NCERT Class 11 Chemistry Chapter 11 Some P Block Elements Short Answer Questions

NCERT Solutions For Class 11 Chemistry Chapter 11 Some P Block Elements Short Question And Answers

Question 1. Boron shows anomalous behaviour and differs from j the rest of the members of its family —why?
Answer:

Boron shows anomalous behaviour because of

1. Exceptionally small atomic size,
2. High ionisation enthalpy and
3. Absence of d -d-orbitals in its valence shell

Question 2. Give reasons for which carbon differs from the rest of the members of its family
Answer:

• Exceptionally small atomic size,
• Higher electro¬ negativity,
• Higher ionisation enthalpy and
• Absence of f-orbitals in the valence shell.

Read and Learn More NCERT Class 11 Chemistry Short Answer Questions

Question 3. Diamond is a non-conductor of electricity but a good conductor of heat—why?
Answer:

Due to the absence of free electrons, it is a non-conductor of electricity. It has the highest known thermal conductivity because thermal motion is distributed in its 3D -structure very effectively

Question 4. Explain why the melting and boiling points of boron are much higher.
Answer:

1. Boron exists as a giant covalent polymer having a three-dimensional network structure both in the solid and the liquid states.
2. For this reason, its melting and boiling points are very high;

Question 5. pπ-pπ back bonding occurs in the case of boron halides but not in the case of aluminium halides —why?
Answer:

• The tendency of pπ-pπ back bonding decreases with an increase in the size of the central atom.
• Since Al is larger than B, pn-pn back bonding does not take place in the case of Al

Question 6. (SiH₃)₃N is weaker base than (CH₃)₃N —why?
Answer:

• pπ-dπ Back bonding occurs in (SiH₃)₃N but not in (CH₃)₃N.
• Therefore, the unshared pair ofelectrons is more available in (CH₃)₃N and for this reason, it is more basic

Question 7. N(CH₃)₃ is pyramidal but N(SiH₃)₃ is planar—explain.
Answer:

• Because of d -orbital resonance, the N-atom in N(SiH₃)₃ molecule is sp² – hybridised and therefore, the molecule is planar
• No such d -d-orbital resonance occurs in N(CH₃)₃ and the central N-atom is sp³ -hybridised. For this reason, this molecule is pyramidal.

Question 8. CO gets readily absorbed in ammoniacal silver nitrate solution but CO₂ does not—explain.
Answer:

Due to the presence of an unshared electron pair on carbon, CO acting as a Lewis base, combines with ammoniacal cuprous chloride to form a stable and soluble complex. No such reaction takes place in the case of CO₂ because in it carbon has no unshared electron pair,

Question 9. Which out of anhydrous and hydrous AlCl₃ is more soluble in ether and why
Answer:

Anhydrous AlCl₃ is electron deficient but hydrous AlCl₃ is not Because of this, anhydrous AlCl₃ combines with ether through the formation of a coordinate bond and gets dissolved in

Question 10. Explain why the B—X bond distance in BX3 is shorter than the theoretically expected value.
Answer:

This is due to the pπ-pπ backbonding of the filled p orbital of halogen (X) into the empty p -orbital of boron.

As a result, the B— X bond possesses some double-bond character and hence B —X bond is shorter than expected

Question 11. Although aluminium lies above hydrogen in the electrochemical series, it is quite stable in water and air. Why?
Answer:

Aluminium is a highly reactive electropositive metal. In the presence of water and air, a thin layer of aluminium oxide (Al₂O₃) forms over the metal surface. Consequently, this protective layer prevents further reaction of aluminium with water or air.

So, aluminium is quite stable in water and air, despite being situated above hydrogen in the electrochemical series.

Question 12. Using chemical reactions show that amphoteric. aluminium is
Answer:

Aluminium reacts with acids as well as with bases. So, it is amphoteric.

