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CBSE Class 11 Chemistry Notes For Chapter 10 S Block Elements Group 1 Elements (Alkali Metals) Introduction

All the alkali metals have one valence electron ( ns¹ ) outside the noble gas core. The loosely held s -electronin the outermost valence shell makes them the most electropositive metals.

• To get the stable electronic configurations of noble gases, they readily lose the valence electron to generate the monovalent (M⁺) ions. Hence, they are never found in a free state but in the combined state of nature.
• Since the last electron enters ns -orbital, these are called s -block elements.
• Since all these elements have similar valence shells or outer electronic configurations, all the alkali metals exhibit a striking resemblance in their physical and chemical properties and they are placed in a definite group (Gr-1).
• Lithium shows some abnormal behavior as its electronic configuration is slightly different from the rest of the members of Gr-1 and also because of its extremely small atomic and ionic radii.

Again, lithium shows some similarities with magnesium present in the group- 2 of the third period

Electronic configuration of alkali metals: 

Occurrence Of Alkali Metals

• Since the alkali metals are highly reactive, they do not exist in a free state. In nature, they mostly occur as compounds like halides, oxides, silicates, borates, and nitrates.
• According to abundance, lithium is placed at the 35th position. It mainly occurs in nature in the tire form of silicates,
• For example: Spodumene: LiAl(SiO₃)₂ and Lepidolite: Li₂Al₂(SiO₃)₃(F, OH)₂
• Sodium and potassium are respectively placed at 7th and 8th position in order of their abundance. Sea water is a major source of NaCl and KCl.
• Sodium is abundantly present in the form of rock salt (NaCl). Other important minerals are Chile salt petre: NaNO₃, borax: Na₂B₄O₇.10H₂O, mirabilite: Na₂SO₄, and trona: Na₂CO₃-NaHCO₃-2H₂O.
• Important ores of potassium are sylvite: KCl, carnallite: KCl-MgCl₂-6H₂O, and feldspar: (K₂O-Al₂O₃-6SiO₂).
• Rubidium and cesium are much less abundant than lithium. Radioactive francium does not occur appreciably in nature. It is obtained from the radioactive decay of actinium
• ²²⁷Ac ₈₉ → ²²³Fr₈₇  +⁴He₂  Its longest-lived isotope ²²³Fr₈₇  has a half¬life period of only 21 minutes.
• Since most of the compounds of alkali metals are water soluble, they are found in adequate amounts in seawater.

CBSE Class 11 Chemistry Notes Chapter 10 S Block Elements

General Trends In Atomic And Physical Flv Properties Of Alkali Metals

The alkali metals show regular trends in their physical and chemical properties with an increase in atomic number. Some important atomic and physical properties of alkali metals are given in the following table:

Atomic and physical properties of alkali metals:

General trends in different atomic and physical properties of alkali metals and their explanations:

1. Atomic and ionic radii 

The atomic and ionic radii of alkali metals are the largest in their respective periods and these values further increase on moving down the group from Li to Cs.

Atomic and ionic radii  Explanation:

On moving from left to right in a period, the number of electronic shells remains the same but the nuclear charge increases with each succeeding element Thus, the valence shell electrons experience a greater pull towards the nucleus and this results in successive decreases in atomic and ionic radii with an increase in atomic number

• Thus, the atomic and ionic radii of alkali metals are the largest in their respective periods.
• On moving down the group, a new electronic shell is ) added to each element and the nuclear charge increases with an increase in atomic number.
• The addition of an electronic shell tends to increase the size of the atom but the increase in nuclear charge tends to decrease the atomic radii by attracting the electron cloud inward. Thus, the two factors oppose each other.
• However, the increase in the number of shells increases the screening effect of the inner electrons on the outermost s -electron and as the screening effect is, quite large, it overcomes the contractive effect of the increased nuclear charge.
• The net result is an increase in atomic and ionic radii down the group from Li to Cs.

2. Ionization enthalpy

The first ionization enthalpies of alkali metals are the lowest in their respective periods. Explanation: Since the alkali metal atoms are largest in their respective periods, their outermost electrons being far away from the nucleus experience less force of attraction and hence, can be removed easily.

1. Ionization enthalpy of group-1 alkali metals decreases down the group.

Explanation:

Since the alkali metal atoms are largest in their respective periods, their outermost electrons being far away from the nucleus experience less force of attraction and hence, can be removed easily.

2. Ionisation enthalpy7 of group-1 alkali metals decreases down the group

Explanation:

On moving down the group from Li to Cs, the distance of the valence s-electron from the nucleus progressively increases due to the addition of a new shell with each succeeding element With an increase in the number of inner shells, the screening effects progressively increase and as a result, the effective nuclear charge experienced by the valence electron progressively decreases and hence, the ionization enthalpies decrease down the group.

3. The second ionization enthalpies of alkali metals are very high.

Explanation:

The monovalent cation formed by the removal of an electron from the alkali metal atom has a very stable noble gas configuration,

For example – 

1. Li⁺: 1s² or [He], Na⁺: 1s²2s²2p⁶ 
2. [Ne], k⁺: 1s²2s²2p⁶3s²3p⁶ or [Ar] etc.

Removal of another electron from the monovalent ion having a stable noble gas configuration is very difficult and requires a huge amount of energy. For this reason, the second ionization enthalpies of alkali metals are very high.

3. Hydration of ions, hydrated radii, and hydration enthalpy

The salts of alkali metals are generally ionic and are soluble in water because the cations get hydrated in water to form hydrated cations: M⁺ + aq —> [M(aq)]⁺.

1. The degree of hydration of ions and the hydrated radii decrease as we move down the group

Explanation:

The smaller the cation, the greater its degree of hydration. Since ionic radii increase down the group, the degree of hydration decreases, and consequently, the radii of die-hydrated ions decrease from Li⁺ to Cs⁺.

2. The order of mobilities of die alkali metal ions in aqueous solution is: Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

Explanation:

Smaller ions are more easily hydrated. As Li+ is the smallest ion among the given ions, it is most easily hydrated and has the least ionic mobility in an aqueous solution whereas Cs+ is the largest and is least hydrated. So its mobility is die highest.

3. Ionic conductance of the hydrated ions increases from [Li(m7)]+ to [Cs(aq)]+.

Explanation:

The ionic conductance of these hydrated ions increases from [Li(aq)]⁺ to [Cs(ag)]⁺ because die size decreases and mobility increases in this order. Hydration of ions is an exothermic process. The energy released when 1 gram-mol of an ion undergoes hydration is called hydration energy or hydration enthalpy,

4. Hydration enthalpy of alkali metal ions decreases from Li⁺ to Cs⁺.

Explanation:

The hydration enthalpy of an ion depends upon the ratio of charge to radius (q: r). Since the radii of alkali metal ions increase down the group, the hydration enthalpies decrease from Li⁺ to Cs⁺. Li⁺ ion has the maximum degree of hydration and for this reason, most of the lithium salts are found to be hydrated

For example: LiCl-2H₂O, LiClO₄-3H₂O etc.

4. Oxidation State

Alkali metals exhibit a +1 oxidation state in their compounds and it remains restricted in a +1 state only.

Oxidation State Explanation: 

Alkali metals have low ionization enthalpies and by losing their valence s -electrons they acquire the stable electronic configurations of the nearest noble gases. Thus, they have a strong tendency to form M⁺ ions and exhibit a +1 oxidation state in their compounds.

The second ionization enthalpies required to pull out another electron from M⁺ ions having very unstable noble gas electronic configuration are very high indeed and are not available under the conditions of chemical bond formation. Hence,v the alkali metals do not form M²⁺ ions, i.e. their oxidation state remains restricted to +1 state.

5. Metallic character

The elements of this group are typical metals that are soft (can be easily cut with a knife) and light. When freshly cut, they are silvery white but on exposure to air, they turn tarnished. The metallic character, which refers to the level of reactivity of a metal, increases on moving down the group.

Metallic character Explanation:

As the ionization enthalpy decreases down the group,  the tendency to lose the valence electron increases, and consequently, the metallic character increases.

6. Photoelectric effect

Alkali metals (except Li) exhibit a photoelectric effect. The emission of electrons from the surface of a metal exposed to electromagnetic radiations of suitable wavelength is called the photoelectric effect.

S Block Elements Class 11 NCERT Notes

Photoelectric effect Explanation:

Due to low ionization enthalpies, the alkali metals exhibit a photoelectric effect. It is to be noted that lithium having the highest ionisation enthalpy does not exhibit a photoelectric effect. Cesium having the lowest ionisation enthalpy possesses the highest tendency to exhibit a photoelectric effect.

Potassium anti-cesium, rather than lithium is used in photoelectric cells:

The ionization enthalpies of potassium and cesium are much lower than that of lithium. For this reason, these two metals on exposure to light easily emit electrons from their surface but lithium does not. Hence, potassium and cesium rather than lithium are used in photoelectric cells.

7. Electronegativity

The alkali metals have low electronegativity which further decreases down the group.

Electronegativity Explanation:

The alkali metals having ns¹ electronic configuration preferably show electron releasing tendency rather than electron accepting. Thus, they have low electronegativities. Since the atomic sizes increase down the group, the tendency of atoms to hold their valence electrons decreases down the group, and consequently, electronegativity decreases down the group.

8. Conductivity

Alkali metals are good conductors of heat and electricity.

Conductivity Explanation:

Due to the presence of loosely bound valence electrons (ns¹) which are free to move throughout the metal structure, the alkali metals are good conductors of heat and electricity.

9. Melting and boiling points 

Melting and boiling points: Melting and boiling points of alkali metals are low and decrease down tire group.

Melting and boiling point Explanation:

The cohesive energy that binds the atoms in the crystal lattices of these metals is relatively low (weak metallic bonding) due to the presence of only one valence electron (ns¹) which can take part in bonding. Hence, their melting and boiling points are low. These further decrease down the group as the strength of the metallic bonds and cohesive energy decrease with increasing atomic size.

10. Nature of bonds formed

Alkali metals form ionic compounds and the ionic character of compounds increases down the group from Li to Cs.

Nature of bonds formed Explanation:

For low ionization enthalpies, alkali metals readily form monovalent cations by losing their valence electrons. As ionization enthalpies decrease down the group, the ionic character of the compounds increases down the group.

11. Density

The densities of alkali metals are quite low and increase down the group from Li to Cs.

Density Explanation:

Due to their large atomic size and weak metallic bond, alkali metals have low density. Both the atomic volume and the atomic mass increase down the group but the corresponding increase in atomic mass is not balanced by the increase in atomic volume. As a result, the densities of alkali metals increase down the group. However, the density of K is less than that of Na because the atomic size and atomic volume of potassium are quite higher than that of sodium. As a result, the ratio of mass/ volume decreases.

Li is the lightest metal having a density of 0.53 g. cm^-3. It cannot be preserved in kerosene because it floats over it. Generally kept wrapped in paraffin wax.

12. Flame coloration

Alkali metals or their salts on heating in the flame of the bunsen burner, impart characteristic colors and they can be easily identified from the color of the flame

Flame coloration Explanation:

Ionization enthalpies of alkali metals are not much higher. Thus, when an alkali metal or its salt (especially chloride due to its more volatile nature) is heated in a Bunsen burner flame, the electrons in the valence shell get excited and jump to higher energy levels by absorbing energy. When the excited electron drops back to its ground state, the emitted radiation falls in the visible region and as a result, alkali metals or their salts impart color to the flame.

Alkali metals can be detected by flame tests and can be estimated by flame photometry or atomic absorption spectroscopy.

13. Softness

Alkali metals are soft (can be cut easily with die help of a knife) and their softness increases down the group.

Softness Explanation:

The softness of alkali metals is due to their low cohesive energy and weak metallic bonding. Further, on moving down the group, the strength of metallic bonding decreases due to an increase in atomic size and as a result, the softness of the metals increases down the group.

Chemical Properties Of Alkali Metals

Alkali metals are highly reactive. Such reactivity may be attributed to their large atomic size, low ionization enthalpies, and low heats of atomization. –

Action of air and moisture

Alkali metals, being highly reactive, react readily with atmospheric oxygen to form oxides. These oxides further react with moisture to form hydroxides which in turn produce carbonates by reacting with atmospheric CO₂

Metal oxides also react with CO₂ to form carbonates.

These metals lose their glossiness and become tarnished due to the formation of carbonate layers on their surface. To protect from atmospheric oxygen and moisture, these metals are always stored in inert hydrocarbon solvents such as kerosene, petroleum ether, etc.

Reaction with oxygen

When the alkali metals are heated with oxygen or excess air, they form different types of oxides depending upon the nature of the metal involved. Lithium mainly forms monoxide (Li₂O), sodium forms peroxide (Na₂O₂), and the other alkali metals (K, Rb, and Cs) mainly form superoxides having the general formula MO₂. The temperature required for the reaction decreases down the group from Li to Cs.

4Li + O₂ → 2Li₂ O₂ (Lithium monoxide)

2Na + O₂ → Na₂O₂ (Sodium peroxide)

M +O₂ → MO₂ (Superoxide) [here, M = K, Rb, Cs]

 Oxygen Explanation

A smaller cation can stabilize a smaller anion while a larger cation can stabilize a larger anion. If both the ions are similar in size, the coordination number will be high and this results in higher lattice energy.

A cation having a weak positive electric field can stabilize an anion having a weak negative electric field. Li⁺  ion and oxide ion (O^2-) have small ionic radii and high charge densities. Hence, these small ions combine to form a very’ stable lattice of Li₂O.

Sodium forms peroxide (Na₂O₂) but potassium forms superoxide (KO₂), even though the peroxide ion,   is larger in size than the superoxide ion, This can be explained in terms of charge densities. Due to the bigger size

Na+ ion has a weaker positive field around It and therefore it can stabilise peroxide Ion which also has a weaker negative field around it. Thus, Na⁺ forms peroxide. K⁺ ion is still bigger in size and the magnitude of the positive field around it is much weaker so it can’t stabilize superoxide ion which also has a much weaker negative field around it. Hence, K forms superoxide.

According to valence bond theory, two O-atoms in superoxide ion (O₂^–) are attached by a 2- 2-electron bond (common covalent bond) and a 2-electron bond. As the unpaired electron is present in the 2-electron bond, the superoxide ion is paramagnetic and all tire superoxides are colored (LiO₂ and NaO₂ are yellow, KO₂ & CsO₂ are orange, RbO₂ is brown).

According to MO theory, there is an unpaired electron in one of the n antibonding orbitals and because of this, the superoxide ion is paramagnetic. The electronic configuration of O₂^–

O₂^–  ion present in common oxides [For example,  Li₂O, NaO₂etc.) and the 02~ ion in peroxides (For example,  Na₂O₂), contain no unpaired electron and for this reason, these are diamagnetic and colorless.

Reaction with water

The alkali metals having high negative reduction potentials (E°) can act as a better-reducing agent than hydrogen. Hence, they react with water to form water-soluble hydroxides and liberate hydrogen gas.

2M + 2H₂O → 2MOH + H₂T [M = alkali metal]

Reactivity with water increases down the group as the electropositive character of the metals increases clown the group. Lithium decomposes water slowly. Sodium reacts with water vigorously. K, Rb, and Cs react with water explosively and the evolved hydrogen gas catches fire.

Alkali metals also react with compounds containing acidic H-atoms

For example:  halogen halides (HX), alcohols (ROH), acetylene HC=CH, etc., to form their corresponding salts and H₂

2Na + 2HX → 2NaX  (Sodium halide) + H₂ ↑

Li + 2C₂H₅OH → 2 C₂ H₅ OLi (Lithium ethoxide) + H₂ ↑

2Na + 2HC=CH →  2NaC = CH (Sodium acetylide)+ H₂ ↑

The standard electrode potential of Li is most negative while that of sodium is least negative i.e., in the reaction with water, Li releases a greater amount of heat than sodium. Despite that, Li reacts less vigorously with water than Na. j This can be explained concerning chemical kinetics. Na has a low melting point and the heat of the reaction is sufficient j to melt it. Molten metal spreads out and exposes a relatively large surface to water and as a result, it reacts with water 1 readily and violently. On the other hand, the melting point of Li is much higher and the heat of the reaction is not sufficient to melt it. Hence, its surface area does not increase and it reacts slowly with water.

NCERT Solutions Class 11 Chemistry Chapter 10 S Block Elements

Reaction with dihydrogen

All alkali metals react with dihydrogen at about 673 K (Li at 1073 K) to form colorless, crystalline hydrides (MH). These hydrides have a high melting point.

1. The reactivity of the alkali metals towards dihydrogen decreases down the group.

Explanation:

As the size of the metal cation increases down the group, the lattice energy of the hydrides decreases down the group. Consequently, the reactivity of the alkali metals towards hydrogen decreases down the group,

2. The ionic character of the alkali metal hydrides increases from Li to Cs.

Explanation:

Since the ionization enthalpy of alkali metals decreases down the group, the tendency to form cations as well as the ionic character of the hydrides increases.

Reaction with halogens

All alkali metals react vigorously with halogens to form crystalline halide compounds having the general formula MX. Lithium halides are covalent due to the very high polarising power of the Li⁺ ion. Halides of other alkali metals are ionic in nature’

⇒ \(2 \mathrm{M}+\mathrm{X}_2 \rightarrow 2 \mathrm{M}^{+} \mathrm{X}^{-}\) ( X = F, Cl, Br or I)

The reactivity of the alkali metals towards a particular halogen increases down the group.  Due to a decrease in ionization enthalpies or an increase in the electropositive character of the metals down the group, the reactivity increases down the group.

The reactivity of halogens towards a particular alkali metal decreases in the order:

F₂> Cl₂ > Br₂ > I₂

Reducing nature

The alkali metals act as strong reducing agents because of their low ionization enthalpies. Since the ionization enthalpies decrease on moving down the group, therefore, in the free state the reducing power also increases in the same order, i.e., Li  < Na < K < Rb < Cs.

The tendency of a metal to lose an electron in solution is measured by its standard electrode potential (E°). The alkali metals have low values (higher negative values) of E° and so they have a strong tendency to lose electrons and can act as strong reducing agents. Lithium, although, has the highest ionization enthalpy, is the strongest reducing agent in solution (E° = -3.04 V). On the other hand, Na is the weakest reducing agent and the reducing character increases from Na to Cs, i.e., Na < K < Rb < Cs.

Explanation of anomalous behavior of lithium

The anomalous behavior of lithium can be explained because the ionization enthalpy is the property of an isolated atom in the gaseous state while the standard electrode potential is concerned when the metal atom goes into solution.

The ionization enthalpy involves the change:

M(g) → M⁺+(g) + e, while the standard electrode potential involves the change: M(s) → M⁺(aq) + e.

The latter change occurs in three steps as follows:

1. M(s)→ M(g) – sublimation enthalpy
2. M(g)→ M+(g) + e – ionisation enthalpy
3. M⁺ (g) + H₉O → M⁺(aq) + hydration enthalpy

The overall tendency for the change depends on the net effect of these three steps. Among the alkali metal cations, Li⁺ ion has the maximum tendency to get hydrated due to its very small size. The high hydration enthalpy compensates the energy required in the first two steps to a large extent and the overall energy required to convert M(s) to M⁺ (aq) is minimum for lithium.

Thus, small size and high hydration enthalpy are responsible for the strong reducing character of lithium.

The solution in liquid ammonia

Alkali metals dissolve in liquid ammonia to give highly conducting deep blue solutions which are highly reducing and paramagnetic. As the concentration increases (> 3M), the color of the solution changes to copper-bronze. These concentrated solutions are diamagnetic.

Solution in liquid ammonia Explanation:

1. When an alkali metal is dissolved in liquid v ammonia, ammoniated cations, and ammoniated electrons are formed as shown below:

M + (x+ y)NH₃ →  [M(NH₃)_x]⁺ (Ammoniated cation) + [e(NH₃)_y]^–(Ammoniated electron)

2. The blue color of these solutions is due to the excitation of the free ammoniated electrons to higher energy levels by absorbing energy corresponding to the red region of visible light.  The transmitted light is blue which imparts a blue color to the solutions.

3. With the increase in the concentration of the alkali metal, the formation of clusters of metal ions starts and because of this, at a much higher concentration (> 3M) the solutions possess metallic luster and attain the color of copper-bronze.

4. These blue solutions are highly conducting because of the presence of ammoniated electrons and ammoniated cations but the conductivity decreases with increasing concentration as the ammoniated cations get attached to the free unpaired electrons.

5. These blue solutions are paramagnetic due to the presence of unpaired electrons. However, tire paramagnetism decreases with increasing concentration due to the association of ammoniated electrons to yield diamagnetic species.

6. The free ammoniated electrons make these solutions very powerful reducing agents.

7. These solutions when kept, form metal amides and release H₂. However, these solutions can be stored in anhydrous conditions in the absence of impurities like Fe, Pt, Zn, etc.

M⁺(am) + e(am) + NH₃(l)→ MNH₂(am) + ½H₂(g)

Where ‘am’ stands for ‘solution in ammonia

2M + 2NH₃→ 2MNH₂ (metal amide)+ H₂

Extraction of alkali metals

Alkali metals cannot be extracted by applying common processes used for the extraction of other metals.

Alkali metals Explanation:

• The alkali metals are strong reducing agents. Hence, they cannot be extracted by reduction of their oxides or other compounds.
• Since they are highly electropositive, the method of displacing them from their salts by any other element is not possible.
• The aqueous solution of their salts cannot be used for extraction by electrolytic method because hydrogen, instead of the alkali metal is discharged at the cathode (discharge potentials of alkali metals are much higher).
• However, by using Hg as a cathode, the alkali metals can be deposited but in that case, the alkali metals readily combine with mercury to form amalgams from which the recovery of metals becomes quite difficult.
• The electrolysis of their fused salts (usually chlorides) is the only successful method for their extraction, Another metal . ‘ chloride is generally added to lower its fusion temperature

General Characteristics Of The Compounds Of Alkali Metals

The compounds of alkali metals are predominantly ionic. Some of the general characteristics of these compounds are described below.

1. Oxides and hydroxides

1. Typical oxides or monoxides of alkali metals

For example: Li₂O and Na₂O are white ionic solids and basic. These oxides react with water to form strong alkalis (MOH).

Example:  Na₂O + HO → 2NaOH

2. All peroxides are strong oxidizing agents. They react with water or acid to give hydrogen peroxide (H₂O₂) and the corresponding metal hydroxide. Na₂O₂ is widely used as an oxidizing agent in inorganic chemistry.

M₂O₂ + 2H₂O →  2MOH + H₂O₂

Example: Na₂O₂ + 2H₂O₂→ 2NaOH + H₂O₂

3. Superoxides are stronger oxidizing agents than peroxides and react with water or acid to give both H₂O₂ and O₂ along with metal hydroxide.

2MO₂ + 2H₂O→  2MOIH + H₂O₂+ O₂

Example: 2KO₂ + 2H₂O₂ →2KOH + H₂O₂ + O₂

The alkali metal hydroxides (MOH) are all white crystalline solids and corrosive. They are the strongest of all bases and readily dissolve in water. Due to excess hydration, a large amount of heat is released. These hydroxides are thermally stable except Li OH. The basic strength of alkali metal hydroxides increases on moving down the group from Li to Cs.

Explanation:

The ionization enthalpies of alkali metals decrease on moving down the group and this causes a weakening of the bond between the alkali metal and the hydroxyl group (M —OH). This results in an increase in the concentration of hydroxyl ions in the solution, i.e.,  the basic character of the solution increases on moving down the group.

Thus, the basic strength of the hydroxides follows the order:

CsOH > RbOH > KOH > NaOH > LiOH

2. Halides

The alkali metal halides can be prepared by combining metals directly with halogens or by reacting appropriate oxides, hydroxides or carbonates with aqueous halogen acids (HX).

2M + X₂  →  2MX ; M₂ O + 2HX↓ 2MX + H₂ O

MOH + HX→ MX + H₂ O

M₂ CO₃+ 2HX→ 2MX + HO₂ + CO₂

The enthalpy of formation (ΔH°_f) of alkali metal halides is highly negative. For a given metal, AHj values decrease from fluoride to iodide. These halides are colorless crystalline solids having high melting and boiling points.

1. The melting point of halides of a particular alkali metal decreases as:

Fluoride > Chloride > Bromide > Iodide.

Explanation:

For a particular alkali metal ion, the lattice enthalpies decrease as the size of the halide ion increases.

Lattice enthalpies of NaP, NaCl, Nalir, and Nal are 893, 770,730, and 685 kJ. mol^-1 respectively, A.s the lattice enthalpy decreases, the energy required to break the crystal lattice decreases, and consequently, the melting points decrease. Thus, the melting points of NaF, NaCl, NaBr, and Nal are found to be 1201K, 10IMK, 1028 K, and 944 K respectively.

2. For a particular halide ion, the melting point of IJX is less than that of

NaX and thereafter the melting points decrease on moving down the group from Na to Cs.

Explanation:

The melting point of LiCl (887K) is less than that of NaCl (I084K), because LiCl is covalent (for smaller atomic size of Li compared to that of Na), but NaCl is ionic. Thereafter, the order of melting point is:

NaCl(1084K)>KCl(1039K)>RbCI(988K)>CsCl(925K) This is observed because the lattice enthalpies decrease as the size of the alkali metal atom increases.

3. Solubilities of the alkali metal halides (except fluorides) decrease on moving down the group since the decrease in hydration enthalpy is more than the corresponding decrease in the lattice enthalpy.

For example, the difference in lattice enthalpy between NaCl and KCl is 67kJ. mol^-1 whereas the difference in hydration enthalpy between Na⁺ and K⁺ ion is 76 kj -mol^-1 Thus, KCl is relatively less soluble in water compared to NaCl.

Explanation:

The solubility of a salt in water depends on its lattice enthalpy as well as its hydration enthalpy. In general, if hydration enthalpy > lattice enthalpy, the salt dissolves in water but if the hydration enthalpy < lattice enthalpy, the salt does not dissolve.

Further, the extent of hydration depends on the ionic size. The smaller the size of the ion, the more it will get hydrated and the greater will be its hydration enthalpy. LiF, for example, is almost insoluble in water because of its higher lattice enthalpy (-1005 kJ . mol^-1 ).

On the other hand, the low solubility of Csl in water is due to smaller hydration enthalpies of the two large ions [-276(Cs⁺)-305(I^–) = -581 kJ.-mol^-1]. o Due to the smaller size and relatively higher electronegativity of Li, lithium halides except LiF are predo¬minantly covalent and hence, are soluble in organic solvents such as acetone, alcohol, ethyl acetate etc.

In contrast, sodium chloride, being ionic, is insoluble in organic solvents.

3. Soils of oxoacids

Alkali metals react with c to acids such as carbonic acid (H₂CO₃), nitric acid (HNO₃), sulphuric acid (H₈SO₄), etc., to form corresponding salts and release H₂. Due to the high polarising power and lattice energy of small Li ions, lithium salts behave abnormally.

4. Nature of carbonates and bicarbonates

All alkali metals form carbonates of the type M₂CO₃. Since the alkali metals are highly electropositive, their carbonates are remarkably stable up to l000°C above which they first melt and then decompose to form oxides. These salts are readily soluble in water. As electropositive character increases down the group, the stability of carbonates increases in the same order:

Cs₂CO₃ > Rb₂CO₃ > K₂CO₃ > Na₂CO₃ > Li₂CO₃

Li₂CO₃ is insoluble in water and unstable towards heat. It decomposes readily to give Li₂O and CO₂.

Explanation:

1. The very small L ion exerts a strong polarising power on the large carbonate (CO₃^2-) ion and distorts the electron cloud of its nearby oxygen atom.

This results in the weakening of the C—O bond and the strengthening of the Li — O bond. This eventually facilitates the decomposition of Li₂CO₃ leading to the formation of Li₂O and CO₂. %

2. The crystal lattice formed by a smaller Li⁺ ion with a smaller O₂ ion is more stable than that 2  formed by a larger CO₃ ion and a smaller Li+ ion. This also favors the decomposition of Li₂CO₃

3. The aqueous solution of carbonates is alkaline. This is because carbonates being the salts of strong bases and weak acids (H₂CO₃) undergo hydrolysis.

M₂CO₃ + 2H₂O  ⇌  2MOH (strong base) + H₂CO₃ (weak acid)

4. Bicarbonates or hydrogen carbonates (MHCO₃) of the alkali metals except LiHCO₃ are obtained in the solid state. These bicarbonates are soluble in water and stable towards heat. On strong heating, all the bicarbonates undergo decomposition to yield carbonates with the evolution of carbon dioxide.

2MHCO₃ (heat)→ M₂CO₃ + CO₂ + H₂O

As the electropositive character of the metals increases down the group from Li to Cs, the stability of the bicarbonates increases in the same order.

5. Nature of nitrates

The alkali metal nitrates (MNO₃) are prepared by the action of HNO₃ on the corresponding carbonates or hydroxides. They are ionic crystalline solids having low melting points and are highly soluble in water. On strong heating, they (except LiNO₃ ) decompose into nitrites and at higher temperatures oxides.

S Block Elements Chapter 10 NCERT Notes

For example:

LiNO₃ decomposes readily on heating to give

6. Nature of sulphates

The alkali metals form sulfates of the type M₂SO₄. All the sulfates except Li₂SO₄ are soluble in water. The sulfates when fused with charcoal, form sulphides.

M₂SO₄.+ 4C→  M₂ S + 4CO

Sulfates of alkali metals form double salts with the sulfates of trivalent metals like Fe, Al, Cr, etc. These double salts crystallize with a large number of water molecules to form alum. A typical example is potassium aluminum

[K₂SO₄→ Al₂(SO₄)₃ -24H₂O].

Lithium sulfate (Li₂SO₄) is not known to form alum.

Anomalous Behaviour Of Lithium (Li) And Similarity Between Li And Mg

Although lithium, the first element of group 1, exhibits most of the characteristic properties of this group, yet it differs from other members of this group in several respects.

Reasons for anomalous behavior of lithium

1. Both Li- atom and Li⁺ ion have very small sizes.
2. Much higher polarising power of very small Li⁺ ion results in increased Points of difference between lithium and other alkali metals covalent character of its compounds.
3. Lithium has the lowest electropositive character, the highest ionization enthalpy, and the highest electronegativity compared to the rest of the members.
4. Non-availability of d -d-orbital in its valence (outermost) shell.
5. Strong intermetallic bonding (cohesive force) due to its small size. On the other hand, lithium shows a diagonal relationship with magnesium

Points of difference between lithium and other alkali metals:

Reasons for the similarities between lithium & magnesium:

Lithium exhibits a diagonal relationship with the 3rd-period group-2 element, magnesium. Reasons for the similarities between lithium and magnesium

• The atomic as well as ionic radii of Li and Mg are almost the same (Li⁺ = 76 pm and Mg²⁺ = 72 pm).
• Both lithium and magnesium have almost similar electronegativities (Li = 0.98 and Mg = 1.2).

Similarities between lithium and magnesium:

Uses Of Alkali Metals

Preparation, Properties, And Uses Of Some Important Compounds Of Sodium

1. Sodium carbonate (washing soda), (Na₂CO₃-10H₂O)

1. Manufacture: Ammonia-soda or Solvay process

Sodium carbonate is commonly known as washing soda. It is generally manufactured by the Solvay process or ammonia-soda process.

Principle: When carbon dioxide is passed through an aqueous solution of NaCl (brine, 28% NaCl solution) saturated with ammonia, sodium bicarbonate is formed.

NH₃ + CO₂ + H₂O > NH₄HCO₃

NH₄HCO₃ + NaCl  ⇌ NaHCO₃ + NH₄Cl

Due to the common ion effect of Na⁺ ion, sodium bicarbonate so formed gets precipitated. Such removal of solid NaHCO₃ shifts the reaction more and more towards the right. This results in a greater yield of NaHCO₃.  In this way, a nearly two-thirds portion of NaCl is converted into NaHCO₃. The precipitated NaHCO₃ is then filtered off, dried, and heated at 150°C to get sodium carbonate.

Evolved CO₂ is reused to saturate the ammoniated brine.

Raw materials:

1. Brine solution (28% aqueous solution of NaCl),
2. Limestone or calcium carbonate (CaCO₃) it is the source of CO₂ and
3. Ammonia.

2. Description of the process

Preparation of ammoniated brine: 

1. This process is carried out in the absorption tower made of iron

2.  From an overhead tank, brine is allowed to trickle down slowly along the tower and ammonia gas from the ammonia recovery tower which is mixed with a small amount of CO₂ is allowed to pass through a tube situated near the bottom of the tower. As a result, the brine solution gets saturated with ammonia while calcium chloride and magnesium chloride are present as impurities in commercial.

3. Sodium chloride gets precipitated as their corresponding insoluble carbonates.

2NH₃ + CO₃ + H₂O → (NH₄)₂CO₃

CaCl₂ + (NH₄)CO₃ → 2NH₄Cl + CaCO₃↓

MgCl₂ + (NH₄)2CO₃ → 2NH₄Cl + MgCO₃↓

4. The ammoniated brine is then filtered to remove the precipitated calcium and magnesium carbonates and the filtrate thus obtained is passed into the carbonation tower.

Carbonation of ammoniated brine:

1. This operation is carried out in a long cast iron tower (carbonation or Solvay tower). The tower is fitted with several horizontal plates.

2. The ammoniated brine solution is trickled down from the top of the tower while CO₂ gas from the lime kiln is introduced into the tower under high pressure through a pipe fitted at the base of the tower.

3. In this way, CO₂ comes in contact with the descending stream of ammoniated brine and they react with each other to form ammonium bicarbonate which subsequently combines with NaCl to produce sodium bicarbonate and ammonium chloride.

NH₃ + CO₂ + H₂O →  NH₄HCO₃

NaCl + NH₄HCO₃ →  NaHCO₃↓ + NH₄Cl

Separation of sodium bicarbonate:

1. The solution coming out of the carbonation tower contains crystals of NaHCO₃. These are separated by passing the solution through vacuum filters.

2. The separated sodium bicar¬bonate is washed with water to remove any sodium or ammonium chloride that may adhere to it and then dried.

3. The filtrate containing NH₄Cl and a small amount of NH₄HCO₃ is taken to the ammonia recovery tower where it comes in contact with Ca(OH)₂.

Calcination:

When the dry NaHCO₃ is heated strongly in a furnace at 180°C, it decomposes to form anhydrous Na₂CO₃. It is called soda ash. It is nearly 99.5 % pure.

Evolved CO₂ is reused in the carbonation tower or absorption tower

Recovery of ammonia:

The filtrate from the carbonation tower which contains ammonium chloride and a little ammonium bicarbonate is made to flow down the ammonia recovery tower. NH₄HCO₃ is decomposed by the heat of steam and NH₄Cl reacts with calcium hydroxide to form ammonia, carbon dioxide, and CaCl₂. The mixture of NH₃ and CO₂ is used for tire saturation of brine while calcium chloride is obtained as a by-product.

1.

2.

Potassium carbonate (K₂CO₃) cannot be prepared by the Soh’ay process. This is because unlike sodium bicarbonate, potassium bicarbonate (KHCO₃) which is fairly soluble in water does not get precipitated when CO₂ is passed through the ammoniated solution of KCl.

 Properties of Sodium carbonate: 

1. State:

Sodium carbonate is available either as anhydrous salt or as hydrated salt. The hydrated salts are white crystalline substances and are mainly of two types

1. Decahydrate (Na₂CO₃-10H₂O) and
2. Monohydrate (Na₂CO₂-H₂O).

The decahydrate is also called washing soda. The anhydrous salt commonly known as soda ash, is a white powder. When sodium carbonate is crystallized from water, the decahydrate is obtained as white transparent crystals. These crystals are efflorescent. When exposed to air for a long time, crystals of decahydrate partially lose their water of crystallization and is converted to monohydrate, a powdery substance which is known as crystal carbonate.

⇒ \(\mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3 \cdot \mathrm{H}_2 \mathrm{O}+9 \mathrm{H}_2 \mathrm{O}\)

2. Action of heat:

When the decahydrate is heated up to 100°C, it slowly loses nine molecules of water of crystallization and gets converted into monohydrate. When the monohydrate is heated above 100°C, the anhydrous salt (Na₂CO₃) is produced as a white powder which melts at high temperatures but never undergoes decomposition.

1.

2.

Anhydrous Na₂CO₃ or soda ash melts at higher temperatures (melting point 852°C) but does not decompose. It turns to monohydrate when kept in the air.

3. Hydrolysis:

It dissolves in water with the evolution of a considerable amount of heat. Being a salt of a weak acid (HCO₃) and a strong base (NaOH), it undergoes hydrolysis in water to give an alkaline solution.

⇒\(\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2\left[\mathrm{Na}^{+}+\mathrm{OH}^{-}\right]+\mathrm{H}_2 \mathrm{CO}_3\)

4. Reaction with acid:

At ordinary temperature, Na₂CO₃ reacts with dilute mineral acids to form the corresponding sodium salts and water along with the evolution of CO₂

Na₂CO₃ + 2HCl→2NaCl + CO₂↑ + H₂O

Na₂CO₃ + 2CH₃COOH→2CH₃COONa + CO₂↑ + H₂O

Reaction with slaked lime: When a solution of Na₂CO₃ is heated with slaked lime (milk of lime) at 80°C, sodium hydroxide with insoluble calcium carbonate is obtained.

Na₂CO₃  + Ca(OH)₂ → CaCO₃↓+ 2NaOH

Uses of sodium carbonate:

• Sodium carbonate is mainly used for softening hard water and for washing clothes.
• It is used in fire extinguishers.
• It is largely used in the manufacture of soap, glass, borax, and caustic soda.
• It is used in the paper, paint, and textile industries.
• A mixture of Na₂CO₃ and K₂CO₃ is used as a fusion mixture.
• It is used as an important laboratory reagent both in qualitative and quantitative analysis.

