NCERT Class 11 Chemistry Chapter 11 Some P Block Elements Long Question And Answers

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

Question 1. Anhydrous aluminum chloride is used as a catalyst and fumes in moist air. Explain these observations.
Answer:

The Al-atom in AlCl₃ has only six electrons in its valence shell, and it requires two more electrons to complete its octet. Therefore, it can act as a Lewis acid catalyst.

For example:  In Friedel-Crafts alkylation and acylation reactions, AlCl₃ acts as a Lewis acid catalyst to generate the electrophile (R⁺ or RC⁺O).

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In moist air, AlCl₃ undergoes partial hydrolysis to form HCl gas and because of this, it fumes in moist air.

AlCl₃(S) + 3H₂O→Al(OH)₃+ 3HCl(g)

Question 2. Out of anhydrous AlCl₃ and hydrated AlCl₃, which] one is more soluble in diethyl ether and why?
Answer:

Anhydrous AlCl₃ is an electron-deficient compound (Lewis acid) while hydrated. AlCl₃, i.e. AlCl₃ -6H₂O, is not because H₂O has already donated a pair of electrons to Al. Therefore, the O-atom of diethyl ether donates a pair of electrons to the Al-atom of anhydrous. AlCl₃, forming a coordinate bond. For this reason, anhydrous AlCl₃ is soluble in diethyl ether.

Question 3. When lead nitrate solution is added to an aqueous solution of H2S, a black precipitate is obtained. However, when lead nitrate solution is added to the filtrate obtained by passing the solution of H₂S through activated charcoal, no black precipitate is obtained. Explain these observations
Answer:

When lead nitrate solution is added to an aqueous precipitate solution of H₂S, a reaction leading to the formation of a black precipitate of lead sulfide (PbS) occurs

H₂S +Pb(NO₃)₂  → PBS↓ (black)+2HNO₃

Activated charcoal, because of its porous structure, adsorbs gaseous substances. So, when aqueous solution of H₂S is filtered through activated charcoal, H2S is adsorbed by it. Due to the absence of H₂S in the filtrate, the addition of lead nitrate solution does not give any precipitate.

Question 4. CO is a poisonous gas while C02 is not—why?
Answer:

Carbon monoxide (:C^– ≡ ⁺O: ) combines with the Fe- atom present in the hemoglobin of blood through the lone pair of electrons on carbon to form a highly stable complex known as carboxyhemoglobin (HbCO). Due to the formation of this complex, hemoglobin cannot further act as an oxygen-carrier. As a result, the body cells get slackened due to the deficiency of oxygen and this ultimately results in death.

Thus, CO is poisonous: Hb + CO→ HbCO.

(Hb – Haemoglobin) On the other hand, the structure of carbon dioxide is \(: \ddot{O}=\mathrm{C}=\ddot{\mathrm{O}}:\).

Although the oxygen atoms in the CO₂ molecule contain lone pairs of electrons, unlike CO it cannot combine with the Fe -atom of hemoglobin because of its larger size. Therefore, CO₂ cannot prevent hemoglobin from carrying oxygen so it is not poisonous.

Question 5. what is a foaming mixture? How can it extinguish the fire?
Answer:

Foam-type fire extinguishers are nowadays used to extinguish petroleum and other oil fires. To produce stable f m of CO₂, a mixture consisting of a concentrated solution of aluminum sulfate and sodium bicarbonate along with

Al₂(SO₄)₃ on hydrolysis produces Al(OH)₃ and H₂SO₄. Sulphuric acid thus obtained reacts with NaHCO₃ to Liberate bubbles Of CO. These CO bubbles form stable sticky foam with AlOH in the presence of licorice extract.

This foam, when applied to the fire, forms a layer and prevents the oil or petrol from coming in further contact with air or oxygen. Consequently, the fire gets extinguished.