For example:

Question 13. car acts as a better lubricant on the moon compared to that on earth” Justify the validity ofthe statement.
Answer:

The statement is not correct. Different substances like air, water vapour and other gaseous materials enter into the layers of graphite when it is on earth, thus enhancing its lubricating property. However, the moon is devoid of atmosphere. Thus, due to the absence of water vapour and gaseous substances, the lubricating property of graphite is quite less on the moon

Question 14. Explain why PbCl₄ is a good oxidising agent.
Answer:

Due to the inert pair effect, the Pb²⁺ ion is relatively more stable than the Pb⁴⁺ ion and so the Pb⁴⁺ ion readily gets reduced to Pb²⁺ ion by accepting two electrons. That is why PbCl₄ is a good oxidising agent.

Question 15. Why do nitrogen and carbon monoxide show similarities in their physical properties?
Answer:

Nitrogen (N₂) and carbon monoxide (CO) exhibit structural similarities because both molecules contain the same number of valence electrons (10). Because of structural similarities (similar distribution of electrons in the bonding orbital), they show striking resemblance in their physical properties like vapour density, solubility in water, boiling point, melting point, etc

Question 16. Explain why graphite is used as a solid lubricant for heavy machinery.
Answer:

• Since any two successive layers in graphite are held together by weak forces of attraction, one layer can slip over the other.
• This makes graphite soft and a good lubricating agent for heavy machinery

Question 17. Diamond is a bad conductor of electricity but a very good conductor of heat—explain.
Answer:

• There is no free electron left in the structure of the diamond made up of sp³ hybridised C-atoms and so, the diamond is unable to conduct electricity.
• On the other hand, it is a very good conductor of heat because its structure distributes thermal motion in three dimensions very effectively.

Question 18. Despite being a covalent substance, the melting point of diamond is very high—why?
Answer:

• In diamonds, there is a three-dimensional network of strong covalent bonds.
• Since a large amount of thermal energy is required for the cleavage of these bonds, the melting point of the diamond is very high.

Question 19. CO is an inflammable gas while CO₂ is not—why?
Answer:

The oxidation states of carbon in CO, and CO are +4 and +2 respectively. In any compound, the maximum oxidation state of carbon is +4. Consequently, the carbon atom in CO tends to increase its oxidation number, i.e., CO has the inherent urge to be oxidised. For this reason, it reacts readily with oxygen, i.e.,

It is a combustible gas. 2CO +O₂ →2CO₂ On the other hand, the oxidation state of carbon in CO₂ is maximum (+4) so it does not tend to be oxidised and unlike CO, it is not combustible.

Question 20. [SiF₆]^2- is known to exist whereas [CF₆]^2- does not exist. Explain.
Answer:

• Silicon can extend its coordination number beyond four because it possesses vacant d-orbitals. Hence, [SiF₆]^2-– exists.
• On the other hand, C has no vacant d-orbitals in its valence shell and thus it cannot extend its coordination number beyond four.
• Hence, [CF₆]^2-does not exist

Question 21. An aqueous solution of sodium hydroxide is added dropwise to the solution of gallium chloride in water. A precipitate is initially formed. The precipitate dissolves on further addition of NaOH solution. Explain the observation using suitable chemical reactions
Answer:

On the addition of NaOH solution to a solution of GaCl₃, a gelatinous white precipitate of Ga(OH)₃ is formed which dissolves on adding excess NaOH by forming a soluble complex.

Question 22. Define buckyball. How is it made?
Answer:

C₆₀ -fullerene is called buckyball. It contains 20 six-membered rings and 12 five-membered rings.  It is prepared by heating graphite in an electric arc in the presence of an inert gas like argon or helium.  The C₆₀ and C₇₀ fullerenes hence formed can be separated readily by extraction with benzene or toluene followed by chromatography in the presence of alumina.

Question 23. CO is readily absorbed by ammoniacal cuprous chloride solution but CO₂ is not. Explain.
Answer:

CO has a lone pair ofelectrons on C-atom. Thus, it acts as a Lewis base and forms a soluble complex with an ammoniacal CuCl solution.