2. Sodium bicarbonate or sodium hydrogen carbonate (baking soda), NaHCO₃

Preparation of sodium bicarbonate:

Sodium hydrogen carbonate is obtained as the intermediate product in the Solvay process of manufacturing sodium carbonate.

It can also be prepared by passing CO₂ through a saturated solution of sodium carbonate. Being less soluble, the white crystals of sodium hydrogen carbonate can be filtered out and dried at room temperature.

Na₂CO₃ + H₂O+ CO₂ ⇌ 2NaHCO₃↓

Properties of sodium bicarbonate:

1. State: It is a white crystalline solid and is sparingly soluble in cold water. It is also stable in air.

2. Hydrolysis: Being a salt of weak acid (H₂CO₃) and strong base (NaOH), it hydrolyses to give a faintly alkaline solution.

NaHCO₃ + H₂O ⇌  [Na⁺ + OH^–] + H₂CO₃

3. Action of heat: On heating, it decomposes to form CO₂, water, and sodium carbonate.

4. Reaction with acids: At ordinary temperature, it reacts with mineral acids to form CO₂, water, and the sodium salt of the acid:

NaHCO₃ + HCl→ NaCl + CO₂ ↑ + H₂O

Uses of sodium bicarbonate:

• It is used as an antacid (known as soda bi-carb). It is also used as a mild antiseptic for skin infections.
• It is the chief ingredient of ‘baking powder’ which is used in preparing breads, biscuits, cakes etc.
• It is used in the preparation of soft drinks like soda- water, lemonades, etc.
• It is also used in fire extinguishers.

3. Sodium hydroxide (caustic soda), NaOH

Manufacture of sodium hydroxide:

Sodium hydroxide is industrially prepared by the electrolysis of an aqueous solution of NaCl (brine) in a specially designed cell called the Castner-Kellner cell or mercury cathode cell.

Class 11 Chemistry Chapter 10 S Block Elements Notes PDF

Sodium hydroxide Principle:

When a brine solution is electrolyzed in a cell using a mercury cathode and graphite anode, metallic sodium discharged at the cathode combines with mercury to form sodium amalgam. Now, electrolysis of slightly alkaline water in the cell using sodium amalgam as anode and iron rod as cathode produces NaOH. The reaction between sodium amalgam and water also produces NaOH.

Sodium hydroxide  Procedure:

1.  The cell consists of a large rectangular iron tank divided into three compartments by two slate partitions which do not touch the bottom of the tank but remain suspended in mercury placed in the grooves

2. The graphite anodes are fixed in the two outer compartments and the cathode which consists of several iron rods is fitted in the central compartment.

3. The layer of mercury at the bottom serves as an intermediate electrode as a cathode in the outer compartment and as an anode in the central compartment by induction.

4. The brine solution is taken in the two outer compartments and a very dilute NaOH solution is taken in the central compartment.

5. The mercury layer is made to flow from one compartment to another by rocking the cell with the help of an eccentric wheel.  On passing electric current, the following reactions take place in the outer and central compartments.

6. In the outer compartment, NaCl undergoes electrolysis. Cl₂ gas formed at the anode comes out from the outlet tube while sodium liberated at the cathode combines with mercury to form sodium amalgam.

NaCl → Na⁺+ Cl^– ; H₂ O → H⁺ + OH^–

• At cathode: Na⁺ + e →Na; Na + Hg→ Na/Hg
• At anode,: Cl^–→ Cl + e; Cl + Cl→ Cl₂ ↑

7. In the central compartment, sodium amalgam (Na/Hg) acts as an anode by induction.

⇒ \(\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}, \mathrm{NaOH} \rightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}\)

• At cathode: \(\mathrm{H}^{+}+e \longrightarrow \mathrm{H} ; \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2 \uparrow\)
• At anode: \(\mathrm{Na} / \mathrm{Hg} \longrightarrow \mathrm{Na}^{+}+e+\mathrm{Hg}\)

Net reaction: \(2 \mathrm{Na} / \mathrm{Hg}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2\left(\mathrm{Na}^{+}+\mathrm{OH}^{+}\right)+2 \mathrm{Hg}+\mathrm{H}_2 \uparrow\)

The concentration of NaOH in the central compartment gradually increases with the progress of electrolysis and when it becomes 20%, the solution is withdrawn, evaporated and converted into pellets or flakes of NaOH.

Properties of sodium hydroxide:

1. State: It is a white, crystalline hygroscopic solid having a melting point of 318°C.

2. Solubility: It dissolves in water with the evolution of heat, giving a strong alkaline solution. It also dissolves in alcohol.

3. Hygroscopic and corrosive nature:

The crystals of NaOH are deliquescent (hygroscopic). When exposed to air they absorb moisture from air and dissolve in the absorbed water. Moist caustic soda generally absorbs CO₂ from air to form sodium carbonate which forms a coating over the surface of the material. As Na₂CO₃ is non-hygroscopic, wet sodium hydroxide becomes dry again.

⇒ \(2 \mathrm{NaOH}+\mathrm{CO}_2 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}\)

It is corrosive. When its concentrated solution comes in contact with the skin it produces a burning sensation. It breaks down the proteins of the skin and because of this property, it is commonly called caustic soda,

4. Reaction with acids, acidic oxides, and amphoteric oxides: Being a strong alkali, it reacts with acids, acidic oxides, and amphoteric oxides to form corresponding salts.

NaOH + HCl→ NaCl +H₂O

2NaOH + SO₂→ Na₂SO₃ +H₂O

Al₂O₃ + 2NaOH →2NaAlO₂ (Sodium aluminate) + H₂O

ZnO + 2NaOH → Na₂ZnO₂ (Sodium zincate) + H₂O

Uses of sodium hydroxide:

It is used

• In the manufacture of soap, paper, artificial silk, dyes, and several chemicals
• In the refining of [etroleum and vegetable oil,
• In the purification of bauxite,
• As a cleaning agent for greasy machines and metal NItoots,
• As a laboratory reagent etc.

4. Sodium chloride (common salt), NaCI

Preparation of sodium chloride:

• The main source of sodium chloride Is seawater which contains 2.7-2,9% of the salt by mass. In tropical countries like India, common salt is generally obtained by the evaporation of seawater.
• Crude sodium chloride obtained by this process contains calcium sulfate (CaSO.), sodium sulfate (Na₂SO₄), calcium chloride (CaCl₂), magnesium chloride (MgCl₂), etc as impurities.
• Since MgCI₂ and CaCl₂ are deliquescent (absorb moisture from the air), impure common salt gets wet in the rainy season.
• For purification, a saturated solution of crude NaCI is prepared and the insoluble impurities are removed by filtration.
• The filtrate is then saturated with hydrogen chloride gas and crystals of pure NaCI separate out due to the common ion effect.
• Chlorides of Ca and Hg being more soluble remain dissolved in the solution. NaCI can also be prepared from rock salt.

Properties of sodium chloride:

• NaCI is a white crystalline solid that melts at 1081K.
• 36 g of NaCI is soluble in 100g of water at 373K. However, solubility does not increase much with an increase in temperature.

Uses of sodium chloride:

• It is used as common salt or table salt for domestic purposes.
• It is used in the manufacture of sodium, caustic soda (NaOH), chlorine, washing soda, sodium peroxide, sodium sulfate, etc.
• It is used in soap industry, in softening hard water, in freezing mixtures, and for regenerating ion exchange resins.

Biological Importance Of Sodium And Potassium

Sodium and potassium ions are the most common cations present in biological fluids. A person weighing 70kg contains about 90g of Na and 170g of K along with 5g of Fe and 0.06 g of Cu.

The daily requirement of Na and K for the human body is about 2 g each.

1. The Na⁺ ions are mainly found outside the cells, in blood plasma, and in the interstitial fluid that surrounds the cells. These ions take part in the transmission of nerve signals, in regulating the flow of water across the cell membranes, and in the transportation of various amino acids and sugars into the cells.
2.  K⁺ ions are the most abundant cations in the cell fluids, where they activate a variety of enzymes, and promote the oxidation of glucose into ATP (adenosine triphosphate), and Na⁺ ions promote the transmission of nerve signals.
3. The Na⁺ and K⁺ ions differ considerably in concentration on the opposite sides of the cell membrane. In blood plasma, for example, the concentrations of Na+ and K+ ions are 143 million-L^-1 and 5 million-L^-1 respectively.
4. Within the blood cells, however, the concentrations of these ions are 10 millimol-L^-1 and 105 millimol-L^-1 respectively.
5. The activities in a nerve cell depend upon the sodium-potassium ion gradient. These ionic gradients are maintained by an ion transport mechanism that operates the active inclusion of K⁺ ions and active exclusion of Na⁺ ions across the cell membrane.
6. The transportation of ions requires energy which is obtained by hydrolysis of ATP. The hydrolysis of one ATP molecule to ADP provides enough energy to move three Na⁺ ions out of the cell two K⁺ ions and one H+ ion back into the cell.

CBSE Class 11 Chemistry Notes For Chapter 10 S Block Elements Group-2 Elements (Alkaline Earth Metals) Introduction

The outermost shell of these elements contains two electrons and the penultimate shell contains eight electrons except forthe first member beryllium which contains two electrons.0 Since the last electron enters the ns orbital, these are also called s-block elements.

Their outermost electronic configuration may be represented as ns², where n- 2 to 7. Due to their similarity in electronic configuration, they are placed in the same group (Group- 2) of the periodic table and closely resemble each other in physical and chemical properties. Two valence electrons are always involved together giving rise to uniform bivalency of the elements.

Beryllium shows some abnormal properties as its electronic configuration is slightly different from the rest of the members. The main reason is that both the beryllium atom and Be²⁺ ion are extremely small. Beryllium also shows some similarities with aluminum of group 13. Like alkali metals, the compounds of these metals are also predominantly ionic. The electronic configurations of alkaline earth metals are given in the following table

Electronic configuration of alkaline earth metals

CBSE Class 11 Chemistry S Block Elements Summary

Occurrence Of Alkaline Earth Metals

Due to low ionization enthalpies and high electropositive character, the alkaline earth metals are chemically very reactive and hence, do not occur in the free state but are widely distributed in nature as silicates, carbonates, sulfates, and phosphates.

• Relative abundance of Be, Mg, Ca, Sr, Ba, and Ra in the earth’s crust is 2, 27640, 46600, 384, 390, and 1.3 x 10-6 ppm respectively. 0 Beryllium, the fifty-first most abundant element by mass in the earth’s crust, is found as silicate minerals like beryl (Be₃Al₂Si₆O₁₈) and phenacite (Be₂SiO₄).
• Magnesium, the sixth most abundant element is found as carbonate, sulphate, and silicate. Its two important minerals are magnesite (MgCO₃) and dolomite [MgCO₃.CaCO₃].
• It is also found in seawater at 0.13% as MgCl₂ and MgSO₄.
• Calcium, the fifth most abundant element by mass found in the earth’s crust, occurs mainly as CaCO₃ in the form of limestone, marble and chalk. Its other important minerals are fluorspar (CaF2), fluorapatite, [3Ca₃(PO₄)2-CaF₃], gypsum (CaSO₄-2H_2O) anhydride, (CaSO₄).
• Strontium and barium are respectively the fifteenth and sixteenth most abundant element. Strontium occurs principally as the mineral celestite (SrSO₄) and strontianite (SrCO₃) while barium occurs mainly as the mineral barytes (BaSO₄).
• Radium is radioactive and extremely scarce. It occurs in very small amounts (1 gin 7 ton) in pitchblende as the decay product of uranium.

General Trends In Atomic And Physical Properties Of Alkaline Earth Metals

The alkaline earth metals show regular trends in their physical and chemical properties with an increase in atomic number. Some important atomic and physical properties of alkaline earth metals are given in the

Atomic and physical properties of alkaline earth metals:

General trends in different atomic and physical properties of alkali metals and their explanations

1. Atomic and ionic radii

The atomic and ionic radii of alkaline earth metals are fairly large but smaller than those of the corresponding alkali metals and these increase on moving down a group.

Atomic and ionic radii Explanation:

The electrons of alkaline earth metals having a higher nuclear charge are more strongly attracted towards the nucleus.  On moving down the group, the atomic as well as ionic radii increase. The addition of new shells and the increasing screening effect jointly overcome the effect of increasing nuclear charge down the group

2. Ionization enthalpy

1. The first and second ionization enthalpies of alkaline earth metals are quite low and decrease down the group from Be to Ra

Explanation:

The low ionization enthalpies of alkaline earth metals are due to their smaller nuclear charge and larger atomic size (compared to the other succeeding elements of the same period) which result in weaker forces of attraction between valence electrons (ns2) and nucleus. On moving down the group, atomic size increases and the screening effect of the inner shell electrons also increases

Since the alkali metal atoms are largest in their respective periods, their outermost electrons being far away from the nucleus experience less force of attraction and hence, can be removed easily. These two effects jointly overcome the effect of increasing nuclear charge down the group. Thus, first and second ionization enthalpies decrease down the group.

2. The first ionization enthalpies of alkaline earth metals are higher than those of the corresponding alkali metals but their second ionization enthalpies are lower than those of the corresponding alkali metals.

Explanation:

The alkaline earth metals have higher values of first ionization enthalpy than those of the corresponding alkali metals because they have smaller size and higher nuclear charge which result in stronger forces of attraction between the valence electrons and the nucleus.

The second ionization enthalpy values of alkaline earth metals are much lower than those of the corresponding alkali metals because the loss of the second electron from an alkaline earth metal cation (M⁺) leads to the attainment of a stable noble gas configuration (ns²np⁶) while the loss of the second electron from an alkali metal cation (M⁺) causes loss ofits stable noble gas configuration

Example:

3. Electropositive or metallic character

The alkaline earth metals are highly electropositive and possess high metallic character. However, they are less electropositive than the alkali metals. Their electropositive or metallic character increases on moving down the group.

Electropositive Explanation:

• Due to their relatively low ionization enthalpies, alkaline earth metals have a strong tendency to lose both valence electrons to form dipositive ions. Thus, they exhibit high electropositive or metallic character.
• As their atoms have smaller sizes and higher ionization enthalpies compared to those of the corresponding alkali metals, their tendency to lose valence electrons is less than that of alkali metals. Hence, alkaline earth metals have less electropositive or metallic character as compared to tine alkali metals.
• On moving down the group from Be to Ra, ionization enthalpies decrease due to an increase in atomic radii. Therefore, the tendency to lose electrons increases and so does the electropositive character

4. Hydration enthalpy

Hydration enthalpies of alkaline earth metal ions are much greater than that of the alkali metal ions & decrease down the group from Be²⁺ to Ba²⁺

Be²⁺ > Mg²⁺ > Ca²⁺ > Sr2+ > Ba²⁺

Hydration enthalpy Explanation:

Due to the smaller size of alkaline earth metal ions, their hydration enthalpies are much greater than those of the alkali metal ions. Therefore, the compounds of alkaline earth metals are found to be more extensively hydrated than those of alkali metals. Magnesium chloride and calcium chloride, for example, exist as hexahydrates (MgCl₂-6H₂O and CaCl₂-6H₂O) while sodium chloride and potassium chloride do not form such hydrates.

The ionic conductance of hydrated alkaline earth metal ions increases from [Be(H₂O)_x]²⁺ to [Ba(HO)_2x]²⁺ due to a decrease in the extent of hydration. The hydration enthalpy of an ion is directly proportional to its charge/radius ratio {q/r). On moving down a group, the radii of the alkaline earth metals increase. As a result, the hydration enthalpies of these metals decrease.

5. Oxidation State

Alkaline earth metals exhibit an oxidation in their compounds. Although the second date of +2 in their compounds. Although the second ionization enthalpy of these elements is nearly double that of the first ionization enthalpy, yet they exist as divalent ions (M²⁺) in most of their compounds.

NCERT Solutions for S Block Elements Class 11 Chemistry

Oxidation State Explanation:

1. The divalent ions (M²⁺) of alkaline earth metals have stable noble gas configurations. Thus, M²⁺ ion is more stable than M+ ion.

M ([Noble gas] ns²) → M²⁺ [Noble gas] + 2e

2. Due to greater charge and smaller size, the divalent cations lead to the formation of very stable lattices, and hence, a huge amount of energy is released. The high lattice enthalpy easily compensates for the high second ionization enthalpy.

3. Divalent cations for their smaller size get hydrated in water to a greater extent and the energy thus released (hydration enthalpy) is large enough to compensate for the second ionization enthalpy

The ΔH_i(3) values of alkaline earth metals are very high because the electron now has to be removed from the stable noble gas configuration. For this reason, the alkaline earth metals do not exhibit an oxidation state of more than +2.

6. Melting & boiling points

Alkaline earth metals have higher melting & boiling points than that of alkali metals. However, on moving down the group, no regular trend is observed.

Melting & boiling points Explanation: 

• Due to their smaller size, the atoms of alkaline earth metals form a more close-packed crystal lattice. Moreover, alkaline earth metals have two electrons in their valence shell whereas alkali metals have only one.
• The larger number of valence electrons leads to the formation of stronger metallic bonds.
• No regular trend in melting and boiling point is observed down the group because the atoms adopt different crystal structures.

7. Nature of bonds formed:

Like alkali metals, alkaline earth metals predominantly form ionic compounds. However, these are less ionic than the corresponding alkali metal compounds. Beryllium, the first member of this group, is an exception as its compounds are covalent. Magnesium also tends to form covalent compounds to some extent. On moving down the group, the tendency to form ionic compounds increases.

Nature of bonds formed Explanation:

Alkaline earth metals form ionic compounds because they have low ionization enthalpies. Their compounds, however, are less ionic because their ionization enthalpies are higher than those of the corres¬ ponding alkali metals. Due to its much smaller size and much higher ionization enthalpy, beryllium forms compounds that are predominantly covalent. Down the group, the tendency to form ionic compounds increases because ionization enthalpy decreases.

8. Density and hardness

The alkaline earth metals are denser and harder than the corresponding alkali metals. However, on moving down the group, no regular trend is observed. It initially decreases from Be to Ca and then increases from Ca to Ba.

Density and hardness Explanation:

The extent of cohesive energy determines the density and hardness of metals and this depends on the number of electrons involved in metallic bonding and the size of the atom. In alkali metals, one electron per atom (the valence electron) is involved in metallic bonding while in alkaline earth metals, two electrons per atom (the valence electrons) are involved. Moreover, the atoms of alkaline earth metals are heavier and smaller in size.

Therefore, the extent of cohesive energy is relatively higher in the case of alkaline earth metals & consequently, the atoms in alkaline earth metals are packed more closely in their lattices. Cohesive energy decreases from Be to Ca due to a gradual increase in size while it is found to increase from Ca to Ba due to the formation of different crystal lattices.

9. Conductivity

The Gr-2 metals are good conductors of heat and electricity.

Conductivity Explanation:

Due to the presence of two loosely bound valence electrons (per atom) which can move freely throughout the crystal lattice, the alkaline earth metals are good conductors of heat and electricity.

10. Flame coloration

When the alkaline earth metals and their salts, except beryllium and magnesium, are heated in the flame of a bunsen burner, they impart characteristic color to the flame.

These colors are as follows:

• Ca: Brickred
• Ba: Apple green
• Sr & Ra: Crimson red

Flame coloration Explanation:

• When the alkaline earth metals or their salts are put into a flame, the electrons of their valence shell absorb energy and get excited to higher energy levels.
• When they drop back to the ground state, the absorbed energy is emitted in the form of visible light having characteristic wavelengths.
• Depending upon the wavelength of light emitted, different colors are imparted to the burner flame.
• Due to their smaller size, the valence electrons in Be and Mg are too strongly bound to get excited by the energy available from the flame. Therefore, they do not impart any color to the flame.

Alkaline earth metals (except Be and Mg) can easily be identified by flame test in qualitative analysis. Further, they can be estimated by flame photometry or atomic absorption spectroscopy.

11. Magnetic property

The alkaline earth metals and their salts are diamagnetic.

Magnetic property Explanation:

Since the divalent ions (M²⁺) of alkaline earth metals have noble gas configurations with no unpaired electrons, their salts are diamagnetic. The metals are also diamagnetic as all the orbitals are filled up with paired electrons.

Chemical Properties Of Alkaline Earth Metals (Group-2 Metals)

Due to their low ionization enthalpies and high electropositive character, alkaline earth metals have a strong tendency to lose their valence electrons. Therefore, they are highly reactive and do not exist in the free state in nature.

1. Reducing nature

The alkaline earth metals are strong reducing agents. However, they are weaker reducing agents than alkali metals. Again, like alkali metals, their reducing strength increases down the group.

Reducing nature Explanation:

The alkaline earth metals except Be, have a fairly strong tendency to lose two valence electrons to form dipositive ions (M→ M²+  2e) i.e. they possess low ionization enthalpies and hence, they are strong reducing agents.

This is indicated by their high negative values of reduction potentials (E°). Their reducing strength, however, is less than the alkali metals as their atomization enthalpies and ionization enthalpies are relatively higher. Reducing strength increases on moving down the group as their ionization enthalpies decrease & electrode potentials become progressively more negative from Be to Ba.

2. Action of air

• Being fairly reactive, the alkaline earth metals are oxidized by the oxygen of the air and get tarnished due to the formation of a fine layer of oxide on their surface. With increasing atomic numbers, the effect of air on the metals gradually increases.
• Be and Mg being less reactive are not much affected by air. Ca and Sr get easily tarnished in air while Ba readily burns when exposed to air. Hence, Ca, Ba, and Sr are usually stored in paraffin.

3. Reaction with oxygen

Alkaline earth metals burn in oxygen to form oxides. Be, Mg, and Ca form monoxides while Sr and Ba form peroxides when they react with oxygen. This is because larger cation stabilizes a larger anion and hence the tendency to form peroxide increases as the size of the metal ion increases

 (M = Be, Mg or Ca)

 (M = Ba, Sr)

4. Reaction with water

• Alkaline earth metals except beryllium react with water to form the corresponding hydroxides along with the liberation of H₂ gas.
• Beryllium having the lowest negative standard electrodepotential (E° of Be²⁺/Be = -1.97V) among all

The group-2 metals is the least electropositive and hence, do not react with water or steam even at red hot conditions.

Ca, Sr, and Ba have relatively higher negative standard electrode potentials similar to those of the corresponding Gr-1 metals and hence, react even with cold water.

Mg has an intermediate value of E° and does not react with cold water but decomposes in boiling water.

M + 2H₂O→M(OH)₂ + H₂↑ (M = Mg, Ca; Sr or Ba)

Thus, the reactivity of the alkaline earth metals towards water increases on moving down the group. However, they are less reactive towards water as compared to the corresponding alkali metals.

5. Reaction with nitrogen

1. All alkaline earth metals burn in nitrogen to form nitrides of the type M3N2. However, Li forms Li3N.

3M + N₂ →M₂N₂ (M = Be, Mg, Ca, Sr and Ba)

2.  The ease of formation of nitrides decreases from Be to Ba. Since N₂ molecule is very stable, it requires very high energy to form nitride ions (N^3-). This large amount of energy is supplied from the lattice enthalpy evolved when crystalline solids containing ions with high charges (M²⁺ and N^3-) are formed.

Be₃N₂ is volatile because it is covalent. Other nitrides of this group are not volatile as they are ionic crystalline solids.

6. Reaction with halogens

The alkaline earth metals directly combine with halogens at higher temperatures to form halides having the general formula, MX₂

Halides can also be obtained by the action of halogen acids on metals, their oxides, hydroxides, or carbonates

M + 2HX→ MX₂ + H₂; MO + 2HX→MX₂ + H₂O

M(OH)₂ + 2HX→MX₂ + 2H₂O

MCO₃ + 2HX→MX₂ + CO₂ + H₂O

BeCl₂ is, however, conveniently prepared by heating BeO with Cl₂ in the presence of charcoal at 1073K

7. Reaction with hydrogen:

All the elements of group-2 except Be, form metal hydrides of the general formula MH₂ when heated with hydrogen

Beryllium hydride can be prepared indirectly by reducing beryllium chloride with lithium aluminum hydride.

2BeCl₂ + LiAlH₄→ 2BeH₂ + LiCl + AlCl₃

Both beryllium hydride (BeH₂) and magnesium hydride (MgH₂) are covalent compounds. In these molecules, both Be and Mg have four electrons in their valence shell. Therefore, these molecules are electron deficient. To make up for their electron deficiency, these two compounds exist as polymers, (BeH₂)_n and (MgH₂)_n in which each Be or Mg -atom forms four three-centre two-electron (3c-2e) bonds or hydrogen bridge bonds or banana bonds.

The structure of polymeric beryllium hydride is shown below:

CaH₂, SrH_2, and BaH₂ are ionic compounds in which a hydride ion (H^–) exists as an anion. Calcium hydride (CaH₂ ) which is also called hydrolith is used for the production of H₂ by the action of HaO on it.

8. Reaction with carbon

When the alkaline earth metals except for Be, are heated with carbon in an electric furnace or when their oxides are heated with carbon, carbides of the type MC₂ are obtained. These carbides are also called acetylides (containing discrete C₂ ions) as on hydrolysis they form acetylene.

(M = Mg, Ca, Sr or Ba)

At much higher temperatures ( ~ 1700°C), beryllium reacts with carbon to form Be₂C. This carbide is called methanide (containing discrete C^4- ion) as on hydrolysis it produces methane. On heating, MgC₂ forms Mg₂C₃, which is called allylide (containing discrete C₃^4- ion) as hydrolysis yields allylene (methyl acetylene).

CaC₂ + 2H₂O → HC≡ CH + Ca(OH)₂

Be₂C + 4H₂O →  2Be(OH)₂ + CH₄

Mg₂C₃+ 4H₂O →  CH₃C ≡ CH + 2Mg(OH)₂

When calcium carbide (CaC₂), an important chemical intermediate, is heated in an electric furnace with atmospheric nitrogen at 1375K, it produces calcium cyanamide (CaNCN).

The mixture of CaNCN and carbon is called nitrolim. It is used as a slow-acting nitrogen fertilizer as it undergoes very slow hydrolysis and evolves NH3 gas for a long period.

CaC₂ + 3H₂O → CaCO₃ + 2NH₃

9. Reaction with acids

The alkaline earth metals react with dilute acids to form the corresponding salt with the liberation of H₂ gas.

M + H₂SO₄ → MSO₄ + H₂↑ (M = Be, Mg, Ca, Sr or Ba)

Beryllium is the only group-2 metal which reacts with alkali to form H₂ and beryllate salt.

Be + 2NaOH + 2H₂O → Na₂ [Be(OH)₄] (Sodium beryllate)+ H₂ ↑

This is observed due to the diagonal relationship between aluminum and beryllium.

10. Solutions in liquid ammonia

Like alkali metals, alkaline earth metals dissolve in ammonia to give deep blue-colored solutions containing ammoniated cations and ammoniated electrons.

M + (x+ 2y)NH₃ → [M(NH₃)_x]²⁺ + 2[e(NH₃)_y ]^–

Evaporation of ammonia from these solutions leads to the formation of hexammoniates M(NH₃)₆ which slowly decompose to yield the corresponding metal amides, M(NH₂)₂ and H₂

11. Tendency to form complexes

The group-2 elements tend to form stable complexes and it is found to be greater than that of alkali metals because their ions have smaller size and higher charge. The tendency to form complexes decreases down the group and this is due to the decrease in ion-dipole interaction with increasing size of the metal ion. Be and Mg have the maximum tendency to form complexes.

Examples of two stable complexes of Be and Mg are \(\left[\mathrm{BeF}_4\right]^{2-}\&\left[\mathrm{Mg}\left(\mathrm{NH}_3\right)_6\right]^{2+}\) respectively

1. Complexation of Ca²⁺ by EDTA and polyphosphates plays an important role in the removal of the metal in water softening.
2.  In chlorophyll, the complex formed by the combination of Mg and the tetrapyrrole system (porphyrin) is very crucial in photosynthesis.

12. Extraction of alkaline earth metals

Like alkali metals, alkaline earth metals are also very reactive and strong reducing agents. So they cannot be extracted by ordinary chemical reduction methods. These metals also cannot be prepared by electrolysis of aqueous solutions of their salts because in that case, hydrogen is discharged at the cathode instead of the metal which has a much higher discharge potential. However, electrolysis can be carried out using a Hg-cathode, but in that case, recovery of the metal from amalgam becomes difficult. These metals are best isolated by electrolysis of their fused salts, usually chlorides.

S Block Elements Class 11 Important Topics

General Characteristics Of The Compounds Of Alkaline Earth Metals

Compounds of group-2 elements are predominantly ionic but are less ionic than the corresponding compounds of group 1 elements and this is due to their increased nuclear charge and smaller size. The general characteristics of some of the compounds of alkaline earth metals are discussed below.

1. Oxides of alkaline earth metals

1. Crystal structure:

Except for BeO (covalent solid), the oxides of the remaining alkaline earth metals are crystalline ionic solids and possess a rock-salt (NaCl) structure with coordination number 6. BeO though covalent, is an extremely hard solid because of its polymeric nature. BeO possesses a covalent lattice with coordination number 4. Both BeO and MgO have several properties that make them useful as refractory materials (for lining furnaces).

These properties are:

• They have high melting points (BeO-2500°C and MgO – 2800°C),
• They have very low vapor pressures,
• They are good conductors of heat, O they are chemically inert and
• They can act as electrical insulators.

2. Stability:

Due to much higher lattice enthalpies, the oxides are very stable towards heat. The lattice enthalpies decrease with an increase in the size of the metal ion.

3. Basic character:

Beryllium oxide, BeO reacts with both acids and alkalis, i.e., it is amphoteric while the oxides of other group-2 metals are basic.

BeO + 2HCl→BeCl₂ + H₂O

BeO + 2NaOH → Na₂ BeO₂ (Sodium beryllate) + H₂O

The basic strength of the oxides increases on moving down the group.

BeO (Amphoteric) < MgO (Weakly basic) < CaO (Basic ) < SrO, BaO( Strongly basic)

4. Reaction with water:

All these oxides except BeO and MgO, react with water to form sparingly soluble hydroxides. These reactions are exothermic.

MO + H₂O→M(OH)₂ + heat, M = Ca, Sr or Ba

2. Hydroxides

1. Basic character:

All the alkaline earth metal hydroxides are basic except Be(OH)₂ which is amphotericin nature. Their basic strength increases on moving from Be(OH)₂ to Ba(OH)₂

The alkaline earth metal hydroxides are, however, less basic than the alkali metal hydroxides.

Basic character Explanation:

Due to low ionization enthalpies of the alkaline earth metals, the M — O bond present in their hydroxides is weak and breaks up easily to give OH^– ions. For this reason, their hydroxides exhibit basic character. On moving down the group, the tendency of the M — OH bond to break heterolytically increases because ionisation enthalpies decrease and consequently, the basic character of the hydroxides increases.

Due to larger ionic sizes and lower ionization enthalpies of alkali metals, the M — OH bonds in their hydroxides are still weaker than those in alkaline earth metal hydroxides. Thus, the alkali metal hydroxides are more basic than the alkaline earth metal hydroxides.

2. Solubility in water:

Hydroxides of alkaline earth metals are less soluble in water than the hydroxides of alkali metals. Again, the solubility of these hydroxides increases markedly on moving down the group

Solubility in water Explanation:

On moving down the group, both the lattice enthalpy and the hydration enthalpy decrease with increasing ionic size. However, the lattice enthalpy decreases more rapidly than the hydration enthalpy, and consequently, their solubility increases down the group.

3. Thermal stability:

The alkaline earth metal hydroxides decompose on heating to give the metal oxide and water.

Thermal stability Explanation:

Thermal stability of these hydroxides increases down the group as the polarising power of the M2+ ion and the lattice enthalpy of the oxide formed decreases with increasing ionic size down the group.

3. Halides

• Due to the high polarising power of the Be²⁺ ion, beryllium halides have a covalent nature having low melting points.
• All other alkaline earth metal halides are ionic and their ionic character increases as the size of the metal ion (M²⁺) increases down the group. These ionic halides are non-volatile solids having high melting points.
• Due to its covalent nature, beryllium halides are sparingly soluble in water but readily soluble in organic solvents. The halides of other group-2 alkaline earth metals are readily soluble in water.
• Except for BeCl₂, all other anhydrous halides of the alkaline earth metals are hygroscopic in nature and form hydrates.
• For example: MgCl₂-6H₂O, CaCl₂-6H₂O, SrCl₂-2H₂O and BaCl₂-2H₂O
• The tendency to form hydrate decreases down the group. Thus, anhydrous calcium chloride is used as a dehydrating agent in the laboratory.
• The dehydration of the hydrated chlorides, bromides, and iodides of Ca, Sr, and Ba can be achieved by heating. However, the corresponding hydrated halides of Be and Mg on heating suffer hydrolysis.
• BeF₂ is highly soluble in water due to the much higher hydration enthalpy of the very small Be²⁺ ion.
• All other fluorides (MgF₂ > CaF₂, SrF_2, and BaF₂ ) are almost insoluble in water because their lattice enthalpies are higher than their hydration enthalpies.
• Except for BeCl₂ and MgCl₂, all other alkaline earth metal chlorides impart characteristic color to a flame.
• For example: CaCl₂: is brick red, SrCl₂:  Is crimson red, BaCl₂: Is grassy green, etc.

Structure of BeCl₂ In the solid state, beryllium chloride has a polymeric chain structure with chlorine bridges as given below:

Which are bonded by two covalent bonds while the other two by coordinate bonds. The Be -atoms in (BeCl₂)n sp³ -hybridized.

In the vapor phase, BeCl₂ exists as a chlorine-bridged dimer which dissociates into linear triatomic monomer at about 1200K. In the dimer, Be is sp² -hybridized while in the monomeric is sp² -hybridized

CaF₂ an industrially important compound, is the main source of both F₂ and HF

CaF₂ + H₂SO₄→ 2HF + CaSO₄

• CaF₂ is also used for making prisms and cell windows for spectrophotometers, an important instrument used in the spectroscopic analysis of compounds.
• In cold countries, CaCl₂ is Used for treating ice on roads, because 30% eutectic mixture of CaCl₂ +H₂O freezes at  – 55%C.

4. Salts of Oxoacids

Trends in the properties of some salts of group-2 elements are discussed below

1. Carbonates:

Solubility in water:

The carbonates of the alkaline earth metals are practically insoluble in water. Their solubilities decrease on moving down the group. BeCO₃ is sparingly soluble in water while BaCO₃ is insoluble in water.

Solubility in water Explanation:

On moving down the group, lattice enthalpy of carbonates remains almost unchanged” (the size of the metal ion is much smaller compared to CO₃^2-  ion) but hydration enthalpies of cations (M²⁺) decrease. Consequently, the solubilities of carbonates decrease down the group.

The extremely low solubility of alkaline earth metal carbonates in water is very important in the precipitation of Ba²⁺, Sr^2+, and Ca²⁺ ions as their carbonates, in Gr-IV qualitative analysis of basic radicals.

Thermal stability: Carbonates of gr.-2 metals decompose on heating to give metal oxide and carbon dioxide

Thermal stability of the carbonates increases down the group with increasing cationic size

Thermal stability Explanation:

The trend can be explained in terms of the stability of the monomeric is sp –hybridized. of the resulting metal oxides. With increasing stability of  Cl the metal oxide, the carbonate becomes more unstable BeCl₂ [monomer] towards heat. The stability of metal oxides decreases down the group due to a decrease in lattice enthalpy with increasing cationic (M²⁺) size. Hence, the stability of the carbonates towards heat increases down the group.

2. Bicarbonates:

The bicarbonates of alkaline earth metals do not exist in the solid state but are found in solutions. When such solutions are heated, bicarbonates decompose to form carbonates with the evolution of CO₂

3. Sulfates

1. Solubilityin water:

The sulfates of alkaline earth metals are relatively less soluble in water than the corresponding sulfates of alkali metals. Further, their solubilities decrease down the group. For example, BeSO₄ and MgSO₄ are more soluble in water, CaSO₄ is less soluble in water and SrSO₄, BaSO_4, and RaSO₄ are practically insoluble in water.

Solubilityin water Explanation:

On moving down the group, the hydration enthalpies decrease with increasing cationic size but the lattice enthalpy remains almost unchanged because the anion, SO₄ is much larger as compared to the metal ions (M²⁺). For this reason, the solubilities of sulfates decrease down the group.

2. Thermal stability: The alkaline earth metal sulfates white solids which dissociate on heating to give the metal oxides and sulfur trioxide.

The thermal stability of the sulfates increases down the group due to an increase in ionic character. This is due to a decrease in the polarising power of the metal ions with increasing ionic size. It becomes evident from the dissociation temperatures of the sulfates given below

4. Nitrates

Nitrates of alkaline earth metals are prepared by heating corresponding metal carbonate with dilute HNO₃

MCO₃  + 2HNO₃→ M(NO₃) + H₂ O + CO₂ (M = Be, Mg, Ca, Sr or Ba)

Magnesium nitrate crystallizes as Mg(NO₃)₂. 6H₂O whereas barium nitrate crystallizes as an anhydrous salt. This again shows that the tendency to form hydrates decreases with increasing ionic size and decreasing hydration enthalpy as we move down the group. Upon heating, all nitrates decompose to give the corresponding oxides with the evolution of NO₂ and O₂.

The nitrates of all these metals are soluble in water. Beryllium is unusual for the fact that it forms a basic nitrate in addition to the normal salt.

1.

2.

Anomalous Behaviour Of Be And Similarities Between Be And Al

Beryllium, the first member of group 2, shows some anomalous behavior, i.e.,it differs from the rest of the members of its family.

