Question

Which one of the following has maximum number of hybrid orbitals ?

• A. $\mathrm{C}{}_6\mathrm{H}{}_6$
• B. $(\mathrm{CH}{}_3){}_4\mathrm{C}$
• C. ${\left(C{H}_3\right)}_{2}C=O$
• D. $\mathrm{CH}{}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CN}$

Answer 1

Answer: (B)

Hydrogen atom does not show hybridisation, thus number of hybrid orbitals in :

(a) $C_6{H}_6^{-}$ All six C-atoms show $s{p}^2$ -hybridisation (i.e. 3 -orbitals by each $C$ -atom )

$\therefore$ Total number of hybrid orbitals $=6\times 3=18$

(b) ${\left(C{H}_3\right)}_4$ C All five C-atoms show

$s{p}^3$ -hybridisation (i.e. 4-orbitals by each C-atom).

$\therefore$ Total number of hybrid orbitals $=5\times 4=20$.

(c) ${\left(C{H}_3\right)}_{2}C=O$ Two $C$ -atoms belong to $C{H}_3$

group show $s{p}^3$ -hybridisation (i.e. 4 -orbital by each $C$ -atom $)$.

One C-atom, bonded with O-atom and ${\left(C{H}_3\right)}_2$ -groups, show $s{p}^2$ -hybridisation (i.e.3-hybrid orbitals).

One $O$ - atom also show $s{p}^2$ -hybridisation Thus, total number of hybrid orbitals $=8+3+3=14$



$•C-1$ show sp-hybridisation (i.e. 2 -hybrid orbitals)

$•C-2$ and 3 show $s{p}^2$ -hybridisaton (i.e. 3 -hybrid orbital by each C-atom)

$•$ C-4 show $s{p}^3$ -hybridisation (i.e. 4 -hybrid orbitals.

$•$ N-atom show sp-hybridisation (i.e. 2 -hybrid orbitals) Thus, total number of hybrid orbitals

$=2+6+4+2=14$