CuCl + NH₃ + CO → [Cu(CO)NH₃]⁺Cl^– (Soluble)

CO₂, on the other hand, does not possess a lone pair of electrons on the C-atom. Hence, it does not act as a Lewis base. Thus, it does not dissolve in ammoniacal CuCl solution

Question 24. What is the chemical composition of the borax bead?
Answer:

Borax forms a glassy mass on heating called a borax bead

Thus, the borax bead is a mixture of sodium metaborate (NaBO₂) and boric anhydride (B₂O₃)

Question 25. Silicon in elemental form does not form a graphite-like structure. Explain.
Answer:

Silicon is larger than carbon. Thus, pn-pn bonding is not as effective in the case of Si as in the case of C-atom.  In graphite, pn-pn bonding is effective due to the smaller size of carbon.  Thus, Si does not resemble graphite. Rather, it resembles a diamond and is a poor conductor of electricity

Question 26. Anhydrous aluminium chloride cannot be prepared by heating hydrated aluminium chloride. Why?
Answer:

When hydrated aluminium chloride (AlCl₃– 6H₂O) is heated, it gets hydrolysed by the water of crystallisation and aluminium oxide is formed. Therefore, anhydrous aluminium chloride cannot be prepared by heating hydrated aluminium chloride

Question 27. Why are the dihalides of carbon unstable but the dihalides of tin and lead stable?
Answer:

All the elements of group 14 (except C and Si) form stable dihalides. Due to the inert pair effect, these elements are highly stable in the +2 oxidation state. Hence, tin and lead form stable dihalides. On the other hand, carbon is stable in the +4 state. Therefore, dihalides of carbon are unstable

Question 28.

1. Why PbO₂ is oxidising?
2. Which of the following is the thermodynamically most stable form of carbon? Coke, diamond, graphite, fullerenes.

Answer:

1. Among the group-13 elements B and Al exhibit +3 oxidation state only. On the other hand, Tl shows +3 as well as +1 oxidation states but due to the inert pair effect, it is more stable in +1 oxidation state. So, TlCl exists but AlCl does not

2. Graphite.

Question 29. PbCl₂ is less stable than SnCl4 while PbCl₂ is more stable than SnCl₂. Justify or contradict
Answer:

In the case of group-14 elements the number of d- or f- electrons increases down the group from Ge to Pb. Hence, 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 down the group. Consequently, Pb is more stable in the +2 state whereas Sn is more stable in the +4 state. Therefore, PbCl₄ is Jess stable than SnCl₄ while PbCl₂ is more stable than “SnCl₂

Question 30. What happens when at first lesser amount and then an excess amount of NaOH solution is added to the Al₂(SO₄) solution?
Answer:

When a lesser amount of NaOH is added to the solution of Al₂(SO)₃, a white precipitate of Al(OH)₃ forms. In the presence of excess NaOH, the solution becomes clear due to the formation of soluble sodium aluminate (NaAlO₂)

Question 31. Explain with reason: SnCl₂ is a solid ionic compound whereas SnCl₄ is a covalent liquid.
Answer:

SnCl₂, Sn(II): [Xe]₄d¹⁰5s¹⁰

SnCl₄, Sn(IV): [Xe]₄d10

Here with an increase in oxidation state from Sn(II) to Sn(IV) the ionisation potential of the central atom (here Sn) increases which makes the Sn—Cl bonds more covalent in SnCl₄ compared to SnCl₂ — Fajan’s rules.

Hence the SnCl₂ molecules are closely packed due to greater ionic character whereas in SnCl₄ the molecules show weaker London forces of interaction due to their covalent nature. This explains the given observation

Question 32. How can you explain the higher stability of BCl₃ as compared to TlCl₃?
Answer:

Because of the poor shielding effect on s-electrons of the valence shell by the inner d -and f-electrons (i.e., 3d, 4d, 5d, 4f-electrons), the inert pair effect is maximum in Tl. As a result, for Tl only the 6p¹ -electron becomes involved in bond formation. Hence the most stable oxidation state of TI is +1 and not +3. Therefore, TlCl is stable but TlCl₃ is unstable.