Reasons for anomalous behavior of beryllium:

• The extremely small size of the Be atom and Be²⁺ ion,
• Much higher polarising power of Be²⁺ ion,
• Higher ionization enthalpy and electronegativity as compared to the other members, absence of vacant d -d-orbitals in its valence shell.
• Again, beryllium resembles its diagonally placed element aluminum, the second typical element of group 13 and period 3, in several properties

Difference between beryllium and other alkaline earth metals

Reasons for the similarities between beryllium and aluminum:

• They have approximately the same polarising power. The polarising power of Be²⁺ is 0.064, while that of Al³⁺is 0.060.0
• The standard electrode potentials of Be and A1 are much closer i.e. Be²⁺/Be = -1.85V and Al³⁺+/Al = -1.66V.0
• The electronegativity of both the elements i.e., beryllium and aluminum are the same (1.5in the Pauling scale).

Similarities between beryllium and aluminium:

Class 11 Chemistry Chapter 10 S Block Elements Overview

Uses Of Alkaline Earth Metals

• and different milk products.

CBSE Class 11 Chemistry Notes For Chapter 11 Some P Block Elements Group-13 Elements (Boron Family) Introduction

The valence shell electronic configuration of the elements of group-13 is ns²np¹ where n = 2-6.  It becomes clear from the electronic configurations that boron (B) and aluminium (Al) have noble gas cores, gallium (Ga) and indium (In) have noble gas cores plus 10 d-electrons and thallium (Tl) have noble gas cores plus 14 F -electrons and 10 d-electrons. Thus electronic configuration of the elements of group 13 is more complex compared to those of groups and 2.

This difference in electronic configuration affects the chemistry of the elements of this group.

Electronic configuration of group 13 – elements:

Occurrence Of Group-13 Elements

1. The elements present in group 13 of the periodic table are boron (B), aluminium (Al), gallium (Ga), indium (In) and thallium (Tl). Except for horon, which is a non-metal, all other elements of this group are metals. The non-metallic character of boron is due to its small atomic size, high ionization enthalpy and comparatively high electronegativity.

2. Boron is a fairly rare element which occurs to a very small extent (0.0001% by mass) in the earth’s crust Natural boron consists of two isotopes: ¹⁰B (19%) and ¹¹B (81%). Boron does not occur in a free state as it is highly reactive. It occurs mainly as orthoboric acid & as minerals like

Borax –  Na₂(B₄O₅(OH₄).8H₂O -8H₂O

Kernite – Na₂(B₄O₅(OH₄).2H₂O

Colemanite – Na₂(B₃O₄(OH₃)₂.2H₂O

3. Aluminium is the most abundant metal, and the third most abundant element (8.3%by mass) in the earth’s crust after oxygen (45.5%) and silicon (27.7%). The important minerals of aluminium are:

Bauxite – (Al₂O₃ – 2H₂O)

Cryolite -(Na₃AlF₆)

Orthoclase (feldspar) – KAlSi₃O₈, 

Mica (Muscovite)-  KAl ₂(AlSi₃O₁₀)(F, OH)₂ etc.

4. Gallium, indium and thallium are quite less abundant and occur in traces in sulphide minerals.

5. The highest concentration of Ga (0.1-1%) is found in a rare mineral known as germanite (a sulphide complex of Zn, Cu, Ge and As).

6. Traces of and Tl are available in sulphide ores of zinc and lead respectively

CBSE Class 11 Chemistry Notes Chapter 11 Some P Block Elements

General Trends In Atomic And Physical Properties Of Group-13 Elements

Some important atomic and physical properties of group-13 most elements are given in the following table

Some atomic and physical properties of group-13 elements:

Trends in different atomic and physical properties of group-13 elements with explanations:

Atomic and ionic radii

1. Atomic and ionic radii of group-13 elements are smaller as compared to the corresponding elements of group-2.

Atomic and ionic radii Explanation 1:

On moving from left to right in the periodic table, i.e., on moving from group-2 to group-13in a given period, the magnitude of nuclear charge increases but the new electron is added to the same shell. Since the electrons in the same shell do not screen each other and the effective nuclear charge increases, the outermost electrons experience greater nuclear charge and are pulled more strongly towards the nucleus. As a result, atomic size decreases. The same is true in the case of ionic radius.

2. On moving down the group, both atomic and ionic radii are expected to increase due to the addition of new electronic shells. However, the observed atomic radius of Ga = 135pm is slightly lesser than that of Al = 143pm.

Atomic and ionic radii Explanation 2:

On moving from Al (Z = 13) to Ga (Z = 31), the d-orbitals are filled by electrons. Since the d-orbitals are larger, these intervening electrons in d-orbitals do not screen the nucleus effectively. As a result, the effective nuclear charge experienced by the electrons in Ga is greater than that experienced by the electrons in Al. Hence, the atomic radius of Ga is slightly less than that of Al. The Ionic radii, however, increase regularly on moving down the group

Ionisation enthalpy

1. First ionisation enthalpies (Δ_iH₁) of group-13 elements are lower than the corresponding elements ofgroup-2.

Ionisation enthalpy Explanation 1:

• The first electron, in the case of group-13 elements (ns²np¹), is to be removed from a p-orbital, while in the case of group-2 elements, it is to be removed from an s-orbital.
• Due to greater penetration of the s-orbital, the s-electron is nearer to the nucleus and is more tightly held by the nucleus than a p-electron of the same principal shell.
• The removal of the s-electron requires a greater amount of energy compared to p- the electron and because of this, the values of first ionisation enthalpies (Δ_iH_i) of the elements of group-13 are low as compared to the corresponding elements of group-2.
• The second and third ionisation enthalpies of these elements are, however, quite high because the second and third electrons are to be removed from ns-orbital.

2. On moving down the group from B to Al, the first ionisation enthalpy, (Aÿ) decreases sharply. However, the value of (Δ_iH₁) of Ga is slightly higher than that ofAl, while that ofTl is much higher than that of.

Ionisation enthalpy Explanation 2:

1. The sharp decrease in (Δ_iH₁) value from B to Al is expected because an increase in atomic size and screening effect (caused by to addition of a new shell) outweighs the effect of ‘increased nuclear charge.

2. The element Ga has ten 3d-electrons which do not screen as  N, much as s- and p-electrons. Therefore, due to poor shielding of 3d-electrons, the effective nuclear charge acting on Ga is slightly higher than that on Al.  Due to this, the (Δ_iH₁) value of Ga is slightly higher than that of Al, even though a new shell has been added on going from Al to Ga.

3. The same explanation can be offered on going from Into Tl. Tl has fourteen 4 f-electrons having a very poor screening effect and because of this, there occurs an unexpected increase in tyre effective nuclear charge, for which (Δ_iH₁) of Tl becomes much higher than that of In.

4. The order of (Δ_iH₁) values of group-13 elements is B > Al < Ga > In < Tl. However, this trend is not observed in the (Δ_iH₂) and (Δ_iH₃) values of these elements and this is because once the outermost p-electron is removed, it is not easy to remove the second and third electrons due to a large increase in effective nuclear charge. As expected, the first three ionisation enthalpies of these elements follow the order: Δ_iH₁< Δ_iH₂ < Δ_iH₃

Oxidation states

The atoms of group-13 elements have three valence electrons, two in the s -s-subshell and one in the p -subshell. Therefore, it becomes clear from their electronic configurations that +3 is expected to be the most common oxidation state of these elements. Therefore, the group oxidation state of the group-13 element is +3.

Due to the small size of the boron, the sum of its first three ionisation enthalpies is very high. Therefore, it cannot lose its valence electrons to form B³⁺ ion rather it forms covalent bonds with other atoms, for example, BH₃ or B₂H₆.

The sum of the first three ionisation enthalpies of Al is much lower than that of B so, it can form an Al³⁺ ion, for example, AlCl₃. Al is a highly electropositive metal.

B and Al exhibit only a +3 oxidation state but Ga, In and Tl show a +3 as well as a +1 oxidation state and on moving down the group the stability of the +3 oxidation state decreases while that of the +1 oxidation state progressively increases.

The stability of the +1 oxidation state follows the order:

Al < Ga <In < Tl. In the case of Tl, the +1 oxidation state is very much more stable than the +3 oxidation state.

Oxidation states Explanation:

The stability of the +1 oxidation state that increases down the group can be explained in terms of the inert pair effect. On moving down the group, the tendency of electrons of the valence shell to participate in bond formation decreases due to poor shielding of these electrons from the attraction of the nucleus by the intervening d- and f- electrons.

This reluctance or inertness ofthe s-electrons to participate in bond formation is called the inert pair effect. Since the magnitude of this effect increases down the group, the +1 oxidation state becomes more and more stable down the group as compared to the +3 oxidation state. The inert pair effect is maximum in the case of Tl and therefore, it shows mainly a +1 oxidation state. Due to lesser stability, Tl3+ salts act as strong oxidising agents. This is evident from its electrode potential data:

Tl³⁺(aq) + 2e → Tl+(aq);E° = +1.25V

The inert pair effect may also be explained by the fact that as the size of the atom increases from Al to Tl, the energy required to unpair the ns² -electrons is not compensated by the energy released due to the formation of two additional bonds.

Electropositive and metallic character

The elements of group- 13 are less electropositive or metallic as compared to the elements of group 2. On moving down the group, the electropositive character of the elements first increases from B to Al and then decreases from Al to Tl.

Electropositive and metallic Explanation:

1. Elements of group-13 are smaller in size and the sum of the three ionisation enthalpies Δ_iH₁+ Δ_iH₂ + Δ_iH₃ needed to form M³⁺ ions is much higher than the sum of two ionisation enthalpies, Δ_iH₁+ Δ_iH₂ < Δ_iH₃ for the corresponding bigger-sized elements belonging to alkaline earth metals needed to form M²⁺ ions. For this reason, the elements of group 13 are less electropositive than the elements of group 2.

2. Boron has the highest sum of the first three ionisation enthalpies among the elements of group 13. Because of this, it has very little tendency to lose electrons and hence it is the least electropositive among group-13 elements. It is a non-metal and a poor conductor of electricity.

3. On moving from B to Al, the sum of the first three ionisation enthalpies decreases considerably (6887 to 5137kJ-mol^-1 ) due to an increase in atomic size and hence, Al has a much higher tendency to lose electrons, i.e., Al is sufficiently electropositive. All is a metal and a good conductor of electricity.

4. Because of the increasingly poor shielding effect of 3d -electrons in Ga, 4d -electrons in and 4 f-electrons in Tl, the effective nuclear charge gradually increases and as a consequence, they exhibit lesser electropositive and metallic character.

Density

Because of smaller atomic and ionic radii, the elements of group 13 have a higher density as compared to the elements of group 2. On moving down the group, density increases.

Density Explanation:

On moving down the group, the density of these elements increases because the extent of the increase in atomic mass is greater than the extent of the increase in atomic size.0 On moving from B to Tl, both atomic mass and no. of electrons in the inner d- and f- subshell increases. Due to the lower shielding effect of d-and f— electrons, the effective nuclear charge increases from B to Tl. As a result, from B to Tl, the atomic size does not increase much.

P Block Elements Class 11 NCERT Notes

Melting and boiling points

Elements of group 13 do not show a regular trend in their melting points. The melting points decrease from B to Ga and then increase from Ga to Tl.

Melting and boiling points Explanation:

This irregular trend is probably due to unusual crystal structures of B and Ga. The much higher melting point of B is due to its giant covalent polymeric crystal structure consisting of icosahedral units with B-atoms at all 12 corners and each B-atom is bonded to five equidistant neighbours resulting in much stronger attractive forces. In contrast, Ga consists of discrete Ga₂ molecules so its melting point is exceptionally low (303K). However, the boiling points of these elements decrease regularly on moving down the group.

Gallium remains liquid over a vast range of temperatures and no other low-melting metal can compare with it. Molten Ga begins to boil only when heated to a temperature of 2276K. Due to this unusual property, gallium is used in thermometers required for measuring very high temperatures (>1000°C).

Electronegativity

Elements of group-13 are more electronegative than the elements of group-1 (alkali metals) and group-2 (alkaline earth metals). On moving down the group, the electronegativity first decreases from B to Al and then increases marginally.

Electronegativity Explanation:

•  Because of the smaller atomic size and higher nuclear charge, the electronegativities of group-13 elements are higher than the corresponding elements of group-1 and 2.0 On moving down the group from B to Al, the atomic size increases considerably and as a result, the attraction of the nucleus for the electrons decreases and hence the electronegativity decreases.
• On moving from Al to Tl, the atomic size increases but at the same time effective nuclear charge increases due to poor shielding of the inner d and f-electrons. As a result, the attractive force of the nucleus for the electrons increases and hence the electronegativity increases

Chemical Elements Properties Of Group-13

The members of group-13 elements have three electrons in their valence shells. Except for the last member Tl, all other members use these electrons to form three bonds and thus exhibit an oxidation state of +3. In the +3 oxidation state, the members of the boron family are expected to form covalent bonds for the following reasons:

• Small size and high charge (+3) cause high polarisation of the anions leading to the formation of covalent bonds.
• The large value of the sum of the first three ionisation enthalpies, Δ_iH₁ + Δ_iH₂ + Δ_iH₃ of these elements also suggests that the bonds will be largely covalent.
• The difference in electronegativity between the elements
  of group 13 and those ofthe higher groups is not very high.

This fact also agrees with the formation of covalent bonds. Because of its small size and high ionisation enthalpies, it is not possible for boron to form B3+ ions by losing its three valence electrons.

Therefore, 113 Boron does not form ionic compounds. It always forms covalent compounds by showing its valence electrons. The sum of the first three ionisation enthalpies, Δ_iH₁ + Δ_iH₂ + Δ_iH₃ is also higher but less than that of B. So A1 also has a strong tendency to form covalent compounds,

Example:  AlCl₃ , AlBr3 and AlCl₃. Like Al, compounds of the rest of the members such as GaCl₃, In Cl₃ etc. are covalent when anhydrous.

However, all the members except B form metal ions in solution. This change from covalent to ionic nature may be explained by the fact that in aqueous solutions these ions undergo hydration and the amount of hydration enthalpy exceeds the ionisation enthalpy.

Ga, In and TI show two oxidation states of +1 and +3 due to the inert pair effect. The compounds in the +1 oxidation state are more ionic than the compounds in the +3 oxidation state. In a trivalent state, the number of electrons in the valence shell of the central atom in a molecule of these elements is only six (two electrons less than the octet) and therefore, such electron-deficient molecules behave as Lewis acids. For example, BCl3 (Lewis acid) readily accepts an unshared pair of electrons from ammonia (Lewis base) to form the adduct, BCl₃-NH₃.

Reaction with dioxygen or air

1. All the members of group 13 react with dioxygen at higher temperatures to form trioxides of the general formula M₂O₃. T1 forms both T1203 and some amount of Tl₂O

The reaction of Al with 02 is known as a thermite reaction which is highly exothermic (AH0 =-1670kJ mol-1 ). A very strong affinity of Al for oxygen is used in the extraction of other metals from their oxides (thermite process). For example, Mn and Cr can be extracted from Mn₃O₄ and Cr₂O₃ respectively by this process.

2. The reactivity of group-13 elements towards dioxygen increases on moving down the group. Pure crystalline boron is almost unreactive towards air at ordinary temperature. Al does not react with dry air.

However, it gets tarnished readily in moist air even at ordinary temperatures due to the formation of a thin oxide (Al₂O₃) layer on the surface which prevents the metal
from further reaction. When amorphous boron and aluminium metal are heated in air, they form boron trioxide and aluminium trioxide (Al₂O₃) respectively.

Ga and In are not affected by air but T1 forms an oxide on
its surface in the presence of air.

3. B and Al react with dinitrogen at high temperatures to form
the corresponding nitrides.

Ga, In and T₁ do not react with N₂ to form the corresponding nitrides.

Boron nitride is a white slippery solid which melts under pressure at 3246K. It is chemically inert towards the air, oxygen, hydrogen, chlorine, etc. even on heating. The total number of valence electrons of one B and one N- -atom is equal to the number of valence electrons of two C-atoms.

Therefore, the structure of boron nitride is almost the same as that of graphite having a layer lattice. In each layer, alternate B and N-atoms (both sp2 -hybridised) form
a planar hexagon. The layers are stacked over one another in such a way that the N-atom of one layer is directly over the B-atom of another layer. Because of its structural similarity with graphite, boron nitride is also called inorganic graphite.

When boron nitride is heated at 1800°C under very high pressure, it gets converted to a cubic form comparable to diamond. This extremely hard variety known as borazon is used for cutting diamonds.

The acid-base character of oxides and hydroxides:

1. Trioxides of the elements of the boron family react with water to form their corresponding hydroxides.

M₂O₃ + 3H₂O→  2M(OH)₃

2. The nature of these oxides and hydroxides changes on moving down the group. Both B203 and B(OH)3 are weakly acidic. They dissolve in alkali to form metal borates.

B₂O₃ + 2NaOH → 2NaBO₂(sodium metaborate) + H₂O

B(OH)₃ + 3NaOH → Na₃BO₃(sodium borate) + 3H₂O

Aluminium oxide and hydroxide are amphoteric. Both of them dissolve in alkalies as well as acids.

1. Al₂O₃(s) + 3H₂SO₄(s)→ Al₂(SO₄)₃(aq) + 3H₂O(l)

2.

Similarly, Al(OH)₃(s) + NaOH(s)→Na[Al(OH)₄](aq)

Al(OH)₃(aq) + 3HCl(aq)→ AlCl₃(aq) + 3H₂O(l)

The oxide and hydroxide of Ga are also amphoteric while those of In and TI are basic.

Therefore, the basic character of oxides and hydroxides increases down the group

3. Thallium forms two types of hydroxides:

Thallic hydroxide [Tl(OH)3] and thallous hydroxide (TlOH). Al(OH)3 is insoluble in H20 but TlOH is soluble and is a
strong base like alkali metal hydroxides.

Thallium Explanation:

On moving down the group, the magnitude of ionisation enthalpy decreases. As a result, the strength of the M— O bond also decreases and therefore, its cleavage becomes progressively easier resulting in the increased basic strength down the group

An extremely hard crystalline form of aluminium oxide called corrundum is used as an abrasive. It can be made by heating amorphous aluminium oxide at about 2000K. Aluminium forms a series of mixed oxides with other j metals, some of them occurring naturally as semi-precious stones. These include ruby (Cr³⁺) and blue sapphire
(CO²⁺, Fe²⁺, Tl⁴⁺).

Reaction with hydrogen

1.  Group-13 elements form hydrides of the type MH₃. The members of the boron family do not combine directly with hydrogen. However, several hydrides are known
which can be prepared indirectly. Boron forms several stable covalent hydrides which are collectively called boranes.

The two most important types of boranes are as follows:

1. Boraness with general formula B_nH_n+4 are called nido-boranes 
   • Examples: Diborane (B₂H₆) pentaborane-9 (B₅H₉).
2. Boranes with the general formula B_nH_n+6 are called arachnoid-boranes
   • Examples:  Tetraborane (B₄H₁₀), and pentaborane-11 (B₅H₁₁).

2. The simplest hydride is diborane (B₂H₆) which is prepared by the reaction of BF₃ with lithium hydride industrially.

3. The other members of group 13 also form several hydrides which are polymeric,

Example:

(AlH₁)_n, (GaH₃)_n, (InH₃)_n–

The stability of these hydrides decreases down the group and thallium hydride is quite unstable. Boron, aluminium and gallium also form complex anionic hydrides such as NaBH₄ (sodium borohydride), LiAlH₄ (lithium aluminium hydride) and LiGaH₄ (lithium gallium hydride). These complex hydrides act as powerful reducing agents

4. The hydrides are weak Lewis acids and readily form adducts with strong Lewis bases to form compounds of the type MH₃:B (B = base).

For example, AlH₃:NMe₃, GaH₃:NMe₃ etc.

NMe₃ + AlH₃→[Me₃N:→AlH₃]

Reaction With Acids And Alkalies

The action of acids:

1. Boron remains inert in the presence of non-oxidising acids such as HCl. However, it undergoes oxidation by strong oxidising acids such as a mixture of hot concentrated H₂SO₄ and HNO₃ (2: 1) to form boric acid (H₃BO₃) at very high temperatures.

2.  The remaining elements of this group react with both oxidising and non-oxidising acids. For example, Al dissolves in dilute HCl and liberates dihydrogen.

2Al(s) + 6HCl(aq)→ 2Al3+(a<7) + 6Cr(a<jr) + 3H2(g)

3. Concentrated nitric acid renders aluminium passive by forming a protective layer of its oxide (AlO₃) on the surface ofthe metal. Thus aluminium vessels can be used
to store concentrated HNO₃.

2Al + 6HNO₃→Al₂O₃+ 6NO₂ + 2H₂O

Ga, In and TI react with dilute acids to liberate H₂

Action of alkalies:

1. When boron is fused with alkalies (NaOH or KOH) at a temperature greater than 775K, it forms borates and liberates dihydrogen

2. Boron dissolves in a fused mixture of Na₂CO₃ and NaNO₃ at 1123K to produce borate and nitrite salt and liberate carbon dioxide.

Al and Ga also react with aqueous alkalies with the evolution of dihydrogen

In and TI does not react with alkalis.

Reaction with halogens

Elements of group-13 react with halogens at high temperatures to produce trihalides ofthe general formula, MX₃. However, thallium (III) iodide does not exist.

Trihalides of boron:

Due to its small atomic size and high ionisation enthalpy, boron forms covalent trihalides

BX₃. BF₃ is a gas, BCl₃ and BBr₃ are liquids and BI3 is a solid. All these are trigonal planar molecules in which the central B -atom is sp² -hybridised. The three unpaired electrons of p -orbitals of three halogen atoms overlap with the three sp² -orbitals of boron to form three sp²-p, B—X cr -bonds. The unhybridised empty p-orbital remains perpendicular to the plane of the molecule.

Since there are only six electrons in the valence shell of the central boron atom in boron trihalides, they can accept two more electrons to acquire a stable octet, i.e., boron trihalides can behave as Lewis acids.

NCERT Solutions Class 11 Chemistry Chapter 11 P Block Elements

The Lewis acid character, however, decreases in the order:

BI₃ > BBr₃ > BCl₃ > BF₃

Explanation: This order of relative Lewis acid strength of boron trihalides, is just the reverse of what may be expected based on the electronegativities of the
halogen atoms can well be explained based on the tendency of the halogen atom to donate its lone pair of electrons to the boron atom through pn-pn back bonding.

Since the vacant 2p -orbital of B and the 2p-orbital of Fatom containing a lone pair of electrons are equal in size, therefore, the tendency of the F -atom to donate the unshared
pair by pπ-pπ back bonding is maximum. BF₃ can well be represented as a resonance hybrid of four resonating structures. As a result of resonance involving pn-pn back bonding, the electron density on the boron atom increases effectively and so its strength as a Lewis acid decreases considerably.

As the size ofthe halogen atom increases on going from Cl to I, the extent of overlap between the 2p -orbital of boron and a large p -orbital of halogen (3p of Cl, 4p of Br and 5p of I] decreases. As a consequence, the electron deficiency of boron increases and thus, the Lewis acid strength decreases on going from BF₃ to BI₃.

Halides of aluminium:

The halides of aluminium in the vapour state as well as in an inert solvent such as benzene exist ns dimers.

For example, Al₂Cl₃ exists as Al₂Cl₃.

Halides of aluminium Explanation:

In AlCl₃, there are six electrons (two electrons less than tyre octet) around the central Al-atom. In the dimeric structure, each Al completes its octet by accepting a lone
pair of electrons from the Cl-atom of another AlCl₃ molecule. The dimeric form exists lit vapour state at < 473K.

However, at higher temperatures, it dissociates to trigonal planar AlCl₃ molecule. In polar solvents such as water, the dimer dissociates and it is the high hydration enthalpy which helps this dissociation leading to the formation of Al³⁺ ion

Al₂Clg + 6H₂O ⇌ 2[Al(H₂O)6]³⁺(aq) + 6Cl^–(aq)

Therefore, anhydrous AlCl₃ is covalent but, hydrated aluminium chloride is ionic. Some important points of distinction between boron and
the other members of its family (especially the next member Al) are discussed in the given table. Anomalous properties of boron

Unlike aluminium halides, boron halides exist as monomers and this is because the boron atom is so small that it cannot accommodate four large-sized halogen atoms around it.

Boron

• Boron is the first member of group 13 of the periodic table. There are three electrons in its valence shell (ls²2s²2p¹).
• It exhibits anomalous behaviour and differs from the other members of its family. The reasons behind its exceptional behaviour can be attributed to the
• Exceptionally small atomic size as compared to other elements of its group,
• Much higher ionisation enthalpy and absence of d orbitals in its outermost or valence shell. Boron forms electron-deficient compounds which act as Lewis acids

Anomalous behaviour of boron:

Some important points of distinction between boron and the other members of its family (especially the next member Al) are discussed in the given table

Anomalous properties of boron:

1. Occurrence

Boron does not exist in a free state in nature. It is always found in the combined state as boric acid and borates. Boron occurs in two isotopic forms, ¹⁰B (19%) and ¹¹B (81%). Its abundance in the earth’s crust is very low (about 0.001%)

Important minerals of boron:

Boron may be obtained from the jets of steam which erupt from the volcano as boric acid and also from the water of the hot spring of Tuscany in small amounts as boric acid.

2. Properties of boron

Boron Physical properties:

1. Boron is an extremely hard solid (next to diamond) having a much higher melting point (2450K) and this is because of its three-dimensional network structure. Its boiling point is 3923 K.

2. Boron exists in two allotropic forms namely:

1. Amorphous,
2. Crystalline.

Crystalline boron is of three types:

1. α – Rhombohedral
2. β – Rhombohedral and
3. ϒ – Tetragonal.

The building units of all these forms are B₁₂ icosahedral units with 20 faces and boron atoms at all the 12 comers or vertices.

The melting and boiling point of boron is 2450K and 3923K respectively. The reason behind such high melting and boiling points is attributed to very strong attractive forces among the B₁₂ units as well as its closely packed stable crystal structure.

Chemical properties of crystalline boron

• Crystalline boron is chemically very inert. It is not oxidised even when heated with oxygen Crystalline boron is not attacked by HCl or HF. It is not affected by various oxidising acids such as shot and cone. HNO₃, H₂SO₄ etc.
• When it is fused with Na₂O₂ or Na₂CO₃ and KNO₃ at high temperatures, sodium borate (Na₃BO₃) is obtained.

Chemical properties of amorphous boron

1. Reaction with air:

When amorphous boron is heated in air at 700°C, it bums with a red flame and undergoes oxidation to form boron trioxide. Boron nitride is also formed by its reaction with N₂ gas of air

4B + 3O₂ → 2B₂O₃; 2B + N₂ → 2BN

2. Reaction with strong alkali:

When amorphous boron is fused with NaOH or KOH at a temperature greater than 773K, it forms borate salts and liberates H₂ gas

3. Reaction with oxidising adds:

Boron is not affected by non-oxidising acids such as hydrochloric add. However, it reacts with oxidising adds like cone. H₂SO₄ and HNO₃ form boric add.

3H₂SO₄ + 2B → 2H₂BO₂ + 3SO₂

6HNO₃ + 2B→2H₃BO₃ + 6NO₂

4. Reaction with halogens:

Boron bums in fluorine gas to form boron trifluoride. Boron reacts with chlorine at high temperatures to form boron trichloride

2B + 3F₂→ 2BF₃ ;2B + 3Cl₂→2BCl₃

An aqueous solution of BC13 is addicting because it undergoes hydrolysis to form a mixture of HCl and boric acid.

BCl₃+ 3H₂O →  3HCl + H₃BO₃

5. Reaction with metals (oxidising property):

The binary compounds of boron with elements having electronegativity lower than boron itself (For example: Metal) are called borides. When boron is heated with a metal at high temperature in an electric arc furnace, borides are obtained (B acts as an oxidant). Borides are hard, inert and have special properties

3Mg + 2B→ Mg₃B₂; 3Ca + 2B→ Ca₃ B₂

6. Reaction with carbon:

When boron is heated with carbon having comparable size and electronegativity at high temperatures in an electric arc furnace, very hard covalent boron carbide (B₄C) is obtained. It is even harder than diamonds. It is used as an abrasive. 4B + C→ B₄C It can also be prepared by reducing B₂O₃with coke at high temperature (2500°C) in an electric furnace

2B₃O₃ → B4C + 6CO↑

Reducing property: When boron is heated strongly with SiO₂ and CO₂, it reduces these oxides to give Si and C respectively.

2B₂O₃ + 7C →  B₄C + 6CO↑

7. Reducing property:

When boron is heated strongly with SiO₂ and CO₂, it reduces these oxides to give Si and C respectively.

8. Reaction with water: Red hot boron reduces steam to yield B₂O₃ and dihydrogen

Uses of boron

• Boron, an extremely hard refractory solid with a high melting point, low density and very low electrical conductivity, finds many applications which are as follows:
• Boron fibres having enormous tensile strength are used in making bullet-proof vests and as reinforcement materials in space shuttles and aircraft.
• Because of the high tendency of isotopes to absorb neutrons, metal borides are used in nuclear reactors as protective shields and control rods.
• It is used in the steel industry (instead of using expensive metals like Mo, Cr and W) for manufacturing special types of hard steel
• Its compounds such as borax and boric add are used for making heat-resistant glass (i.e., p a mild antiseptic.
• Boron compounds are used as rocket fuels because of their high energy/mass ratio.
• Boron carbide (B₄C) is used as an abrasive for polishing or grinding.
• Boron is used as a semiconductor for making electronic devices glass), glass-wool and gÿre glass
• An aqueous solution of orthoboric acid is used as

Some P Block Elements Chapter 11 NCERT Notes

Some Important Compounds Of Boron

1. Borax, Na₂B₄O₇.10H₂O or Na₂[B₄O₅(OH)₄] 8H₂O

Borax or sodium tetraborate decahydrate which occurs naturally as tincal (sugar) in certain dried-up lakes is the most important compound of boron. Borax contains the tetranuclear units

Therefore, its correct formula is Na₂[B₄O5(OH)₄]-8H₂O.

Borax Preparation

1. From tincal:

Naturally occurring borax or tincal, which contains about 50% borax is boiled with water, concentrated and then filtered to remove insoluble impurities. The filtrate is then concentrated and cooled when crystals of borax separate.

2. From colemanite:

Finely powdered mineral, colemanite (Ca₂ B₆O₁₁) is boiled with sodium carbonate solution and CaCO₃, Na₂B₄O₇ and NaBO₂ are obtained

Precipitate of CaCO₃ is filtered off and the filtrate is then concentrated and cooled to get the crystals of borax. A current of CO₂ is passed through the mother liquor when sodium metaborate presentient gets converted into borax

4NaBO₂ + CO₂→ Na₂B₄O₇(Borax) + Na₂CO₃

3. From boric acid:

Borax may also be obtained by neutralising boric acid with sodium carbonate. Crystals of Na₂B₄O₇.10H₂O separate on cooling.

4H₃BO₃ + Na₂CO₃ →  Na₂B₄O₇ + 6H₂O + CO₂↑

Borax Physical properties:

1. It is a white crystalline solid.
2. It is less soluble in cold water but more soluble in hot water.
3. When ordinary borax is recrystallised from water at a higher temperature (≈60°C), crystals of sodium tetraborate pentahydrate (Na₂B₄O₇.5H₂O) separate. This is called ‘goldsmith’s sugar.

Borax Chemical properties

1. Nature of aqueous solution:

The aqueous solution of borax is alkaline in nature and this is because borax undergoes hydrolysis to form the strong alkali, NaOH and the weak acid, boric acid. It acts as a buffer.

As the aqueous solution of borax is alkaline, it can be titrated against an acid using an orange indicator

Na₂B₄O₇ + 2HCl + 5H₂O→ 4H₃BO₃ + 2NaCl

When phenolphthalein is added to an aqueous solution of borax, the solution becomes pink in colour. However, when glycerol (a polyhydroxy compound) is added to the solution, it becomes colourless again.

Aqueous solution Explanation:

Since the aqueous solution of borax is alkaline in nature, it becomes pink when phenolphthalein is added to it. When glycerol is added to that solution, it combines with B(OH)₄ and removes it from the equilibrium by forming a stable chelate complex. As a consequence, the equilibrium shifts to the right making boric acid a strong acid. Because of the increased concentration of H⁺ ions, complete neutralisation of OH^– ions occurs and the solution becomes colourless again.

H₃BO₃ + H₂O  ⇌  H⁺ B(OH)₄^–; K_a = 6 × 10^-10

2. Reaction with caustic soda:

When a calculated amount of NaOH is added to borax, sodium metaborate is obtained.

Na₂B₄O₇ + 2NaOH→4NaBO₂ + H₂O

Reaction with sulphuric acid: When a calculated quantity of concentrated sulphuric acid is added to a hot concentrated solution of borax, boric acid is produced

3. Reaction with ethanol and sulphuric acid:

When borax is heated with ethanol and concentrated H₂SO₄, vapours of triethyl borate are formed which on ignition bum with a green-edged flame.

Na₂B₄O₇+ H₂SO₄ + 5H₂O →  Na₂SO₄ + 4H₃BO₃

H₃BO₃ + 3C₂H₅OH→B(OC₂H₅)₃ (Triethyl borate)+  3H₂O

This reaction is used as a test for the detection of borate ion (BO^3-₃) in qualitative analysis.

4. Action of heat:

When borax is heated strongly in the flame of a Bunsen burner, it loses its water of crystallisation and swells up to form a puffy mass. On further heating, the

mass turns into a transparent liquid which Nolldllics to form a bead that consists of sodium metaborate (NaBO₂) and boric anhydride (B₂O₃)

Preparation til boron from borax

A hot and concentrated solution of borax reacts with concentrated H₂SO₄ to form I boric acid (H₃BO₃). Boric added when heated tit high temperature, successively dissociates to form boron d ioxide, (B₂O₃). Boron trioxide when heated with Mg-powdor produces boron.

Na₂B₄O₇ + H₂SO₄+ 5H₂O→Na₂SO₄ + 4HaBO₃

Borax bead test

The borax bead test is very Important In a qualitative analysis for the detection of coloured metal ions like Cu²⁺, Ni²⁺, Co²⁺, Cr²⁺ etc.

• At first, a hot platinum loop is touched with borax and then heated in a Bunsen burner’s flame.
• Borax at first swells up and finally melts to form a colourless bead in the loop.
• The hot loop is touched with the salt under investigation and heated at first in the oxidising flame and then in the reducing flame.
• The metal ion is identified from the colour of the bead. This test is called the borax bead test.

Borax Bead Test:

Reactions:

1. Metallic compounds undergo decomposition on heating to form metallic oxides.

2M(NO₃)₂ → 2MO + 4NO₂ + O₂

2MSO₄ → 2MO + 2SO₂ + O₂

(M = Cu, Fe, Co, Ni, Mn, Cr)

2. The basic metallic oxides dissolve in the acidic diborane trioxide(BO) of the borax bead and form coloured metal metaborate salts

MO + BoO₃ → M(BO₂)₂

3. Copper Iron and other metallic salts form -ic metaborate in oxidising flame and -metaborates in reducing flame

Examples:

1. Reactions with copper salt:

In oxidising flame, cupric metaborate (blue) is formed

In reducing flame cuprous metaborate is formed

1.

2.

2. Reactions with iron salt:

In oxidising flame, ferric metaborate (yellow) is formed

In reducing flame ferrous metaborate (green) is formed

3. Reactions with cobalt salt: Both in oxidising and reducing flame, cobalt metaborate (blue) is formed

As the oxidation states of Co (+2) and Cr (+3) remain unchanged, the colour of the bead obtained from them is the same for both reducing and oxidising flame.

Uses of borax

Borax is used:

• In the manufacture of heat-resistant borosilicate (pyrex) glass,
• For preparing medicinal soaps,
• As a flux in soldering metals.
• In the candle industry as a stiffening agent,
• In softening water
• For the borax bead test,
• In the manufacture of perborate
• Na₂(OH)₂B(O — O)2B(OH)₂ 6H₂O, is an important cleansing and bleaching agent used in washing powders.

Orthoboric acid or boric acid, H₂BO₃o r B(OH)₃

The trivial name of orthoboric acid is boric acid

Orthoboric acid Preparation:

1. From colemanite:

Sulphur dioxide is passed through a hot concentrated solution of the mineral cole, Win’ll the resulting solution is concentrated and cooled, and crystals of boric add separate out. Calcium bisulphite being highly soluble In water remains dissolved In the mother liquor.

Ca₂ B₆O₁₁  (Colemanite)+ 4SO₂ + 11H₂ O →  2Ca(HSO₃)₂ (Calcium bisulphite)+ 6H₃BO₃ ( Boric acid)

2. From borax:

When a hot concentrated solution of borax is treated with hydrochloric acid or sulphuric acid, boric acid Is obtained. The resulting solution is concentrated and then cooled when crystals of boric acid separate out

Na₂B₄O₇ + 2HCl + 5H₂O →4H₃BO₃ + 2NaCI

Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + Na₂SO₄

3. From boron compounds by hydrolysis:

Certain boron compounds such as halides, hydrides and nitrides on boiling with water (hydrolysis) produce boric add.

BCI₂ + 3H₂O → H₃BO₃+ 3HCl

B₂H₆(dlborane) + 6H₂O→2H₃BO₃ + 6H₂↑

BN(boron nitride) + 3H₂O→ H₃BO₃ + NH₃

Orthoboric acid’s Physical properties

• It Is a white needle-like crystalline solid with a soft soapy touch.
• It Is sparingly soluble in cold water but highly soluble in hot water.
• It Is steam volatile.

