Question

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

• A. $\ce{C_6H_6}$
• B. $\ce{(CH_3)_4C}$
• C. $\left( CH _{3}\right)_{2} C = O$
• D. $\ce{CH_3 - CH= CH - CN}$

Answer 1

Answer: (B)

$\ce{(CH_3)_4C}$

Answer 2

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) $\left( CH _{3}\right)_{4}$ 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) $\left( CH _{3}\right)_{2} C = O$ Two $C$ -atoms belong to $CH _{3}$

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

One C-atom, bonded with O-atom and $\left( CH _{3}\right)_{2}$ -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 $sp ^{3}$ -hybridisation (i.e. 4 -hybrid orbitals.

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

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