Al₂(SO₄)₃ + 6NaHCO₃→2AI(OH)₃ + 3Na₂SO₄ + 6CO₂↑

Question 6. AIF₃ does not dissolve in anhydrous HF but dissolves in KF. When BF3 is added to the above solution containing KF, aluminum trifluoride is precipitated. Explain.
Answer:

AlF₃ does not dissolve in anhydrous HF because it is not available for coordination with AIF₃ due to the presence of intermolecular hydrogen bonding. It dissolves in KF because F_ ion can coordinate with AIF₃ to form a salt

3KF +AlF₃→K₃[AIF₆]

The salt is decomposed by BF₃ which is a Lewis add (electron deficient) and AlF₃ is precipitated.

K₃[AlF₆] + 3BF₃→ AIF₃↓ + 3KBF₄

Question 7. Explain why carbon dioxide is a gas at room temperature but silicon dioxide is a solid substance,
Answer:

As the atomic sizes of carbon and oxygen are almost equal, carbon and oxygen atoms can easily form a double bond between them by effective pπ-pπ overlapping (O=C=O), and because of this, carbon dioxide exists as discrete molecules. The linear CO₂ molecules are non-polar, and so the intermolecular forces in carbon dioxide are too weak to allow the formation of molecular aggregates. That is why carbon dioxide is a gas at room temperature.

On the other hand, Si-atom is much larger than O-atom, and the 3p -orbital of silicon and 2p -orbital of oxygen differ quite appreciably in their sizes and energies. So formation of a double bond by effective pπ-pπ overlapping does not take place. Therefore, silicon dioxide cannot have a molecular structure of the type O=Si=O, similar to carbon dioxide. In other words, silicon dioxide does not exist as discrete SiO₂ molecules.

Instead, silica possesses a giant three-dimensional structure [(SiO₂)] in which each silicon atom is linked to four O-atoms tetrahedrally and each O-atom is linked to two silicon atoms. Because of its giant three-dimensional polymeric structure, silicon dioxide is a solid at room temperature.

Question 8. Why does Ga (+1) undergo a disproportionation reaction?
Answer:

Due to the inert pair effect, gallium exhibits both +1 and +3 oxidation states. However, the +3 oxidation state of gallium is more stable than its +1 oxidation state. For this reason, Ga(+1) undergoes a disproportionation reaction (selfoxidationreduction) to form gallium metal & the more stable Ga3+ ion in aqueous solution.

Question 9. Unlike In⁺, Tl⁺ does not undergo a disproportionation reaction—Explain.
Answer:

Both In and T1 can exhibit oxidation states of +1 and +3. However, because of the prominent inert pair effect in Tl, the +1 oxidation state of Tl is more stable than its +3 oxidation state while the +3 oxidation state of In is more stable than its +1 oxidation state. As a consequence, in an aqueous solution, the less stable In+ undergoes a disproportionation reaction to form a more stable In³⁺ but ⁺ being more stable, does not undergo a disproportionation reaction.

Question 10. Discuss the pattern of variation in the oxidation states of B to Tl.
Answer:

1. Since B and Al have no d -or f-electrons, they do not exhibit an inert pair effect. Therefore, they exhibit only one oxidation state of +3 due to the presence of two electrons in the s -and one electron in the p-orbital of the outermost shell.

2. The remaining elements from Ga to Tl contain d or both d and electrons in their inner shells and hence exhibit +3 as well as +1 oxidation states due to the inert pair effect.

3. As the number of d -and /-electrons increases on moving down the group, the inert pair effect becomes more and more prominent. As a consequence, the stability of the +1 oxidation state increases (i.e., Ga < In < Tl ), while that of the +3 oxidation state decreases (i.e. Ga > In > Tl ). The +1 oxidation state of Tl is more stable than its +3 oxidation state

Question 12.BF₃ is a weaker Lewis acid than BCl₃, even though F is more electronegative than Cl. Explain
Answer:

The B -atom in BF3 or BCl₃ has only sue electrons in its valence shell, and hence it is capable of accepting a pair of electrons to complete its octet. Therefore, both BF₃ and BCl₃ act as Lewis acids. Because of equal sizes of the empty 2p orbital of B and filled 2p -orbital of F, the lone pair of electrons of F is donated to the empty 2p-orbital pπ-pπ back bonding) to a considerable extent and as a result, the electron deficiency of B decreases in BF₃

In BCl₃, on the other hand, the size of the 3p -orbital of Cl containing the lone pair of electrons is much bigger than the empty 2p -orbital of B, and hence the donation of a lone pair of electrons from Cl to B does not take a place significantly. Therefore, the B atom in BCl₃ is a stronger Lewis acid than BF₃.

