Page No 324: - Chapter 11 P Block Elements class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 11.9:

What are electron deficient compounds? Are BCl₃ and SiCl₄ electron deficient species? Explain.

Answer:

In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet.

(i) BCl3

BCl₃ is an appropriate example of an electron-deficient compound. B has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6. However, it is still short of two electrons to complete its octet.

(ii) SiCl4

The electronic configuration of silicon is ns² np². This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, SiCl₄ is not an electron-deficient compound.

Question 11.10:

Write the resonance structures of  and .

Answer:

(a) 

(b) 

There are only two resonating structures for the bicarbonate ion.

Question 11.11:

What is the state of hybridisation of carbon in (a)   (b) diamond (c) graphite?

Answer:

The state of hybridisation of carbon in:

(a) 

C in  is sp² hybridised and is bonded to three oxygen atoms.

(b) Diamond

Each carbon in diamond is sp³ hybridised and is bound to four other carbon atoms.

(c) Graphite

Each carbon atom in graphite is sp² hybridised and is bound to three other carbon atoms.

Question 11.12:

Explain the difference in properties of diamond and graphite on the basis of their structures.

Answer:

Rationalise the given statements and give chemical reactions:Question 11.13:

• Lead(II) chloride reacts with Cl₂ to give PbCl₄.

• Lead(IV) chloride is highly unstable towards heat.

• Lead is known not to form an iodide, PbI₄.

Answer:

(a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4. On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable. This is because of the inert pair effect. Hence, PbCl₄ is much less stable than PbCl₂. However, the formation of PbCl₄ takes place when chlorine gas is bubbled through a saturated solution of PlCl₂.

(b) 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).

(c) Lead is known not to form PbI₄. Pb (+4) is oxidising in nature and I^– is 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).

Question 11.14:

Suggest reasons why the B–F bond lengths in BF₃ (130 pm) and  (143 pm) differ.

Answer:

The B–F bond length in BF₃ is shorter than the B–F bond length in  . BF₃ is an electron-deficient species. 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.

Question 11.15:

If B–Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.

Answer:

As a result of the difference in the electronegativities of B and Cl, the B–Cl bond is polar in nature. However, the BCl₃ molecule is non-polar. This is because BCl₃ is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole-moments of the B–Cl bond cancel each other, thereby causing a zero-dipole moment.

Question 11.16:

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 reasons.

Answer:

Hydrogen fluoride (HF) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride (AlF) does not dissolve in it. Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AlF dissolves. This is because of the availability of free F^–. The reaction involved in the process is:

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. Therefore, when BF₃ is added to the solution, B replaces Al from the complexes according to the following reaction:

Question 11.17:

Suggest a reason as to why CO is poisonous.

Answer:

Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The CO–Hb complex is more stable than the O₂–Hb complex. The former prevents Hb from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the CO–Hb complex is about 300 times more stable than the O₂–Hb complex.

Question 11.18:

How is excessive content of CO₂ responsible for global warming?

Answer:

Carbon dioxide is a very essential gas for our survival. However, an increased content of CO₂ in the atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sunrays. Higher the level of carbon dioxide, higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby causing global warming.

Question 11.19:

Explain structures of diborane and boric acid.

Answer:

(a) Diborane

B₂H₆ is an electron-deficient compound. 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:

2 boron and 4 terminal hydrogen atoms (H_t) lie in one plane, while the other two bridging hydrogen atoms (H_b) lie in a plane perpendicular to the plane of boron atoms. Again, of the two bridging hydrogen atoms, one H atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-centre two-electron (2c – 2e^–) bonds, while the two bridging (B–H–B) bonds are three-centre two-electron (3c – 2e^–) bonds.

(b) Boric acid

Boric acid has a layered structure. Each planar BO₃ unit is linked to one another through H atoms. The H atoms form a covalent bond with a BO₃ unit, while a hydrogen bond is formed with another BO₃ unit. In the given figure, the dotted lines represent hydrogen bonds.

Question 11.20:

What happens when

(a) Borax is heated strongly,

(b) Boric acid is added to water,

(c) Aluminium is treated with dilute NaOH,

(d) BF₃ is reacted with ammonia?

Answer:

(a) 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) When boric acid is added to water, it accepts electrons from ^–OH ion.

(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

(d) BF₃ (a Lewis acid) reacts with NH₃ (a Lewis base) to form an adduct. This results in a complete octet around B in BF₃.

Question 11.21:

Explain the following reactions

(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;

(b) Silicon dioxide is treated with hydrogen fluoride;

(c) CO is heated with ZnO;

(d) Hydrated alumina is treated with aqueous NaOH solution.

Answer:

(a) 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) When silicon dioxide (SiO₂) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF₄). Usually, the Si–O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.

The SiF₄ formed in this reaction can further react with HF to form hydrofluorosilicic acid.

(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.

Question 11.22:

Give reasons:

(i) Conc. HNO₃ can be transported in aluminium container.

(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.

(iii) Graphite is used as lubricant.

(iv) Diamond is used as an abrasive.

(v) Aluminium alloys are used to make aircraft body.

(vi) Aluminium utensils should not be kept in water overnight.

(vii) Aluminium wire is used to make transmission cables.

Answer:

(i) 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) 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.

(iii) 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) In diamond, carbon is sp³ hybridised. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason, diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.

(v) 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) The oxygen present in water reacts with aluminium to form a thin layer of aluminium oxide. This layer prevents aluminium from further reaction. However, when water is kept in an aluminium vessel for long periods of time, some amount of aluminium oxide may dissolve in water. As aluminium ions are harmful, water should not be stored in aluminium vessels overnight.

(vii) Silver, copper, and aluminium are among the best conductors of electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminium is a very ductile metal. Thus, aluminium is used in making wires for electrical conduction.

Question 11.23:

Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon?

Answer:

Ionisation enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy (786 kJ). This is because of an appreciable increase in the atomic sizes of elements on moving down the group.

Question 11.24:

How would you explain the lower atomic radius of Ga as compared to Al?

Answer:

Question 11.25: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 very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al.

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?

Answer:

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.

Diamond:

The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest naturally-occurring substances. It is used as an abrasive and for cutting tools.

Graphite:

It has sp² hybridised carbon, arranged in the form of layers. These layers are held together by weak van der Walls’ forces. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.