Orthoboric acid Chemical properties

1. Addle nature:

It Is a very weak monobasic acid (K_a = 6 × 10^-10). It does not donate protons like most protonic acids. In fact, due to the small size of B and the presence of only six electrons Its valence shell B(OH)₃ behaves as a Lewis acid and accepts a pair of electrons from OH” Ion water thereby releasing a proton.

H(OH)₃ + 2H₂O ⇌  [B(OH)₄]^–+ H₃O⁺

B(OH)₃ behaves as a very weak acid (pK_a  = 9.2) because it only partially reacts with water to form [B(OH)₄]^–and H₃O⁺ Ions. So, B(OH)₃ or H₃BO₃ cannot be titrated satisfactorily with NaOH solution because no sharp end point Is obtained.

If some polyhydroxy compound such as glycerol, mannitol or catechol is added to the titration mixture, then boric acid behaves as a strong monobasic add.  This occurs due to the Removal of [B(OH)₄]^– Ion from the equilibrium mixture by the formation of a stable complex with the polyhydroxy compound.

It can then be titrated with NaOH solution and the endpoint can be detected using phenolphthalein as an indicator. ‘

2. Action of heats:

When orthoboric acid is heated, it loses molecules of water in three stages at different temperatures thus forming different products

3. Reaction with ethyl alcohol:

Boric acid reacts with ethyl alcohol In the presence of concentrated sulphuric acid to form triethyl borate

Vapours of triethyl borate burn with a green-edged flame. This test is used for detecting boric acid in qualitative analysis.

It is to be noted that this test can also be performed without using H₂SO₄ However, for detecting borate ions, the presence of H₂SO₄ is required. Therefore, boric acid and borate ions can be distinguished by this test.

4. Reaction with fluoride salt:

Boric acid reacts with fluoride salt in the presence of concentrated H₂SO₄ to form volatile boron trifluoride (BF₃). This compound burns with a green-edged flame.

2H₃BO₃ + 3CaF₂ + 3 H₂SO₄ → 3CaSO₄ + 2BF₃ + 6H₂O

5. Reaction with ammonium bi fluoride:

When boric acid is heated with ammonium bi fluoride, no residue is obtained because all the resulting compounds are gaseous

B₂O₃3(s) + 6NH₄BF₄(S)→  8BF₃(G) + 6NH₃(g) + 3H₂O(g)

6. Reaction with potassium fluoride:

When the aqueous solutions of two acidic compounds, boric acid and potassium fluoride (KHF2) are mixed, an alkaline solution is obtained due to the formation of potassium tetrafluoroborate (KBF₄) and potassium hydroxide (KOH).

B(OH)₃+ 2KHF₂→  KBF₄ + KOH + 2H₂O

Being a Lewis acid, B(OH)₃ has a strong tendency to combine with relatively smaller (F^–) ions to form fluoroborate ion (BF^–) and for this reason, this unbelievable reaction takes place.

Uses of boric acid

Boric acid is used

• As a mild antiseptic for washing eyes under the name Boric lotion
• In the manufacture of heat-resistant borosilicate glass,
• As a preservative for milk and foodstuffs,
• In the manufacture of enamels and glazes for pottery.

Structure Of boric acid:

The shining white crystals of boric acid contain B(OH)₃ units linked by H -bonds in infinite layers of nearly hexagonal symmetry. Since the adjacent layers in the boric acid crystal are held together with weak attractive forces, one layer can easily slide over the other and hence, boric acid is soft and slippery touch.

Diborane, B₂H₆

Boron hydrides are binary compounds of B and H. Although boron does not combine directly with hydrogen, several boron hydrides collectively called boranes, (in analogy with alkanes) are known. Depending upon their general formulae, these hydrides

Can be divided into several categories of which the following two are the most important:

1. Nido-boranes: General formula: \(\mathrm{B}_n \mathrm{H}_{n+4} \text {, example} \mathrm{B}_2 \mathrm{H}_6\) (diborane), B₅H₉ (pentaborane-9), B₆H₁₀ (hexaborane- 10), B₈H₁₂(octaborane-12),B₁₀H₁₄ (decaborane-14) etc. 
2. Arachno-boranes: General formula: \(\mathrm{B}_n \mathrm{H}_{n+6} \text {, example }\) B₅H₁₄ (pentaborane-11), BgH₁₂ (hexa-borane-12), BgH₁₄(octaborane-14) etc. The mostimportant hydride ofboron is diborane (B₂Hg).

Preparation of diborane

1. Laboratory preparation:

Diborane is prepared by the oxidation of sodium borohydride (NaBH₄) with I₂ in a diglyme solution.

Diglyme is a polyether whose formula is CH₃OCH₂CH₂OCH₂CH₂OCH₃

2. From boron trifluoride etherate:

It may be prepared by the reduction of boron trifluoride etherate with lithium aluminium hydride (LiAlH₄) in diethyl ether or sodium borohydride (NaBH4) in diglyme

3. Industrial preparation:

On an industrial scale, diborane is prepared by reducing BF₃ with LiH or NaH. 450K

Diborane Physical properties

Diborane is a colourless, foul-smelling, highly toxic gas having a boiling point of 180K.

Diborane Chemical properties

1.  Thermal stability:

It is stable only at low temperatures. When it is heated at 373-523K in a sealed tube, several higher boranes are obtained

However, by controlling the temperature, pressure and reaction time, various individual boranes can be prepared.

2. Combustibility:

When it is exposed to air, it spontaneously catches fire because of the strong affinity of boron towards oxygen. This reaction forming boric anhydride and water is highly exothermic.

B₂H₆ + 3O₂ →B₂O₃ + 3H₂O; ΔH = – 1976 kJ -mol^-1

The higher boranes also spontaneously in the air.

3. Hydrolysis:

It undergoes ready hydrolysis to produce boric acid.

B₂H₆(g) + 6H₂O(aq)→2H₃BO₃(aq) + 6H₂(g)↑

It reacts with methanol to form trimethyl borate.

B₂H₆ + 6CH₃OH → 2B(OCH₃)₃ + 6H₂↑

4. Reaction with Lewis bases:

When diborane is treated with Lewis base, it undergoes cleavage to form monoborane which then reacts with Lewis base to form an adduct.

B₂H₆ + 2NMe₃→ 2BH₃-NMe₃; B₂H₆ + 2CO → 2BH₃-CO

5. Reaction with ammonia:

When diborane is treated with ammonia, an additional compound is formed. The compound on further heating at about 473 K decomposes to give a volatile compound called borazine (or borazole).

Borazine is isosteric (i.e., the same number of atoms) and isoelectronic (i.e., the same number of electrons) with benzene and its structure is similar to that of benzene. Like benzene, all the atoms in borazine are sp² -hybridised. The n -n-bonding of borazine is dative and it arises due to sideways overlapping of filled p-orbitals of N and empty p-orbitals of B. Because of its similarity with benzene, borazine is also called inorganic benzene.

Class 11 Chemistry Chapter 11 Some P Block Elements Notes PDF

6. Formation of complex borohydrides:

Diborane reacts with several metal hydrides to form borohydrides containing tetrahedral [BH₄]_ ion.

Both sodium and lithium borohydrides are used as very good reducing agents in the synthesis of organic compounds in the laboratory. These two compounds may also be used as starting material for the preparation of other borohydride compounds.

7. Reaction with alkalis:

Diborane dissolves in strong alkalies such as NaOH or KOH solution to form metaborates and H₂ gas.

B₂H₆ + 2KOH + 2H₂O→ 2KBO₂ + 6H₂(Potassium Metaborate)↑

8. Reaction with halogen acids:

Diborane reacts with halogen acids to form halodiborancs and hydrogen gas. The order of reactivity of halogen acids is: HI > HBr > HCl

9. Reaction with halogens:

Diborane reacts with halogens to form corresponding halodiboranes. The order of reactivity of halogens is Cl₂ > Br₂ > I₂. Thus, chlorine reacts with diborane explosively at room temperature, bromine reacts rapidly at 373 K but iodine reacts slowly at higher temperatures

Uses of diborane:

1. Diborane is used in the preparation of several borohydrides such as LiBH₄ NaBH₄, etc.
2. It is used as a reducing agent in organic synthesis.
3. It is also used as a fuel for supersonic rockets.

Structure Of diborane, B₂H₆:

The structure and bonding of diborane seem to be very interesting. In the excited state, the B atom has the electronic configuration 2s¹2p_x¹ 2p_y¹ and therefore, it has only three electrons available for sharing. Now, 14 electrons (for six B—H and one B—B bond) are required if boron forms all conventional covalent bonds in ethane (C₂H₆)

But there are only 12 electrons (six from two B atoms and six from six H-atoms). Thus, the molecule is short of two electrons and its structure can not be similar to that of ethane (C₂H₆)

Based on electron diffraction study:

• Diborane has a bridged structure as given in There are two types of hydrogen atoms in this bridged structure. The two boron atoms and four terminal hydrogen atoms (shown by thick lines) lie in the same plane, while the remaining two hydrogen atoms (shown by dotted lines) lying above and below the plane form bridges and these are called bridge hydrogen atoms.
• The two B-H-B bridges lie in a plane which is nearly perpendicular to the plane containing the terminal B—H bonds.
• There are two bonds in the diborane molecule:
  1. The four terminal B—H bonds are normal covalent bonds, each Being formed by sharing a pair of electrons between boron and hydrogen atoms. These are quite strong bonds and called two-centre electron pair bonds or two-centre two-electron bonds (2c-2e bonds),
  2. The two bridge bonds B …. H……B are quite different from the normal electron pair bonds. Each bridge H-atom is bonded to two boron atoms by sharing only one pair of electrons.
• Such bridge bonds are called three centre electron pair bonds or three centres two-electron bonds (3c-2e bonds). Three-centre electron pair bonds or three-centre two-electron bonds are very weak bonds and are often called banana bonds as they resemble bananas in shape.
• Molecules like diborane (B₂H₆) which do not have a sufficient number of electrons to form normal covalent bonds are called electron-deficient molecules.

Based on hybridisation:

Boron atoms (excited state electronic configuration: 2s¹2p_x¹ 2p_y¹ in diborane undergo sp³ -hybridisation.

1. The two half-filled sp³ -hybrid orbitals of each boron atom overlap with the half-filled orbitals of hydrogen atoms to form normal covalent bonds.

2. The third half-filled hybrid orbital of one of the two boron atoms and the vacant orbital of the remaining boron atom overlap simultaneously with the half-filled Is -orbital of a hydrogen atom to form a B……H….B bridge bond

3. This bond involves three atoms (two boron atoms and one hydrogen atom) and contains only two electrons because one overlapping orbital of boron is empty. Hence, this B–‘H-‘-B bond is called three centre electron pair (3c-2e) bonds. Because of its typical shape resembling a banana, it is also called a banana bond

Aluminium

Aluminium, the second member of the boron family (group-13), is the most abundant metallic element in the earth’s crust. It is found in a variety of aluminosilicate compounds such as clay, mica and feldspar. The only ore of aluminium from which the metallic aluminium can be extracted profitably (in industry) is bauxite. Bauxite is hydrated aluminium oxide whose molecular formula is Al₂O₃-2H₂O

Aluminium Physical properties:

• Aluminium is a bluish-white metal with a brilliant lustre. But aluminium easily gets tarnished by the formation of a thin layer of oxide on the surface.
• It is a light metal whose density is 2.73g-cm^-1. Aluminium possesses high tensile strength, yet it is malleable and ductile.

Aluminium is a very good conductor of heat and electricity.

Aluminium Chemical properties:

It is not as reactive as its high negative electrode potential (E° = -1.66V) would imply and this is because there is a very thin layer of oxide on its surface.

1. Action of air:

Al remains unaffected in dry air but in the presence of moist air, a thin film of oxide is formed over its surface. Hence, the metallic of disappears. When burnt in oxygen it produces brilliant light.

4Al + 3O₂ → 2Al₂ O₃+ 772 kcal

The reaction is highly exothermic and the heat evolved is used for the reduction of oxides of Cr, Fe, Mn etc. (known as the thermite process).

2. Action of water: Aluminium decomposes boiling water thereby evolving hydrogen gas.

2Al + 6H₂O→2Al(OH)₃ + 3H₂↑

3. The action of non-metals:

Besides oxygen, aluminium reacts with other non-metals such as nitrogen, sulphur and halogens to form nitride, sulphide and halides respectively

4. Action of acids:

It dissolves both in dilute and concentrated hydrochloric acid and dilute sulphuric acid along with the evolution of hydrogen gas.

2Al + 6HCl→ 2AlCl₃ + 3H₂↑

2Al + 3H₂SO₄→Al₂(SO₄)₃ + 3H₃↑

The reaction with dilute sulphuric acid is very slow probably due to the insolubility of tyre oxide film over the metal in the acid. Hot and concentrated sulphuric acid dissolves aluminium with the evolution of sulphur dioxide (SO₂) gas.

2Al + 6H₂SO₄→Al₂(SO₄)₃ + 3SO₂ + 6H₂O

Dilute and concentrated nitric acid have no action on aluminium and this is due to the formation of an impenetrable oxide layer on its surface. Nitric acid may, therefore, be kept in the aluminium vessel.

5. Action of alkalis:

Aluminium dissolves in hot and cone. NaOH or KOH solutions form sodium aluminate with the evolution of hydrogen gas.

2Al + 2NaOH +2H₂O → 2NaAlO₂  (Sodium aluminate) (Soluble) +  3H₂ ↑

Aluminium reacts with hot and concentrated sodium carbonate (Na2C03) solution to form sodium aluminate, carbon dioxide and hydrogen.

2Al + Na₂CO₂ + 3H₂O  → 2NaAlO₂ (Sodium aluminate) (Soluble) + CO₂ + 3H₂↑

Uses of aluminium:

• Aluminium alloys (duralumin: Al, Mg, Cu and magnalium: Al, Mg) are light and strong and thus, are used in the construction of aircraft, ships and cars.
• It is a better conductor than copper and is used for making electric power cables.
• It is used for making doors, windows, building panels, mobile homes and household utensils.
• Finely divided Al powder is used in preparing aluminium paint and as an ingredient in solid fuels in rockets.
• Aluminium foils are used In wrapping soaps, cigarettes and confectioneries.
• Al is used to extract metals such as Cr, Mn etc., from their ores (thermite process).
• A mixture of ammonium nitrate and Al dust (ammonal) is used to make bombs and crackers.

CBSE Class 11 Chemistry Some P Block Elements Summary

CBSE Class 11 Chemistry Notes For Chapter 11 Some P Block Elements Group-14 Elements (Carbon Family)

1. The valence shell electronic configuration of the elements of group 14 is ns²np², where n = 2-6.
2. It becomes clear from these electronic configurations (given in the table below) that carbon and silicon have noble gas cores, germanium and tin have noble gas plus 10 d-electron cores and lead has noble gas cores in addition to 14/ and lOd -electron cores.
3. Thus, the electronic configurations of group-14 elements are similar to that of group-13 elements. However, they contain one more p-electron as compared to group-13 elements.

Electronic configurations of a group of  elements

Occurrence Of Group-14 Elements

1. The members of group 14 are carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb).

2. Carbon is the seventeenth most abundant element by mass in the earth’s crust It is widely distributed in nature in free as well as in combined states. In a free state, it occurs in coal, graphite and diamond. These are the main allotropes of carbon.

Carbon in the form of coal and coke is used mainly as fuel. In a combined state, it is present widely as metal carbonates, hydrocarbons (petroleum), carbohydrates and carbon dioxide (0.03%) in the air. Gases like propane and butane are the major constituents of LPG.

Moreover, the main constituent of all organic compounds is carbon. Two stable isotopes of carbon are present in nature namely ₆C¹² and ₆C¹³ Another isotope of carbon (₆C¹⁴) is radioactive. The age of antique articles is determined by the ratio of ₆C¹² and ₆C¹⁴ present in them. This process is called radiocarbon dating.

3. Silicon is the second (about 27.7% by mass) most abundant element (next to oxygen) in the earth’s crust and is present in the form of silica and silicates. Germanium occurs only in traces (1.5 ppm). Both germanium and silicon in very pure form find applications as semiconductors.

4. The natural abundance of tin and lead is very low (2 and 13 ppm respectively). The principal ore of tin is tinstone or cassiterite (Sn02) and that of lead is galena (PbS). Both tin and lead form several alloys. Tin is also used for tin plating while some lead-containing compounds are used as the constituents of paints.

5. The first two elements of this group, carbon and silicon are non-metals, germanium is a semi-metal (metalloid) while tin and lead are metals

General Trends In Atomic And Physical

Properties Of Group 14 Elements

Some important physical and chemical properties of group-14 elements are given in the following table. The trend in properties may be largely understood from their electronic configurations. The effect of inadequate shielding by d – and f- electrons is prominent in the case of Ge and Pb.

Some atomic and physical properties of group-14 elements

Trends in various atomic and physical properties of group -14 elements and their explanation

1. Atomic radii

1. The atomic radii of group-14 elements are smaller than the corresponding elements of group-13.

Atomic radii  Explanation 1:

On moving from a group-13 to a group-14 element within the same period, the magnitude of nuclear charge increases. As a result, the outermost electrons experience greater nuclear charge and are pulled more strongly towards the nucleus. Consequently, the atomic radius decreases.

2. On moving down the group, the atomic radii of the group- 14 elements increase regularly.

Atomic radii  Explanation 2:

The increase in atomic radii down the group is due to the addition of new electron shells. However, the increase in atomic radii from Si to Pb is small and this is because the effective nuclear charge somewhat increases due to ineffective shielding ofthe intervening d and /-electrons.

2. Ionisation enthalpy

1. First ionisation enthalpies of group-14 elements are higher than those of corresponding members of group-13.

Ionisation enthalpy Explanation 1:

This is because of the greater nuclear charge and smaller atomic size of group-14 elements as compared to the corresponding members of group-13.

2. In group 14, ionisation enthalpy decreases from C to Sn and then increases from Sn to Pb. The overall order is

Ionisation enthalpy Explanation 2:

Because of the increase in atomic size and the screening effect of inner electrons which outweigh the impact of increased nuclear charge, the ionisation enthalpies decrease down the group from C to Sn. However, a small increase in ionisation enthalpy from Sn to Pb is because the effect of increased nuclear charge (82-50 = 32 units), in this case, outweighs the inadequate shielding effect of intervening d and /-electrons.

3. Electropositive or metallic character

The group-14 elements are less electropositive and hence less metallic than the elements of group-13. Again, on moving down the group, the metallic character increases from C to Pb.

Electropositive Explanation:

The less electropositive character of group-14 elements is due to the smaller size of their atoms and higher ionisation enthalpies as compared to those of the corresponding group-13 elements. On moving down the group, the electropositive character increases as the size of the atom increases and ionisation enthalpy decreases.

4. Electronegativity

Because of the smaller size, the elements of group 14 are more electronegative than the corresponding elements of group 13. Values of electronegativity, however, do not decrease regularly down the group.

Carbon is the most electronegative element (2.5) while the remaining elements possess almost die same value (1.8 to 1.9) because of poor shielding by d -and f-electrons.

5. Melting and boiling points

The elements of group 14 have higher melting and boiling points as compared to the corresponding elements of group 13.

This is because the atoms of group-14 elements form a greater number of covalent bonds (four) with each other as compared to group-13 elements (which form only three bonds) and hence there exist strong binding forces between their atoms both in the solid as well as in the liquid states. Further, the melting and boiling points decrease down the group due to an increase in atomic size and consequent decrease in the interatomic forces of attraction. However, the melting point of Pb is higher than that of Sn.

6. Catenation

The self-inking property of the elements by which their atoms mutually combine to form long open chains (straight or branched) and rings of different sizes is called catenation. Carbon has the maximum tendency to catenate and because of this carbon forms a vast number of organic compounds. The catenation property of group- 14 elements follows the order: C>>Si > Ge = Sn. Lead does not exhibit the property of catenation.

Catenation Explanation:

The tendency of an element to undergo catenation increases with an increase in the strength of the M—M bond, where M represents an atom of the given element. As the size of the carbon atom is very small, the C—C bond is quite strong and the formation of this bond is thermodynamically very favourable.

However, on moving down the group, the atomic size increases and the strength of the bond decreases and consequently, the tendency to exhibit the property of catenation decreases

C —C (348 kj-mol^-1) > Si—Si  (297kJ-mol^-1) > Ge —Ge  (260kJ-mol^-1)> Sn—Sn (240kJ-mol^-1)

7. Oxidation state

1. Carbon and silicon exhibit a +4 oxidation state. Due to the inert pair effect, the other three elements ofthe group (Ge, Sn and Pb) show +2 and +4 oxidation states. This is due to the weak shielding effect of- and- electrons.

2. On moving down the group from Ge to Pb, the number of d- and f- electrons increases due to which the inert pair effect increases gradually. As a result, the stability of the +4 oxidation state decreases. The increasing order of the +2 oxidation state is Ge < Sn < Pb. Lead is most stable at +2 oxidation state.

3. The compounds in which the group-14 elements are in a +4 oxidation state, are covalent since the charge on the element is high and its size is small. However, those compounds in which the oxidation state is +2, are ionic, since the charge of the element is small and its size is large. For example, SnCl₂ is ionic whereas SnCl₄ is a covalent liquid. On moving down the group, the tendency of forming covalent compounds decreases and that of forming ionic compounds increases.

8. Multiple bonding

Carbon, the first member of the carbon family, forms multiple bonds with carbon (C=C, C ≡ C), oxygen (C=O) and nitrogen (C=N, C≡ N ) and this occurs because of comparable sizes and energies of overlapping orbitals (2p ). However, Si and other members possess p -orbitals having comparatively bigger sizes and higher energies (3/7, 4p, 5p etc.) as compared to that of carbon.

Therefore, sideways overlap is not effective and consequently, multiple bonding does not take place. However, in some compounds, silicon becomes involved in multiple bonds with oxygen and nitrogen atoms and these are called pn-in bonds.

Example:

This can be illustrated by the structure of trisilylamlne, N(SiH3)3. The central N-atom of this molecule is sp² – hybridised. It is, therefore, a trigonal planar molecule. A lone pair of electrons present in the 2p -orbital of the N-atom becomes involved in sideways overlap with the vacant 3d orbitals of Si leading to the formation of pπ-dπ back-bonding (shown by dotted lines). Since the availability of the unshared pair of electrons on N (engaged in pn-dn backbonding) is reduced, the compound becomes weakly basic.

The central N-atom of trimethylamine, N(CH₃)₃, on the other hand, is sp3 -hybridised. The three hybrid orbitals are involved in the formation of cr -bonds with three carbon atoms and the unshared electron pair exists in the fourth sp³ -hybrid orbital. Carbon does not possess any low-energy vacant d -d-orbital suitable for -bonding. Therefore, the shape of the molecule is pyramidal. Since the availability of the unshared electron pair on N is much higher, the molecule is strongly basic

NCERT Solutions for P Block Elements Class 11 Chemistry

Chemical Properties Of Carbon Family

1. Nature of compounds

1. The general valence shell electronic configuration of the elements of the carbon family is ns²np². This indicates that their atoms have four valence electrons.

2. Compounds of these elements are normallynotionicin nature as the formation of both M⁴⁺ and M^4- ions require a large amount of energy, thus making it unfavourable.

3. Compounds formed by the elements of this family are generally covalent, in which they complete their octets by sharing valence electrons with atoms of other elements.

4. Carbon and silicon exhibit a +4 oxidation state. The remaining three elements of this group, i.e., Ge, Sn and Pb, however, exhibit +4 as well as +2 oxidation states. This is due to the inert pair effect arising out of poor shielding of valence s -electrons by intervening d – or f-electrons.

5. As the number of d or f-electrons increases down the group from Ge to Pb, the inert pair effect becomes gradually more prominent. As a result, the stability of the +4 oxidation state decreases and the +2 oxidation state increases.

The stability of the +2 oxidation state follows the order:

1. Ge < Sn < Pb. Lead is most stable in the +2 oxidation state. vii] The compounds of group-14 elements in which these elements exhibit a +4 oxidation state are expected to be covalent because of their extremely high charge. However, the compound in which these elements exhibit an oxidation state of +2 is expected to be ionic because of its large size and a small charge

2. SnCl₂, for example, is anionic compound while SnCl4 is a covalent compound. Again, on moving down the group, the tendency of the elements to form covalent compounds decreases whereas the tendency to form ionic compounds increases

3. Since carbon has no d -d-orbital, it cannot expand its valence shell and hence its maximum covalency or coordination number is four. However, due to the availability of vacant d -d-orbitals, Si, Ge, Sn and Pb can form hexa-coordinated complexes by increasing their coordination number from 4 to 6. For example, [SiF₆]^2-, [GeF₆]^2-, [Sn(OH)₆]^2-, [PbCl₆]^2-etc

2. Trends in oxidising and reducing properties

1. Because of the inert pair effect, elements like Ge, Sn and Pb exhibit +2 and +4 oxidation states. Therefore, these elements in the +2 oxidation state act as reducing agents (M→ M⁴⁺+2e) while in the +4 oxidation state, they act as oxidation agents (M⁴⁺+2e →M²⁺).

2.  The +4 oxidation state of get is most stable followed by Sn and Pb. It thus follows that in group -14, Ge(2) salts are the strongest reducing agents followed by Sn(2) salts. However, germanium is much less abundant in nature than tin and therefore, Sn(II) salts such as SnCl₂ are largely used as reducing agents.

For example:

⇒  \(2 \mathrm{FeCl}_3+\mathrm{SnCl}_2\rightarrow 2 \mathrm{FeCl}_2+\mathrm{SnCl}_4 \)

⇒  \(2 \mathrm{HgCl}_2+\mathrm{SnCl}_2\rightarrow \mathrm{Hg}_2 \mathrm{Cl}_2+\mathrm{SnCl}_4 \)

⇒ \(\mathrm{Hg}_2 \mathrm{Cl}_2+\mathrm{SnCl}_2\rightarrow 2 \mathrm{Hg}+\mathrm{SnCl}_4\)

3. The +2 oxidation state of Pb is most stable followed by Sn and Ge. Therefore, Pb(IV) salts such as lead tetraacetate, Pb(OCOCH3)4 and Pb02 are largely used as oxidising agents.

For example:

Pb02 + 4HC1 (cone.)- PbCl2 + Cl2 + 2H20

Formation of oxides

When the elements of group 14 are heated in oxygen,

They form two types of oxides:

1. Monoxides of the formula MO and
2. Dioxides ofthe formula MO.

Lead also forms another oxide called trialled tetraoxide or red lead (Pb304) which can be obtained by heating litharge (PbO) more than O₂ or air at 673 K

The acid-base character of the oxides:

1.  The oxides of these elements in higher oxidation states are generally more acidic than those in lower oxidation states.

2. Again, the acidic character of the oxides decreases on moving down the group. The dioxides such as CO₂ and SiO₂ are acidic, GeO₂ is less acidic than SiO₂ while SnO₂ and PbO₂ are amphoteric. Being acidic, the dioxides of mC, Si and Ge react only with bases.

CO₂ + 2NaOH→Na₂CO₃+ H₂O

CO₂ + Ca(OH)₂→CaCO₃ + H₂O

SiO₂ + 2NaOH→Na₂SiO₃ + H₂O

GeO₂+ 2NaOH→Na₂GeO₃ + H₂O

3. Being amphoteric, the dioxides of Sn and Pb react with both acids and bases

SnO₂+ 2NaOH→Na₂SnO₃ (Sodium stannate)+ H₂O

SnO₂ + 4HCl→ SnCl₄ + 2H₂O

PbO₂ + 2NaOH→Na2PbO₃ (Sodium plumbate)+ H₂O

4. Among the monoxides of these elements, CO is neutral, GeO is acidic while SnO and PbO are amphoteric.

Oxidising and reducing the power of oxides:

1. Among the monoxides of group-14 elements, CO is the strongest reducing agent because the +4 oxidation state is the most stable for carbon. CO is used in the extraction of many metals from their oxides.

For example:

2. Among the dioxides of group-14 elements, Pb02 is the strongest oxidising agent because of the inert pair effect. It oxidises HC1 to Cl2 and reacts with concentrated H2S04 or HN03 to liberate dioxygen.

3. Formation of halides

The elements of group-14 combine with halogens to form halides of the formula MX4 and MX2 (X = F, Cl, Br or I). All the elements of this except carbon react directly with halogens under suitable conditions.

Tetrahalides ofgroup-14 elements:

1. All the elements of this group form tetrahalides of the formula MX4.

2. These tetrahalides are mostly covalent. The central atom of these halides is sp3 -hybridised and the molecules are tetrahedral shape. Thein two exceptions are SnF4 and PbF4 which are more.

3. IoThe ionic character and thermal stability of these halides decrease with an increase in n atomicnumbatomic number of halogen atoms. Example:

PbCl₄ is a stable compound but PbBr₄ is an unstable compound while Pbl4 is not known. Because of the strong oxidising power of Pb(+4) and the strong reducing power of I^–, Pbl₄ does not exist. The unstable nature of PbBr₄ is because Pb(+4) is strongly oxidising and Br is weakly reducing.In both the cases, the halide ion(I^– or Br^– ) reduces Pb⁴⁺ to Pb²⁺ (Pb⁴⁺ + 2X^–→Pb²⁺+ X₂)

Then on-existence of Pbl₄ can be explained alternatively as follows— The amount of energy released by the initial formation ofthe Pb—I bond is not sufficient to unpair 6s² electrons and excite one of them to 6p -orbital to have four unpaired electrons around the Pb-atom

4.  Carbon tetrachloride (CCl₄) does notundergo hydrolysis. However, the tetrahalides of other elements undergo ready hydrolysis. For example:

SiCl₄ + 4H₂O→Si(OH)₄ + 4HCl

SnCl₄+ 2H₂O→SnO₂ + 4HCI

Explanation:

That CCl₄ does not undergo hydrolysis can be explained by the fact that carbon having no vacant d -d-orbital cannot expand its coordination number beyond 4. However, silicon can undergo ready hydrolysis because it can expand its octet (i.e., coordination number beyond four) due to the availability ofvacant d -orbitals. reaction

CCl₄+ 2H₂O → Noreaction

SiCl₄ + 4H₂O → Si(OH)₄ (Silicic acid) + 4HCl

The reaction proceeds through the following steps:

1. In the first step, a lone pair of electrons from the oxygen atom of the H20 molecule is donated to an empty d -d-orbital of Si, forming a five-coordinate intermediate which has a trigonal bipyramidal shape.
2. In the second step, the intermediate loses a molecule of HCl and in this way, one Cl-atom of SiCl₄ is displaced by one OH group.
3. The remaining three Cl-atoms are displaced successively by OH groups yielding silicic acid.

5. Since carbon has no vacant d -orbitals in its valence shell, its tetrahalides do not form complexes. However, tetrahalides of other elements of this group form complexes due to the availability of vacant d -d-orbitals in their valence shells. These can, therefore, increase their coordination number up to six.

In other words, the tetrahalides of carbon do not act as Lewis acids but the tetrahalides of the other elements act as strong Lewis acids. The tetrahalides of Si, Ge, Sn and Pb, for example, can form hexahalo complexes like [SiF₆]^2-_, [GeF₆]^2-, [SnCl₆]^2-_ and[PbCl₆]^2- with corresponding halide ions.

SiF₄+ 2HF→H₂SiF₆ (hydrofluorosilicic acid) (or hexafluorosilicic acid)

SnCl₄ + 2Cl→SnCl^2-₆ [Hexachlorostannate (IV) ion]

Dihalldes of group-14 elements:

All the elements of group-14 except carbon and silicon form dihalides of the formula MX2. On moving down the group, the stability of these dihalides increases steadily due to the inert pair effect, i.e., the stability follows the order:

GeX₂<<SnX₂ < PbX₂

Anomalous Behaviour Of Carbon

Carbon, die first member of group 14, differs from the remaining members of its family in many properties.

Reasons for anomalous behaviour of carbon: 

1. Its atomic size is exceptionally small.
2. It has higher ionisation enthalpy and electronegativity.
3. It has no vacant dorbital in its valence shell.
4. It tends to form multiple bonds

Anomalous properties of carbon:

Allotropic Forms Of Carbon

The phenomenon of the existence of an element in two or more forms having different physical but similar chemical properties i? called allotropy and the different forms are called allotropes.

Carbon exists in several allotropic forms which may be classified as:

1. Crystalline and
2. Amorphous.

The four crystalline allotropic forms of carbon are:

1. Diamond
2. Graphite,
3. Fullerene
4. Carbon nanotubes.

The four amorphous allotropic forms of carbon are:

1. Charcoal
2. Soot or lamp black
3. Coke and
4. Gas carbon.

Amorphous carbon is not pure and remains mixed with various elements and compounds. Finer X-ray studies have shown that the amorphous varieties of carbon are composed of very minute crystalline units like graphite which are distributed throughout their masses in a most disordered fashion.

A synopsis of various allotropes of carbon is given in the following chart:

Crystalline Carbon Diamond

Diamond is a very precious substance which is availablein South Africa, New South Wales, Brazil, the Ural mountains and at Golconda in India. Two varieties of natural diamonds are available. One is the lustrous and colourless (or slightly coloured) variety which is generally used as precious gem¬ stones and the other is the black or deep-coloured opaque variety, known as the carbonado robot.

The weight of precious diamond is expressed in carats (1 carat = 200mg).

Some well-known diamonds are:

Cullinan (3032 carats), Kohinoor (present weight 106 carats), Pitt (136.25 carats), Regent (193 carats), Orloff (193 carats) and Great Mogul (186 carats). Diamond can also be prepared artificially but because of high cost and poor quality is seldom made artificially

Diamond Physical properties

1. Diamond is transparent, lustrous and crystalline. It may be colourless or slightly yellow coloured although some black or dark coloured varieties are also available.
2. It is the hardest naturally occurring substance known and it has a very high melting point (3843 K).
3. It is the heaviest among all the allotropic forms of carbon; its density is 3.51 gem^-3
4. Diamond has a very high refractive index (2.417) and thus light passing through it, suffers total internal reflection innumerable times. For this reason, diamonds appear to be extremely bright and lustrous.
5. It is transparent to X-rays and this property helps to distinguish a real diamond from an artificial one (made of glass).
6. Diamond is a non-conductor of electricity but good conductor of heat

Chemical properties:

At ordinary temperatures, diamond is chemically inert. It does not react with acids, alkalies, chlorine, potassium chlorate etc. However, it reacts with certain substances at much higher temperatures.

1. It is oxidised by oxygen at 800-900°C to produce pure carbon dioxide:

C + O₂ → CO₂

2. Diamond is converted into graphite in the absence ofair at much higher temperatures

Molten diamond can be converted into graphite by applying heat but graphite cannot be converted into diamond by heating to a very high temperature. The change is, therefore, unidirectional and this is because graphite is thermodynamically more stable than diamond. This type of allotropy is known as monotropy.

3. When a diamond is reacted with molten sodium carbonate, sodium monoxide and carbon monoxide are produced

C + Na₂CO₃ → Na₂O+2CO

4. It undergoes oxidation by fluorine at 700°C to form carbon tetrafluoride

C+ 2F₂→ CF₄

5. At about 250°C, diamond gets oxidised by a mixture of K₂Cr₂O₇ and concentrated sulphuric acid (i.e., chromic acid) to CO₂

6. Diamond reacts with sulphur vapour at 1000°C to form carbon disulphide: C + 2S→ CS₂

Structure of diamond:

1. In diamond, each C-atom is sp³ – hybridised and linked to four other C-atoms tetrahedrally by covalent bonds.

2. The value of each C—C—C bond angle is 109°28′ and each C —C bond distance is 1.54Å

3. An innumerable number of such tetrahedral units are linked together to form a three-dimensional giant molecule containing very strong bonds extended in all directions. Because of such a three-dimensional network of strong covalent bonds, diamond is extremely hard.

4. Since a huge amount of thermal energy is required to break a large number of strong covalent bonds, its melting point is very high.

5. All 4 valence electrons of each sp3 -hybridised C-atom in diamond crystal participate in forming covalent bonds and there is no free electron on any carbon atom. Thus, a diamond is an anon-conductor of electricity. vi] Diamond has the highest known thermal conductivity because its structure distributes thermal motion in three dimensions very effectively.

6. Unlike graphite in which the C-atoms are arranged in different distant layers, the C-atoms in diamond are placed at a covalent bond distance (1.54A). Because of this, the density of diamond is higher than that of graphite

Diamond Uses:

1. Because of its transparency, dazzling lustre and beauty, diamond is extensively used as precious gemstone.
2. Because of its extreme hardness, it is used for cutting glass, polishing hard surfaces and drilling purposes. Black or dark-coloured diamonds are generally used for this purpose.

Graphite

Graphite is available as minerals in Sri Lanka, Mexico, Italy, California (U.S.A), Siberia, Korea, Spain and India. The word graphite originates from the Greek word ‘graph’ which

Preparation of artificial graphite: Acheson process:

• In this process, coke dust mixed with silica is heated to a temperature of 3000-3500°C with the help of electrodes in an electric furnace made of fire bricks for 25-30 hours.
• The mixture is kept covered by sand.
• In the first stage of the reaction, silica reacts with carbon to form silicon carbide (SiC) and carbon monoxide (CO).
• Silicon carbide thus formed decomposes to yield graphite and silicon. Q
• At higher temperatures, silicon, on being vapourised, escapes from the furnace and graphite is left.

SiO₂ + 3C→ SiC + 2CO↑; SiC→Si + C [graphite]

Graphite Physical Properties 

1. Graphite is a dark greyish-coloured opaque, soft and slippery crystalline substance possessing metallic lustre.
2. It is lighter (density 2.25 g-cm-3 ) than diamond.
3. It is a good conductor of heat and electricity.