Question 13. Discuss the pattern of variation in the oxidation states of B to Tl.
Answer:

Since B and Al have no d -or f-electrons, they do not exhibit an inert pair effect. Therefore, they exhibit only one oxidation state of +3 due to the presence of two electrons in the s -and one electron in the p-orbital of the outermost shell.

The remaining elements from Ga to Tl contain d or both d and f electrons in their inner shells and hence exhibit +3 as well as +1 oxidation states due to the inert pair effect.

As the number of d -and /-electrons increases on moving down the group, the inert pair effect becomes more and more prominent. As a consequence, the stability of the +1 oxidation state increases (i.e., Ga < In < Tl ), while that of the +3 oxidation state decreases (i.e. Ga > In > Tl ). The +1 oxidation state of Tl is more stable than its +3 oxidation state.

Question 14. Which out ofdil. H₂SO₄, HCl, and HNO₃ can be used in the preparation of carbon dioxide from PbCO₃.
Answer:

PbC03 reacts with dilute H₂SO₄ and dilute HCl to form insoluble PbS04 and PbCl₂ respectively. A protective coating of the salt is formed over the surface of the marble pieces and this cuts off the contact between the acid and the marble. So the reaction stops and no more CO₂ is produced. It is for this reason, that dilute H₂SO₄ and HCl cannot be used in the preparation of CO₂ from PbCO₃. When dilute HNO₃ is used, highly soluble Pb(NO₃)₂ is formed and so, the evolution of CO₂ takes place smoothly. Therefore, dilute HNO₃ can be used in the preparation of CO₃ from PbCO_3.

Question 15. Explain why BF₃ exists whereas BH₃ does not. 
Answer:.

Because of pn-pn backbonding, the lone pair ofelectrons of F is donated to the B -atom. This delocalization reduces the deficiency of electrons on the B -atom and as a consequence, the stability of the BF₃ molecule increases.

Due to the absence of a lone pair of electrons on the H -atom, similar compensation does not take place in BH₃. In other words, deficiency of electron on B retains, and therefore, to reduce its electron deficiency, BH₃ dimerizes to form B₂H₆ (diborane).

Question 16. Consider the compounds, BCI₃ and CCl₄. How will they behave with water? Justify.
Answer:

BCl₂ is an electron-deficient molecule as the central Batom has only sue electrons in the valence shell. Therefore, it can accept a pair of electrons donated by water and undergo hydrolysis to form boric acid (H₂BO₃) and HCl.

On the other hand, the C-atom in CCl₄ has 8 electrons in its valance shell and it has no vacant d -d-orbitals to extend its octet. So it cannot accept a pair of electrons from H₂O and hence CCl₄ does not undergo hydrolysis.

Question 17. Is boric acid a protic acid? Explain.
Answer:

Boric acid is not a protic acid because it does not ionize in water to give a proton:

However, due to the presence of only 6 electrons in the valence shell of boron, B(OH)₃ accepts a lone pair ofelectrons from Oatom of H₂O to form a hydrated species. The positively charged O-atom pulls the O —H bonding electrons towards itself thereby facilitating the release of proton. Therefore, B(OH)₃ acts as a weak monobasic acid

Question 18. Describe the shapes of BF₃ and BH^–₄. Assign the hybridization of boron in these species.
Answer:

In the BF₃ molecule, the B -atom is sp² -hybridized. Thus, the BF₃ molecule contains three bond pairs and exhibits trigonal planar geometry.