On the other hand, due to the absence of d -and f-electrons, B does not exhibit an inert pair effect and all three valence electrons become involved in bond formation. Hence, B exhibits an oxidation state of +3 and thus forms BCl₃. So, BCl₃ is more stable than TlCl₃

Question 33. Why does boron trifluoride behave as a Lewis acid?
Answer:

The B-atom in the BF₃ molecule has only she electrons in its valence shell and thus two more electrons are required to complete its octet. Therefore, BF₃ can easily accept a pair of electrons from basic substances such as NH₃, (C₂H₅)₂O etc. and thus behaves as a Lewis acid.

Question 34. BF₃ is reacted with ammonia?
Answer:

Being a Lewis acid, BF₃ accepts a pair of electrons from NH₃(a Lewis base) to form a complex

Question 35. Aluminium chloride exists as a dimer, but boron trichloride does not. Explain
Answer:

The boron atom is so small that it cannot accommodate four large-sized Cl-atoms around it so it cannot complete its octet by forming a dimer. However, the Al-atom being larger can accommodate four Cl-atoms around it. For this reason, AlCl₃ exists as a dimer in which each Al-atom accepts an unshared pair of electrons from the Cl-atom of another molecule to complete its octet.

Question 36. Sn(II) is a reducing agent but Pb(II) is not—why?
Answer:

Because of inert pair effect, both tin and lead show two oxidation states of +2 and +4. But this effect is more prominent in the case of Pb than in Sn and consequently, +2 oxidation state of Sn is less stable than its +4 oxidation state. Therefore, Sn(II) acts as a reducing agent and gets converted to the more stable Sn(IV) by losing two electrons.

In contrast, the +2 oxidation state of Pb is more stable than its +4 oxidation state due to the prominent inert pair effect. Therefore, Pb(II) does not lose electrons easily and does not act as a reducing agent

Question 37. CO is stable but SiO is not—why?
Answer:

Since electronegativity, has a strong tendency to form pn-pn multiple bonds, it combines with oxygen to form CO which is stabilised by resonance as follows:

⇒ \(: \mathrm{C}=\ddot{\mathrm{O}}: \leftrightarrow: \overline{\mathrm{C}} \equiv \stackrel{+}{\mathrm{O}}:\)

Silicon, on the other hand, due to its bigger size and lower electronegativity, does not tend to form pn-pn multiple bonds. Thus, it does not combine with oxygen to form stable SiO

Question 38. [SiF₆]^2-  is known but [SiF₆]^2- is not. Why?
Answer:

Possible reasons for the non-existence of [SiF₆]^2- are:

1. Six fluorine atoms can be easily accommodated around silicon atoms due to smaller size while six larger chlorine atoms cannot be accommodated around silicon atoms.
2. The unshared pair of electrons present in a relatively small 2p -orbital of F interacts with the d -orbitals of Si better than the unshared pair ofelectrons present in a relatively large 3p -orbital of Cl.

Question 39. Explain why CCI₄ is resistant to hydrolysis but SiCl₄undergo ready hydrolysis.
Answer:

Carbon does not undergo hydrolysis because carbon cannot extend its coordination number beyond four due to the absence of a vacant d-orbital in its valence shell. On the other hand, SiCl₄ can undergo ready hydrolysis because Si has a vacant d-orbital in its valence shell and can extend its coordination number beyond four

Question 40. No visible reaction is observed when Al metal is left in contact with concentrated HNQ3. Explain.
Answer:

Al is a reactive metal and hence it initially reacts with a cone. HNO₃ to form Al₂O₃. The oxide forms a protective layer No the surface of the metal and it becomes positive Thus no visible reaction is observed

Question 41. Thermite reaction cannot be stopped by pouring water. Explain.
Answer:

In a thermite reaction, the oxygen needed for the reaction is supplied by the metal oxide. Thus, stopping the oxygen supply (by pouring water) has no effect.

