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NCERT Class 11 Chemistry Chapter 11 The s-Block Elements
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The p-Block Elements
Chapter: 11
| Part – II |
1. Discuss the pattern of variation in the oxidation states of:
(i) B to Tl.
Ans: The electric configuration of group 13 elements is ns2 np1. Therefore, the most common oxidation state exhibited by them should be +3. However, it is only boron and aluminium which practically show the +3 oxidation state. The remaining elements, i.e., Ga, In, Tl, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, Tl (+1) is more stable than Tl (+3). The inert pair effect causes the s-electrons to remain non-bonding as we move down a group. This makes Ga(+1) unstable, In(+1) fairly stable, and Tl(+1) highly stable.
| Group 13 element | Oxidation state |
| B | +3 |
| AI | +3 |
| Ga, In, Tl | +1, +3 |
(ii) C to Pb.
Ans: The electronic configuration of group 14 elements is ns2 np2. Therefore, the most common oxidation state exhibited by them should be +4. However, the +2 oxidation state becomes more and more common on moving down the group. Carbon and silicon predominantly show the +4 oxidation state, but as we move down to lead (Pb), the +2 state becomes more stable due to the inert pair effect, while the +4 state becomes less stable.
| Group 13 element | Oxidation state |
| B | +3 |
| AI | +3 |
| Ga,In,TI | +1,+3 |
2. How can you explain higher stability of BCl3 as compared to TlCl3?
Ans: Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. BCl3 is more stable than TlCl3 because the +3 oxidation state of B is more stable than the +3 oxidation state of Tl. In Tl, the +3 state is highly oxidising and it reverts back to the more stable +1 state. Therefore, BCl3 is stable but TlCl3 is comparatively unstable.
3. Why does borontrifluori.de behave as a Lewis acid?
Ans: The electronic configuration of boron is ns2 np1. It has three electrons in its valence shell. Thus, it can form only three covalent bonds. Hence, boron trifluoride remains electron-deficient and acts as a Lewis acid. This means that boron has only six electrons around it, leaving its octet incomplete. When a boron atom bonds with three fluorine atoms, it still lacks two electrons to complete its octet. This electron deficiency is what makes boron trifluoride (BF₃) a Lewis acid, as it readily accepts electron pairs from other atoms or molecules.
4. Consider the compounds, BCl3, and CCl4. How will they behave with water justify?
Ans: Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result.
BCl3 + 3H2O ⟶ 3HCl + B(OH)3
CCl4 completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When CCl4 and water are mixed, they form separate layers.
CCl4 + H2O ⟶ No reaction
5. Is boric acid a protonic acid? Explain.
Ans: Boric acid B(OH)3 is not a protonic acid. It is a Lewis acid.
Boric acid accepts electrons from hydroxyl ions of H2O molecules.
B(OH)3 + 2HOH → [B (OH)4]– + H3O+
6. Explain what happens when boric acid is heated.
Ans: On heating orthoboric acid (H3BO3) at 370 K or above, it changes to metaboric acid (HBO2). On further heating, this yields boric oxide B2O3.
7. Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.
Ans: BF3
The BF3 molecule is formed by bonding between three sp2 orbitals of B (Boron) and 3p orbitals of 3 F (Fluorine) atoms. The halides of the boron form trigonal planar geometry.
BH4–
Boron-hydride ion (BH4–) is formed by the sp3 hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.
8. Write reactions to justify the amphoteric nature of aluminium.
Ans: A substance is called amphoteric if it displays characteristics of both acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour
2Al(s) + 6HCl(dil.) → 2AlCl3(aq) + 3H2(g)
Aluminium also dissolves in aqueous alkali and liberates dihydrogen.
2Al + 2NaOH + 6H2O → 2Na + [Al(OH)4]− + 3H2
These reactions demonstrate the amphoteric nature of aluminium.
9. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient species? Explain.
Ans: Electron-deficient compounds are those that do not have a complete octet of electrons in their valence shell, making them capable of accepting electron pairs from nucleophiles.
BCl3 is an electron deficient compound because boron has only 6 valence electrons and is short of 2 electrons. Therefore, it can accept a pair of electrons from nucleophiles.
SiCl₄ is not electron-deficient Because Silicon has 4 valence electrons. Forming 4 bonds with chlorine gives it 8 valence electrons.They are also known as Lewis acid. BCl3 and SiCl4 both are electron deficient species.
Therefore, only BCl₃ is an electron-deficient compound. SiCl₄ is not.
