Question

The hybridisation of $C$ in diamond, graphite and ethyne is in the order

• A. $sp ^{3}, sp , sp ^{2}$
• B. $sp ^{3}, sp ^{2}, sp$
• C. $sp, sp^{2}, sp^{3}$
• D. $sp ^{2}, sp ^{3}, sp$

Answer 1

Answer: (B)

$sp ^{3}, sp ^{2}, sp$

Answer 2

(i) As diamond has tetrahedral structure in which all the C-atoms are bonded by $4 \sigma$ -bonds with 4 neighbouring C-atoms. Thus, it has $s p^{3}$ hybridisation.

(ii) Graphite forms an hexagonal sheet like structure in which each C-atom is bonded by $3 \sigma$ -bonds with $3- C$ atoms in the same plane, 4th electron of C-atom is free to move thus, show $s p^{2}$ -hybridisation.

(iii) Ethyne has $CH = CH$ structure, in which one C-atom is bonded with other C-atom through only one $\sigma$ -bond. (as $\pi$ -bonds do not take part in hybridisation)

Thus, it shows $sp$ -hybridisation.

Hence, correct order is: $s p^{3}, s p^{2}, s p$