NCERT Class 11 Chemistry Chapter 10 S Block Elements Long Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 10 S Block Elements Long Question And Answers

Question 1. How can anhydrous magnesium chloride be prepared from magnesium chloride hexahydrate?
Answer:

Anhydrous magnesium chloride cannot be prepared by simply heating MgCl₂-6H₂O because it gets hydrolyzed by its water of crystallization.

However, when hydrated MgCl₂ is heated at 650K in the presence of HCl, its hydrolysis is prevented; and it loses its water of crystallisation to form anhydrous MgCl₂

Read and learn More NCERT Class 11 Chemistry

However, when hydrated MgCl₂ is heated at 650K in the presence of HCI, its hydrolysis is prevented, and it loses its water of crystallization to form anhydrous MgCl₂.

Question 2. Which out of BeCl₂ and CaCl₂ would give an acidic solution when dissolved in water?
Answer: 

Being a covalent compound and a good Lewis acid, BeCl₂ forms a hydrated salt, Be(H₂O₂)4Cl₂. The hydrated ion undergoes hydrolysis in solution producing H₃O⁺. This occurs because the Be — O bond is very strong and so in the hydrated ion this weakens the O —H bond. Hence, there is a strong tendency to lose protons. For this reason, the aqueous solution of BeCl₂ is acidic.

[Be(H₂O)₄]²⁺ + H₂O ⇌ [Be(H₂O)₃OH]⁺ + H₃O⁺

On the other hand, CaCl₂ is a strongly ionic compound and does not behave as a Lewis acid (the size of Ca is relatively large and its octet is filled up). Moreover, since it is a salt of strong acid and strong base, it does not undergo hydrolysis and therefore, its aqueous solution is neutral.

Question 3 Explain The below Observation

1.

2.

Answer:

1. The smaller Li⁺ ion exerts strong polarising power and distorts the electron cloud of the nearby oxygen atom of the OH^– ion. This results in the formation of a strong Li —O bond and the weakening of the O —H bond. This ultimately facilitates the decomposition of LiOH into Li₂O and H₂O. The polarising power of the large Na+ ion is much lower and thus, NaOH remains unaffected by heating.

2. The smaller Li⁺ ion exerts a strong polarising power on highly polarisable H^– ion and as a result, the two atoms remain strongly attached by a covalent bond. On the other hand, due to the low polarising power of Na⁺ ion, NaH is essentially ionic & so it dissociates on heating to yield metallic sodium and dihydrogen.

Question 4. Discuss the roles of Na₂O₂, KO_2, and LiOH in the purification of air.
Answer:

Sodium peroxide (Na₂O₂) is used to purify the air in submarines and confined spaces as it removes carbon dioxide (CO₂) and produces O₂.

Potassium superoxide (KO₂) is used to purify the air in space capsules, submarines, and breathing masks because it can absorb carbon dioxide (CO₂) thereby removing it and producing O₂ Both functions are important life support systems.

Lithium hydroxide (LiOH) is used for the absorption of carbon dioxide (CO₂) in space capsules and submarines.

Question 5. Lithium forms monoxide while sodium forms) peroxide in the presence of excess oxygen why
Answer:

Larger cations can be stabilized by larger anions because if both the ions are comparable in size, the coordination number will be high and this gives rise to a high lattice enthalpy. Lithium-ion, Li⁺ as well as oxide ion, O^2-, have small ionic radii and high charge densities.

Hence these small ions combine and form a very stable lattice of lithium monoxide (Li₂O). Similarly, the formation of sodium peroxide (Na₂O₂) can be explained based on the stable lattice formed by the packing of relatively large Na⁺ ion and peroxide ion O₂

Question 6. A freshly cut piece of sodium metal appears shiny but its metallic lustre soon gets tarnished when exposed to air. Give reason
Answer:

When a freshly cut piece of metallic sodium is exposed to moist air, it readily reacts with oxygen to form sodium monoxide (Na₂O). The resultant sodium monoxide and also the metal itself readily react with the moisture of the air to form sodium hydroxide (NaOH).

In the subsequent step, both Na₂O and NaOH combine with CO₂ of air to form sodium carbonate (Na₂CO₃). Thus, a coating of sodium carbonate is formed on the surface of the metal and as a result of this, the metallic luster is tarnished.