10. Write the resonance structure of CO32- and HCO3–.
Ans:
11. What is the state of hybridisation of carbon in:
(a) CO32-
(b) diamond
(c) graphite?
Ans: The state of hybridisation of carbon in:
(a) C in CO32− is sp2 hybridised and is bonded to three oxygen atoms.
(b) Diamond is each carbon in diamond that is sp3 hybridised and is bound to four other carbon atoms.
(c) Each carbon atom in graphite is sp² hybridised and is bonded to three other carbon atoms.
12. Explain the difference in properties of diamond and graphite on the basis of their structures.
Ans: Following are the difference in properties of diamond and graphite on the basis of their structures:
| DIAMOND | GRAPHITE |
| Diamond is a solid form of pure carbon with its atoms arranged in a crystal. | Graphite is an allotrope of carbon which is used for making moderator rods in nuclear power plants. |
| Since diamond exists as a three dimensional network solid, it is the hardest substance known with high density and high melting point. | In graphite, any two successive layers are held together by weak forces of attraction. This makes graphite soft. |
| The C–C bond length in diamond is 154 pm. | The C–C bond length in graphite is 141.5 pm. |
| In diamond, the carbon atom is sp3 hybridised. | In graphite, carbon atom is sp2 hybridised |
| It acts as an electrical insulator. | It is a good conductor of electricity. |
| It is made up of tetrahedral units. | It has a planar geometry. |
13. Rationalise the given statements and give chemical reactions:
(a) Lead (II) chloride reacts with Cl2 to give PbCl4.
Ans: Lead is known not to form PbI4. Pb (+4) is oxidising in nature and reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises I – to I2 and itself gets reduced to Pb(II). However, the formation of PbCl4 takes place when chlorine gas is bubbled through a saturated solution of PlCl2.
PbCl2(s) + Cl2(g) ⟶ PbCl4(l)
(b) Lead (IV) chloride is highly unstable towards heat.
Ans: On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb (II).
PbCl4(l) → PbCl2(s) + Cl2(g)
(c) Lead is known not to form an iodide Pbl4.
Ans: Lead is known not to form PbI4. Pb (+4) is oxidising in nature and Iis reducing in nature.The combination of Pb(IV) and iodide ions is unstable because iodide is a strong reducing agent. In this reaction, Pb(IV) oxidises I⁻ to I₂, while itself gets reduced to Pb(II).
Pbl4 ⟶ Pbl2 + l2
14. Suggest reason why the B-F bond lengths in BF3 (130 pm) and BF– (143 pm) differ.
Ans: The B–F bond length in BF₃ is shorter than in BF₄⁻. This is because BF₃ is an electron-deficient species, leading to partial double-bond character in its B–F bonds, while BF₄⁻ has purely single bonds.With a vacant p-orbital on boron, the fluorine and boron atoms undergo pπ – pπ back-bonding to remove this deficiency. This imparts a double-bond character to the B–F bond.
This double-bond character causes the bond length to shorten in BF3 (130 pm). However, when BF3 coordinates with the fluoride ion, a change in hybridisation from sp2 (in BF3) to sp3 (in) occurs.Boron now forms 4σ bonds and the double-bond character is lost. This accounts for a B–F bond length of 143 pm in ion.
15. If B-Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
Ans: The BCl3 molecule has a zero dipole moment because of polarity. In BCl3 since the molecule is symmetrical (planar). Thus the polarities cancel out.
16. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reason.
Ans: Aluminium trifluoride (AlF₃) is insoluble in anhydrous HF because it forms a stable, insoluble lattice in the absence of additional fluoride ions. However, when sodium fluoride (NaF) is added, AlF dissolves. This is because of the availability of free F–. The reaction involved in the process is:
AIF3 +3NaF ⟶ Na3[AiF6]
Sodium hexafluoroaluminate.(III)
When boron trifluoride (BF3) is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. When BF₃ is added, boron replaces aluminium in the complexes, resulting in the formation of [BF4]− and precipitating AlF₃.
Na3[AiF] + 3BF3 ⟶ 3Na[BF4] + AIR 3
17. Suggest a reason as to why CO is poisonous.
Ans: CO is a poisonous gas that reduces the amount of heme-bound oxygen, so less oxygen is available in the bloodstream. CO reacts with haemoglobin to form carboxyhemoglobin which can destroy the oxygen carrying capacity of haemoglobin and the man dies of suffocation.