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Question 19. What are electron-deficient compounds? Are BCl₃ and SiCI₄ electron-deficient species? Explain
Answer:

Electron-deficient compounds are those in which the central atom either does not have 8 valence electrons or in which the central atom has 8 valence electrons but can expand its valency beyond 4 due to the presence of volatile d -d-orbitals. BCl₃ is an electron-deficient molecule as the central B atom has six electrons in its valence shell. So, it can accept a pair of electrons from Lewis bases like NH₃ to form an adduct.

The central Si atom in SiCl₄ has 8 electrons in its valence shell but it can expand its covalency beyond 4 due to the presence of vacant d -orbitals. Therefore, in principle, SiCl₄ should also act as an electron-deficient compound. However, it cannot form [SiCl₆]^2- by accepting two Cl^– ions and this is because

1. Small Si -atoms cannot accommodate the large-sized Cl^– atoms around and
2. The interaction between lone pairs of Cl^– atom and empty p -orbitals of  Siatom is weak.

Question 20. Explain the difference in properties of diamond and graphite based on their structures.
Answer:

Diamond and graphite:

Question 21. Rationalize the statements and give reactions:

1. Lead (II) chloride reacts with Cl₂ to give PbCl₄.
2. Lead (IV) chloride is highly unstable towards heat.
3. Lead is known not to form an iodide, Pbl₄.

Answer:

1. Pb shows +2 and +4 oxidation states in its compounds. However, due to the inert pair effect, the +2 oxidation state is more stable than the +4 oxidation state. As a result, lead combines with Cl₂ to form lead (II) chloride, But, Cl₂ is an oxidizing agent and oxidizes Pb²⁺ to Pb⁴⁺. Thus, PbCl₂ reacts with Cl₂ to form PbCl₄.

PbCl₂ + Cl₂→PbCl₄

2. Due to the inert pair effect, the +2 oxidation state of lead is more stable than the +4 oxidation state. Thus, PbCl₄ on heating decomposes to form PbCl₂.

3. Pbl₄ does not exist because the amount of energy released by the initially formed two Pb—I bonds is not sufficient enough to unpair the 6s² electrons and promote one of them to the higher energy 6p orbital. The strong oxidizing power of Pb⁴⁺. Ion and the strong reducing power of I^– ion is also responsible for the non-existence of Pbl₄

Question 22. Suggest reasons why the B— F bond lengths in BF₃ (130 pm) and BF4 (143 pm) differ
Ans:

The sp² -hybridized B atom in trigonal planar BF₃ molecule has an empty 2p -orbital. Because of similar sizes the vacant and filled p -orbitals, pn-pn back bonding involving the transfer of a pair of electrons from F to B occurs. As a result, the B B — F bond acquires some double bond character. On the other hand, in BF₄ ion, the B atom is sp³ -hybridized and so, it has no empty p -orbital to accept the electrons donated by the F atom. As a consequence, in BF₄ ion, the B— F bond is a purely single bond. Since a double bond is shorter than a single bond, the B —F bond in BF₃ is shorter in length (130 pm) than the B —F bond (143 pm) in BF^–₄

Question 23. If the B—Cl bond has a dipole moment, explain why the BCl₃ molecule has zero dipole moment.
Answer:

The B-atom in the BCl₃ molecule is sp² – hybridized and the molecule possesses trigonal planar geometry. As the molecule is symmetrical, the resultant dipole moment of the three B —Cl bonds is zero. Thus BC13 molecule has zero dipole moment.

Question 24. Aluminum trifluoride is insoluble in anhydrous HF but dissolves with the addition of NaF. Aluminum trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons
Answer:

Anhydrous HF being a strongly H -bonded covalent compound does not give F^– ions for complexing with AlF₃and so AlF₃ does not dissolve in HF. In the presence of the ionic compound NaF, F^– ions combine with AlF₃ to form a soluble complex

3NaF + AlF₃→Na₃[AlFg] Sodium hexafluoroaluminate (III) (Soluble complex)

Due to its small size and higher electronegativity, boron has a much higher tendency to form a complex than aluminum. Because of this, when gaseous BF is bubbled through this solution, aluminum trifluoride precipitates out.