4Na + O₂→ 2Na₂O; Na₂O + H₂O →2NaOH

2Na + 2H₂O→2NaOH + H₂↑; Na₂O+ CO₂→Na₂CO₃

2NaOH + CO₂ →Na₂CO₃ + H₂O

Question 7. What happens when each of the following compounds is heated? 

1. Li₂CO₃ 
2. Na₂CO₃
3. LiNO₃
4. KNO₃

Answer:

1. Lithium carbonate (Li₂CO₃ ) decomposes readily on heating to give lithium monoxide (Li₂O) and carbon dioxide (CO₂).

2. Na₂CO₃ does not decompose on heating.

3. Lithium nitrate (LiNO₃) decomposes on heating to form lithium monoxide (Li₂O), nitrogen dioxide (NO₂), and dioxygen (O₂).

4.  Potassium nitrate (KNO₃) decomposes on heating to give potassium nitrite (KNO₂) and dioxygen

Question 8. What happens when

1. HCl gas is passed through a concentrated solution of common NaCl-containing impurities like Na₂SO₄, CaSO₄, CaCl₂, MgCl_2, etc.
2. Caustic soda beads are exposed to air for a long time.
3. How will you convert Na₂CO₃ into NaHCO₃ and vice versa?

Answer:

1. When HCl gas is passed through a concentrated solution of common NaCl with impurities, crystals of pure NaCl separate out because of the common ion effect,  Caustic soda (NaOH) is a deliquescent substance and becomes wet on exposure to air.

On long exposure, the solid beads dissolve in the absorbed water Moist caustic soda then absorbs carbon dioxide (CO₂) from the air to form sodium carbonate (Na₂CO₃) which forms a coating over the surface of the material. As sodium carbonate (Na₂CO₃) is not a deliquescent substance the wet sodium hydroxide becomes dry again.

2NaOH + CO₃→ Na₂CO₃+ H₂O

2.  Sodium bicarbonate (NaHCO₃) can be obtained by passing carbon dioxide (CO₂) through a saturated solution of sodium carbonate. Sodium bicarbonate, being less soluble gets separated from the solution as a white crystalline substance.

Na₂ CO₃ + CO₂+ H₂O→2NaHCO₃

When sodium bicarbonate (NaHCO₃) is heated, it decomposes to give sodium carbonate (Na₂CO₃) and carbon dioxide (CO₂)

Question 9. Why are the group-2 metals harder and have higher melting and boiling points than group-1 metals?
Answer:

The magnitude of the cohesive energy determines the hardness as well as melting and boiling points of the metals and it depends on the number of electrons involved in metallic bonding. In the case of group 1 metals, one electron per atom (valence electron, ns² ) is involved in metallic bonding while in group 2 metals, two electrons per atom (valence electrons, ns²) are involved in metallic bonding. Moreover, atoms of group 2 metals are smaller in size than those of group 1 metals.

Consequently, stronger metallic bonding exists in group 2 metals which results in higher cohesive energy and close packing of the atoms. This accounts for the greater hardness and higher melting and boiling points of group 2 metals as compared to group 1 metals.

Question 10.

1. Give some common tests used for the detection of calcium compounds. 

2. A white solid When heated liberates a colorless gas that does not support combustion. The residue is dissolved in water to form (B) which can be used for whitewashing. When excess CO₂ gas is passed through the solution of (B), it gives a compound (C) which on heating forms (A). Identify (A), (B) and (C). Give the reactions
Answer:

The following tests may be performed for the detection of Ca -compounds:

1. Calcium salts give a brick-red color in the flame test,
2. When ammonium oxalate solution is added to a solution of a calcium salt, a white precipitate of crystalline calcium oxalate is obtained which is insoluble in acetic acid but soluble in mineral acids.
3. In addition to a solution of sodium carbonate to a neutral (or ammoniacal) solution of a Ca-salt, white calcium carbonate is precipitated, which is soluble in acids.
4. The observations suggest that the compound (A) is limestone, i.e., CaCO₃, the compound (B) is calcium hydroxide & the compound (C) is calcium bicarbonate.