18. How is excessive content of C02 responsible for global warming?
Ans: Carbon dioxide is a very essential gas for our survival. However, an increased content of CO2 in the atmosphere poses a serious threat. An increase in fossil fuel combustion, the decomposition of limestone, and a reduction in tree populations have resulted in higher levels of carbon dioxide. Carbon dioxide is a very essential gas for our survival. However, an increased content of CO2 in the atmosphere poses a serious threat. Some of it is dissipated into the atmosphere while the remaining part is radiated back to the earth. As a result, temperature of the earth increases. This is the cause of global warming.
19. Explain structures of diborane and boric acid.
Ans: Diborane is a chemical compound that consists of boron and hydrogen atoms and has a molecular formula B2H6. B2H6 has only 12 electrons – 6 e– from 6 H atoms and 3 e– each from 2 B atoms. Thus, after combining with 3 H atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:
Boric acid has a layered structure. Each planar BO3 unit is linked to one another through H atoms.Boric acid is an acid containing one atom of boron, three atoms of hydrogen and three atoms of oxygen. As we know it is a monobasic lewis acid. Now we draw the structure of boric acid:
20. What happens when:
(a) Borax is heated strongly
Ans: When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.
(b) Boric acid is added to water
Ans: When boric acid is added to water, it accepts electrons from –OH ion.
B(OH)3 + 2HOH ⟶ [B(OH)4]− + H3O+
(c) Aluminium is treated with dilute NaOH
Ans: Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.
2AI + 2NaOH + H2O ⟶ 2NaAIO2 + 3H2
(d) BF3 is reacted with ammonia?
Ans: BF3 (a Lewis acid) reacts with NH3 (a Lewis base) to form an adduct. This results in a complete octet around B in BF3.
BF3 + NH ⟶ [H3N ⟶ BF]
21. Explain the following reactions.
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper.
Ans: When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicon polymers called methyl-substituted chlorosilanes (MeSiCl3, Me2SiCl2, Me3SiCl, and Me4Si) are formed
(b) Silicon dioxide is treated with hydrogen fluoride.
Ans: When silicon dioxide (SiO2) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF4). The Si–O bond is typically very strong and resists attacks from halogens and most acids, even at high temperatures. However, it can be attacked by hydrofluoric acid (HF).
SiO2 + 4HF⟶ SiF4 + 2H2O
The SiF4 formed in this reaction can further react with HF to form hydrofluorosilicic acid.
SiF4 + 2HF ⟶ H2SiF6
(c) CO is heated with ZnO.
Ans: When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.
ZnO + CO → Zn + CO2
(d) Hydrated alumina is treated with aqueous NaOH solution.
Ans: Alumina dissolves to form sodium meta-illuminate
22. Give reasons:
(i) Cone. HNO3 can be transported in aluminium container.
Ans: Concentrated HNO3 can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open the drain.
Ans: Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.
2Al(s) + 2NaOH(aq) + 2H2O(l) ——-> 2NaAlO2(aq) + 3H2(g)
(iii) Graphite is used as lubricant.
Ans: Graphite has a layered structure and different layers of graphite are bonded to each other by weak van der Waals’ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.
(iv) Diamond is used as an abrasive.
Ans: Diamond is very hard and hence can be used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
Ans: Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as Cu, Mn, Mg, Si, and Zn. It is very malleable and ductile. Therefore, it is used in making aircraft bodies.
(vi) Aluminium utensils should not be kept in water overnight.
Ans: Generally, aluminium metal does not react with water quickly but, when it is kept overnight, it reacts slowly with water in presence of air.
2Al(s) + O2(g) + H2O(l) ——–> Al2O3(S) + H2(g)
a very small amount of (in ppm) Al3+ produced in the solution is injurious to health if the water is used for drinking purposes.
(vii) Aluminium wire is used to make transmission cables.
Ans: Aluminium wire is used to make transmission cables: Aluminium is a good conductor of electricity and is much lighter than copper. Its low density and cost, combined with good electrical conductivity, make it suitable for use in transmission cables for long-distance power lines.
23. Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon.
Ans: Ionisation enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol).This is expected due to its small size. However, as we move down the group to silicon, there is a sharp decrease in the enthalpy to 786 kJ. Because there is increase in atomic size on moving from carbon to silicon, the screening effect increases. Thus the force of attraction of nucleus for the valence electron decreases as compared to carbon. Thus the ionisation enthalpy decreases from carbon to silicon.