Question 25. How is excessive content of COz responsible for global warming?
Answer:

The visible and some ultraviolet radiations (which are Due to small size and higher electronegativity, boron has a much higher tendency to form complex than aluminum. Because of this, when gaseous BF3 is bubbled through this solution, aluminum trifluoride precipitates short wavelengths) from the sun and reaches the earth by passing through CO₂ present in the atmosphere, and as a result, the earth becomes heated. However, when the earth becomes cool, the energy is emitted from the earth’s surface in the form of infrared radiations (which have longer wavelengths and have a heating effect).

CO₂ does not allow these radiations to pass through itself and becomes heated by absorbing them. Some of this heat is dissipated into the atmosphere while the remaining part is radiated back to the earth. As a result, the temperature of the earth increases. In this way, CO₂ helps to maintain the temperature on the earth required for the existence of living organisms. However, if the amount of CO₂ in the air increases due to some human activity, the temperature of the earth increases more than required. This phenomenon is called global warming.

Question 26. Explain the following reactions—

1. Silicon is heated with methyl chloride at high temperatures in the presence of copper.
2. Silicon dioxide is treated with hydrogen fluoride.
3. CO is heated with ZnO.
4. Hydrated alumina is treated with aqueous NaOH.

Answer:

1. Si reacts with methyl chloride (CH₃Cl) at a high temperature in the presence of Cu as a catalyst to form mono, di, and trimethylchlorosilane along with a small amount of tetramethylsilane

2. SiO reacts with HF to form silicon tetrafluoride which dissolves in HF to form hydroflurosilicic acid

SiO₂+4HF → SiF₄ +2H₂O

SiF₄ +2hF → H₂SiF₆

3. ZnO is reduced to Zn by CO at high temperatures.

ZnO + CO→Zn + CO₂

4. Hydrated alumina dissolves in aqueous NaOH solution to form sodium meta aluminate or sodium aluminate

Question 27. Give reasons:

1. Cone. HNO₃ can be transported in an aluminum container.
2. A mixture of dilute NaOH and aluminum pieces is used to open the drain.
3. Graphite is used as a lubricant. 
4. Diamond is used as an abrasive.
5. Aluminum alloys are used to make aircraft bodies.
6. Aluminum utensils should not be kept in water overnight.
7. Aluminum wire is used to make transmission cables.

Answer:

1. When Al reacts with the cone. HNO₃, a very thin film of aluminum oxide (A1203) is formed on its surface. This oxide layer protects Al from further reaction and hence aluminium containers can be used to transport cones. HNO₃

2. The H₂ gas liberated in the reaction of Al with dilute NaOH is used to open clogged drains.

2Al+ 2NaOH + 2H₂O→2NaAlO₂ + 3H₂

3. Graphite possesses a hexagonal layered structure. These layers slip over one another because they are held by weak van der Waals forces. Hence it acts as a lubricant.

4. The C -atoms in diamond are sp3 -hybridized. Each C atom is linked to the other four C -atoms tetrahedrally by single covalent bonds. Thus, diamond is very hard and is hence used as an abrasive.

5. Aluminium alloys such as duralumin (Al-95%, Cu-4%, Mg-0.5%, Mn-0.5%) is light, tough and corrosion resistant. Hence it is used for making aircraft bodies.

6. Aluminium reacts with water and dissolved oxygen to form a thin film of aluminum oxide (Al₂O₃).

2Al(s) + O₂ (g) + H₂O(l)→Al₂O₃(s) + H₂(g)T

A very small amount of Al2O33 may dissolve in water to give a solution containing a few ppm of Al³⁺ ions. Since Al³⁺ions are toxic to health, drinking water should not be kept in an aluminum vessel overnight.