Further at high temperatures (1270-1300K), Al reacts with H₂O to form H₂ gas which spreads the fire rather than extinguishing it

Question 42. Why were lead sheets used on the floors in the Hanging Gardens of Babylon
Answer:

To prevent water from escaping, the lead sheet was extensively used on the floors in the Hanging Gardens of Babylon (one of the wonders ofthe ancient world that was built during the Egyptian civilisation).

Question 43. Explain why HF is not stored in glass containers the visible reaction is observed
Answer:

SiO₂ present in glass reacts readily with hydrofluoric acid (HF) to form H₂SiF₆ which is soluble.

Hence, HF is not stored in glass containers.

Question 44. What is the state of hybridisation of carbon in

1. CO₃^2-
2. Diamond
3. Graphite?

Answer:

The hybridisation state of C

• In CO₃^2- is sp²
• In diamond is sp³
• In graphite is sp²

Question 45. AICI₃ is covalent but ionizes in water—why?
Answer:

AICI₃ ionizes in an aqueous solution because the amount of hydration enthalpy released exceeds the ionization enthalpy. 5137kJ mol^-1 energy is required to convert Al to Al³⁺, hydration Al³⁺

ΔH hydration – 4165kJ mol^-1& ΔH _hydration for Cl^-1 is -381 kj. mol^-1. Since this exceeds ionization enthalpy (-5137k J.mol^-1), hence, AlCl undergoes ionization in water

NCERT Solutions For Class 11 Chemistry Chapter 11 Some P Block Elements Warm Up Exercise Question And Answers

Question 1. Give reactions to justify the amphoteric nature of Ga.
Answer:

Question 2. Why does BF₃ form an adduct with ammonia?
Answer:

In the NH molecule NH₃ -atom has a lone pair ofelectrons and in BF₃ molecule 2 electrons are required to complete the octet of the B-atom. Thus, NH₃ (a Lewis base) reacts with BF₃ (a Lewis acid) to form an adduct

H₃ N: + BF₃→ [H₃ N→ BF₃ ]

Question 3. Boron is distinctly non-metallic—why?
Answer:

Boron is distinctly non-metallic because of its small atonfWsize, high ionisation enthalpy high electronegativity.

Question 4. Using chemical reactions shows that boron acts as an oxidising agent as well as a reducing agent.
Answer:

B Is heated with Mg in an electric arc furnace magnesium boride Is formed. I lore, B acts as an oxidising agent.

Question 5. Metal borides having 10B are used in nuclear reactors why?
Answer: 10_B has a greater tendency to absorb a high-energy neutron

Question 6. BO₃^3- has a trigonal planar structure—why?
Answer:

The three half-filled orbitals (2s, 2p_x and 2py) of boron in its excited state undergo sp² -hybridisation. The resulting three sp² -hybridised orbitals overlap with 2p – orbitals of three O^– forming three B – O^– bonds. Thus, BO₃^3- ion has a trigonal planar structure

Question 7. Although boric acid [B(OH)₃] contains three -OH groups, it is sparingly soluble in water—why?
Answer:

Boric acid molecules form cyclic two-dimensional associated giant molecules through intermolecular hydrogen bonds. So it finds little or no opportunity to form hydrogen bonds with water and is hence, sparingly soluble in water

Question 8. Among group-14 elements which one exhibits pπ-pπ multiple bonding?
Answer:

Among all the group-14 elements carbon exhibits pn-pn multiple bonding.

Question 9. Account for the anomalous behaviour of carbon from other group-14 elements.
Answer:

Due to small size, high ionisation enthalpy, high electro- * negativity and unavailability of d-orbitals, the behaviour of carbon is different from other elements of group-14

Question 10. The shape of (SiH₃)₃P is pyramidal. Comment.
Answer:

A larger 3p -orbital of P is unable to participate in efficient pπ-dπ bonding. Thus, the central P-atom undergoes sp³-hybridisation and forms pyramidal (SiH)₃P

Question 11. Which element among the group-14 elements is a metalloid?
Answer: Germanium (Ge)

Question 12. Which compound of lead is used as “Sindoor”
Answer: Red lead (Pb₃O₄)

Question 13. Among the dioxides of group-14 elements. PbO₂ is the strongest oxidising agent— explain.
Answer:

Due to the inert pair effect, the +2 oxidation state of Pb is more stable. Hence, PbO₂ is the strongest oxidising agent among the dioxides of group-14 elements.