The corresponding reactions are as:

Question 11. Compare the alkali metals and alkaline earth metals concerning their

1. Basicity of oxides
2. Solubility of hydroxides and
3. Solubility of nitrates.

Answer:

1. Basicity of oxides:

The ionization enthalpy of alkali metals is less than the corresponding alkaline earth metals. So the alkali metal oxides are more basic than the corresponding alkaline earth metal oxides.

2. Solubility of hydroxides:

Due to of small size and higher ionic charge, the lattice enthalpies of alkaline earth metal hydroxides are much higher than those of alkali metal hydroxides and hence, the solubility of alkali metal hydroxides is much higher than that of alkaline earth metal hydroxides.

3. Solubility of nitrates:

Nitrates of alkali and alkaline earth metals are soluble in water. However, the solubility of alkali metal nitrates increases down the group because their lattice enthalpies decrease more rapidly than their hydration enthalpies. Nitrates of alkaline earth metals follow the reverse trend i.e., their solubility decreases down the group and this is because their hydration enthalpies decrease more rapidly than their lattice enthalpies.

Question 12. What are the properties that make oxides of MgO and  BeO useful for lining furnaces?
Answer:

The given properties make MgO and BeO useful for lining furnaces

• They have high melting points [melting point of Beo is 2500°C (approx) and MgO is 2800°C (approx)].
• They are very good conductors of heat.
• They have very low vapor pressure.
• They are chemically inert.
• They are insulators

Question 13. The halides of alkali metals are soluble in water except for LiF. Why?
Answer:

The solubility of a salt in water depends on its lattice enthalpy as well as its hydration enthalpy. A salt dissolves in water when its hydration enthalpy exceeds its lattice enthalpy value. The lattice enthalpy of LiF is very high and its hydration enthalpy value does not exceed this value. As a result, LiF is insoluble in water.

However, the other halides of alkali metals possess higher hydration enthalpy values compared to their corresponding lattice enthalpies. Hence, they are soluble in water.

Question 14. Why Is LiCO₃ decomposed at a lower temperature whereas Na₃ CO₃ at a higher temperature
Answer: 

Li+ ion being smaller in size forms a more stable lattice with the smaller oxide ion (O^2-) than the larger carbonate ion (CO₃ ) and consequently, Li₂CO₃  decomposes into Li₂ O at a much lower temperature. The high polarising power of very small Li⁺ ions also facilitates the decomposition of Li₂CO₃.

On the other hand, the larger Na+ ion forms a more stable lattice with the larger CO₃ ion than with the smaller O^2- ion. Therefore, Na₂ CO₃ is quite stable and decomposes only at very high temperatures.

Question 15. Compare the solubility and thermal stability of the given compounds of the alkali metals with those of the alkaline earth metals. 

1. Nitrates
2. Carbonates Sulphates.

Answer:

1. Nitrates:

The nitrates of alkali metals as well as alkaline earth metals are highly soluble in wOn heating, nitrates of both alkali and alkaline earth metals undergo decomposition.

Nitrates of alkali metals decompose to form metallic nitrite and oxygen. On the other hand, nitrates of alkaline earth metals decompose to form the corresponding oxides with the evolution of NO₂ and O_2.

Due to the diagonal relationship, lithium nitrates behave similarly to magnesium nitrate.

2. Carbonates:

1. Except for Li₂CO₃, other carbonates of alkali metals readily dissolve in water. However, carbonates of alkaline earth metals are practically insoluble in water. Their solubilities decrease on moving down the group. So, BeCO₃ is sparingly soluble in water while BaCO₃ is insoluble in water.

2. Carbonates of alkali metals except Li₂CO₃ are stable and do not decompose on heating, but carbonates of alkaline earth metals decompose on heating to give metal oxide and carbon dioxide.

The thermal stability of the carbonates increases down the group. Like MgCOg, Li₂CO₃ decomposes on heating.

3. Sulphates:

1. Except Li₂SO₄, the remaining sulfates of alkali metals are water-soluble. 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,

2. The sulfates of alkali metals are stable compounds and do not decompose on heating. On the other hand, alkaline earth metals dissociate on heating to give metal oxides and sulfur trioxide.

The thermal stability of the sulfates increases down the group. Li₂SO₄ dissociates on heating just like MgSO_4.

Question 16. Starting with sodium chloride how would you proceed to prepare

1. Sodium metal
2. Sodium peroxide

Answer:.