Graphite Chemical properties

1. Graphite is more reactive than diamond

2. When graphite is heated in air at 700°C, it is oxidised to carbon dioxide:

C + O→CO₂

3. At 500°C, fluorine reacts with graphite to produce carbon tetrafluoride (CF4). The compound is a non-conductor of electricity. It is also called graphite fluoride.

4. Graphite is not attacked by dilute acid or alkali. However, when it is subjected to react with molten sodium carbonate, carbon monoxide is formed.

5. Graphite is oxidised to carbon dioxide with a mixture of K₂Cr₂O. and concentrated H₂SO₄ (chromic acid).

6. When graphite is heated in the presence of a mixture of cones. nitric acid and sulphuric acid containing a small amount of potassium chlorate, greenish-yellow-coloured solid graphitic acid (CnH₄O₅) is obtained. Its exact structural formula is not known yet. Graphite can be identified by this test (diamond does not respond to this test).

7. On complete combustion, graphite produces mellitic acid
[C₆(COOH)₆].

Structure of graphite

1. Each carbon atom in graphite is sp2 -sp2-hybridised and is linked to three other carbon atoms directly in the same plane forming a network of planar hexagons and these two-dimensional layers exist in different parallel planes.

2. In each layer, the C—C bond length is 1.42Å and the distance between two adjacent layers is 3.35Åwhich is greater than the C—C covalent bond distance. So, the layers are supposed to be held together by relatively weak van der Waals forces of attraction

Some P Block Elements Class 11 Important Topics

3. As the distance between two parallel layers is sufficiently large, graphite is less dense than diamond.

4.  Since the layers are weakly held together, on application of pressure, one layer easily slides over the other. Thus, graphite is found to be soft and lubricating

5. In the formation of hexagons in a layer of graphite, only three of its four valence electrons are used to form three sigma bonds (Csp²-Csp²). The remaining electrons of each carbon atom present in an unhybridised p -p-orbital are utilised to form n -n-bonds.

6. The n -n-electrons are mobile and can move freely through the graphite crystal. Because of the presence of free mobile electrons, graphite is a good conductor of electricity and heat.

7. Of all the crystalline allotropes of carbon, graphite is thermodynamically the most stable one. Its standard enthalpy of formation (AfH°) is taken as zero.

Graphite Uses:

• Graphite is largely used for lining and making electrodes for electric furnaces.
• When mixed with oil and water, graphite is used as a lubricant in machinery.
• It finds use in making crucibles resistant to high temperatures.
• By mixing with desired quantities ofwax or clay, graphite is used for making cores of lead pencils
• Graphite is used as a moderator in nuclear reactors

Fullerenes

Fullerenes or Buckminsterfullerene (named after the famous American designer of the geodesic dome, Robert Buckminster Fuller) is the latest allotrope of carbon discovered in 1985 collectively by three scientists namely R. E. Smalley, R. F. Curl and H. W. Kroto. It is a crystalline allotrope of carbon in which the carbon atoms exist in a cluster form. It is also known to be the purest form of carbon because, unlike diamond and graphite does not have surface bonds that are to be attracted by other atoms.

Structure of fullerenes C₆₀:

Fullerenes are expressed by the general formula C_n, where n is an even number between 30-600, for example, C₆₀, C₇₀, C₈₀….etc.

All these are cage-like spheroidal molecules having polyhedral geometry containing pentagonal and hexagonal planes. Number of hexagons in a C„ molecule =(n/2- 10),

Example:

In fullerene C₆₀ molecule, number ofhexagons =(60/2-10) = 20. Structure of fullerene C₆₀; The C₆₀ molecule consists of twenty-six-membered rings and twelve five-membered rings of sp² -hybridised carbon atoms fused into each other

Each carbon atom forms three or -bonds with the other three carbon atoms and the remaining electron on each carbon is involved in the formation of the n -bond and as a result, the system is expected to be aromatic. However, it is not aromatic because the molecule is not planar and it does not have (4n + 2) electrons.

It is a non¬ aromatic system. This fusion pattern provides a marvellous symmetry to the structure in which the fused ring system bends around and closes to form a soccer ball-shaped molecule (“buckyball”). Of all the fullerenes, C60 is the most stable one.

Structure of fuUerene C₇₀:

The molecule acquires the shape ofa rugby ball. It consists of twelve five-membered rings and twenty-five six-membered rings and their arrangement is the same as that of a C₆₀ molecule.

C₇₀ Preparation:

• The preparation of fullerenes involves heating of graphite in an electric arc in the presence of an inert gas such as He or Ar.
• A sooty material is recovered which consists mainly of C₆₀ with a small amount of C₇₀ and traces of other with an even number of C atoms up to 350 and above.
• The C₆₀ and C₇₀ fullerenes can further be separated from the sooty material by extraction with benzene or toluene followed by chromatographic separation using alumina (Al₂O₃) as the adsorbent.
• In Russia, America, Canada and New Zealand C₆₀ and C₇₀ fullerenes are isolated from natural sources. Fullerenes of this type are formed by the red giant star Antares.

Properties and applications: 

• Fullerenes are solids with high melting points.
• Being covalent, they are soluble in organic solvents.
• They react with alkali metals to form solid compounds such as K₃C₆₀. This compound acts as a superconductor even at temperatures of the order of 10-40K.
• Because of their spherical shapes, they exhibit wonderful lubricating property

Carbon Nanotubes

Carbon nanotubes are crystalline allotropes of carbon with cylindrical nanostructure. This allotrope was discovered by Sumio Iijima (Japan) in 1991. A carbon nanotube consists of a two-dimensional array of hexagonal rings of carbon just as in a layer of graphite or a chicken wire.

The layer is then rolled. into a cylinder and capped at each end with half of a C₆₀  fullerene

Properties and applications: 

The nanotubes are approximately 50000 times thinner than a human hair. These are very tough, about 100 times as strong as steel. They are electrically conducting along the length of the tube.

These cylindrical carbon molecules having unusual properties are valuable for nanotechnology, electronics, optics and other fields of material science and technology. They are also being used as probe tips for the analysis of DNA and proteins by atomic force microscopy (AFM). Many other applications have been envisioned for them as well, including molecular-size test tubes or capsules for drug delivery

Amorphous Carbon Charcoal

Vegetable charcoal

1. Wood charcoal: When wood is subjected to destructive distillation in an iron retort, the volatile organic compounds present escape and the residue left in the retort is called wood charcoal.
2. Sugar charcoal: It can be prepared by heating pure sugar in a closed vessel or by eliminating water from sugar by reacting it with concentrated sulphuric acid

Animal charcoal

1. Bone charcoal:

• Small pieces of animal bones are treated with superheated steam to remove the adhering fat and marrow.
• The dried bones are then subjected to destructive distillation in an iron retort when volatile substances are distilled out and a black residue containing carbon and about 90% impurities like calcium phosphate and calcium carbonate are left behind.
• This is known as bone black or bone charcoal. These impurities are removed by dissolving the black material in dilute HCl.
• The insoluble deep black powder thus obtained is almost pure charcoal and is called ivory black.

2. Blood charcoal:

• Charcoal obtained by destructive distillation of blood is known as blood charcoal.

Charcoal Physical properties

Charcoal is black, soft and porous. It is a bad conductor of heat and electricity. Its specific gravity lies between 1.4 and 1.9. But because of its porosity, air enters in its pores.

• As a result, its specific gravity gets reduced to 0.2 and hence charcoal floats on water. When porous charcoal b freed from entrapped air, it can retain any other ga In its pores, This phenomenon is known as adsorption.
• The adsorbed gas escapes on heating and is much more reactive than the ordinary gas.

Activated charcoal:

Activated charcoal has high adsorption power as compared to ordinary charcoal.

Activated charcoal Preparation:

When the charcoal obtained by destructive distillation of coconut shell is heated to about 800-900°C in a limited supply of air or steam, activated charcoal is obtained.0 besides this, activated charcoal may also be obtained by the destructive distillation of sawdust soaked In aqueous solution of ZnCl₂or MgCl,

Activated charcoal Properties:

Activated charcoal is not only a good adsorbent for gases but it also has the power of decoloursing a coloured substance and absorbing the taste of a substance. Moreover, a catalyst accelerates the rates of many chemical reactions.

Charcoal Chemical properties

1. At higher temperatures, charcoal burns in air or oxygen to form C02 gas. However, in a limited supply of oxygen, it’s combustion produces carbon monoxide.

C +O₂ → CO₂; 2C + O→2CO

2. Charcoal forms different compounds with sulphur, nitrogen and hydrogen at high temperatures.

2C + N₂ → (CN) (Cyanogen)

2C + H₂ → C₂H₂ (Acetylene)

In the first reaction, two solids (C and S) react together to form a liquid (carbon disulphide

3. Charcoal combines with heated Ca, Al, Fe etc. to form their corresponding carbides.

Ca + 2C→CaC₂; 3Fe + C→Fe₃C; 4Al + 3C→Al₄C₃

4. When a mixture of silica (SiO₂) and coke dust is heated at 1500’2000°Cin an electric furnace, silicon carbide (SiC) is formed. It is a black, bright and very hard solid. It is known as carborundum which is used for polishing metals

SiO₂ +3C → SiC+2CO

5. When steam is passed over white-hot charcoal or coke, a mixture containing equal volumes of CO and H₂, called water gas is produced

6. Reducingproperty: Charcoal is a good reducing agent.

At higher temperatures, various metal oxides are reduced by charcoal to their corresponding metals.

CuO + C→ Cu + CO; PbO + C→ Pb + CO

Fe₂O₃ + 3C→2Fe + 3CO

At higher temperatures, charcoal reduces carbon dioxide to CO and sodium sulphate to sodium sulphide.

CO₂ + C→2CO;

⇒ \(\mathrm{Na}_2 \stackrel{+6}{\mathrm{~S}} \mathrm{O}_4+4 \stackrel{0}{\mathrm{C}} \rightarrow \mathrm{Na}_2S{ }^{-2}+4 \stackrel{+2}{\mathrm{~C}} \mathrm{O}\)

Charcoal in the burning condition is oxidised by concentrated nitric acid or sulphuric acid to CO₂

Charcoal Uses: 

• Wood charcoal is used as fuel and as a reducing agent in the extraction of metals.
• It is used in decolourising sugar syrup and refining oils, fats, glycerine etc.
• It is used in the treatment of drinking water as it adsorbs excess chlorine after chlorination.
• It is used in gas masks as it adsorbs poisonous gases.
• It is also used for preparing gunpowder and black paint (Black Japan).

Lamp Black or carbon black

When organic liquids rich in carbon such as kerosene, petrol, turpentine oil, benzene etc. are subjected to bum in controlled air, a black sooty smoke is produced. This smoke on condensation in a cold container forms soot. This soot is called lamp black or carbon black. It may also be obtained when natural gas (methane) is subjected to albumin-controlled air. It is the purest form of all the amorphous allotropes of carbon.

Lamp Black Properties:

It is amorphous, black and non-conductor of heat and electricity.

Lamp Black Uses:

It is used in the preparation of printing ink, shoe polish and black paints

Coke and gas carbon

When anthracite coal (96% carbon) is subjected to destructive distillation, the solid residue left in the iron retort is called coke. At higher temperatures (1000 -1200°C), hard coke is called coke. At higher temperatures (1000 -1200°C), hard coke is obtained whereas at 600-650°C, we get soft coke. The black hard dense residue deposited on the relatively cooler upper part Qf returns as gas carbon. it possesses thermal water gas is produced. and electrical conductivity and electrical conductivity.

Coke Uses:

• Hard coke is used as fuel and as a reducing agent for domestic fuel
• Gas carbon is widely used as electrodes in batteries, arc lights and during electrolysis.

All the allotropic modifications of carbon consist of the same element:

When a fixed weight of a pure allotropic form of carbon is heated in a long hard combustion tube in the presence of pure oxygen, CO₂ and CO are produced.

CuO, kept in the tube, converts CO into CO₂. CO₂ so obtained is absorbed in a previously weighed potash bulb attached to the exit end of the tube. An increase in the weight of the bulb increases the amount of CO₂ formed.

When the experiment is performed separately with the same fixed weight of any other allotrope, the same amount of CO₂ is obtained in each case. This experiment, thus, proves that the different allotropes consist ofthe same element, i.e., carbon.

Carbon Monoxide

Laboratory preparation:

1.  In the laboratory, carbon monoxide is prepared by dehydrating formic acid or oxalic acid after heating with concentrated sulphuric acid.

2. When potassium ferrocyanide is heated with an excess of cone, sulphuric acid, pure carbon monoxide is obtained

CO cannot be fired by concentrated sulphuric acid:

Concentrated sulphuric acid is a strong oxidising agent. Thus when CO ( a reducing agent )is passed through concentrated H₂SO₄, it is oxidised by sulphuric acid to CO₂

Other methods of preparation

From carbon:

When steam is passed over red hot coke, water gas or synthesis gas (CO + H₂) is obtained. CO is separated from the mixture by liquefaction.

When air is passed over hot coke, producer gas (CO + N₂) is formed. CO is separated by liquefaction

From carbon dioxide:

When CO₂ is passed over red hot carbon, zinc, Iron etc., it is reduced to CO.

CO₂ + C→2CO; CO₂ + Zn→CO + ZnO

CO₂+ Fe→FeO + CO

From metal oxides:

Carbon reduces die oxides of zinc, lead or iron to produce CO.

ZnO + C→ Zn + CO; Fe₂O₃+ 3C→2Fe + 3CO

From nickel tetracarbonyl:

Pure CO is obtained when nickel tetracarbonyl vapour is heated above 150°C

Carbon monoxide Physical properties:

1. Carbon monoxide is a colourless, tasteless, odourless gas which is lighter than air.

2. It is slightly soluble in water. It is a neutral oxide.

3. It is a highly poisonous gas. If only a volume of CO is present in 10,000 volumes of air, then that air is considered to be poisonous. Carbon monoxide molecule.

4.

with a lone pair of electrons on carbon combined with Fe-atom present in the haemoglobin of the blood to form a very stable complex compound named carboxyhaemoglobin.

5. Hb + CO → HbCO; (Hb =Haemoglobin) As CO is almost 100 times more rigidly bonded to Fe-atom than O₂, O₂ can no longer combine with haemoglobin.

6. In other words, haemoglobin fails to act as an oxygen-carrier. As a consequence, the body tissues become slackened due to lack of ofoxygen ultimately causing death

7. In case of CO poisoning, the patient should immediately be taken to an open area and artificial respiration with carbogen (a mixture of oxygen and 5-10% CO₂) should be started.

Carbon monoxide Chemical properties:

1. Combustion:

Carbon monoxide is itself a combustible gas but does not support combustion. It burns in the air with a blue flame and is oxidised to C02. Because of the evolution of a large amount of heat, CO is used as fuel.

2CO + O₂ → 2CO₂ + 135.2 kcal

The two important fuels containing carbon monoxide are water gas and producer gas. Water gas contains 50% of H₂, 40% of CO, 5% of CO₂ and 5% of CH₄ and N₂ while producer gas contains 25% of CO, 4% of CO₂,70% of N2 and traces of H₂, CH₄ and O₂

2. Reducing property:

Carbon monoxide is a powerful reducing agent. The oxidation number of carbonin CO is +2 and the highest oxidation number of carbon is +4. So, CO tends to be oxidised and behaves as a strong reducing agent. Various metal oxides are reduced by CO to the corresponding metal.

CuO + CO→Cu + CO₂; PbO + CO→Pb + CO₂

ZnO + CO→Zn + CO₂; Fe₂O₃ + 3CO→2Fe + 3CO₂

At 90°C, CO reduces iodine pentoxide (I₂O₅) to give violet-coloured iodine. This reaction is called the Ditte reaction.

I₂O₅ + 5CO→I₂ + 5CO₂

3. Reaction with sodium hydroxide:

Being a neutral oxide CO does not react with alkali or base under ordinary conditions. But at 200°C and under high pressure, it reacts with caustic soda solution to yield sodium formate.

4. Absorption of CO:

When CO is passed through an ammoniacal or acidified cuprous chloride solution, it gets absorbed in that solution to give a white crystalline addition compound as a precipitate. CO can be separated from a gas mixture by this process.

Cu₂Cl₂ + 2CO + 4H₂O→2[CuCl.CO. 2H₂O]↓

The addition compound evolves CO on heating.

5. Formation of addition compounds:

1. In the presence of sunlight, CO combines direedy with chlorine gas to form carbonyl chloride or phosgene gas. It is a colourless poisonous gas:

2. CO reacts with sulphur vapour to produce carbonyl sulphide.

3. Combines with many transition metals to form metal carbonyl compounds. For example, CO reacts with nickel powder at 30-40°C under ordinary pressure to form nickel tetracarbonyl. Again, at 200°C and 100 atmosphere pressure, CO reacts with freshly reduced iron to form pentacarbonyl.

Ni + 4CO→Ni(CO)₄; Fe + 5CO→Fe(CO)₅

4. Formation of organic compounds:

Hydrogen reacts with CO at 350°C in the presence of Ni or Pt catalyst to yield methane. If the reaction is carried out at 300°C and 200 atmospheric pressure in the presence of ZnO and Cr₂O₃ catalyst, methyl alcohol is produced. The oxidation number of carbon in CO decreases from +2 to -4 in methane and to -2 in methyl alcohol.

Therefore, in these two cases, CO exhibits its oxidising property.

⇒ \(\stackrel{+2}{\mathrm{CO}}+3 \mathrm{H}_2 \rightarrow \stackrel{-4}{\mathrm{CH}_4}+\mathrm{H}_2 \mathrm{O} ; \stackrel{+2}{\mathrm{CO}}+2 \mathrm{H}_2 \rightarrow \stackrel{-2}{\mathrm{CH}_3^{-}}\mathrm{OH}\)

Identification of carbon monoxide:

1. Carbon monoxide burns in air with a blue flame and the gaseous product turns lime water milky [H₂ also burns with a blue flame but in this case, steam is formed which turns white anhydrous copper sulphate blue.]

2. CO is completely absorbed by the Cu₂Cl₂ solution in a cone. hydrochloric acid or ammonium hydroxide and as a result, a white crystalline addition compound is precipitated.

3. When a filter paper soaked with a solution of platinum or palladium chloride is held in CO gas, the paper turns pink-green or black due to the reduction ofthe metal salts.

PtCl₂ + CO + H₂O→Pt + CO₂ ((pink-green))+ 2HCl

PdCl₂ + CO + H₂O→Pd + CO₂ ((black))+ 2HCl

4. When CO gas is passed through an ammoniacal AgN03 solution, the solution becomes brown

5.  When a dilute solution of blood shaken with CO, is subjected to spectroscopic analysis, the observed band in the spectrum indicates the presence of CO. The presence of traces of the air can be detected by this experiment.

6. The presence of a very small amount of CO in the air can be detected with the help of a halamite tube or colour detector tube. When air containing CO is introduced into this tube I₂ O₅ present in the tube reacts with CO to liberate I₂

Because of the violet colour of evolved I₂, colour of the tube changes and the presence of COin air is indicated

I₂O₅ + 5CO→I₂  (Ditte reaction)+ 5CO₂

Structure of carbon monoxide:

Both the carbon and the oxygen atoms in a CO molecule are sp -hybridised. One of the sp -hybrid orbital of each atom is used to form a C —O cr -bond while the other sp -orbital ofeach contains a lone pair of electrons. The two unhybridised 2p -orbitals of each atom are involved in the formation of two pn-pn bonds. In terms of resonance, the CO molecule can be best represented as a resonance hybrid of the following two I resonance structures(1 and 2 ).

The resonance structure (1) is relatively more stable because of the fulfilment of the octet of both atoms.

Uses Of carbon monoxide: 

• CO is used as fuel in the form of producer gas or water gas.
• It is used as a reducing agent in the extraction of metals.
• It is used for the preparation of pure nickel by Mond’s process.
• It is used for the
• Preparation of methanol, methane, formic acid and synthetic petrol (Fischer-Tropsch process).

Preparation of pure nickel:

Ni(CO)₄ is prepared by the reaction between impure nickel and carbon monoxide. Ni(CO)₄ is then allowed to decompose by heating to 1.50°C to get pure nickel.

Class 11 Chemistry Chapter 11 Some P Block Elements Overview

Carbon Dioxide

Carbon Dioxide Laboratory preparation:

At ordinary temperature, CO₂ is prepared in the laboratory by the action of dilute HCl on calcium carbonate (CaCO₃) or marble.

CaCO₃ + 2HCl→CaCl₂ + CO₂↑ + H₂O

The gas is collected in the gas jar by the upward displacement of air, as it is 1.5 times heavier than air. Carbon dioxide thus produced contains a small amount of HCl and water vapour. The gas is then passed successively through NaHCO₃  solution and cone, sulphuric acid to remove HCl vapour and water vapour respectively.

Dilute sulphuric acid cannot be used for the preparation of CO₂  from marble or limestone:

This is because sulphuric acid reacts with CaCO₃ to produce insoluble; CaSO₄  which forms a layer of CaCO₃. This insoluble layer prevents CaCO₃  from reacting with the acid and as a result, the evolution of CO₂ ceases within a very short time

CaCO₃ + H₂SO₄→CaSO₄+ CO₂ + H₂O

On the other hand, when dilute hydrochloric acid is, used, highly soluble calcium chloride (CaCl₂) is formed. So, the reaction proceeds without any interruption

CO₂ can be prepared by the action of dilute H₂SO₄ on Na₂CO₃:

The salt, Na₂SO₄ produced soluble in water or dilute H₂SO₄

Na₂CO₃ + H₂SO₄→Na2SO₄ + CO₂ + H₂O

At ordinary temperatures, CO₂ is highly soluble in water. Therefore, it is not collected by the downward displacement of water. The solubility of COz in hot water is very low and hence it can be collected over hot

Other methods of preparation:

1. From carbonate salts:

Except for alkali metal carbonates, all other carbonates undergo thermal decomposition to produce CO₂ and the oxides ofthe corresponding metals.

BaCO₃ decomposes only at very high temperatures.

Calcium carbonate or limestone is thermally decomposed (1000°C) for the preparation of carbon dioxide on a commercial scale.

2. From bicarbonate salts:

Bicarbonates of all the elements decompose on heating with the evolution of CO₂

3. From fermentation:

A large amount of CO₂ is obtained as a by-product during the manufacture of ethyl alcohol by fermentation of sugar

4. From water gas:

Water gas is industrially prepared by passing steam through a bed of white-hot coke at about 100°C. C + H₂O→CO + H₂. When a mixture of water gas and excess steam is passed over (Fe₂O₃+ Cr₂O₃) catalyst heated at 400°C, CO is oxidised to CO₂

(CO + H₂) + H₂O→CO₂ + 2H₂

The gaseous product is then passed through a solution of potassium carbonate when CO₂ is completely absorbed and KHC03 is formed. H2 and unconverted CO pass out. When the resulting KHC03 solution is boiled, CO₂ is obtained.

K₂CO₂+ CO₂ + H₂O→2KHCO₃

Carbon Dioxide Physical properties

1. Carbon dioxide is a colourless, odourless and tasteless gas having slightly acidic properties.

2. CO₂ is 1.5 times heavier than air. So, this gas often accumulates in abandoned wells or pits and because of this, severe breathing problems are caused in such places.

3. By the application of pressure (nearly 40 atmospheric pressure and a temperature < 40°C), CO₂ can be easily liquefied. When liquid CO₂ is allowed to vaporise rapidly by releasing the pressure, it further gets cooled down and freezes like ice. This is called dry ice or cardice.

4. When solid carbon dioxide is allowed to evaporate at atmospheric pressure, it gets converted into the vapour state without passing through the intermediate liquid state. Therefore, unlike ordinary ice, it does not wet the surface of the substance and because of this, it is called dry ice.

5. It is highly soluble in water (1.7 cm3 of CO₂ dissolves in 1 cm3 of water). The solubility increases with an increase in pressure. Aerated waters such as soda water, lemonade etc. contain CO₂ under pressure. When the cork of the bottle of aerated water is opened, the pressure is released and excess CO₂ escapes in the form of bubbles. Its solubility in water, however, decreases with a temperature rise.

Carbon Dioxide Chemical properties

1. Combustion:

Carbon dioxide is neither combustible nor helps in combustion. When it (heavier than air) falls on a binning substance, it removes air from the surface of the substance and thereby the substance can no longer remain in contact with air. As a result, the fire is extinguished. A burning jute stick when inserted into a jar of CO₂, extinguishes.

However, when a burning Mgribbon or metallic sodium is inserted into a CO₂ jar, it continues to bum with the separation of black carbon.

⇒ \(\stackrel{+4}{\mathrm{C}} \mathrm{O}_2+2 \stackrel{0}{\mathrm{Mg}} \rightarrow 2 \stackrel{+2}{\mathrm{MgO}}+\stackrel{0}{\mathrm{C}} ; \stackrel{+4}{\mathrm{C}} \mathrm{O}_2+4 \stackrel{0}{\mathrm{Na}} \rightarrow 2 \stackrel{+1}{\mathrm{Na}}_2 \mathrm{O}+\stackrel{0}{\mathrm{C}}\)

During the burning of such metals, the temperature, due to the liberation of a large amount of heat, is so high that CO₂ decomposes into carbon and O₂ and it is the oxygen which helps in the burning ofthe metals.

In these reactions, CO₂ acts as an oxidising agent and itself gets reduced to carbon. These reactions prove the existence of carbon in C02. It is to be noted that the oxidation number of carbon in CO₂ is +4 and this is its highest state of oxidation.

Thus, there is no possibility of an increase in its oxidation number, i.e., CO₂ cannot be further oxidised. That is why CO₂ cannot exhibit any reducing property. For the same basic reason, C02 is not combustible [CO, on the other hand, is combustible because in this case, the oxidation number of carbon may increase from +2 to +4 ].

2. Acidic property:

Carbon dioxide is an acidic oxide. It dissolves in water forming an unstable dibasic acid called carbonic acid (H₂CO₃). CO₂ is, therefore, regarded as the anhydride of carbonic acid.

CO₃+H₂O→H₂CO₃

H₂CO₃ is known only in solution and when the solution is heated, CO₂ is evolved out The solution turns blue litmus red but it cannot change the colour of methyl orange.

H₂CO₃ forms two types of salts, bicarbonates (HCO₃ ) and carbonates (CO^2-₃). Being an acidic oxide, CO₂ combines directly with strongly basic oxides such as CaO, Na₂O etc. to form their corresponding salts.

CaO + CO₂→CaCO₃; Na₂O + CO₂→Na₂CO₃

Reaction with alkali:

When CO₂ is passed through a strong alkaline solution of NaOH, a carbonate salt is first formed. If the passage of CO₂ is continued for a long time, white crystals of sparingly soluble sodium bicarbonate are precipitated. The bicarbonate salt decomposes on heating to form carbonate salt, CO₂ and water.

2NaOH + CO₂→Na₂CO₂ + H₂O

Na₂CO₃ + CO₂ + H₂O→ 2NaHCO₂

Rection with lime water:

When CO₂ is passed through lime water, the solution becomes milky due to the formation of white insoluble calcium carbonate. However, when excess CO₂ gas is passed through this milky solution, its milkiness disappears as insoluble calcium carbonate gets converted into soluble calcium bicarbonate

Ca(OH)₂ + CO₂→CaCO₃↓ (white) +H₂O

CaCO₃+ CO₂ + H₂O→Ca(HCO₃)₂ (soluble)

On heating, calcium bicarbonate decomposes to form calcium carbonate, CO₂ and water and as a result, the clear solution becomes milky again.

Ca(HCO₃)₂→CaCO₃↓ + CO₂ + H₂O

Manufacture of sodium carbonate:

When CO₂ gas is passed through a concentrated solution of sodium chloride (brine) saturated with ammonia at 30-40°C, white crystals of sodium bicarbonate are precipitated. The reaction occurs in two stages

NH₃ + CO₂ + H₂O ⇌ (NH₄)₂CO₃

CaSO₄ + (NH₄)₂CO₃→CaCO₃↓ + (NH₄)2SO₄

Sodium carbonate is prepared by thermal decomposition of sodium bicarbonate. The Solvay process for the manufacture of sodium carbonate is based on this reaction.

Production of ammonium sulphate:

This is carried out by passing CO, and NH₃ gases through a slurry of powdered gypsum (CaSO₄,2H₂O) in water. At first, NH₃ and CO₂ react together in the presence of water to form ammonium carbonate. It then reacts with calcium sulphate (gypsum) to form calcium carbonate and ammonium sulphate by double decomposition.

2NH₃ + CO₂ + H₂O ⇌  (NH₄)₂CO₃

CaSO₄ + (NH₄)₂CO₃→CaCO₃ + (NH₄)2SO₄

The nitrogenous fertiliser ammonium sulphate is manufactured by using this reaction. In this process, (NH₄)₂SO₄ is produced without using H₂SO₄

Production of urea:

At 200-210°C and 150 atm pressure, CO₂ reacts with ammonia to produce urea.

CO₂ + 2NH₃ ⇌(Ammonium carbamate) NH₄ COONH₂ ⇌(Urea) CO(NH₂ )₂ + H₂ O

The important fertiliser, urea is manufactured on a large scale by using this reaction.

Photosynthesis:

Plants absorb atmospheric carbon dioxide. In the presence of chlorophyll and sunlight, the absorbed CO₂ combines with water (absorbed from the soil) to form glucose, water and oxygen. This process is called photosynthesis. In this process, CO₂ is reduced to carbohydrates by water

Reduction of CO₂ :

When CO₂ is passed over heated C, Fe, Zn etc., it is reduced to CO

Identification of carbon dioxide:

• It extinguishes a burning stick.
• Lime water becomes turbid when CO₂ is passed through it. When excess CO₂ is passed through it, the turbidity disappears but when that clear solution is boiled, the turbidity reappears.

N₂gas also extinguishes burning sticks but it does not turn the water milky. Again, SO₂ gas also turns lime water milky but unlike CO₂, it reacts with an acidified solution of potassium dichromate and changes the colour of the solution from orange to green

Uses Of carbon dioxide:

1. CO₂ is used in the manufacture of sodium carbonate by the Solvay process and also for the manufacture of fertilisers such as urea, ammonium sulphate etc.

2. CO₂ is used in fire extinguishers.  It finds extensive use in the preparation of aerated waters such as soda water, lemonade etc… And baking powder.

3. Solid carbon dioxide i.e., dry ice is used as a refrigerant under the commercial name drikold. Dry ice is also used for making cold baths in the laboratory by mixing it with some volatile organic solvents. It is extensively used as a coolant for preserving perishable articles in the food industry, for curing local burns and for surgical operations of sores.

4. Supercritical CO₂ is used as a. solvent to extract organic compounds from their natural sources, for example, caffeine from coffee beans, perfumes from flowers etc.

5. It is used under the name carbogen (a mixture of 95% O₂ and 5% CO₂) for the artificial respiration of patients suffering from pneumonia and affected by poisonous gases (CO poisoning).

Liquid CO₂ is used as a substitute for chlorofluorocarbons in aerosol propellants.

Fife extinguisher:

It is a specially designed metallic pressure vessel having a nozzle at one end. A glass bottle containing dilute sulphuric acid is placed inside it and the remaining portion of the vessel is filled with a concentrated solution of sodium bicarbonate. When required, the glass bottle can be broken by pressing a knob fitted with the vessel at the other end. When the glass bottle is broken, the add comes in contact with sodium bicarbonate solution and reacts to yield copious C02 gas. The gas, ejected under high pressure through the nozzle, falls on the burning substance and as aresult, the fire gets extinguished

Na₂CO₃ + H₂SO₄→Na₂SO₄ + CO₂↑ + H₂O

Baking powder:

The baking powder is used. Fire extinguisher in the preparation of bread consists of a dry mixture of potassium hydrogen tartrate, NaHCO₃, tartaric acid and -starch. When this mixture comes in contact with water present in the bread, a chemical reaction leading to the formation of CO₂ occurs. The resulting CO₂ gas evolved in the form of bubbles making the bread porous and soft. Moreover, NaHC03 and tartaric acid also produce C02 on thermal decomposition

Structure of carbon dioxide:

In a CO₂ molecule, the carbon atom is sp -hybridised whereas the oxygen atoms are sp² – hybridised. Carbon forms two cr -bonds and two pπ- pπ bonds with two oxygen atoms. The shape of the carbon dioxide molecule is, therefore, linear. The molecule is symmetrical (the two bond moments cancel each other) and hence, it is non-polar. The C —O bond length is 1.15Å. CO₂ can be represented as a resonance hybrid of the following three structures:

Compounds Of Silicon

1. Silicon tetrachloride (SiCI₄ )

Silicon tetrachloride Preparation:

Silicon tetrachloride is prepared by heating either silicon or silicon carbide with chlorine

Silicon tetrachloride Properties and uses:

1. Physical state:  It is a volatile liquid (boiling point: 330.5 K).

2. Hydrolysis:

SiCl₄ undergoes ready hydrolysis to produce silicic acid, Si(OH)₄ which on further heating undergoes partial dehydration to yield silica gel (SiO₂  xH₂O).

Silica gel is an amorphous and very porous solid which contains about 4% of water. It is used as an adsorbent in column chromatography and as a catalyst in the petroleum industry. When the hydrolysis of SiCl₄ is carried out at a much higher temperature, finely powdered silica Is obtained instead of silicic acid

The finely powdered silica thus obtained is used as a thixotropic agent (which reduces viscosity temporarily) in polyester, epoxy paints and resins and as an inert filler in silicon rubber.

3.  Reduction: Reduction of SiCl₄ with H₂ gas gives silicon

Ultrapure silicon used for making transistors, computer chips and solar cells is prepared by this method.

4. Reaction with silicon:

When a mixture of SiCl₄ and Si is pyrolysed, a series of perhalosilanes of the general formula, Si_n Cl_{2n+ 2} where n = 2-6, are obtained.

Chains of perhaloslianes are longer than those of silanes and this is due to the formation of pπ-dπ bonding between a lone pair of electrons present on Cl and the empty d-orbitals of Si

2. Silicones

Silicones e Definition:

The synthetic organosilicon polymers containing repeating R₂SiO units held by Si — O — Si linkages are known as silicones The general formula of these compounds is (R₂SiO)n where R = methyl or aryl group. Commercial silicones are generally methyl derivatives and in some cases phenyl derivatives.

Silicones Preparation

Hydrolysis of dichlorodimethylsilane (obtained by the reaction between methyl chloride and silicon in the presence of Cu as catalyst) followed by polymerisation involving intramolecular dehydration yields straight chain polymers, i.e., silicones.

P Block Elements Class 11 Notes PDF

The length of the polymer can be controlled by the reaction of dimethylsiianol with chlorotrimethylsilane. This blocks the terminal end ofthe polymer as follows—

Silicones Properties:

• Silicones containing short chains are oily liquids; those with medium chains are viscous oils, greases and jellies and those with long chains are rubber-like solids.
• They are stable to heat and are also resistant to oxidation, i.e., they are very inert.
• They are water repellents (hydrophobic) & good electrical insulators.

Silicones Uses:

• Silicones are used for making water-proof cloth and paper.
• These are used as electrical insulators.
• Silicon oils are used in high-temperature baths and vacuum pumps.
• Silicon rubbers are very useful as they can retain their elasticity over a wide range of temperatures.
• These are mixed with paints and enamels to make them resistant to the effects of sunlight, high temperatures and chemicals.
• These are used for preparing vaseline-like greases which are used as lubricants in aeroplanes

3. Silicates

Silicates Definition:

Silicates are compounds in which the anions present are either discrete SiO₄^4- tetrahedral units or several such units joined together by corners, i.e., by sharing one oxygen atom but never by sharing edges The negative charge on the silicate structure is neutralised by positively charged metal ions.

Classification Of silicates:

Depending upon the number of comers (0, 1, 2, 3 or 4) of SiO₄^4- tetrahedral unit shared with another tetrahedral unit through oxygen atoms, silicates are following six types:

1. Orthosllicates:

These are simple silicates which contain discrete SiO₄^{4- tetrahedrons}. Some examples are—zircon: Zr₂[SiO₄] , forsterite: Mg₂[SiO₄] , willemite: Zn₂[SiO₄] and phenacite: Be₂[SiO₄]

2. Pyroslllcates:

When two SiO₄^4- – tetrahedra share one corner (i.e., one oxygen atom), Si₂O₇^6-anion is formed. Silicates containing discrete Si₂O units are called pyrosilicates.

The common examples of phyllosilicates are:

1. Thortveitite: Sc₂[Si₂O₇] and
2. Hemimorphite: Zn₄(OH)₂[Si₂O₇]-H₂O

3. Ring or cyclic silicates:

When two O-atoms per tetrahedron are shared to form closed rings, structures with the general formula, (SiO₃)_n^2n-– are obtained. The silicates containing these anions are called cyclic silicates.

Some common examples are :

1. Wollastonite: Ca₃[Si₃O₉] (containing the cyclic ion,  Si₃O₉^6-  and
2. Beryl: Be₃Al₂[Si₆O₁₈] (containing the cycle ion, [Si₆O₁₈]^-12

4. Chain silicates:

If two oxygen atoms of each tetrahedral unit are so shared that a linear single-strand chain of the general formula, (SiO₃)₂ is formed, then the silicates containing these anions are called chain silicates.

Minerals of this type are called pyroxene and these include:

1. Enstatite: Mg₂[(SiO₃)₂],
2. Diopside: CaMg[(SiO₃)₂] and s
3. Spodumene: LiAl[(SiO₃)₂].

P Block Elements Class 11 NCERT Notes

On the other hand, those chain silicates, containing double chain are called amphiboles. Here, two chains are attached through the O-atom. These silicates contain (Si₄O₁₁)_n ^6- ions. Minerals of asbestos are most commonly known as amphiboles.

