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

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

(A)

(B) Diamond

(C) Graphite

Answer:

The state of hybridisation of carbon in:

(A)

is hybridised and is bonded to three oxygen atoms.

(B) Diamond

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

(C) Graphite

Each carbon atom in graphite is 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.
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 hybridised and is bonded to four other carbon atoms through a σ bond.

In graphite, each carbon atom is 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 to give

  • Lead(IV) chloride is highly unstable towards heat

  • Lead is known not form an iodide,

Answer:

(A) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is and . 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, is much less stable than . However, the formation of takes place when chlorine gas is bubbled through a saturated solution of .

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

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

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

Answer

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

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

Q 14 Structure of Double-Bond Character Causes the Bond Length to Shorten in BF3

Character Causes the Bond Length to Shorten in BF3

Q 14 Structure of Double-Bond Character Causes the Bond Length to Shorten in BF3

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

Q 14 1 Structure of Tetrahedral structure

Tetrahedral Structure

Q 14 1 Structure of Tetrahedral structure

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

Q 15 Structure of the respective dipole-moments of the B–Cl bond

Dipole-Moments of the B–Cl Bond

Q 15 Structure of the respective dipole-moments of the B–Cl bond