1. Metallic sodium can be obtained by the electrolysis of a mixture of sodium chloride (40%) and calcium chloride (60 %) in a fused state. The function of calcium chloride is to lower the reaction temperature from 807°C (m.p. of NaCl) to about 577°C. The molten sodium metal thus obtained is liberated at the cathode

Overall reaction: Nacl → Na⁺ + Cl^–

At cathode: Na⁺ + e→ Na;

At anode: Cl^– → Cl + e; Cl + Cl → Cl₂ ↑

2.  Sodium chloride is first converted to sodium by electrolytic reduction. The metal is then heated more than air. The initially formed sodium oxide reacts with excess O₂ to form Na₂O₂.

Question 17. What happens when  

1. Magnesium is burnt in the air 
2. Calcium nitrate is heated?

Answer:

1. When magnesium bums in the air, magnesium oxide (MgO) and magnesium nitride (Mg₃N₂) are obtained as products

2. On heating calcium nitrate, it decomposes to form CaO, NO_2, and O₂ reaction

Question 18. The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:

Alkali metals form monovalent cations (such as Na⁺, and K⁺) while alkaline earth metals form divalent cations (such as Mg²⁺, and Ca²⁺). Due to the increase in charge of the cations, the lattice energy of the corresponding salt increases. For this reason, hydroxides and carbonates of sodium and potassium have lower lattice enthalpy values than the hydroxides and carbonates of magnesium and calcium.

As hydration enthalpies of the hydroxides and carbonates of sodium and potassium are greater than their lattice enthalpies, these salts readily dissolve in water. However, in the case of the hydroxides and carbonates of calcium and magnesium, the lattice enthalpy values are greater than that of hydration enthalpy and consequently, these salts are less soluble in water.

Question 19. Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:

As Li⁺ is the smallest ion among all the alkali metal ions, it can polarise water molecules more easily than the other alkali metal ions. So, numerous water molecules get attached to lithium salts as water of crystallization. However, this is not observed in the case of other alkali metal ions. Thus, lithium salts are commonly hydrated.

Example: LiCl-2H₂O  and those of the other alkali ions are usually anhydrous.

Question 20. Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in acetone?
Answer:

The lattice enthalpy of ionic LiF (formed by small Li⁺ion and F^– ion) is higher than its hydration enthalpy. On the other hand, the lattice enthalpy of LiCl containing small Li⁺ ion and large Cl^– ion is considerably lower than its hydration enthalpy.

Thus, LiF is almost insoluble in water while LiCl is soluble. Furthermore, Li⁺ ions can polarise bigger Cl^– ions more easily than smaller F^– ions. As a result, LiCl has more covalent character than LiF and so, it is also soluble in the organic solvent acetone.

Question 21. What happens when

1. Sodium metal is dropped in water?
2. Sodium metal is heated in a free supply of air
3. Sodium peroxide dissolves in water?

Answer:

Sodium hydroxide is formed. H₂ gas is evolved which catches fire due to the exothermicity of the reaction.

2Na(s) + 2H₂O(l)→ 2NaOH(aq) + H₂(g)

2. Sodium peroxide is formed by heating sodium metal in a free supply of air.

2Na(s) + O₂ (g)→ 2Na₂O₂(s)

3. H₂O₂ is formed when sodium peroxide dissolves in water

Na₂O₂(s) + 2H₂O(l)→2NaOH(aq) + H₂O₂(Z)

Question 22. Comment on each of the given observations:

1. Lithium is the only alkali metal to form a nitride directly.
2.  M²⁺+(aq) + 2e → M(s) (where M = Ca, Sr, or Ba) is nearly constant.

Answer:

1. The lattice energy of lithium nitride (Li₃N) which consists of a small cation (Li⁺) and a small anion (N₃-) is much higher and this energy compensates for the high bond dissociation energy of the N=N bond and the energy to form N₃– ion. Larger alkali metal ions cannot compensate for these energy requirements. Hence, Li⁺ is the only alkali metal that forms nitride directly.

2. M²⁺ (aq) + 2e → M(s) Where M = Ca, Sr, Ba

Standard electrode potential, \(E_{\mathrm{M}^{2+} / \mathrm{M}}^0\) depends on 3

1. Enthalpy of vaporisation
2. Ionization enthalpy and
3. Enthalpy of hydration.

As the combined effect of these factors is almost the same for Ca, Sr, and Ba, their E° values are nearly constant.