For example: Crocidolite or blue asbestos [Na₂Fe₅(OH)₂(Si₄O₁₁)₂], amosite or brown asbestos [(Mg, Fe)(OH)₂(Si₄O₁₁)₂

Asbestos is heat and fire-resistant and thus is used as a shed for houses. Fine asbestos fibres, on entering the lungs cause asbestosis which can even result in lung cancer.

5. Sheet silicates:

Sharing of three comers i.e., three O-atoms of each tetrahedron results in the formation of an infinite two-dimensional sheet structure of the formula (Si₄O₅)_n ^2n-. Silicates containing these anions are called sheet silicates.

Some of the common examples are:

1. Kaolinite: [Al₂(OH)₄Si₂O₅] and
2. Alc: [Mg₃(OH)₂Si₄O₁₀] .

Clay also belongs to this class containing (Si₂O₆) ^2-– anions

Three-dimensional silicates:

If all the four comers i.e., all the four O-atoms of each tetrahedron are shared with other tetrahedra, a three-dimensional network structure is obtained. These have a general formula, (SiO₂)_n. Some common examples are quartz, tridymite and cristobalite

When a few silicon atoms in a three-dimensional network of SiO₂ are replaced by Al³⁺ions, the overall structure thus obtained carries a negative charge and is called aluminosilicate. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge. Such three-dimensional aluminosilicates are called zeolites.

A common example is natrolite:

Na₂[Al₂Si₃O₁₈]-2H₂O. Feldspars and ultramarines are two other types of three-dimensional aluminosilicates.

Many open channels of molecular levels are present in the structure of the zeolites. Depending on the shape and size of these open channels, ions or molecules of different shapes and sizes are adsorbed by the zeolites. Thus, zeolites are used as molecular sieves for separating molecules of different sizes.

Other two types of three-dimensional aluminosilicates are:

1. Feldspar example: Orthoclase, KAlSi₃O₈) and
2. Ultramarine example: Ultramarine blue, Na₈(AlSiO₄)₆S₂ )

NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons Short Question And Answers

Question 1. The hydrolysis of n-propyl magnesium bromide and another alkyl magnesium bromide may obtain propane. Write the name and structure of that alkylmagnesium bromide.
Answer:

Propane can be obtained by hydrolysis of isopropyl magnesium bromide (Me₂CHMgBr)

Question 2. When a concentrated aqueous sodium formate solution is used in Kolbe’s electrolysis method, no alkane is obtained—why?
Answer:

In Koibe’s electrolysis method, two— R groups of two RCOONa molecules combine to form the alkane, and R— R and two CO₂ molecules are obtained from two COONa groups. As, there is no alkyl group in the salt, sodium formate (HCOONa), no alkane is formed on electrolysis of its concentrated aqueous solution.

Question 3. Prepare 2,2-dimethylpropane by Corey-House synthesis.
Answer:

Question 4. How will you convert: \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHBr} \longrightarrow \mathrm{CH}_3 \mathrm{CHDCH}_3\)
Answer:

NCERT Class 11 Chemistry Chapter 13 Hydrocarbons Short Questions

Question 5. Identify RI and R’l in the following reaction \(\mathrm{RI}+\mathrm{R}^{\prime} \mathrm{I} →{\mathrm{Na} / \text { ether }} \text { Butane + Propane + Ethane }\). What is the role of ether in this case?
Answer:

When the Wurtz reaction is carried out with a mixture of HI and R’l, two hydrocarbons with an even number of carbon atoms (R— R and R’—R’) and one hydrocarbon with an odd number of carbon atoms (R—R’) are formed. Among the formed alkanes, butane (CH₃CH₂CH₂CH₃) and ethane (CH₃CH₃) are respectively R— R and R’—R’ whereas, propane (CH₃CH₂CH₂) is R — R’.

So, it is evident that R is an alkyl group containing two carbon atoms, i.e., ethyl group (CH₃CH₂ — ) and Rr is a one carbon atom-containing alkyl group, i.e., methyl group (—CH₃). Therefore, RI is ethyl iodide (CH₃CH₂I) and R’l is methyl iodide (CH₃I). The role ofether in this case is that it acts as a solvent.

Question 6. How will you convert methyl bromide to (1) methane and (2) ethane in one step?
Answer:

1.

2.

Question 7. Arrange the following compounds in order of their increasing stability and explain the reason: 2-butene, propene, 2-methyl but-2-ene
Answer:

The order of increasing stability of the given alkenes is— propene < 2-butene < 2-methyl but-2-ene

The number of hypercoagulable hydrogen in propene, 2-butene, and 2-methylbut-2-ene molecules are 3, 6, and 12 respectively. With an increase in several hypercoagulable hydrogens, stability due to the hyperconjugation effect of the molecules increases.

Question 8. Between the position and chain isomer of but-1ene, which exhibits geometrical isomerism and why?
Answer:

Position isomer of but-1-ene is but-2-ene (CH₃CH=CHCH₃) and chain isomer of but-1-ene is 2-methylpropene [(CH₃)₂C=CH₂] . In a molecule of but-2-ene, different groups are attached to the C-atom which is linked to the double bond.

So, but-2-ene exhibits geometrical isomerism. On the other hand, in a molecule of 2-methylpropene, two similar atoms (H-atom) are attached to the C-atom linked to the double bond. So, 2-methylpropene does not exhibit geometrical isomerism.

Question 9. Write the structures of the two alkenes obtained when 2-butanol is heated with excess ofconcentrated H₂SO₄. Which is obtained predominantly?
Answer:

Stability due to hyperconjugation is higher in the case of but-2-ene compared to that it is obtained predominantly.

Question 10.

Answer:

Question 11. Identify the compounds obtained on heating (CH₃)₄NOHO
Answer:

Compounds formed: trimethyl amine and methyl alcohol.

Question 12. What is the major product formed in the reaction between CH₂=CH—NMe₃Ie and HI? Write its structure.
Answer:

Question 13. What happens when a mixture of ethylene and O₂ gas is passed through a solution of PdCI₂ in the presence of CuCl₂  at high pressure and 50°C?
Answer:

When a mixture of ethylene and O₂  gas is passed through a PdCl₂  solution in the presence of CuCl₂  at high pressure and 50°C, acetaldehyde is formed as the product.

Question 14. Write the IUPAC names of the following compounds:

1. HC=C-CH(CH₃)₂
2. CH₃-C=C-C(CH₃)₃

Answer:

IUPAC name: 3 – Methylpent-1-yen

IUPAC name: 4,4-dimethyl pent-2-yen

Question 15. C = C bond length is shorter than C and C —C —why?
Answer:

The σ -bond In C≡C is formed due to the overlapping of two small sp -hybridized orbitals. In C—C, the σ -bond is formed due to the overlapping of two bigger sp² -hybridized orbitals whereas in C — C.

The σ –bond is formed due to the overlapping of two even bigger sp³ -hybridized orbitals. Again, the multiplicity of the bond between two atoms increases, and the atoms come closer to each other leading to a decrease in bond length. So, bond length follows the order: C=C < C=C < C— C.

Question 16. Write structures and IUPAC names of the alkynes having molecular formula C₅H₈.
Answer:

Structures and IUPAC names of the alkynes having molecular formula C₅H_n are-

⇒ \(\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \mathrm{H}_2 \stackrel{2}{\mathrm{C}} \equiv \stackrel{1}{\mathrm{C}} \mathrm{H} \quad \text { (Pent-1-one) }\)

⇒ \(\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \equiv \stackrel{2}{\mathrm{C}} \stackrel{1}{\mathrm{C}} \mathrm{H}_3 \quad \text { (Pent-2-one) }\) Δ

Question 17. What happens when the ethanolic solution of 1,1,2,2-tetrabromoethane is heated with zinc dust?
Answer:

When the ethanolic solution of 1,1,2,2-tetrabromoethane is heated with Zn dust, acetylene is formed as the product.

Question 18. What happens when the gas obtained by the action of water on calcium carbide is passed through an ammoniacal AgNO₃ Solution? Identify the solution through which acetylene gas is passed to form a red precipitate.
Answer:

Acetylene is formed by die action of water on calcium carbide. When acetylene gas is passed through ammoniacal silver nitrate solution, a white precipitate of silver acetylide (Ag₂C₂) is obtained.

Question 19. Which gas is used in carbide lamps or Hawker’s lamps? How does the gas produce a bright illuminating flame in the lamp?
Answer:

Acetylene gas is used in carbide lamps or Hawker’s lamps for producing bright illuminating flame. The percentage of carbon in acetylene is greater than in saturated hydrocarbons having the same number of carbon atoms.

As a result, incomplete combustion of acetylene gas takes place and the heated carbon particles thus formed produce Hluminating flame and bright light.

Hydrocarbons Class 11 Short Question and Answers NCERT

Question 20. What is Lindlar’s catalyst? Mention its use.
Answer:

Use of Lindlar’s catalyst:

This catalyst is used to add 1 molecule H₂ i.e., for partial hydrogenation of an alkyne. As cis-hydrogenation takes place in this case, a cis-alkene can be prepared from a non-terminal alkyne by using this catalyst.

Question 21. Which out of ethylene & acetylene is more acidic and why?
Answer:

The greater the s -the character of a hybridized carbon atom, the greater will be its electronegativity. The s -character of sp hybridized carbon atom of acetylene is greater than that of the sp² -hybridised carbon atom of ethylene.

So the electronegativity of the carbon atom of acetylene (CH=CH) is greater than the carbon atom of ethylene (CH₂=CH₂).

The tendency of the H -atom attached to the more electro¬ negative carbon atom to be removed as a proton (H+) is relatively higher. So, the acidity of ethylene is less than that of acetylene, i.e., acetylene is more acidic than ethylene.

Question 22. How will you convert ethyne into ethanol?
Answer:

Question 23. Which out of ethyne and propyne is more acidic and why?
Answer:

In ethyne (HC=CH), there are two terminal hydrogen atoms whereas, in propyne (CH₃C=CH) there is only one. Apart from this, one electron-donating —CH₃ group is attached to the carbon atom present on the other side of the triple bond in the propyne molecule.

This decreases the acidity of the alkynyl hydrogen atom. So, ethyne is more acidic than propyne.

Question 24. How will you distinguish between but-1-yne and but-2-yne?
Answer:

But-1-yne (CH₃CH₂C=CH) being a terminal alkyne reacts with a solution of ammoniacal stiver nitrate (AgNO₃) to form a white precipitate of silver 1-butynide.

However, but-2-yne (CH₃C=CCH₃) being a non-terminal alkyne does not undergo this type of reaction. So, but-1-one and but-2-one can be distinguished by observing the result of the above-mentioned test using ammoniacal stiver nitrate solution.

Question 25. How will you convert propylene into propylene?
Answer:

Question 26. What is the expected shape ofa benzene molecule in the absence of resonance?
Answer:

In the absence of resonance, benzene will be considered as the compound, 1,3,5-cyclohexatriene. The structure of this hypothetical compound is—

The structure will be such because each carbon-carbon bond length will be unequal. As, C=C is shorter than C—C, the structure of the molecule will appear as an irregular hexagon instead ofa regular one.

Question 27. Write structures of two aromatic ions in which there is a p-orbital containing 2 electrons and the other in which there is a vacant orbital.
Answer:

Cyclopentadienyl anion has two electrons in one of its p -orbital whereas, cyclopropenyl cation has a vacant p -orbital.

Question 28. Write the name and structural formula of the dibromobenzene which forms three mononitro compounds.
Answer:

Question 29. What happens when each of the following compounds is heated with acidified K₂Cr₂O₇ solution—
Answer:

Ethylbenzene:

Ethylbenzene gets oxidised to benzoic acid.

Question 30. Benzene exhibits a greater tendency towards substitution reactions but a lesser tendency towards addition reactions — Explain.
Answer:

Benzene attains stability by resonance. If benzene undergoes addition reactions, then it no longer participates in resonance. The aromaticity of benzene is no longer retained due to loss of conjugation.

Consequently, the extra stability of benzene is lost. So, benzene has a lower tendency to undergo an addition reaction. However, during the formation of the substitution product, the aromaticity and stability of benzene remain intact. So, benzene has a higher tendency towards substitution reactions.

Question 31. Benzene burns with a luminous sooty flame but methane bums with a non-luminous flame with no black smoke. Why?
Answer:

Due to the high percentage of carbon, elementary carbon is produced during the burning of benzene. As a result, black smoke is formed. The presence of hot carbon particles in the flame makes the flame luminous. In methane, the percentage of carbon is low so no elementary carbon is produced during the burning of methane. Thus, methane burns with a non-luminous flame with no black smoke.

Question 32. Name the electrophiles that participate in the following reactions: nitration, chlorination, Frieclel-Crafts alkylation, Friedel-Crafts acylation, and sulphonation.
Answer:

Electrophiles which participate in nitration, chlorination, Friedel-Crafts alkylation, Friedel-Crafts acylation, and sulphonation reactions are nitronium ion (⁺NO₂), positively charged chlorine ion (Cl⁺) or chlorine-iron (III) Chloride complex (Cl — Cl⁺— ^–FeCl₃), carbocation (R), B: acylium ion (R⁺CO) and sulfur trioxide (SO₃) respectively.

Question 33. Why iodobenzene cannot be prepared directly by combining benzene and iodine in the presence of iron filings? Why is this reaction possible in the presence of nitric acid?
Answer:

Iodobenzene cannot be directly prepared by combining benzene and iodine because the reaction is reversible. However, in the presence of nitric acid, this reaction becomes possible because nitric acid oxidizes the hydrogen iodide as it is formed and so drives the reaction to the right.

Question 34. Write the structure of the compound 4-(1-isopropyl butyl)-3-propyl undecane.
Answer:

Question 35. Write the structure of 2,2,3-trimethylpentane and label the 1°, 2°, 3° and 4° carbon atoms.
Answer:

Question 36. Write the IUPAC names of two different optically active alkanes with the lowest molecular mass.
Answer:

2 optically active alkanes with the lowest molecular masses-

Question 37. What is the state of hybridization of the quaternary carbon atom present in the neopentane molecule? Write the IUPAC name and structure of the alkane formed by the combination of neopentyl and tert-butyl groups.
Answer:

The state of hybridization of the quaternary carbon atom In a nooporitanu molecule is sp³

The alkane formed by the combination of neopentyl and terf-butyl groups Is—

IUPAC name: 2,2,4,4-tetramethylpentane

Question 38. How many chain isomers will be obtained on replacement of different H-atoms of n-pentane? Write their structures and IUPAC names.
Answer:

Three non-equivalent H-atoms are present in a pentane (CH₃CH₂CH₂CH₂CH₃) molecule. Thus, the replacement of these H-atoms by —CH₃ groups results in three isomeric alkanes. These are as follows —

CH₃CH₂CH₂CH₂CH₂CH₃ (hexane)

CH₃CH₂CH₂CH(CH₃)₂(2-methylpentane)

CH₃CH₂CH(CH₃)CH₂CH₃(3-methylpentane)

Question 39. Write the trivial and IUPAC names of the branched chain alkane with the lowest molecular mass.
Answer:

The trivial name of the branched chain alkane with the lowest molecular mass is isobutane and its IUPAC name is 2-methylpropane.

Question 40. Write the IUPAC name and structure of the alkane having formula C₈H₁₈ and containing a maximum number of methyl groups.
Answer:

The alkane of formula C₈H₁₈ containing the maximum number of methyl groups is

IUPAC name: 2,2,3,3-tetramethylbutane

Question 41. Benzene cannot be used as a solvent in ozonolysis of unsaturated hydrocarbons—why?
Answer:

Benzene itself reacts with ozone to form a triozonide. So, benzene cannot be used as a solvent in ozonolysis of unsaturated hydrocarbons

Class 11 Chemistry Chapter 13 Hydrocarbons NCERT Short Q&A

Question 42. Which one of the following three alkyl halides does not undergo Wurtz reaction and why?

1. CH₃CH₂Br
2. CH₃I
3. (CH₃)₃CB

Answer:

1. The second step of the Wurtz reaction is an S_N2 reaction.
2. An S_N2 reaction is very susceptible to steric effect so, a 3° alkyl halide does not take part in an S_N2 reaction.
3. So, (CH₃)₃CBr being a 3° alkyl halide does not undergo Wurtz reaction

Question 43. Which compound is the strongest acid and why?

1. HC ≡CH
2. C₆H₆
3. C₂H₆
4. CH₃OH

Answer:

CH₃OH is the strongest acid because here the H-atom is bonded to oxygen which is more electronegative titan carbon irrespective of its state of hybridisation

Question 44. Give two equations for the preparation of propane from the Grignard reagent.
Answer:

Question 45. How can CH₃D be prepared from CH₄?
Answer:

Question 46. The preparation of which of the following alkanes by Wurtz reaction is not practicable?

1. (CH₃)₃C-C(CH₃)₃
2. CH₃—CH(CH₃)—CH₂CH₃
3. (CH₃)₂CHCH₂CH₂CH(CH₃)₂

Answer:

1. Not practicable. Although it is a symmetrical alkane, prepared from a carboxylic acid? its preparation requires a f-alkyl halide, (CH₃)₃CX, which does not participate in Wurtz reaction.
2. Not practicable. It is an unsymmetrical alkane.
3. Practicable, because it is a symmetrical alkane

Question 47. When a concentrated aqueous solution of a mixture of sodium salts of two monocarboxylic acids is subjected to electrolysis, ethane, propane, and butane are liberated at the anode. Write the structures and names of the two starting sodium salts
Answer:

Ethane may be produced from two CH₃COONa molecules, propane from 1 CH₃COONa molecule and one CH₃CH₂COONa molecule, and butane (CH₂CH₂CH₂CH₃) from 2CH₃CH₂COONa molecules. Evidentlyethane, propane & butane may be obtained from the mixture of sodium ethanoate (CH₃COONa) & sodium propionate (CH₃CH₂COONa)

Question 48. How can an alkane having one carbon atom less be prepared from a carboxylic acid?
Answer:

Question 49. Arrange the following compounds in the increasing order of their acidic character.

1. H₂O, CH₂=CH₂, NH₃
2. HC≡CH, CH₃OH, CH₃CH₃

Answer:

1. Increasing order of acidic character: CH₂=CH₂ < NH₃ < H₂O [Because, the increasing order of electronegativity: C₂ < N < O ]

2. Increasing order of acidic character: CH₃CH₃ < HC = CH < CH₃OH [increasing order of electronegativity: c ,<C. <ol

Question 50. An alkane (molecular mass 72) produces only one monochloro derivative. Give IIJPAC the name of the compound. Give reasons.
Ana.

As the alkane (C_nH_2n+2) on monochlorination produces only one monochloride derivative, so all of its hydrogens are equivalent. The molecular mass of the alkane is 72, i.e.,12 × n +{2n + 2) = 72, or n = 5. Hence, the alkane contains a carbon atom, l.e., it is one of the isomeric pentanes. The pentane in which all the H-atoms are equivalent is neopentane, (CH₃)₄ C. The IUPAC name of the compound is 2,2-dimethylpropane.

Question 51. 
Answer:

A = CH₃CHO (Acetaldehyde);

B= CHCHOH

(Ethyl alcohol); C = CHCl₃

D = HCOONa (Sodium formate)

Question 52. Acetylene does not react with NaOH or KOH, even though it possesses an acidic character —why?
Answer:

Acetylene (HC = CH) is a weaker acid than water (H₂O) and OH- is a weak base than HC=C^–. As a result, weak acid HC=CH does not react with weak base OH^– to form strong acid H₂O and strong base HC≡C. So acetylene fails to react with NaOH or KOH.

Question 53.  Write the products obtained when propyne ion (CH₃C = ^–C:) is allowed to react with

1. H₂O,
2. CH₃OH
3. H₃ (liquid)
4. 1-hexene and
5. Hexane.

Also mention, where no reaction occurs
Answer:

1. CH₃C = CH + OH^–
2. CH₃C = CH + CH₃O^–
3. No reaction
4. No reaction
5. No reaction.

(Propyne is a weaker acid than H₂O and CH₃OH but it is a stronger acid than, NH3 1-hexene and hexane.]

Question 54. Write the names of the products obtained in the following cases
Answer:

Answer:

1. Br CH₂CH₂ COOH (3-bromopropanoic acid).
2. C₆H₅ COCH₃ (Acetophenone).

NCERT Solutions Class 11 Chemistry Chapter 13 Hydrocarbons Short Q&A

Question 55. How can allyl chloride be prepared from 1-propanol?
Answer:

Question 56. trans-pent-2-ene is polar but Frans-but-2-ene is non-polar —why
Answer:

In a trans-but-2-ene molecule, the two C-—CH₃ bond moments, oriented in opposite directions cancel each other and so the net dipole moment in a trans-but-2-ene molecule is zero. On the other hand, in the trans-pent-2-ene molecule, although the C-*-CH₃ and C—C₂H₅ bond moments act in opposite directions, they are not equal in magnitude so they cannot cancel each other. Hence, the molecule possesses a net dipole moment.

Question 57. Convert acetylene into but-2-one.
Answer:

Question 58. Ozonolysis of an alkene leads to the formation of an aldehyde and an isomeric ketone having the molecular formula, C₃H₆O. Identify the alkene.
Answer:

The aldehyde and the ketone having molecular formula, C₃H₆O are CH₃CH₂CHO and CH₃COCH₃ respectively.

Therefore, the starting alkene is 2-methylpent-2-ene. (CH₃)₂C= O + O=CHCH₂CH₃-(CH₃)₂C =CHCH₂CH

Question 59. An alkene having molecular formula, C₄H₈ reacts with HBr to form a tertiary alkyl bromide. Identify the alkene and the alkyl bromide.
Answer:

Question 60. Carry out the following transformation (in two steps): Methyl acetylene →1-bromopropan
Answer:

Question 61. H How can butan-2-one be prepared from acetylene?
Answer:

Question 62. Write the IUPAC names of the products obtained when buta-1,3-diene reacts with bromine in1:1 molar ratio
Answer:

Question 63. Distinguish between buta-1,3-diene and but-1-one.
Answer:

But-1-one (CH₃CH₂C = CH) reacts with ammoniacal cuprous chloride solution to give a red precipitate. Buta-1,3- diene, however, does not respond to this test.

Short Question and Answers for Class 11 Chemistry Hydrocarbons

Question 64. Prove that benzene is insoluble in water and is lighter than water.
Answer:

Benzene is taken in a separating flask and water is added to it. The mixture is shaken and then allowed to stand until the two layers are separated. The upper layer contains benzene while water settles at the bottom. This observation proves that benzene is insoluble in water and is lighter than water.

Question 65. Why does benzene burn with a sooty flame?
Answer:

As the carbon content in benzene molecule (C₆H₆, C: H = 1: 1 ) is relatively higher as compared to the saturated hydrocarbon, hexane (C₆H₁₄ C: H = 1:2.3), elementary carbon is formed during burning of benzene. So, benzene bums with a sooty flame

Question 66. Nitration of aniline with 98% H₂SO₄ and cone. [Anilinium sulfate (water soluble)] HNO₃ occurs very slowly and mainly metasubstitution occurs —why?
Answer:

In the presence of 98% H₂SO₄, the — NH₂ group of aniline takes up a proton (H) and is converted to an electron attracting (-1) and meta-directing — NH₃ group. Electron electrophilic attack of the —NO₂ group occurs very slowly. This accounts for the extremely slow rate of nitration of aniline and the formation of the mainly mega-substituted products under strong acidic conditions.

Question 67. Activating groups are ortho-/para- directing, while; the deactivating groups are meta- directing—why?
An8.

The activating groups by exerting their electron-donating +1 and/or +R effects increase electron densities at the ortho and para- positions to a larger extent than the meta-position. So, the electrophiles (E+) enter preferably the ortho- and parapositions. On the other hand, the deactivating groups by exerting their electron-withdrawing -I and I or -R effects decrease electron densities of meta-positions to a lesser extent than the ortho- and para-positions. So the electrophile enters preferably the relatively more electron-rich meta-position.

Question 68. How will you remove traces of aniline present in benzene?
Answer:

If benzene containing traces of aniline (basic) is shaken with a cone. H₂SO₄ then aniline dissolves in the acid forming a salt.- The acid layer is thus removed. In this way, traces of aniline present in benzene can be removed

Question 69. How can traces of phenol present in a sample of benzene be removed?
Answer:

If the sample of benzene containing traces of phenol (acidic) is shaken with 10% NaOH solution then phenol reacts with NaOH to form sodium phenoxide which dissolves in the aqueous layer. The aqueous layer is then removed. In this way, traces of phenol present in a sample of benzene can be removed.

Question 70. How will you distinguish between benzene and hexane by a simple test in the laboratory?
Answer:

The percentage of carbon in benzene is much higher than the corresponding alkane, hexane. So benzene bums with the formation of elementary carbon and as a result, a sooty flame is produced. During the burning of hexane no such sooty flame is produced. Thus, by observing the nature of the flame produced, the two compounds can be easily distinguished.

Question 71. Aniline does not undergo Friedel-Crafts reaction, though it contains an electron-donating group—why?
Answer:

Aniline (Lewis base) combines with AlCl₃ (Lewis acid) by donating the unshared electron pair on nitrogen, to form a complex. As a result, the N-atom of the — NH₂ group acquires a positive charge and this, being converted into an electron attracting group, decreases the electron density of the ring to such an extent that the Friedel-Crafts reaction does not occur.

Question 72.  How much trisubstituted benzene may be obtained from o-, m-, and p-chlorotoluene and why?
Answer:

Each of o – and m -chlorotoluene will give 4 trisubstituted benzenes, while p -chlorotoluene will give two trisubstituted benzenes. This is because there are 4 types of non-equivalent hydrogens (or positions) in each of the ortho- and meta¬ isomers and two types of non-equivalent hydrogens (or positions) in the para-isomer

Question 73. C₅H₁₂ and C₈H₁₈ are two alkanes that form one monochloride each reacting with chlorine. Write the names of the alkanes and structural formulas of the chlorides.
Answer:

As both the alkanes form one monochloride, it can be said that all H-atoms in these two alkanes are equivalent. Therefore, the alkane with molecular formula C₆H₁₂ is (CH₃)₄C and the alkane with molecular formula C₈H₁₈ is (CH₃)₃C-C(CH₃)₃

Question 74. How can you convert methyl acetylene to acetone?
Answer:

At 60-80°C temperature, when methyl acetylene or propyne is passed through a dilute (20%) solution of sulphuric acid containing 1% H₂SO₄, it combines with one molecule of water to form the unstable compound, 2-propanol. Addition of water molecule to unsymmetrical alkyne (propyne in this case) through Markownikoff’s rule. 2-propenol! Rapidly tautomerizes to acetone

Question 75. Mention two reactions of benzene to show its behavior is different from that of the open-chain unsaturated hydrocarbons.
Answer:

1. Benzene does not decolorise bromine water.
2. Benzene does not react with halogen acids such as HCl, HBr, etc

Question 76. Is it possible to isolate pure staggered ethane or pure eclipsed ethane at room temperature? Explain.
Answer:

The energy difference between eclipsed and staggered conformations of ethane is only 2.8kcal .mol^-1, which is easily achieved by collisions among the molecules at room temperature. So it is not possible to isolate either pure staggered or pure eclipsed form of ethane at room temperature.

Question 77. Explain why rotation about carbon-carbon double bond is hindered?.
Answer:

Carbon-carbon double bond consists of a σ -and a π bond. The π-bond is formed by lateral overlap of unhybridised p -p-orbitals of two carbon atoms above and below the plane of the carbon atoms. If an attempt is made to rotate one of the atoms of the double bond concerning the other, the p orbitals will no longer overlap, thereby causing the fission of the n bond. Since breaking of a π -bond requires a considerable amount of energy, which is not available at room temperature, rotation about the carbon-carbon double bond is hindered.

Question 78. Arrange the following carbanions in order of their Hg+ decreasing stability

Answer:

Due to greater s -character sp -hybridized carbon is more electronegative than sp³ -hybridized carbon and hence can accommodate the -ve charge more effectively. So 1 and 2 are more stable than 3. Again —CH group has an electron-donating +1 effect, therefore it interacts with the -ve charge on carbanion carbon and hedestabilizesises 1 relative to 2. Thus, the stability of the given carbanions decreases in the sequence: B> 1> 3

Question 79. Explain why Tert butylbenzene cannot be oxidized to benzoic acid by treatment with alkaline KMnO₄
Answer:

Alkylbenzenes can be oxidized to benzoic acid provided that the side chain contains one benzylic {hydrogen atom. Since tert-butylbenzene does not contain any benzylic hydrogen, so the alkyl chain cannot be oxidized to the — COOH group.

Question 80. Explain why HF forms hydrogen bonding with acetylene even though it is non-polar
Answer:

Due to the sp -hybridization of carbon, the electrons of the C —H bond of acetylene are attracted considerably toward carbon. Consequently, each carbon carries a partial negative charge and each hydrogen carries a partial positive charge. Owing to the presence of a partial positive charge on hydrogen, acetylene forms an H -bond with the F -atom of the HF molecule

⇒ \(\begin{array}{r}
\delta+\delta-\delta-\delta+\delta-\delta+ \\
\cdots \mathrm{H}-\mathrm{C} \equiv \stackrel{\delta}{\mathrm{C}}-\mathrm{H} \cdots \mathrm{F}-\mathrm{H}
\end{array}\)

Hydrocarbons Chapter 13 Short Question and Answers Class 11

Question 81. One mole ofa symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. Identify the alkene
Answer:

Let, the aldehyde be C_nH_2n+1CHO

The molecular mass of this aldehyde

=[12n + (2n +1) + (12 +1 + 16)]u

= (14n + 30)u

Thus, 14n + 30 = 44

n = 1

So the aldehyde is CH₃CHO.

The alkene is CH₃CH=CH— CH..

Question 82. The addition of HBr to 1 -butene gives a mixture of mechanisms: A and B as the main products together with a small amount of another compound C. Identify A, B, and C
Answer:

Question 83. An alkene, C₆H₁₂, reacts with HBr in the absence as well as in the presence of peroxide to give the same product. Find its structure.
Answer:

Symmetrical alkenes react with HBr in the presence or absence of peroxide to give the same product. Hence the given alkene may have the structure (1) or (2)

Question 84. How will you prepare a sample of propane free from other alkanes using ethyl bromide, methyl bromide, and diethyl ether as the organic compounds, together? The product‘T is iodobenzene. With other inorganic materials
Answer:

It can be prepared by Corey-House synthesis:

Question 85. The catalytic hydrogenation of which of the following is most exothermic?

Answer:

The least substituted alkene, having the lowest number of hyperconjugative structures, has the least thermodynamic stability (i.e., highest energy content) and so it has the highest heat of hydrogenation. Now, out of the given compounds, (C) is the least substituted alkene and so it has the highest heat of hydrogenation.

Question 86. Identify A, B, C & D in the following reaction—

Answer:

1. H —C≡C —H
2.  H —C ≡ CNa
3. H—C ≡ C—CH₃
4. H₂C=CH—CH₃

Question 87. Identify A, B, C Si D in the following reaction

Answer:

1.  CH₂=CH₂
2. CH= CH
3. CH₃CHO
4. CH₃—CH₃

Question 88. What organic compound is obtained when

1. Ethyl iodide is subjected to react with Zn -Cu couple/aqueous ethanol and 
2. Iodoform is heated with Ag powder?

Answer:

1. CH₃CH₃
2. C₂H₂
   

Question 89. Write structures of A and B in the following reactions.

Question 90.  Which one of these two reactions can be used for the identification of ethylenic unsaturation? Why? ½ + ½+½ + 1
Answer:

R-CH(Br)—CH₃

R—CH(Br) —CH₂Br

Question 91.  A hydrocarbon (A) is obtained when 1,2-dibromoethane reacts with alcoholic KOH. (A) decolourises alkaline KMnO₄ solution. (A) contains acidic hydrogen. Identify (A) with reasons.
Answer:

Question 92. Writes structures of A, B, and C 

Answer:

Question 93. How will you convert?

1. HC = CH→H₃C-CH₂-CH₃

Answer:

Question 94. An organic compound (A) composed of C and H contains 85.71 %C. It shows M⁺ at mlz = 42 in the mass spectrum. The compound reacts with HBr in the absence of peroxide to yield an organic compound (B) which is isomeric with the compound (C) obtained when the compound reacts with HBr in the presence of peroxide. Identify A, B, and C.
Answer:

⇒ \(C: H=\frac{85.71}{12}: \frac{14.29}{1}=7.14: 14.29=1: 2\)

Emperical formula = CH₂, Molar formula = (CH₂)_n

According to the problem, n × (12 + 2) = 42.

n = 3. So, the actual molecular formula is C₃H₆. DBE = 1. So, probable structures of CH₆ are—

Structure A is accepted as it can undergo given reactions:

Question 95.

1. Write equations of all the steps of the reaction of methane with chlorine in the presence of diffused sunlight

2. Identify A and B

Answer:

⇒ \(\mathrm{A} \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OSO}_3 \mathrm{H}, \mathrm{B} \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Question 18. Convert benzene into aniline by using the following reagents in the correct order: alkaline KMnO₄ and then HCl NH₃, heat; Br₂/KOH; CH₃Cl /anhydrous AlCl₃.
Answer:

Question 96. Write structures of the organic products obtained in the following reactions :

Answer:

Question 97. Identify (X) and (Y) in the following reactions:

Answer:

Question 98. Identify(M) and (N)in the following reactions

Answer:

Question 99. Write the product of the following reaction:

Answer:

NCERT Class 11 Chemistry Hydrocarbons Short Questions and Answers PDF

Question 100. Benzene in reaction with NOCl in the presence of acid produces an organic compound (1).

1. On treatment with NaNH₂/Liq.NH₃ furnishes another organic compound (15).
2. Treatment with NaNO₂/HBF₄ affords an organic compound
3. Which on heating gives an organic compound.
4. Identify(1), (2), (3) and (4).

Answer:

Question 101. Two different compounds produce only acetaldehyde on ozonolysis. Draw the structures of the two compounds.
Answer:

Question 102. Write the name and the structural formula of the product obtained when hydrogen bromide reacts with propene in the presence of benzoyl peroxide.
Answer:

Question 103. Identify the compound in the following reaction

Answer:

Identify the compound A in the following reaction:

Question 104.

1. Among benzene and toluene, which one will undergo nitration reaction easily and why?

2. Identify A, B, C and D

Answer:

If the electron density of the benzene ring increases, then the reactivity of the ring towards electrophilic substitution also increases. In toluene, the —CH₃ group increases the electron density of the ring and as a result, the reactivity of the ring also increases due to +1 and the hyperconjugation effect of the —CH₃ group. So, nitration occurs more easily for toluene than unsubstituted benzene.

Question 105. Write the name and structural formula of A in the following reaction.

Answer:

Question 106. What happens when What happens ether?
Answer:

Question 107. Two isomeric compounds A and B have the molecular formula C₃H.Br forms the same compound C on dehydrobromination. C on ozonolysis produces acetaldehyde and formaldehyde. Identify A, B, and C
Answer:

⇒ \(A \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}\)

⇒ \(B \Rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)

⇒ \(C \Rightarrow \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2\)

Question 108. An alkene A on ozonolysis gives a mixture of ethanal & pentan-3-one. Write the structure & IUPAC name of A.
Answer:

Question 109. An alkene ‘A’ contains three C —C, eight C —H crbonds and one C — C π -bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of ‘A’.
Answer:

An aldehyde with molar mass 44u is ethanal (CH₃CH=:O). The formula of the alkene ‘A’ which on ozonolysis gives two moles of ethanol can be determined as follows—

⇒ \(\underset{\text { Ethanal }}{\mathrm{CH}_3 \mathrm{CH}}=\underset{\text { Ethanal }}{\mathrm{O}}+\underset{\text { But-2-ene (A) }}{\mathrm{O}}=\underset{\mathrm{CHCH}_3}{\mathrm{CH}_3-\mathrm{CH}}=\underset{\mathrm{CH}}{\mathrm{CH}}-\mathrm{CH}_3\)

There are three C — C cr -bonds, eight C —H tr -bonds, and one C —C 7t -bond in but-2-one.

Question 110. Propanal and pentane-3-one are the ozonolysis products of an alkene. What Is the structural formula of the alkene?
Answer:

Question 111. Write chemical equations of the combustion reaction of the following hydrocarbons: Butane Pantene Hexyne Toluene
Answer:

Chapter 13 Hydrocarbons NCERT Solutions Short Questions Class 11

Question 112. Draw The Cis- and trans-structure of hex- 2 ene which isomer will have higher B>P and Why?
Answer:

The general formula of hex-2-ene is CH₃—CH₂—CH₂—CH=CH—CH₃. Structural formulas of cis-and trans-isomers of this compound are-

The ds-isomer being more polar than the trans-isomer has a higher value of dipole moment than that of the trans-isomer. Intermolecular dipole-dipole interaction in the case of cis-isomers is more than that in trans-isomers. So, the boiling point of the customer is higher.

Question 113. Why is benzene extraordinarily stable though it contains three double bonds?
Answer:

There are (4n + 2) delocalized 7r -electrons (n = 1) in the planar benzene molecule.

Consequently, it attains stability due to aromaticity. So, benzene is extraordinarily stable despite having three double bonds.

Question 114. Out of benzene, m-dinitrobenzene, and toluene which will undergo nitration most easily, and why
Answer:

Nitration of the benzene ring is an electrophilic substitution reaction. In this reaction, the presence of an activating group ( —CH₃) increases the reactivity of the benzene ring, whereas the presence of a deactivating group (—NO₂) decreases the reactivity of the benzene ring. Therefore, order of nitration is toluene > benzene > m-dinitrobenzene.