Question 14. Give the Lewis acidity order: SiI₄, SiCl₄, SiBr₄ > SiF₄
Answer:

Lewis acidity order: Sil₄ > SiBr₄ > SiCl₄ > SiF₄

Question 15. What is the state of hybridisation of carbon in fullerene?
Answer: In fullerene, carbon is sp² -hybridised.

Question 16. How can you decolourise a sample of slightly brown-coloured impure sugar?
Answer:

When an aqueous solution of brown-coloured impure sugar is heated with activated charcoal, a mixture is obtained, which on filtration gives a colourless solution. The solution is concentrated by heating and then cooled. As a result, colourless crystals of pure sugar are obtained.

Question 17. Give two differences between diamond and graphite.
Answer:

Two characteristic differences between diamond and graphite are:

1. Diamond is hard but graphite is soft
2. Diamond is an insulator while graphite is a good conductor of electricity.

Question 18. CO forms an additional compound but CO₂ does not—why?
Answer:

In the CO molecule, the C-atom has a lone pair of electrons. By donating this lone pair of electrons CO forms an addition compound with metals or non-metals. Also in the CO molecule, the C-atom exhibits an oxidation no. of +2.

So, it can increase its oxidation no. from +2 to +4 by forming additional compounds. On the other and C-atom in the CO₂ molecule does not have any lone pair of electrons. Besides, the C-atom exhibits +4 oxidation state in CO₂. So, it has no opportunity to increase its oxidation number. Thus, CO forms an additional compound but CO₂ does not

Question 19. Explain why blue flame is seen in a coal oven.
Answer:

At the bottom section of the coal oven, carbon burns in the presence of excess oxygen producing CO₂ This CO₂ while moving upwards, is reduced by red hot carbon (coke or coal) to CO in the middle section ofthe oven.

The CO bums in the open air at the top of the oven with a blue flame to form CO₂.

1. At the bottom : C + O₂ →C02
2. At the middle: CO₂ + C→2CO
3. At the top: 2CO +O₂ →2CO₂

Question 20. How will you separate CO and CO₂ from a mixture?
Answer:

When a mixture of CO₂ and CO is passed through Cu₂Cl₂ solution acidified with HCl, CO is absorbed in the die solution but CO₂ escapes without participating in a chemical reaction. The solution thus obtained liberates CO on heating

2CO + Cu₂Cl₂ + 4H₂O→2[CuCl-CO. 2H₂O]

Question 21. How will you confirm that a gas is CO₂ but not SO₂?
Answer:

Both the gases are passed through the K₂Cr₂O₇ solution, SO₂ turns the orange colour of the K₂Cr₂O₇ solution green but CO₂ does not.

Question 22. Write the formula of white asbestos. What type of silicate is it?
Answer:

Formula of white asbestos is Mg₃(OH)₄[Si₂O₅] . It is a type of sheet silicate

Question 23. How can ultrapure silicon be prepared from impure silicon?
Answer:

At first, impure Si is treated with chlorine gas to produce impure SiCl₄, which on distillation forms pure SiCl₄. Pure SiCl₄ thus produced is reduced by H₄ gas to ultrapure silicon

Question 24. Explain why silicones are water-repelling in nature.
Answer:

Silicon chains are surrounded by non-polar organic groups. Thus, they are water-repelling in nature (water is a polar solvent)

Question 25. What are zeolites? Give two important uses of zeolites
Answer:

Zeolites are used

1. For softening of hard water
2. As a molecular sieve to separate molecules of different size