Chemistry Class 11 NCERT Solutions: Chapter 11 the p Block Elements Part 3

Q: 11. What is the state of hybridisation of carbon in

(A) Equation

(B) Diamond

(C) Graphite

Answer:

The state of hybridisation of carbon in:

(A) Equation

Equation is Equation hybridised and is bonded to three oxygen atoms.

(B) Diamond

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

(C) Graphite

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

Q: 12. Explain the difference in properties of diamond and graphite on the basis of their structure.

Answer:

Q.12- table showing difference between diamond and graphite.

Diamond

Graphite

It has a crystalline lattice

It has a layered structure.

In diamond, each carbon atom is Equation hybridised and is bonded to four other carbon atoms through a σ bond.

In graphite, each carbon atom is Equation hybridised and is bonded to three other carbon atoms through a σ bond. The fourth electron forms a π bond.

It is made up of tetrahedral units.

It has a planar geometry.

The C–C bond length in diamond is 154 pm.

The C–C bond length in graphite is 141.5 pm

It has a rigid covalent bond network

which is difficult to break.

It is quite soft and its layers can be separated easily.

It acts as an electrical insulator.

It is a good conductor of electricity.

Q: 13. Rationalise the given statements and give chemical reactions:

  • Lead(II) chloride reacts with Equation to give Equation

  • Lead(IV) chloride is highly unstable towards heat

  • Lead is known not form an iodide, Equation

Answer:

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

Equation

(B) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Equation is highly unstable and when heated, it reduces to Equation

Equation

(C) Lead is known not to form Equation . Equation is oxidising in nature and Equation is reducing in nature. A combination of Equation and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises Equation to Equation and it self gets reduced to Equation .

Equation

Q: 14. Suggest reasons why the Equation bond lengths in Equation (130 pm) and Equation (143 pm) differ.

Answer

The Equation bond length in Equation is shorter than the Equation bond length in Equation . Equation is an electron - deficient species. With a vacant p-orbital on boron, the fluorine and boron atoms undergo Equation back bonding to remove this deficiency. This imparts a doublebond character to the Equation bond.

double-bond character causes the bond length to shorten in BF3

This double-bond character causes the bond length to shorten in Equation (130 pm). However, when Equation coordinates with the fluoride ion, a change in hybridisation from Equation (in Equation ) to Equation Equation occurs. Boron now forms Equation bonds and the double – bond character is lost. This accounts for a Equation bond length of 143 pm in Equation ion.

Q: 15. If Equation bond has a dipole moment, explain why Equation 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 Equation molecule is non – polar. This is because Equation is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole – moments of the Equation bond cancel each other, thereby causing a zero – dipole moment.

Explore Solutions for Chemistry