Question 115. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Answer:

Ethylation of benzene means the introduction of an ethyl group in the benzene ring. This reaction is carried out by Friedel-Crafts reaction of benzene with ethyl halide (chloride or bromide), ethene, or ethanol. Lewis acids, other than anhydrous AlCl₃, that can be used in this reaction are anhydrous FeCl₃, SnCl₄, BF₃, HF, etc.

Question 116. Why does an iline not participate in the Friedel-Crafts reaction?
Answer:

Aniline reacts with AlCl₃ complexing \(\mathrm{C}_6 \mathrm{H}_5-\stackrel{\oplus}{\mathrm{N}} \mathrm{H}_2-\stackrel{\ominus}{\mathrm{AlCl}} \mathrm{Cl}_3\) which makes —NH₂ group electron-withdrawing nature. Consequently, the benzene ring becomes highly deactivated so aniline does not participate in the Friedel-Crafts reaction;

Question 117. The reaction of CH₂=CH—N(CH₃)₃I takes place contrary to Markownikoff’s rule—why?
Answer: 

The carbocation formed due to the addition of H+ to the carbon atom containing a higher number of hydrogen atoms becomes unstable because of the electron-attracting —NMe₃ group.

So, the reaction takes place contrary’ to Markownikoff’s rule;

Question 118. Mention the limitations of the Wurtzreaction.
Answer:

Limitations:

• Tertiary alkyl halides do not respond to this reaction,
• Methane cannot be prepared by this reaction and
• Preparation of unsymmetrical alkanes cannot be done by this method;

Question 119. What product is formed when the given compound reacts with HBr and why?
Answer: The product obtained

 is because the carbocation formed in the first step

Is, resonance stabilized due to —OCH₃ group

NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons  Warm-Up Exercise Question And Answers

Question 1. What are the chief constituents of LPG?
Answer: The chief constituents of LPG are n-butane and isobutane.

Question 2. Why do C—C bonds instead of C—H bonds of alkanes dissociate due to the effect of heat?
Answer: The bond energy of the C— C bond (ΔH = 83 kcal. mol^-1) is less than that of the C—H bond (ΔH= 99 kcal. mol^-1). So, the C—C bond dissociates more easily than the C—H bond.

Question 3. Write the IUPAC name of the straight-chain hydrocarbon consisting of 20 carbon atoms.
Answer: IUPAC’s name of the straight-chain hydrocarbon consisting of 20 carbon atoms is eicosane.

Question 4. Give the structures of the isomers of molecular formula C₅H₁₂–
Answer: CH₃CH₂CH₂CH₂CH₃ (n -pentane) CH₃CH(CH₃)CH,CH₃ (isopentane) and (CH₃)₄C (neopentane)

Question 5. Explain why dry ether is used in the Wurtzreaction.
Answer:

Dry ether is used because it is present in ether, then it may react with metallic sodium thereby rendering it ineffective

2Na + 2H₂O→2NaOH + H₂

Question 6. Predict whether Me3CBr will take part in the Wurtz reaction or not
Answer:  Wurtz reaction proceeds through the S_N2 pathway. As tertiary alkyl halides do not participate in the S_N2 reaction (due to steric effect), Me₃CBr does not participate in the Wurtz reaction

Question 7. Explain why methane does not react with chlorine in the dark.
Answer:  The reaction does not take place because in the dark Cl —Cl bond does not dissociate to form Cl free radical;

Question 8. One molecule of a hydrocarbon produces one molecule each of acetone, methyl glyoxal, and formaldehyde on ozonolysis. Identify the hydrocarbon.
Answer:

The hydrocarbon is 3, 4-dimethylpenta-l, 3-diene [CH₃—C(CH₃)=C(CH₃)—CH=CH₂] or, 2,4-dimethylpenta-1,3-diene [CH₂=C(CH₃)—CH=C(CH₃) —CH₃];

Short Answer Questions on Hydrocarbons Class 11 NCERT

Question 9. Explain why 1-butyne reacts with ammoniacal silver nitrate to produce a white precipitate, but 2-butyne does not
Answer:

1-butyne (CH₃CH₂C=CH) being a terminal alkyne reacts with ammoniacal AgNO₃  solution to produce a white precipitate but 2-butyne (CH₃C=CCH₃) being a non-terminal alkyne does not react with ammoniacal AgNO₃ solution;

Question 10. How will you detect the presence of acetylene in a gas mixture?
Answer:

If the gas mixture when passed through ammoniacal AgNO₃ solution or ammoniacal Cu₂Cl₂ solution forms a white or red precipitate, then the gas mixture contains acetylene.

Question 11. Explain why the carbon-carbon bond in acetylene is shorter than the carbon-carbon bond in ethylene.
Answer:

Cr -bond in acetylene (HC=CH) is formed due to the overlapping of two small sp-hybridized orbitals whereas in ethylene (H₂C=CH₂) it is formed by overlapping of two bigger sp² hybridized orbitals. So, the bond length of HC=CH <H₂C=CH₂;

Question 12. How will you distinguish between ethylene and acetylene?
Answer:

Acetylene reacts with ammoniacal AgNO₂ solution to form a white precipitate of silver acetylide (AgC=CAg) but ethylene does not give a similar reaction with ammoniacal AgNO₃ solution.

Question 13. The population of which conformation increases with the temperature rise?
Answer: The population of the less stable conformation Increases with the increase in temperature.

Question 14. What are the carbides that react with water to form methane commonly known as?
Answer: The carbides which react with water to form methane are commonly known as methanldes.

Question 15.

Question 16. Why are hydrocarbons insoluble in water but highly soluble in solvents like petroleum ether, benzene, carbon tetrachloride, etc?
Answer:

An important principle regarding dissolution is ‘like dissolves like’. It means that polar molecules dissolve in polar solvents while non-polar molecules dissolve in nonpolar solvents. This dissolution process is thermodynamically favorable. Water is a highly polar solvent whereas, petroleum, ether, benzene, and carbon tetrachloride are non-polar solvents. As hydrocarbons are non-polar compounds, they are insoluble in water but soluble in petroleum ether, benzene, and carbon tetrachloride.

Question 17. Why are the alkanes called paraffins?
Answer: Alkanes are called paraffin as their chemical reactivity is quite low (Latin: parum = little, affinis = affinity).

Question 18. What are the typical reactions of alkanes?
Answer: Typical reactions of alkanes are substitution reactions.

Question 19. Mention the type of mechanism through which halogenation of alkanes occurs.
Answer: Free-radical mechanism.

Question 20. What happens when methane is heated at 1000°C in the absence of air?
Answer:

Methane when heated at 1000°C in the absence of air, decomposes to form a fine powder of carbon which is known as carbon black:

⇒ \(\mathrm{CH}_4 →{1000^{\circ} \mathrm{C}} \mathrm{C}+2 \mathrm{H}_2 \uparrow\)

Question 21. What is the main constituent of natural gas which is used as a fuel?
Answer: The main constituent of natural gas which is used as a fuel is methane (90%).

Question 22. Why is light or heat essential for the chlorination of alkanes?
Answer:

Cl free radical is required for the initiation of the reaction between an alkane and chlorine, i.e., homolysis of the Cl—Cl bond is necessary. The energy required for this hemolysis is derived from light or heat. So, light or heat is essential for the chlorination of alkanes.

Question 23. Which gas is responsible for explosions in coal mines?
Answer: Methane is responsible for explosions in coal mines.

Question 24. Write the IUPAC name of freon – 113.
Answer: IUPAC name off neon-113, i.e., Cl₂FC— CClF₂ is 1,1,2-trichloro-1,2,2-trifluoroethane

Question 25. Which reaction helps locate the position of the double bond in alkenes?
Answer:

The reaction which helps locate the position of the double bonds in alkenes is ozonolysis.

Question 26. An alkene (C₄Hg) reacts with HBr in the presence or in the absence of peroxide to give the same compound. Identify the alkene.
Answer:

As tire given alkene (molecular formula: C₄HO) reacts with HBr to give the same product in the presence and absence of peroxide, the alkene is symmetrical. So, a symmetrical alkene with molecular formula C₄H₈ is but-2-ene (CH₃CH=CHCH₃).

Question 27. Calculate the number of sigma (or) and pi (n) bonds in methyl acetylene.
Answer:

In methyl acetylene (CH₃-C CH), there are 6 a-bonds and 2 bonds.

Question 28. Which of the following compounds will react with metallic sodium to produce H₂ gas?

1. C₂H₄
2. C₆H₆
3. C₂H₂
4. CH₃CH₂CH₃

Answer: 3. Acetylene (C₂H₂) reacts with metallic sodium to produce H2 gas:

⇒ \(\mathrm{HC} \equiv \mathrm{CH}+2 \mathrm{Na} \rightarrow \mathrm{NaC} \equiv \mathrm{CNa}+\mathrm{H}_2 \uparrow\)

Question 29. The C₂ — C₃ bond 1,3-butadiene possesses some double bond characteristics.
Answer:

The C₂—C₃ bond in 1,3-butadiene possesses some double bond character because of the delocalization of n -electrons.

Question 30. An arena when oxidised forms 1,3-dicarboxylic acid. Write the numbers of side chains and their position in the arena.
Answer:

As the arena gets oxidized to a dicarboxylic acid, it has two side chains. It can be said that the two side chains are at 1,3- or meta-position of each other because a 1,3-dicarboxylic acid forms in the oxidation.

Question 32. Distinguish between benzene and toluene with the help of a chemical reaction.
Answer:

Toluene on oxidation by alkaline KMn04 and subsequent acidification produces shining white crystals of benzoic acid. Benzene, on the other hand, does not undergo oxidation with alkaline KMnO₄ to form any white precipitate.

Question 33. Between — NH₂ and —NO₂, which group facilitates nucleophilic substitution reaction in the benzene ring?
Answer:

The group that facilitates nucleophilic substitution reaction in the benzene ring is — NO₂ because it decreases the electron density of the benzene ring.

Class 11 Hydrocarbons Short Questions and Answers NCERT Solutions

Question 34. Arrange in order of increasing reactivity towards electrophilic substitution: benzene, nitrobenzene, toluene, chlorobenzene.
Answer:

The order of increasing reactivity towards electrophilic substitution of the compounds is :

Nitrobenzene< Chloro¬ benzene < Benzene < Toluene.

Question 35. Name the halogen carrier in the chlorination of benzene.
Answer: The compound that acts as the halogen carrier in the chlorination of benzene is either AlCl₃ or FeCl₃.

Question 36. Benzene undergoes de-sulphonation but not denitration. Why?
Answer:

Since sulphonation is a reversible reaction, benzene can undergo a desulphonation reaction. However, nitration is an irreversible reaction. So, benzene cannot undergo a nitration reaction.

Question 37. If the calculated and the experimental heats of combustion of benzene are 824.1 and 789.1 kcal mol^-1 respectively, then calculate the value of resonance energy of benzene.
Answer:

Resonance energy = calculated heat of combustion experimental heat of combustion = (824.1 – 789.1)kcal. mol^-1= 35 kcal .mol^-1

NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons Long Question And Answers

Question 1. How can an eclipsed conformation of ethane be converted into a staggered conformation?
Answer:

In an ethane molecule, if one carbon atom is kept fixed around the C—C bond axis and the other carbon atom is rotated at a minimum angle of 60°, then the eclipsed conformation is converted to the staggered conformation.

Question 2. Give examples of a chiral conformation and an achiral conformation of n-butane.
Answer:

Gauche-staggered conformation of n-butane is chiral because it cannot be superimposed on its mirror image.

However, the fully eclipsed conformation of n-butane is achiral as it can be superimposed on its mirror image.

Question 3. Arrange the following conformations of n-butane according to their increasing stability:

1. Gauche-staggered
2. Fully eclipsed
3. Eclipsed and
4. Ante-Mggered

Answer:

When 2-iodopropane is used as the alkyl halide in the Wurtz reaction, the alkane obtained is 2,3-dimethylbutane.

Question 4. Which of the following alkanes cannot be prepared by the Wurtz reaction in good yield?

1. (CH₃)₂CHCH₂CH(CH₃)₂
2. (CH₃)₂CHCH₂CH₂CH(CH₃)₂
3. (CH₃)₃CCH₂CH₂CH₂CH₃
4. CH₃CH₂C(CH3)₂CH₂CH₃
5. (CH₃)₃C-C(CH₃)₃

Answer:

(1), (3) and (4) are three unsymmetrical alkanes. So, these cannot be prepared by Wurtz reaction in good yield. Again, for preparing alkane (5), a 3° alkyl halide is required. So, despite being a symmetrical alkane, (5) cannot be prepared by the Wurtz reaction.

NCERT Class 11 Chemistry Chapter 13 Hydrocarbons Long Question and Answers

Question 5. How will you prepare methane and ethane starting from ethanoic acid?
Answer:

Question 6. How many monochloro derivatives are obtained on chlorination of n-pentane, isopentane and neopentane? Write down their structures.
Answer:

There are three and four types of non-equivalent hydrogen atoms in n-pentane (CH₃CH₂CH₂CH₂CH₃) and isopentane [CH₃CH(CH₃)CH₂CH₃] respectively. Whereas, in neopentane [(CH₃)₄C], all H -atoms are equivalent.

Therefore, chlorination of n-pentane, isopentane and neopentane form three, four and one monochloride derivatives respectively.

Question 7. Chlorination of cyclohexane to prepare chlorocyclohexane is more practicable than the chlorination of methylcyclohexane to prepare l-chloro-l-methylcyclohexane— explain.
Answer:

There are five types of non-equivalent H -atoms in methylcyclohexane

When it undergoes chlorination, four other monochloride derivatives are formed along with 1-chloro-1-methylcyclohexane

As a result, the yield of the desired product is low’—’ and Low it is difficult to separate the product from the mixture. On the other hand, all H -atoms in cyclohexane are equivalent and thus, only chlorocyclohexane

Is formed as the product for this reason, Achlorination of cyclohexane to prepare chlorocyclohexane is more feasible than the chlorination of methylcyclohexane to prepare 1-chloro-l-methylcyclohexane.

Question 8. Write the IUPAC names of the following compounds:

Answer:

Question 9. Write the structural formula:

1. 3-(1- methyl ethyl) hex-2-ene; 
2. 4-ethyl- 2, 4- dimethyl kept-1- ene

Answer:

Question 10. Write the IUPAC names and structures of the alkenes having the molecular formula C₅H₁₀.
Answer:

IUPAC names and structures of the alkenes having the molecular formula C₅H₁₀ are as follows

Hydrocarbons Class 11 Long Question and Answers NCERT

Question 11. Write the mechanism of acid-catalysed dehydration of butyl alcohol.
Answer:

Dehydration of isopropyl alcohol in the presence of concentrated H₂SO₄ is an El reaction. The reaction occurs in three steps. The second step of the reaction is the slowest, i.e., it is the die rate-determining step of the reaction.

Step 1: Protonation of the alcohol.

Step 2: Elimination of water molecules and formation of carbocation

Step 3: Elimination of proton from carbocation

Question 12. Write the structures of A and B obtained from given reactions
Answer:

Answer:

A is R —CHBr —CH₃ and B is RCHBrCH₂Br. The alkene, HBr and the formed alkyl bromide (A) are all colourless. So, the left reaction cannot be used to detect ethylenic unsaturation.

On the other hand, the alkene and the formed dibromoalkane (B) are colourless but bromine has a reddish-brown colour.

So, the right reaction can be used to detect ethylenic unsaturation because decolourisation of bromine takes place in this reaction.

Question 13. How can a double bond be created in a molecule of a compound which has a carbon-carbon single bond?
Answer:

A double bond is created in a molecule of a compound containing a carbon-carbon single bond by the can-given method.

Question 14. Which reaction is used to detect ethylenic unsaturation and why? Write the structures and IUPAC names of the compounds expected to be obtained in the given reactions:

⇒ \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HCl}→{\text { Peroxide }}\)

⇒ \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr}→{\text { Peroxide }}\)

Answer:

Question 15. Write the product of the given reaction. Explain its formation:

Answer:

Due to the -R and -I -effect of the — NO₂ group, carbocation [I] is less stable than carbocation [II]. So the reaction proceeds through carbocation [II] and the major product formed is

Question 16. State Markownikoffs rule. Explain with an example. How would you convert ethylene to acetylene? Identify the compound in the reaction—

Answer:

Markownikoffs rule and example: Conversion of ethylene to acetylene

Question 17. Ethane can be dried by passing through concentrated H2S04 but not ethylene—why?
Answer:

Ethane being a saturated hydrocarbon does not react with concentrated H₂SO₄

CH₃ —CH₃  (Ethane) + cone. H₂SO₄ . So, ethane can be dried by passing through concentrated H₂SO₄.

On the other hand, ethylene being an unsaturated hydrocarbon, when passed through concentrated H₂SO₄ gets absorbed by the acid and forms ethyl hydrogen sulphate. So, ethylene cannot be dried by passing through concentrated H₂SO₄.

Question 18. Identify the compounds A, B and C in the following reaction and write their names:

Answer:

Carbonyl compounds B and C contain three and two carbon atoms respectively. There are also three carbon atoms on one side of the double bond and two carbon atoms on the other side of the double bond in the alkene.

Therefore, two alkenes with molecular formula C₅H₁₀ are: 2-methylbut-2-ene [CH₃ —C(CH₃)=CHCH₃] and pent-2-ene (CH₃CH₂CH=CHCH₃) . If A (C₅H₁₀) is 2-methylbut-2-ene, thenB (C₃H₆O) &C(C₂H₄O) areacetone (CH₃COCH₃) and acetaldehyde (CH₃CHO) respectively.

If A is pent-2-ene, then B and C are propanal (CH₃CH₂CHO) and acetaldehyde (CH₃CHO) respectively.

Question 19. An alkene on ozonolysis produces propanone and propanal. The alkene is—

1. 2-methyl pent-2-ene
2. 3-methyl pent-2- ene
3. A-methyl pent-2-ene
4. Hex-3-ene.

Answer:

The products formed on ozonolysis are propanone and propanal. Therefore, the alkene can be determined as—

The alkene is (1) 2-methylpent-2-ene.

Question 20. How can cis- and trans-hydroxylation of cis-2-butene be carried out? Comment on the optical activity of the formed products.
Answer:

Osmium tetroxide adds to the double bond of cis-2- butene to form osmic ester which is hydrolysed by an aqueous ethanolic solution of sodium bisulphite. In this case, the two —OH groups get attached to the doubly bonded carbon atoms from the same side of the double bond and form 1,2-diol.

So, cis-hydroxylation takes place in case of this reaction. On the other hand, cis-2-butene reacts with peracids to form the corresponding epoxide. The resulting epoxide on hydrolysis with dilute acid or alkali yields 1,2-diol.

Epoxidation followed by hydrolysis causes the addition of two —OH groups from the opposite sides of the double bond. So, trans-hydroxylation takes place in case of this reaction.

Question 21. What product would you expect from the following reaction?

Answer:

An alkyne should be formed when a vicinal dihalide is refluxed with ethanolic KOH. However, in the given case, an alkyne does not form because a six-membered ring system cannot accommodate a linear portion like C—C=C —C. So, the compound formed is 1,3-cyclohexadiene.

Question 22. Identify and B in the following reactions

Answer:

Question 23. Give IUPAC names of the following compounds:

1. CH₃CH₂-CH=C=CH₂

2.

3.  CH₂=CH-CH(CH₃)-(CH₃)C=CH₂

Answer:

IUPACname: 1,2-pentadiene.

IUPAC name: 1-methyl-1,4-cyclohexadiene.

IUPAC name: 2,3-dimethyl-1,4-pentadiene

Question 24. What are the two planar conformations of 1,3-butadiene? Which conformation is less stable and why?
Answer:

The two planar conformations of1,3-butadiene are—

Due to steric hindrance or strain, s-ds-conformation is less stable.

Question 25. Between 1,3- and 1,4-cyclohexadiene, which compound has a lower value of heat of hydrogenation and why?
Answer:

1,3-cyclohexadiene being a conjugated diene is more stable than 1,4-cyclohexadiene which is a nonconjugated diene. So, the heat of hydrogenation of 1,3- cyclohexadiene has a lesser value than that of 1,4- cyclohexadiene.

Question 26. Calculate the double bond equivalent of benzene from its molecular formula.
Answer:

Double bond equivalent (DBE) of compound,

⇒ \(\mathrm{DBE}=\frac{\sum n(v-2)}{2}+1\)

Where n is the number of different atoms present in the molecule and v is the valency of each atom. The molecular formula of benzene is C₆H₆.

So, DBE of benzene = \(\frac{6(4-2)+6(1-2)}{2}+1=4\)

Class 11 Chemistry Chapter 13 Hydrocarbons NCERT Long Q&A

Question 27. What knowledge about the carbon-carbon bond length in benzene may be obtained from valence bond theory?
Answer:

The benzene molecule is a resonance hybrid of Kekule structures (1) and (2) and the contribution of each hybrid structure is 50% i.e., equal.

The single bonds (C—C) and the double bonds (C—C) in structure (1) become double and single bonds respectively in structure (2). As the two equally stable resonance structures (1) and (2) contribute equally to the hybrid.

It may be said that any two adjacent carbon atoms of a benzene molecule are linked by a bond intermediate between a single and a double bond. So, all the carbon-carbon bonds of benzene are equivalent and their lengths are equal (1.39A).

Again, the bond order of each bond is the same (1.5). So, it can be said that all carbon-carbon bonds are equal in length.

⇒ \(\text { Bond order }=\frac{\text { Double bond }+ \text { Single bond }}{2}=\frac{2+1}{2}=1.5\)

Question 28. Which of the following representations is correct and why?

Answer:

According to representation

(1), it seems that structures (1) and (2) have a separate existence. There is no separate existence of structures (1) and (2).

So, (1) and (2) cannot be related by ’. Thus, the representation (i) is incorrect. Since (1) and (2) are two resonance structures which have no separate existence. So, (1) and (2) can be related by’ •*-»ÿ ‘. Thus, the representation

(2) is correct.

Question 29. What is the basic difference between aromatic and anti¬ aromatic compounds?
Answer:

Monocyclic planar conjugated polyene systems containing (4n + 2) delocalised; π-electrons (n = 0, 1,2,3, are called aromatic compounds.

Monocyclic planar conjugated polyene systems containing 4n delocalised π- electrons (n = 1,2,3,…) are called antiaromatic compounds.

Question 30. Will cyclooctatetraene exhibit aromatic character? Explain.
Answer:

Since cyclooctatetraene does not contain (4n + 2)n electrons, it does not exhibit an aromatic character.

As cyclooctatetraene has 4n – electrons (n=2), it should be an antiaromatic compound. However, the ring of this compound is very large so it does not exist in the unstable planar shape, rather it forms a tub-shaped structure. As a result, conjugation is lost and so cyclooctatetraene is a non-aromatic compound.

Question 31. Using the theory of aromaticity, explain the finding that A and B are different compounds, but Cand D are identical.
Answer:

As A is an antiaromatic compound (4nπ -electron system, n = 1 ), it becomes unstable due to the delocalisation of π electrons. As delocalisation of π -electrons does not take place for A, B is not the resonance structure of A.

B is the structural isomer of A. So, A and B are two different compounds. Again, C is an aromatic compound [(4n + 2)π electron-system, n = 1 ] which attains stability due to the delocalisation of electrons. So, delocalisation of electrons takes place for C. D is the resonance structure of C, i.e., Cand D are same compound.

Question 32. Classify each of the given species as aromatic, antiaromatic and nonaromatic

Answer:

1. Is an antiaromatic compound because the B atom contains a vacant p -orbital,
2. Is a non-aromatic compound because one carbon atom of the ring does not have a p -orbital.
3. Behaves as an aromatic compound with (4n + 2)π-electrons (n = 1) because of the vacant d -d-orbital and lone pair of electrons of the sulphur atom

Is an aromatic ion with (4n+2)π- electrons, n = 0

Question 33. Which is the smallest aromatics species?
Answer:

The smallest aromatic species is cyclopropenyl cation.

Question 34. Write structural formulae of isomeric nitrotoluenes.
Answer:

Structural formulae ofisomeric nitrotoluenes are

Question 35. Write structural formulae of isomeric dibromophenols.
Answer:

Structural formulae of isomeric dibromo phenols are

Question 36. More than three dibromobenzenes are not possible—explain.
Answer:

Considering the resonance structures of benzene, it is easy to understand that positions 1,2- and 1,6- are indistinguishable. Similarly, positions 1,3- and 1,5- are indistinguishable. Thus, in the case of bromobenzene, only three isomers are possible which are as follows

Question 37. Write the IUPAC names of the given compounds

Answer:

1. 1,2-dihydroxybenzene
2. 1-phenylpropanoid-l-one
3. 2-hydroxybenzoic acid
4. Al-phenylethylamine
5. l-bromo-3-chlorobenzene
6. 3-phenylpropanoid acid
7. 2,4,6-trinitrotoluene
8. 4-hydroxy-3-methoxy benzaldehyde

Question 38. Write structures and IUPAC names:

1. Mesitylene
2. Styrene
3. Pyrogallol
4. Picric acid
5. Salicylaldehyde
6. Benzophenone
7. TNT
8. Phthalic acid
9. Anthranilic acid.

Answer:

Question 39. Classify the following groups as o-/p-or m- directing group and activating or deactivating group:

1. -NO₂
2. -Cl
3. -C₂H₅
4. -CP₃
5. OH
6. — NHCOCH₃
7. —NH₃
8. — O
9. —COCH₃

Answer:

1. — NO₂ (deactivating and m -directing),
2. —Cl (deactivating and o/p -directing),
3. —C₂H₅ (activating and o-lp- directing),
4. — CF₃ (deactivating and m -directing),
5. —OH (activating and o-lp- directing),
6. — NHCOCH₃ (activating and o-lp- directing),
7. — NH₃ (deactivating and m directing),
8. — Oe (activating and o-lp-directing
9. — COCH₃ (deactivating and m -directing).

Question 40. Explain each of the following observations:

1. Although —Cl is a deactivating group, it is o-lp-directly.
2. The —CH₃ group is an o-/p- directing group, even though the carbon atom contains no unshared pair of electrons.
3. The —OCH₃ group is an activating and o-/p directing group.
4. The — CCl₃ group is a m -m-directing group, even though the carbon atom is not bonded to a more electronegative atom by a double or triple bond.

Answer:

3. —OCH₃  group is an o-/p-directing group because an unshared pair of electrons on O-atom participate in resonance (+R -effect) and increase the electron density of the ring at ortho- and para-positions.

So, electrophiles (E) preferably enter the ortho- and para-positions. Due to an increase in electron density, the ring becomes more activated than the unsubstituted benzene towards an electrophilic substitution reaction. Thus, — OCH₃ is an activating group.

4. —CCl₃ is an electron-withdrawing group because of its -I effect which is attributed to the presence of three highly electronegative Cl -atoms. Consequently, it decreases the electron density of the benzene ring, especially at the ortho- and para-positions.

So, —CCl₃ is a deactivating group which makes the ring less reactive towards electrophilic substitution and substitution occurs preferably at meta-position.

NCERT Solutions Class 11 Chemistry Chapter 13 Hydrocarbons Long Q&A

Question 41. How will you prepare benzene from the given compounds?

1. C₆H₅COOH
2. C₆H₅CMe₃ 
3. C₆H₅CH₂Cl
4. C₃H₅Br

Answer:

Question 42. Write two processes to convert C₆H₆ into C₆H₅D.
Answer:

Question 43. Write the formed.

Answer:

Question 44. Why groups like — CHO, — NO₂, — B(OR) — PBr₃ and — SR₂ act as meta-directing groups?
Answer:

Since the carbon atom of the:

1. — The CHO group is bonded to the oxygen atom by a double bond, the nitrogen atom of the
2. — The NO₂ group is linked with the oxygen atom by a double bond, the boron atom of the
3. —B (OR)₃ group contains a vacant p -orbital and the phosphorus and sulphur atoms of the groups

—⁺ PBr₃ and — ⁺SR₂-, have vacant d-orbitals, all of these groups reduce the electron densities of ortho- and para- positions by their -R effect. Consequently, the electron density at the metaposition becomes relatively higher and the electrophile preferably enters the meta-position. Thus, these groups behave as meta-directing groups

Question 45. Arrange the compounds in increasing order of their rate of nitration and give reason: Benzene, Toluene, Nitrobenzene, Hexadeuterobenzene (C₆D₆).
Answer:

The increasing order of rate of nitration of given compounds is nitrobenzene < benzene = hexadeuterobenzene < toluene. The electron-attracting nitro (— NO₂) group decreases the electron density of the nitrobenzene ring and as a result, its nitration occurs at a rate slower than that of benzene.

On the other hand, the electron-repelling methyl (— CH₃) group increases the electron density of the toluene ring and as a result, its nitration proceeds at a rate faster than that of benzene. Benzene and hexadeuterobenzene (C₆D₆) undergo nitration at the same rate because, in an aromatic electrophilic substitution reaction, cleavage of the C — H or C — D bond does not occur at the rate-determining step.

Question 46. A mixture of benzene and bromine solution remains unchanged for an indefinite period, but if an iron nail is added to the solution, bromination of benzene occurs rapidly—explain.
Answer:

Benzene is an aromatic compound having no ethylenic unsaturation. So benzene does not participate in additional reactions with bromine. Again, the substitution reaction of benzene does not take place with the poor electrophile bromine alone.

So a solution of bromine in benzene remains stable (f.e., unchanged) for an indefinite period. However, when an iron nail is added to the solution, bromination of benzene occurs to yield bromobenzene because iron then acts as a halogen carrier. The red solution of bromine becomes colourless

2Fe + 3Br₂→2FeBr₃; Br₂ + FeBr₃ → Br⁺ FeBr^–

C₆H₆ + Br⁺ FeBr^–₄ →C₆H₅Br + HBr + FeBr₃

Question 47. Write the monosubstituted compounds formed in each of the following reactions and state whether each reaction is faster or slower than that of benzene.

1. Nitration of C₆H₅NHCOCH₃,
2. Bromination of C₆H₅CBr₃
3. Chlorination of C₆H₅CMe₃
4. Nitration of C₆H₅—C₆H₅
5. Nitration of C₆H₅—COOCH₃
6. Sulphonation of C₆H₅CHMe₂
7. Nitration of C₆H₅CN,
8. Bromination of C₆H₅I,
9. Nitration of C₆H₅-C₆H₄C₆H₅

Answer:

1. p-O₂NC₆H4NHCOCH₃ (for this compound nitration occurs faster than benzene).

2. m-BrC₆H₄CBr₃ (for this compound bromination occurs slower than benzene)

3. p-ClC₆H₄CMe₃ (for this compound chlorination occurs faster than benzene)

4. p-O₂NC₆H₄C₆Hg (for this compound nitration occurs faster than benzene)

5. m-O₂NC₆H₄COOMe (for this compound nitration occurs slower than benzene)

6. p-HSO₃C₆H₄CHMe₂ (for this compound sulphonation occurs faster than benzene)

7. m-O₂NC₆H₄CN (for this compound nitration occurs slower than benzene)

8. p- BrC₆H₄I (reaction occurs slower than benzene)

9.

(reaction occurs faster than benzene in the middle ring because it is attached to two activating — CgH5 groups on both sides.)

Question 48. Write three methods by which alkyl side chains can be introduced into the benzene ring
Answer:

The: methods by which alkyl side chains, can be introduced into the benzene ring are—

1. By using (a) CH₃CH₃X, AlCl₃,  CH₂=CH₂ HF and CH₃CH₂OH, BF₃ or concentrated H₂SO₄ in Friedel-Crafts alkylation reaction.

2. By acylation of benzene using CH₃COCI or (CH₃CO)₂O, AlCl₃ followed by . Clemmensen reduction  the formed ketone.

4. By reacting CH₃CH₂Br with Ph₂CuLi according to Corey-House synthesis.

Question 49. Classify the following groups based on their orientation and reactivity:

Answer:

1,4 and 5 are activating and ortho-/para- directing groups. 3 and 6 are deactivating and meta-directing groups. 2 is a deactivating group (due to — NO₂ ) and is ortho-/para directing >C=C<

Question 50. 1-butyne and 2-butyne are allowed to react separately with the reagents given below:

1. Na, liquid NH₃ ; 
2. H₂ (1 mole), Pd-BaSO₄, quinoline,
3. H₂SO₄,H₂O, H₂SO₄ ; 
4. H₂/Pt.

Which reagent(s) will produce the same product in both cases? Write the structures, of products formed in these cases
Answer:

Both 1-butyne & 2-butyne react separately with reagents 1 and 2 to produce 1-butene and 2-butene respectively

However, reagents(3 & 4 react with 1-butyne & 2-butyne separately to yield same products (2-butanone & butane)

Question 51. Ethyne reacts with dil. H2S04 in the presence of Hg2+ salts to give acetaldehyde, but with HC1, under similar conditions, it gives vinyl chloride. Account for such observation.
Answer:

In the first step, ethyne reacts with Hg²⁺ to form a cyclic complex (I). This is then attacked by more nucleophilic H₂O, 2— rather than, weakly nucleophilic SO₄, to form unstable vinyl alcohol which then tautomerism to give acetaldehyde

If HCl is used instead of H₂SO₄ then the complex (I) is attacked by more nucleophilic Cl-, rather than weakly nucleophilic H₂O, to give vinyl chloride

An alkane has a molecular mass of 72. Give the structure of all the possible isomers along with their IUPAC OH names

Let the alkane be C_nH_2n+2. It’s molecular

=12n + (2n + 2) = 14n + 2

.*. 14n + 2 = 72, thus n – 5 and hence the alkane is C₅H₁₂

The isomers of the alkane C₅H₁₂ are—

Long Questions and Answers for Class 11 Chemistry Hydrocarbons

Question 52. Find the number of structural and configurational isomers of a bromo compound C₅H₉Br formed by the addition of HBR to 2-pentyne.
Answer:

Addition of one molar proportion of HBr to CH₃—CH₂—C=C—CH₃ produces two structural isomers 1 and 2

Each of these structural isomers can exist as a pair of geometrical isomers (cis and trans) and hence there are four possible configurational isomers

Question 53. Identify the products P and Q in the following reaction:

Answer: In the absence of light, the reaction occurs via a polar mechanism

Question 54. Identify the product ‘T’ in the following reaction and the major product. Account for its formation.

Answer:

The product‘T is iodobenzene.

Explanation: Since I am less electronegative than Cl, so I+ is the effective electrophile that takes part in the reaction

Question 55. Identify the major product obtained on; monobromination (Br₂/FeBr₃) of meta methyl anisole and account for its formation
Answer:

Both —CH₃ and —OCH₃ are o-/p-directing groups. Therefore, the possible positions of attack which are facilitated by these groups are indicated by arrows as shown below

Attack by the electrophile (Br⁺) is disfavoured at C₂ because this position is most crowded. Again -I effect of — the OCH₃ group does not favour attack at C₆. So most favourable attack occurs at C₄, thereby producing 4-bromo-3-methylanhole as

Question 56. The enthalpy of hydrogenation of cyclohexene is -119.5 kj. mol^-1 ,. If the resonance energy of benzene is 150.4 kj. mol^-1 , estimate its enthalpy of Br2/FeBr3 hydrogenation.
Answer:

Enthalpy of hydrogenation of cyclohexene

=-119.5 kj- mol^-1

So enthalpy of hydrogenation of hypothetical cyclohexatriene

= 3 × -119.5 kj- mol^-1,

In other words, the calculated (or theoretical) enthalpy of hydrogenation of benzene =-3 × -119.5 kj- mol^-1,

Let the actual (i.e., experimental) enthalpy of hydrogenation of benzene = × kj- mol^-1

Now, R. E. of benzene = calculated enthalpy of hydrogenation of benzene- actual enthalpy of hydrogenation of benzene

o,-150.4 = -3 X 119.5 —x

x = -3 ×119.5 + 150.4 = -208.1 kj. mol^-1

Question 57. How low will you prove:

1. Acidic character of acetylene.
2. Presence of terminal =CH2 group in 1-pentene.
3. Presence of acetylenic hydrogen in 1-butyne.
4. 2-butene is a symmetrical alkene. 1-butyne

Answer:

When acetylene is added to water and shaken, the resulting solution does turn blue litmus red. The following reactions in which the H-atoms of acetylene are replaced by metal atoms prove the acidic character of acetylene.

1. HBr (a) Acetylene (HC=CH) reacts with sodium in two steps to form monosodium acetylide (HC = CNa) and disodium acetylide (NaC = CNa) respectively and in each case, H is evolved.

2. When acetylene gas is passed through ammoniacal Cu₂Cl₂ or AgNO₃ solution, metallic acetylide is precipitated in each case

3. Ozonolysis of 1-pentene leads to the formation of formaldehyde (HCHO) as one of the products. This proves that there is a terminal =CH₂ group present in 1-pentene

When 1-butyne is treated with ammoniacal cuprous chloride, a red precipitate of cuprous 1-butynide is obtained. Again, when an aqueous solution of silver nitrate is added to the alcoholic solution of 1-butyne, a white precipitate of silver 1-but nude is obtained.

Ozonolysis of any symmetrical alkene results in the formation of only one carbonyl compound (2 moles). Since 2-butene, on ozonolysis, produces two moles of acetaldehyde (CH₃ CHO), it must be a symmetrical alkene. Its structure is CH₃ CH=CH CH_3.

Question 58. Identify the major product obtained in each of the following reactions and explain its formation

Answer:

1. 