7. Aluminium is a good conductor of electricity (twice that of Cu based on weight) and is not affected by the atmosphere as well. For this reason, it is used in transmission cables.

Question 28. Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon.?
Answer:

• On moving from carbon to silicon
• The size of Si increases due to the addition of a new shell
• Effective nuclear charge on Si increases due to an increase in the number of protons
• The shielding effect increases due to an increase in the number of inner-filled orbitals.
• The decrease in ionization enthalpy due to and
• Is more than the increase in ionization enthalpy due to
• Thus, there is a phenomenal decrease in ionization enthalpy as we move from carbon to silicon

Question 29. How would you explain the lower atomic radius of Ga as compared to Al?
Answer:

The electronic configuration of Ga is as follows

₁₃A1: ls²2s²2p⁶3s²3p¹

₃₁Ga: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p¹

The effective nuclear charge of Ga is greater than that of Al. This is due to the poor shielding of the d-electrons of Ga. Thus, the electrons in Ga experience more attraction towards the nucleus than that experienced by the electrons of Al. Consequently, the atomic radius of Ga is lower compared to A

Question 30. 

1. Classify the following oxides as neutral, acidic, basic, or amphoteric: CO, B₂O₃, SiO₂, CO₂, Al₂O₃, PbO₂, Tl₂O₃.
2. Write suitable chemical equations to show their nature.

Answer:

1. Neutral: CO; Acidic: B₂O₃, SiO₂, CO₂; Basic: Tl₂O₃; Amphoteric: Al₂O₃, PbO₂ .

2.

1. As B₂O₃, SiO_2, and CO₂ are acidic, they react with alkalis to form salts

2. Al₂O₃and PbO₂ reacts with both acids and alkalis as they are amphoteric

3. Tl₂O₃ reacts with acids as it is a basic oxide

Question 31. In some reactions, thallium resembles aluminum, whereas in others it resembles group-I metals. Support this statement by giving some evidence.
Answer:

Both thallium and aluminum belong to group 13 and thus have a general electronic configuration of ns²np¹. Both Al and Tl exhibit +3 oxidation states and form compounds like AlCl₃ and T1Cl₃ respectively. Both of them form octahedral ions like [AlF₆]^3-and [TIF₆]^3-. Again, due to the inert pair effect, T1 shows an oxidation state like the alkali metals of group-1 and forms TlCl, Tl₂O, TlClO₄, etc. TlOH like alkali metal hydroxides, dissolves in water to form strongly alkaline solutions. Again, Tl₂CO₃ is water soluble, and Tl₂SO₄ forms alum like that ofthe alkali metals.

Question 32. When metal X is treated with sodium hydroxide, a water white precipitate (A) is obtained, which is soluble more than NaOH to give a soluble complex (B). Compound (A) is soluble in dilute HC1 to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities
Answer:

AS metal X, on treatment with NaOH forms a white precipitate, which dissolves in excess NaOH to form soluble complex B, the metal X is Al.

Thus X = Al(aluminium)

A = Al(OH)₃(Aluminium hydroxide

B =Na[Al(OH)₄]

C = AlCl₃(aluminiumchloride)

D =Al₂O₃Alumina

Question 33. A certain salt X, gives the following results. O Its aqueous solution is alkaline to litmus.0 It swells up to a glassy material Y on strong heating 0 When cone. H2S04 is added to a hot solution of X, and a white crystal of an acid Z separates. Write equations for all the above reactions and identify X, Y, and Z.
Answer:

Aqueous solution of the salt (X) is alkaline to litmus. Thus, (X) must be salt ofa weak acid and strong base.

When the salt (X) is heated, it swells up to form a glassy material. Therefore, (X) must be borax and (7) must be a mixture of sodium metaborate (NaBO₂) and boric anhydride (B₂O₃).

When cone. H₂SO₄ is added to a hot solution of(X), i.e., borax, white crystals of (Z) separate. Therefore, (Z) must be orthoboric acid (H₃BO₃). The equations of the reactions involved are as follows