24. How would you explain the lower atomic radius of Ga as compared to Al?
Ans:
| Atomic radius (in pm) | |
| Aluminium | 143 |
| Gallium | 135 |
Although Ga has one shell more than Al, its size is lesser than Al. This is because of the poor shielding effect of the 3d-electrons. The shielding effect of d-electrons is quite poor, so the effective nuclear charge experienced by the valence electrons in gallium is significantly greater than in aluminium.
25. What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?
Ans: Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example, oxygen exists in two allotropic forms: O₂ (dioxygen) and O₃ (ozone).
In diamond, carbon is SP3 -hybridised. Since diamond is a three dimensional network solid, it is the hardest substance with high density whereas graphite has a layered structure. The layers in graphite are held together by van der Waals forces of attraction, which is why graphite is soft and slippery.
26. (a) Classify the following oxides as neutral, acidic, basic or amphoteric.
CO, B2O2, SiO2, CO2, Al2O3, PbO2, Tl2O3
Ans: Classification of the given Oxides:
Acidic oxide: B2O3, SiO2, CO2
Basic oxide: Tl2O3
Amphoteric oxide: Al2O3, PbO2
Neutral oxide: CO
(b) Write suitable equations to show their nature.
Ans:
27. In some of the reactions thallium resembles aluminium, whereas in others it resembles group 1 metals. Support this statement by giving some evidence.
Ans: Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is +3. The most prevalent oxidation state for this group is +3. However, the heavier members can also exhibit a +1 oxidation state due to the inert pair effect. Aluminium displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminium and alkali metals.
Thallium, like aluminium, forms compounds such as TlCl3 and Tl2O3. It resembles alkali metals in compounds Tl2O and TlCl.
28. When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl 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.
Ans: The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.
The white precipitate, referred to as compound A, is aluminium hydroxide. When an excess of the base is added, compound B, which forms, is sodium tetrahydroxoaluminate(III).
Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.
Al(OH)3 + 3HCl → AlCl3 + 3H2O
So, compound C is aluminium chloride.
– Now, when the compound A i.e. aluminium hydroxide is heated it undergoes decomposition reaction and yields alumina (D) along with water.
2Al(OH)3 → Al2O3 + 3H2O
29. What do you understand by
(a) inert pair effect.
Ans: As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. This effect is due to the poor shielding of d and f block electrons of the inner shell, which increases the effective nuclear charge.
(b) allotropy. and
Ans: Allotropy is the phenomenon in which an element exists in more than one form having the same chemical properties but different physical properties.
(c) catenation.
Ans: The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in Si and S.
32. Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
Ans: In the laboratory, CO2 can be prepared by the action of dilute hydrochloric acid on calcium carbonate.
The reaction involved is as follows:
Carbon monoxide:
Industrial:
Laboratory:
Carbon dioxide:
industrial:
Laboratory:
CaCO3 + 2HCI(aq) → CaCI2(aq) + CO2(g) + H2O(l)
33. An aqueous solution of borax is:
(a) neutral.
(b) amphoteric.
(c) basic.
(d) acidic.
Ans: Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3), therefore, it is basic in nature, i.e., option (c) is correct. Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3). It is, therefore, basic in nature.
34. Boric acid is polymeric due to
(a) its acidic nature.
(b) the presence of hydrogen bonds.
(c) its monobasic nature.
(d) its geometry.
Ans: Boric acid is polymeric due to the presence of H-bonds. Therefore, option (b) is correct.
Explanation:
Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.
35. The type of hybridisation of boron in diborane is:
(a) sp
(b) sp2
(c) sp3
(d) dsp2
Ans: The type of hybridisation of boron in diborane is sp2. Each boron atom in diborane (B2H6) is sp3 hybridised.
36. Thermodynamically the most stable form of carbon is:
(a) diamond.
(b) graphite.
(c) fullerenes.
(d) coal.
Ans: (b) Graphite is thermodynamically the most stable form of carbon.
37. Elements of group 14
(a) exhibit oxidation state of +4 only.
(b) exhibit oxidation state of +2 and +4.
(c) form M2-and M4+ ion (d) form M2+ and M4+ ions.
Ans: (b) The elements of Group 14 have 4 valence electrons, which typically gives them an oxidation state of +4. However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable.Therefore, this group exhibits +4 and +2 oxidation states:
| Group 14 element | Oxidation state |
| C | +4 |
| Si | +4 |
| Ge, Sn,pb | +2,+4 |
38. If the starting material for the manufacture of silicons is RSiCl3 write the structure of the product formed.
Ans: (i)
(ii)