Although the alkene is an unsymmetrical one, Markownikoff’s rule is not directly applicable here because there are the same number of H-atoms attached to double-bonded carbons. Out of the 2 carbocations ( C₆H₅CHCH₂CH₃ & C₆H₅CH₂CHCH₃ ) obtained in the first step of the reaction, the first one (a benzylic carbocation) is more stable because it is stabilised by resonance involving the benzene ring. So, this carbocation is formed more easily and readily and in the second step, it combines with the Br ion to give 1-bromo-l-phenylpropane as the major product.

2. 

This reaction occurs according to Markownikoff’s rule. Chlorine is more electronegative than iodine. So, in the I^δ+– Cl^δ- molecule, the I-atom with a partial positive charge combines first with the alkene as an electrophile. Between the two carbocations, (CH₃)₂⁺CCH₂I and  (CH₃)₂⁺CICH₂ formed in the first one being a 3° carbocation is relatively more stable. So, it is formed more easily and readily and in the second step, it combines with Cl to give 2-chloro-lido-2-methylpropane as the major product.

3. 

Double bond is more reactive than triple bond towards electrophilic addition reactions. For this reason, bromine (1 mole) is added mainly to the double bond of the compound to produce 4, 5 -dibromo pent-1-one as the major product.

Question 59. Write the names and structures of the two alkenes (molecular formula: C₄H₆) which give the compound when added to HBr in the absence of organic peroxide but different compounds when added to HBr in the presence of peroxide
Answer:

When but-1-ene (CH₃CH₂CH=CH₂) and but-2-ene (CH₃CH=CHCH₃) react with HBr in the absence of an organic peroxide, the same product is obtained.

However, in the presence of an organic peroxide, the addition of HBr to but-1-ene occurs contrary to Markownikoff’s rule and hence differences are obtained from these two cases

1.

2.

Question 60. Mention two reactions in which ethylene and benzene behave differently and two reactions in which they behave similarly.
Answer:

Benzene, unlike ethylene, fails to discharge the red colour of bromine in CCl₄ or the reddish-violet (purple) colour of potassium permanganate solution because, unlike ethylene the π -electron system of benzene possesses extraordinary.
O₂ stabilisation.

So, in these two reactions ethylene and benzene behave differently.

In the following two reactions, ethylene and benzene behave similarly.

1. Both ethylene and benzene bums with sooty flame to produce CO₂ and H₂O

⇒ \(\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒  \(2 \mathrm{C}_6 \mathrm{H}_6+15 \mathrm{O}_2 \longrightarrow 12 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}\)

2. Both ethylene and benzene react with ozone to form ozonide (an additional

⇒ \(\mathrm{C}_2 \mathrm{H}_4+\mathrm{O}_3 \longrightarrow \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_3\) (Ethylene (ozonides)

⇒  \(\mathrm{C}_6 \mathrm{H}_6+3 \mathrm{O}_3 \longrightarrow \mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_9\) (Benzene tri ozonide)

Question 61. Distinguish between:

Answer:

1. Toluene (C₆H₅CH₃), on oxidation by alkaline KMnO₄ solution followed by acidification gives a white crystalline precipitate of benzoic acid (C₆H₅COOH). On the other hand, tert-butylbenzene does not undergo such an oxidation reaction by an alkaline KMnO₄ solution.

2. O -xylene and m -xylene, on oxidation by alkaline KMnO₄ solution, produce phthalic acid and isophthalic acid respectively. Phthalic acid, when heated, forms phthalic anhydride which responds to phthalein test On the other hand, isophthalic acid on heating does not produce any anhydride.

Hydrocarbons Chapter 13 Long Question and Answers Class 11

Question 62. Write the name & structure of the following compounds:

1. An unsaturated aliphatic hydrocarbon which forms monosodium salt.
2. An organic compound which causes depletion of the ozone layer.
3. An alkane which is used as a fuel for household cooking.
4. An alkyl bromide which reacts with alcoholic KOH to form only 1-butene.
5. An alkene which reacts with HBr in the presence or absence of peroxide to give the same product.
6. A compound containing iodine which, when heated with silver powder, produces acetylene.
7. An alkyl bromide (C₄H₉Br) does not participate in the Wurtz reaction.
8. An alkene which on ozonolysis forms glyoxal and formaldehyde.

Answer:

1. Propyne (CH₃C= CH).
2. Dichlorodifluoromethane (CF₂Cl₂).
3. Butane(CH₃CH₂CH₂CH₃).
4. 1-bromobutane (CH₃CH₂CH₂CH₂Br).
5. 2-butene (CH₃CH=CHCH₃).
6. Iodoform (CHI₃).
7. Tert-butylbromide (Me₃CBr).
8. 1,3-butadiene (CH₂=CH—CH=CH₂)

Question 63. Identify A … G in the following reaction sequence

Answer:

Question 64. Write structures and names of the compounds A to Q in the following reaction sequences:

Answer:

Question 65. How will you distinguish between each of the given pairs of compounds by a single chemical test? 

1. Ethylene and acetylene
2. Ethane and acetylene
3. 1-butyne and 2-butyne
4. Ethane and ethylene.
5. Propene and propyne
6. 1-butene and 2-butene
7. 2-pentene and benzene
8. Benzene and cyclohexene.

Answer:

The distinction between two compounds should be written in a tabular form. A reagent which either causes a colour change, or evolution of a gas or the appearance of a precipitate should be selected for this purpose.

Question 66. How will you carry out the following transformations:

1. Acetylene → Acetone
2. Acetylene →Dldeuteroacetylene(C₂D₂)
3. Acetylene → Acetylenedicarboxyllc acid
4. 1-butyne → 2-butyne
5. Propene → 1-propanol  Propyne →  Propanal

Answer:

Question 67. The boiling points of three isomeric pentanes are 36.2°C, 28°C and 9.5°C respectively. Identify the compounds and give a reason.
Answer:

The strength of van der Waals forces depends on the area of products, A and B obtained in the following reactions: contact between molecules. The area of contact between straight-chain n -pentane (CH₃CHCH₂CH₂CH₃) molecules is maximum. So, the extent of van der Waals’ attraction among its molecules is maximum. For this reason, its boiling point is highest (36.2°C). On the other hand, the area of contact between spherical neopentane [(CH₃)₄C] molecules is minimal.

So, _ the extent of van der Waals’ attraction among its molecules is the minimum. For this reason, its boiling point is the lowest (9.5°C). Again, the area of contact between isopentane [(CH₃)₂CHCH₂CH₃] molecules is intermediate between n-pentane and neopentane and so, its boiling point (28°C) is intermediate between the other two isomers.

Class 11 Hydrocarbons Long Questions and Answers NCERT Solutions

Question 68. Write the structure and the name of the monobromoderivative which is obtained as the major product when n-butane reacts with bromine in the presence of light. Why is it produced in larger amounts?
Answer:

N-butane reacts with bromine in the presence of light to give 2-bromobutane as the major product.

The reaction occurs through the free radical mechanism. As 2° free radical (CH₃CH₂CHCH₃) is relatively more stable than 1° free radical (CH₃CH₂CH₂CH₂), displacement of 2° H-atom occurs rapidly to give 2-bromobutane as the major product

Question 68. Possible methods for the preparation of 4-methyl-2- pentyne arc are given. Which method is desirable & why?

Answer:

In both the methods given, the reaction in the second step (reddish-brown) proceeds through the S_N2 pathway and it is known that an S_N2 reaction is very susceptible to steric effect So, the product will be obtained in good yield if, in the second step, methyl or primary alkyl bromide is used. In the second step of the second method, methyl bromide (CH₃Br) has been used. Hence, the second method is desirable.

Question 70.  Three separate cylinders contain methane, and ethylene acetylene respectively. How will you identify them?
Ana.

The three gases are first separately passed through the ammoniacal solution of cuprous chloride. The gas, which gives a red precipitate, is acetylene. The gases in the remaining two cylinders are separately passed through a solution of bromine in CCl₄. The gas, which decolourises the reddish-brown solution of bromine, is ethylene. Hence, the remaining gas in the other cylinder is methane.

Question 71. Give an example alkene which on oxidation by acidic solution of KMn04 or on ozonolysis gives the same compound. Give reason.
Answer:

A terminal =CR₂ group of an alkene gets converted into a ketone when the alkene is heated with an acidic solution of KMnO₄ or subjected to ozonolysis. Hence, an example of such an alkene is 2,3-dimethyl but-2-ene

Question 72. Write the formulas and names of the alkenes which on hydrogenation form 2-methylpentane.
Answer:

The carbon skeleton of the probable alkenes is

As there are four different positions of the double bond in the given carbon skeleton, four alkenes are possible which form 2-methylpentane on hydrogenation. The probable alkenes are:

Question 73. Write two possible methods of preparing 2-methylpropane by Corey-House synthesis. Out of these two methods, which one is better and why?
Answer:

Two possible methods of preparing 2-methylpropane by Corey-House synthesis are as follows-

In Corey-House synthesis, the third step is an S_N2 reaction (sensitive to steric effect). So, this step is highly favourable for methyl or primary halides, and less’ favourable for secondary alkyl halides arid1 does not occur in the case of tertiary alkyl halides. In methods (1) and (2), a secondary halide and methyl halide have been used respectively In ease of the third step. So, method

(1) Is better than the method

(2) for preparing 2-methyl propane by Corey-House synthesis.

Question 74. A or B  A and B are the two geometrical isomers. Identify them.
Answer:

The alkene which gives only acetaldehyde on ozonolysis is 2-butene (CH₃– CH —CH CH₃).

CH₃CH=O+O=CHCI ⇒ CH₃CH=CHCH₃

So, A and B are the two geometrical isomers of 2-butene:

Question 75. Dlazomethane (CH₂N₂), on decomposition forms singlet methylene (: CH₂) which gets attached to different non-equivalent C—H bonds of alkanes to form various alkanes. Name the alkanes formed when pentane (CH₃CH₂CH₂CH₂CH₃) reacts with singlet methylene. Assuming methylene to be highly reactive and less selective, calculate the probable amounts of the formed alkanes.
Answer:

Three alkanes are formed when pentane reacts with singlet methylene because there are three non-equivalent C—H bonds in pentane molecules. So, the alkanes formed are:

As methylene is highly reactive and less selective, its insertion occurs randomly. So, the amounts of the formed compounds have calculated the basis of the probability factor and number of equivalent C— H bonds. For example, Percentage of hexane,

(CH₃CH₂CH₂CH₂CH₂CH₃) = \(\frac{6}{12}\) × 100 = 50

Percentage of 2-methyl pentane

CH₃—CH – CH₃—CH₂CH₂CH₃  =\(\frac{6}{12}\) × 100 = 50

percentage of 3-methyl pentane = \(\frac{2}{12}\) × 100 = 16.7

Question 76. How will you prepare (CH₃)₂CD¹⁴ CH₃ from propane (CH₃CH₂CH₃)?
Answer:

NCERT Class 11 Chemistry Hydrocarbons Long Question and Answer PDF

Question 77. How will you prepare \({ }^{14} \mathrm{CH}_3{ }^{14} \mathrm{CH}_2{ }^{14} \mathrm{CH}_3\) taking \({ }^{14} \mathrm{CH}_3 \mathrm{I}\) as the only source of carbon?
Answer:

Question 78. In the reaction of 2-pentene with HI, the two isomeric iodopentanes are produced in almost equal amounts —why?
Answer:

The two doubly bonded carbon atoms in 2-pentene are bonded to the same number (one) of H-atoms. So, the two isomeric iodopentanes are produced in nearly equal amounts.

The two 2° carbocations (CH₃CH₂CH₂CHCH₃ and CH₃CH₂CHCH₂CH₃ ) obtained on the addition of proton at C-2 or C-3 are almost equally stable. So, the reaction proceeds through the two routes nearly at the same rate and consequently, the two isomeric iodopentanes are formed in nearly equal amounts.

Question 79. From the following two reactions, arrange HC = CH, NH₃ and H₂O in the increasing order of their acidic character.

1. HC = CH + NaNH₂→ HC = CNa + NH₃
2. HC = CNa + H₂O → HC = CH + NaOH

Answer:

In reaction no. , HC ≡ CH exhibits its acidic character and produces NH₃ from NaNH₂. So HC = CH is more acidic than NH₃.

On the other hand, in reaction (2), water exhibits its acidic character and produces HC = CH from HC = CNa. So, H₂O is more acidic than HC = CH. Thus, the increasing order of acidic character: NH₃ < HC=CH < H₂O

Question 80. Unlike acetylene, ethylene dissolves in concentrated sulphuric acid—why?
Answer:

In the first step of the reaction with concentrated H₂SO₄, ethylene forms an ethyl cation (CH₃C⁺H₂) and acetylene forms a vinyl cation (CH₂ =⁺CH) by accepting a H® ion.

Since vinyl cation is less stable than ethyl cation, in the case of acetylene, the first step (rate-determining step) of the reaction does not occur easily. Thus, unlike ethylene, acetylene fails to dissolve in concentrated H₂SO₄

Question 81. Write the structure of the product expected to be formed when CH₂=CH—CH₃(C = 14C) is subjected to free radical chlorination.
Answer:

Question 82. Identify the products obtained when ethylene gas is passed through bromine water in the presence of sodium chloride.
Answer:

Question 83. Which alkenes are formed by dehydrating the following alcohols in the presence of acid? Give the reaction mechanism.

Answer:

Question 84. The conjugated dienes are more reactive than alkenes which in turn are more reactive than alkynes towards electrophilic addition reactions —explain.
Answer:

The reactivity of alkenes, alkynes or conjugated dienes towards electrophilic addition reaction depends on the stability of the intermediate carbocation obtained in the rate-determining step by addition of the electrophile (E⁺).

Out of the three carbocations (la, Ila and IUa), (IlIa) is the most stable because it is stabilised by resonance. Again, out of (a) and (Ila), (IIla) is less stable because the positive charge in it is placed on a more electronegative sp2 -hybfiflis6d carbon atom.

Thus, the stabilities of these carbocations follow the order IIIa> la > Ila. Therefore, the order of activity of these compounds is conjugated diene > alkene > alkyne.

Chapter 13 Hydrocarbons NCERT Solutions Long Questions Class 11

Question 85. Calculate the resonance energy of 1,3-butadiene from the following data

Answer:

The heat liberated due to hydrogenation of one double bond = 30 kcal – mol^-1

The heat liberated due to hydrogenation of two double bonds = 30 × 2 = 60 kcal – mol^-1.

Heat liberated due to hydrogenation of 1,3-butadiene (CH=CH—CH=CH₂) = 57 kcal -mol^-1.

Therefore, resonance energy of 1,3-butadiene = 60- 57 = 3 kcal – mol^-1

Question 86. Dehydration of alcohols to alkene is carried out by treating with a cone. H₂SO₄ but not with cone. HCl or HNO₃. Give reasons.
Answer:

Dehydration of alcohol proceeds via the formation of a carbocation intermediate. If HCl is used as the dehydrating agent then chloride ion (Cl^–), being a good nucleophile, attacks at carbonium ion carbon (Cl⁺) thereby producing alkyl chloride as the substitution product together with the alkene as the elimination product.

If cone. H₂SO₄ Is used as the reagent the H₂SO₄ ion derived from H₂SO₄ does not act as a nucleophile. Instead, the carbocation loses a proton from the β -carbon atom to give alkene (R—CH=CH₂) as the elimination product.

If cone. HNO₃ is used as the reagent then it being a strong oxidising agent, brings about oxidation of the alcohol first to an aldehyde or a ketone and then to a carboxylic acid.

Question 87. How will you prepare ethylbenzene by using ethyne as the only organic substance and any other inorganic substance of your choice?
Answer:

Ethylbenzene (C₆H₅C₂H₅) may be prepared from ethyne (acetylene) through the following steps:

Question 88. Explain why the bromination of benzene requires FeBr₃ as a catalyst, while the bromination of anisole (C₆H₅OCH₃) does not require any catalyst.
Answer:

Since the benzene molecule is not so reactive,© for bromination it requires more reactive bromine cation (Br) or the complex Br— Br—FeBr₃ as the electrophile. Due to the presence of electron-donating (+R) methoxy (— OCH₃) group.

The anisole ring becomes much more reactive towards an electrophilic substitution reaction. When the non-polar bromine molecule comes in contact with the anisole ring, it 6+ 6 — becomes partially polarised (Br— Br)and its positive end (weak electrophile) undergoes easy attack by anisole. Therefore, due to the greater reactivity of anisole, its bromination requires no catalyst.

Question 89. Neither vinyl chloride (CH₂=CH—Cl) nor chlorobenzene (C₆H₅ —Cl) can be used as an alkylating agent in the Friedel-Crafts reaction—why?
Answer:

In vinyl chloride or chlorobenzene, the unshared pair of electrons on the Cl atom is involved in resonance interaction with the σ -electron system and as a result, the C—Cl bond in both cases acquires some double bond character.

The Lewis acid AlCl₃ is incapable of breaking such a strong C—Cl bond. Moreover, even if the C— Cl bond breaks, the carbocations produced would be unstable (due to a positive charge on an sp² – hybridised carbon atom).

Hence such a bond is very much reluctant to undergo cleavage. For this reason, vinyl chloride or chlorobenzene cannot be used as an alkylating agent in the Friedel-Crafts reaction.

Question 90. Two methods for the preparation of propylbenzene are given below-

Question 91. Which one of the two methods is more effective for the preparation of propylbenzene? Give reason.
Answer:

Method 2 is more effective for the preparation of propylbenzene. This is because, in method 1, the alkylating agent containing a chain of three carbon atoms isomerises to give isopropyl benzene as the principal product.

Moreover, the alkyl group activates the benzene ring towards further substitution. So, there is a possibility of polyalkylation of benzene. However, although method 2 involves two steps, the desired propylbenzene is obtained as the only product with a higher yield. In the first step of the reaction.

The CH₃CH₂CO — group is introduced into the ring. Since the acyl group has no possibility of isomerisation, no other isomeric group can enter the ring.

Furthermore, the acyl group being an electron-attracting one deactivates the ring and consequently, polyacylation cannot take place. Hence in the first step, only propiophenone (C₆H₅COCH₂CH₃) is produced and in the second step, it is reduced by Clemmensen method to give only propylbenzene.

Question 92. Which ring (A or B) is in each of the following? Compounds will undergo nitration readily and why?

Answer:

• Ring A is attached to the electron-donating or activating group — :O: COPh, whereas ring B is attached to the electron-attracting or deactivating — COOPh group. So, ring A is more reactive than ring B towards electrophilic substitution reaction. Hence, ring A will undergo nitration readily.
• Since a deactivating — N02 group is attached to ring A, it is relatively less reactive than ring B towards electrophilic substitution. Consequently, ring B undergoes nitration readily.
• Ring A is attached to an electron-donating or activating — CH₃ group while ring B is linked to an electron-attracting or deactivating — CF₃ group. So, ring A is relatively more reactive than ring B towards electrophilic substitution. Hence, ring A undergoes nitration at a faster rate.
• Ring B undergoes nitration readily. The reason is similar to that given in the case of compound (1).

Question 93. Write the names and structures of the compounds formed during the Friedel-Crafts reaction of benzene with

1. CH₂Cl₂
2. CHCl₃ and
3. CCl₄

Answer:

Question 94. Show the formation of the electrophile in each case:

1. Cl₃/AlCl₃
2. Br₂/Fe
3. Conc.HNO₃ + conc.H₂SO₄

Answer:

Question 95. How will you prepare the following compounds from benzene?

1. PhCH₂CH₂Ph,
2. PhCH₂CH₂CH₂Ph and
3. PhCH₂CH₂CH₂CH₂Ph

Answer:

Question 96. An optically active compound A (C₁₀H₄) gets oxidised to benzoic acid (C₆H₅COOH) by alkaline KMnO₄ However, compound B, which is an optically inactive isomer of A does not get oxidised by alkaline KMn04. Identify A and B.

Answer:

As A is oxidised to C₆H₅COOH, A is a substituted benzene which has only one side chain consisting of four carbon atoms. Again, as A is optically active, there must be an unsymmetric carbon atom present in the side chain.

So, the side chain is —CH(CH₃)CH₂CH₃ and A is sec-butylbenzene, C₆H₅CH(CH₃)CH₂CH₃. B, an isomer of A does not get oxidised by alkaline KMnO₄.

Thus, there is no benzylic hydrogen in the compound. So, the side chain is — C(CH₃)₃. The compound B is tert-butylbenzene, C₆H₅C(CH₃)₃

Question 97. Considering the stability of <r -complex, explain why — OCH₃ is o-lp- orienting while —NO₂ is mega-orienting.
Answer:

Electrophilic substitution reaction in anisole proceeds via the following reaction mechanism:

There is an extraordinarily stable (every atom has its octet fulfilled) resonance structure in both ortho- and para-a -complex, but there is no such resonance structure in the meta-σ- complex. So, ortho- and para-cr -complex is more stable than meta-cr complex. Consequently, electrophilic substitution proceeds easily and rapidly via ortho- and para-a -complex resulting in ortho and para-substituted compounds as major products. Thus, — OCH₃ is ortho-/para-orienting group.

Electrophilic substitution reaction in nitrobenzene proceeds via the following reaction mechanism:

Both ortho- and para-complexes are extraordinarily unstable resonance structures (due to the presence of a positive charge on two adjacent atoms). However, in meta-σ -complex there is no such resonance structure and so it is more stable than ortho- and para- σ -complex.

Thus, the reaction proceeds rapidly via the meta- σ -complex and the meta- substituted compound is obtained as the major product Thus, — NO is /nefa-orienting group.

Question 98.  Identify (A)-(F) in the following reaction

Answer:

Question 99.

1. What will be the major product when propyne is treated with aqueous H₂SO₄? Explain the equation.

2. An organic compound (A), C₇H₈O is insoluble in aqueous NaHCO₃ but soluble in NaOH. (A), on treatment with bromine water rapidly forms compound (B), C₇H₅OBr₂. Give structures of (A) & (B). What will be (A) if it does not dissolve in NaOH solution but shows the reaction given above?

Answer:

Propyne does not react with aqueous H₂SO₄ in the absence of Hg²⁺ ion. In the presence of an Hg²⁺ ion, propyne reacts with aqueous H₂SO₄ to give the unstable compound prop-2-enol (according to Markownikoff’s rule) which tautomerism to give acetone.

The problem is solved by assuming that the compound ‘B’ has the molecular formula C₇H₆OBr₂

Long Answer Questions on Hydrocarbons Class 11 NCERT

Question 100.  Write the structural formula of the compounds A to F:

Answer:

Question 101. Both Br₂(g) and NO₂(g) are reddish-brown gaseous substances. How will you chemically distinguish between them?
Answer:

Question 102. Draw the structural formula of the compound from A to F.
Answer:

Question 103. Convert:

1. 2-propanol → 1-propanol
2. 2-butene→Ethane

Answer: 

Question 104. Write the IUPAC names of the following compounds:

Answer:

Question 105. For the Riven compounds write structural formulas and IUFPAC names for all possible isomers having the number of double or triple bonds as indicated:

1. C₄H₈ (one double bond)
2. C₅H₈ (one triple bond)

Answer:

Question 106. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

1. Pent-2-ene
2. 3,4-dimethyIhept-3-ene
3. 2-ethyl but-1-ene
4. I-phenyl but-1-ene

Answer:

Question 107. Explain why the following systems are not aromatic.

Answer:

1.

There are no p-orbitals on one of the H H CH₃H CH₃ carbon atoms forming the ring structure of this system and It is not a cyclic conjugated polyene containing (4n + 2)n -electrons. So, the system is not aromatic.

2.

There are no p-orbitals on one of the carbon atoms forming the ring structure of this system and it is not a cyclic conjugated polyene containing (4n + 2)n -electrons. So, the system is not aromatic.

3.

Cyclooctatetraene has a non-planar structure and there are 8π -electrons in it. So, cyclooctatetraene is a non-aromatic compound.

Question 108.  How will you convert benzene into

1. p-nitrobromobenzene
2. m-nltrochlorobenzene
3. p-nitrotoluene
4. Acetophenone

Answer:

Hydrocarbons Class 11 Long Question and Answers NCERT

Question 109. In H₃C—CH₂—C(CH₃)₂—CH₂CH(CH₃)₂, identify 1°, 2°, and 3° carbon atoms and give the number of H atoms bonded to each one of these.
Answer:

Number of H-atoms attached to 1° carbon atom = 15

Number of H-atoms attached to 2° carbon atom = 4

Number of H-atoms attached to 3° carbon atom = 1

Question 110. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support the Kekule structure for benzene?
Answer:

As the products A, B and C cannot be obtained from any one of the two Kekule structures, this confirms that o-xylene is a resonance hybrid of the two Kekule structures 1 and 2.

Question 111. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also, give a reason for this behaviour.
Answer:

The hybridisation state of carbon in the compounds benzene, n-hexane and ethyne is as follows—

The nucleus. Thus, the correct order of decreasing acidic behaviour is ethyne > benzene > n-hexane.

Question 112. How would you convert the given compounds into benzene?

1. Ethyne
2. Ethene
3. Hexane

Answer:

Question 113. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Answer:

The structural formula of 2-methyl butane is—

The structures of different alkenes by putting double bonds at different positions along with satisfying the tetravalency of each carbon atom which gives 2-methyl butane on hydrogenation are as follows-

Question 114. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E⁺.

1. Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
2. Toluene, p-H₃C —C₆H₄—NO₂, P-O₂N—C₆H₄—NO₂

Answer:

The electron density of the benzene nucleus increases in the presence of an electron-donating group (activating group). Consequently, electrophiles can easily attack the benzene nucleus. On the other hand, the electron density of the benzene nucleus decreases in the presence of the electron-withdrawing group (deactivating group). This makes electrophilic substitution difficult for the benzene nucleus.

Therefore, the order of the different compounds according to their decreasing relative reactivity with an electrophile E+ is—

Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene

Toluene > p-CH₃C₆H₅NO₂ > p-O₂NC₆H₄NO₂

CBSE Class 11 Chemistry Notes For Chapter 2 Structure Of Atom Discovery Of Fundamental Particles

The atomic theory of matter was first proposed by Sir John Dalton (an English scientist) in 1808. His theory, called Dalton’s atomic theory was a landmark in the history of chemistry. According to this theory, the atom is the smallest, indivisible, discrete particle of matter, which takes part in chemical reactions.

However, the researches done by eminent scientists like J. J. Thomson, Goldstein, Rutherford, Chadwick, Bohr, and others towards the end of the 19th century and in the beginning of the 20th century have conclusively proved that atoms were no longer the smallest indivisible particle. Rather, atoms are -3 composed of several smaller particles called subatomic particles. At present J. scientists have identified about 35 different subatomic particles, some of L which are stable while the others are unstable. These particles may be divided under three heads which is shown in the adjacent table.

The three subatomic particles namely electrons, protons, and neutrons are the main constituents of an atom and are regarded as the fundamental particles

Cathode rays: Discovery of electron 

William Crookes, in 1878, studied the conduction of electricity through gases. A discharge tube was filled with a gas at very low pressure (0.01 mm Hg) and the electrodes were connected to a source of high voltage \(\approx 10^4 \mathrm{~V}\).

It was observed that the glass wall behind the anode began to glow with a faint greenish light called fluorescence.

Further investigations revealed that this fluorescence was due to the bombardment of the glass wall by certain invisible rays which were emitted from the cathode surface and moved towards the anode with tremendous speed.

The rays originated from the cathode and were called the cathode ray.

CBSE Class 11 Chemistry Notes Chapter 2 Structure of Atom

Nature of cathode rays: J.J Thomson (in 1897) and others characterized these rays based on different experimental findings.

1. Cathode rays are emitted perpendicularly from the cathode surface and travel towards the anode.
2. They cast a sharp shadow on any opaque object placed in their path. So, like ordinary light, cathode rays also travel in straight lines.
3. When a light paddle wheel (made of mica) is placed in their path, the wheel begins to rotate. This indicates that cathode rays are composed of material particles and possess momentum.
4. As cathode rays possess momentum, they can penetrate thin foil metals like aluminum.

1. Cathode rays are deflected by an electric field (towards the positive plate) as well as by a magnetic field, suggesting that cathode rays consist of negatively charged particles.
2. When these rays strike a metal foil, the foil gets heated indicating that cathode rays produce a heating effect.
3. Cathode rays ionize the gas through which they pass.
4. Like ordinary light, cathode rays affect photographic plates. This is called fogging.
5. Cathode rays produce fluorescence on the glass walls of a tube or a screen coated with zinc sulfide 1 ‘(ZnS) or barium platinocyanide Ba[Pt(CN)6].
6. Cathode rays produce X-rays when they strike against the surface of hard metals like tungsten, molybdenum, etc.
7. The characteristics of cathode rays are independent of the material of the cathode nature of the gas used in the discharge tube.
8. Considering the various properties of cathode rays, J. J. Thomson concluded that cathode rays are made of material particles, cathode rays carry a negative charge.
9. He named these negatively charged particles as negatron. Later, these particles were named electrons by G. J. Stoney (in 1874).

J. J. Thomson (1897) used discharge tubes fitted with electrodes made of different metals and filled different gases in the tube. Every time he found that the ratio of charge to mass of electrons (e/m) was the same.

e/m of electron = \(\frac{\text { charge of cathode ray particle }}{\text { mass of cathode ray particle }}\)

= \(1.76 \times 10^8 \mathrm{C} \cdot \mathrm{g}^{-1}\)

= \(1.76 \times 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1}\)

R. A. Millikan (1917) with the help of his oil drop experiment, determined the charge of an electron. Charge ofan electron (e) = -1.602 x 10^–19 C (or, -4.8 x 10^–1°esu)

No other fundamental particle is known to contain a charge lower than this. This is the minimum measurable quantity of negative charge.

The quantity of electrical charge carried by all negatively charged panicles is an integral multiple of this charge.

Hence, the electronic charge is considered to be die fundamental unit of electricity and is called one unit. The mass of an electron can be calculated from the values of e and e/m.

Mass of an electron = \(\frac{e}{e / m}=\frac{1.602 \times 10^{-19} \mathrm{C}}{1.76 \times 10^8 \mathrm{C} \cdot \mathrm{g}^{-1}} \)

= \(9.11 \times 10^{-28} \mathrm{~g}=9.11 \times 10^{-31} \mathrm{~kg}\)

= 0.000548

1 u = \(1.66 \times 10^{-24}\)

Mass of an electron = \(\frac{1}{1837}\) x mass of a hydrogen atom Thus, a hydrogen atom is 1837 times heavier than an electron. The mass of an electron being very small may be considered as negligible for all practical involving chemical calculations.

Therefore, an electron may be defined as a subatomic particle having a unit negative charge (1.602 × 10^-9C) & negligible mass (9.11 × 10^-28g).

The radius of an electron = 2.8 × 10^-13cm

Electrons are universal constituents of matter:

The elm ratio of negatively charged particles constituting the cathode rays was found to be the same irrespective of the nature of the cathode and the nature of the gas used in the discharge tube, thereby showing that the electrons are the basic constituents of all atoms.

Other experiments showing the existence of electrons:

The following experiments show that the same charge-to-mass ratio exists for the electrons emitted—

1. Spontaneously from radioactive substances in the form of X-rays.
2. when ultraviolet rays are incident on the surface of active metals (Example: Na, K, etc.).
3. When certain metal filaments are heated to a very high temperature.
4. When any form of matter is exposed to X-rays.

Origin of cathode rays in discharge tube:

1. On applying high voltage, electrons are first emitted from the surface of the cathode which travel in straight lines with high speed.
2. During their passage through the gas inside the tube, more electrons are ejected due to the bombardment of the gas molecules by the high-speed electrons.
3. On increasing the voltage in the discharge tube, the speed of the electrons increases, and the electron density in the cathode ray increases.

Applications of cathode ray tube:

In picture tubes of televisions:

The picture tube of a television is a cathode ray tube. Due to fluorescence, a picture is formed on the television screen when an electron beam strikes the screen coated with a fluorescent or phosphorescent material.

In fluorescent tubes:

These types of tubes are filled with gases like argon, nitrogen, etc. along with a small amount of mercury vapor under very low pressure.

• They are cathode ray tubes with their inner walls coated with a suitable fluorescent material.
• On passing electric current, electric discharge occurs inside the tube.
• As a result, electrons from the cathode are transferred to the higher energy state due to collision with the atoms and molecules of the gases or the mercury vapor.
• On their return to the ground state photons are emitted in the form of UV rays. These rays hit the inner walls of the tube and produce visible light.

Anode rays: Discovery of proton

• Since negatively charged electrons are the essential constituents of all atoms and the atom as a whole is electrically neutral, it was thought that some positively charged particles must also be present in the atom.
• Goldstein, in 1886, performed the discharge tube experiment using a perforated cathode. On passing electric discharge at low pressure, he observed that some luminous rays were emitted from the side of the anode which passed through the holes in the cathode and produced fluorescence on the glass wall coated with zinc sulfide.
• These rays were originally called canal rays as they passed through the holes in the cathode.
• As they travel from the side of the anode towards the cathode, these rays are also called anode rays.
• These were named positive rays (since they were found to carry a positive charge) by J. J. Thomson (1907).

Characteristics of anode rays:

• Anode rays travel in straight lines but their speed is much less than that ofthe cathode rays.
• They consist of material particles.
• They are positively charged as indicated by the direction of their deflection in the electric and magnetic fields.
• From the extent of deflection in the electric field, it was proved that the particles constituting the anode rays arc are much heavier than electrons.
• The E/M value of the particles in the anode rays is much smaller than that of the cathode ray particles.
• Furthermore, the e/m value of the particles depends on the nature ofthe gas taken in the discharge tube.
• The E/M value of anode ray particles is maximum when hydrogen gas (the lightest element) is used in the discharge tube.
• The mass of the particle with this maximum e/m value is almost the same as that of an H-atom and its charge is equal to that of an electron but opposite in sign.
• It may therefore be concluded that such anode ray particles with maximum e/m value are none other than H-atoms devoid of electrons. These were called protons by Rutherford (1911).

Anode rays produced using H₂ gas in the discharge tube consist of positively charged particles:

H⁺ [highest e/m ], D⁺, H⁺, HD⁺, and D⁺ [lowest e/m].

Origin of anode rays:

• On applying high voltage between the electrodes in the discharge tube, cathode ray particles (i.e., electrons) move at a high speed toward the anode.
• In the course of their motion, they collide with the gas molecules or atoms and knock out one or more electrons to produce positively charged ions, thereby constituting anode rays.

⇒ \(\text { A (neutral gaseous atom) } \stackrel{\text { ionised }}{\longrightarrow} \mathrm{A}^{+}+e\)

• Thus, the anode rays are not emitted from the anode but produced from the gaseous substance present in the discharge tube.
• So it is clear that the e/m ratio of the ray particles depends on the gaseous substance in the discharge tube. For example, the use of hydrogen to the formation of H+ ions which constitute the orhunode rays.

⇒ \(\mathbf{H} \text { (atom) }-\boldsymbol{e} \stackrel{+}{\longrightarrow} \mathbf{H}^{+} \text {(proton) }\)

E/m of proton = +9.58× 10⁴C.g^-1

Charge of proton = + 1.602×10^-19 C (Or, +4.8 × 10^-10esu)

Mass of proton (m) = \(\frac{\text { charge of a proton }}{c / m \text { of a proton }}\)

= \(\frac{1.602 \times 10^{-19}}{9.58 \times 10^4}\)

= \(1.6725 \times 10^{-24} \mathrm{~g}\)

This mass is nearly the same as that of an H-atom. A proton Is 1836 times heavier than an electron. So, a proton carries 1 unit of +ve charge and it is 1 836 times heavier than an electron.

Radius of a proton \(\approx 1.2 \times 10^{-13} \mathrm{~cm}\)

Proton is a fundamental constituent of all atoms:

• Taking different gases in the discharge tube experiment, it was shown that the mass of anode ray particles is minimal when hydrogen is used as the gaseous substance.
• Masses of other anode ray particles produced in different experiments using different gases are always integral multiples of the mass of a proton.
• Thus a proton, like an electron, is also a fundamental constituent of all atoms.
• Apart from the electrical discharge in gases under low pressure, protons are also emitted in certain nuclear reactions

Example: The bombardment of aluminum metal with a particle

⇒ \(\left({ }_2^4 \mathrm{He}\right)\) or bombardment of nitrogen gas with neutron \(\left({ }_0^1 n\right)\)

⇒ \({ }_{13}^{27} \mathrm{Al}+{ }_2^4 \mathrm{He} \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_1^1 \mathrm{H} ; \quad{ }_7^{14} \mathrm{~N}+{ }_0^1 n \rightarrow{ }_6^{14} \mathrm{C}+{ }_1^1 \mathrm{H}\)

Differences Between Cathode And Anode Rays

Radioactivity:

After the discovery of electrons and protons, it is well established that the atoms are divisible into sub-atomic particles.

• This was further supported by the phenomenon of radioactivity, discovered by Becquerel in 1896.
• The phenomenon of spontaneous emission of active radiations by certain elements like uranium, radium, etc.
• Is called radioactivity, and the elements emitting such radiation are called radioactive elements.

On placing a sample of uranium mineral in a cavity made in a block of lead and allowing the emitted rays to pass through strong electric or magnetic fields, the radiation is resolved in three directions.

• The rays which are deflected slightly towards the negative plate and hence carry + ve charge are called α rays. The particles present in a -rays are called particles, α- rays are called α – particles, and each particle carries 2 units +ve charge and has a mass of 4u. So these are helium atoms with two units +ve charge and are represented as ₂⁴He.
• The rays that are deflected towards the positive plate to a larger extent must carry a -ve charge. These are called α-rays. The particles present in β -rays are called β -P-particles. These particles have the same charge and mass as that of the electrons and are represented as _-1⁰e.
• The rays that remain undeflected are called γ-rays. These are purely electromagnetic radiations.