Question 23. State as to why

1.  A solution of Na₂CO₃ is alkaline?
2. Alkali metals are prepared by electrolysis of their fused chlorides.
3. Sodium is found to be more useful than potassium.?

Answer:

1. Na₂CO₃ is a salt of weak acid H₂CO₃ and strong base NaOH. In an aqueous solution, it ionizes to form Na⁺ and CO^2-₃ – ions.

Na₂CO₃ ⇌  2Na⁺  (aq) + CO^2-₃ (aq)

The formed CO^2-₃ ions hydrolyze in an aqueous solution to produce acetic acid and OH^– ion.

CO^2-₃ (aq) + 2H₂O(Z) ⇌ H₂CO₃ ⇌+ 2OH^– ( aq)

As H₂CO₃ is a weak acid, it remains mostly unionized. Consequently, the concentration of OH^– ions increases in the solution thereby making the solution alkaline.

2.

• As alkali metals are strong reducing agents, they cannot be extracted by chemical reduction from their oxides or other compounds,
• As alkali metals are highly electropositive, these metals cannot be displaced from their salts with the help of other elements,
• Alkali metals cannot be obtained even by the electrolysis of aqueous solutions of their salts.

In this case, H₂, instead of the alkali metal is liberated at the cathode because the discharge potential of alkali metals is higher than that of hydrogen.

Thus, to prepare alkali metals, electrolysis of their fused chlorides is carried out. For example,

NaCl →Na⁺ + Cl^–

During electrolysis, at the cathode,

2Na⁺ + 2e →2Na; and at the anode, 2Cl→ Cl₂ + 2e

3. Sodium is found to be more useful because Na is not as reactive as K. For this reason, reactions of sodium with different substances can be controlled and usage of sodium is far more safe than potassium. Thus, sodium is more useful than potassium

Question 24. Write balanced equations for reactions between 

1. Na₂O₂ & Water,
2. KO₂ & water,
3. Na₂ O& CO₂

Answer: The balanced equations of the given reactions are.

1.

2. 2K O₂ (S) + 2H₂ O(l) → 2KOH(aq) + H₂ O₂ (aq) + O₂ (g)

3. Na₂ O + CO₂→ Na₂ CO₂

Question 25. How would you explain the given observations? 

1. BeO is almost insoluble but BeSO₄ is soluble in water.
2. BaO is soluble but BaSO4 is insoluble in water.
3. Lil is more soluble than KI in ethanol.

Answer:

1. O^2- is smaller in size than SO₄^2-. Consequently, a small Be²⁺ ion is tightly packed with a small O^2- ion, and thus, the lattice enthalpy of BeO is greater than its hydration enthalpy. So BeO is insoluble in water. On the other hand, a small Be²⁺ ion is loosely packed with a large SO^2-. Ion and thus, the lattice enthalpy of BeSO₄ is less than its hydration enthalpy. So, BeSO₄ is soluble in water.

2. Large Be²⁺ ion is tightly packed with large SO₄^2-.ion and thus, the lattice enthalpy of BaSO₄ is greater than its hydration enthalpy. So, BaSO₄ is insoluble in water On the contrary, a large Ba²⁺ ion is loosely packed with a small Oion, and thus lattice enthalpy of BaO is less than its hydration enthalpy. So, BaO is water soluble.

KI is predominantly ionic. On the other hand, due to the high polarising power of very small Li+ ions, Lil is predominantly covalent. For this reason, Lil is more soluble than KI in the organic solvent ethanol.

Question 27. How will you distinguish between:

1. Mg and Ca
2. Na₂SO₄ and BaSO₄
3. Na₂CO₃ and NaHCO₃,
4. LiNO₃ and KNO₃

Answer:

1. Calcium, when heated, imparts brick brick-red color to the flame but magnesium does not.

2. Sodium sulfate (Na₂SO₄) is soluble in water but barium sulfate (BaSO₄) is insoluble.

3. Na₂CO₃ is stable to heat but NaHCO₃ decomposes on heating to produce CO₂ gas which turns limewater milky

4. When LiNO₃ is heated, it decomposes to yield reddish-brown vapors of NO₂. However, when KNO₃ is heated, it decomposes to yield colorless O₂ gas