Question

The molecule of $CO _{2}$ has angle $180^{\circ}$. It can be explained on the basis of :

• A. $sp^{3}$ hydridisation
• B. $sp^{2}$ hybridization
• C. $sp$ hybridisation
• D. $d^{2}sp^{3}$ hybridization

Answer 1

Answer: (C)

$sp$ hybridisation

Answer 2

Carbon (At. no. $=6$ $EC =2, 4 (1 s^{2} 2 s^{2} 2 p^{2})$ in excited state $s p$ hybrids atom orbital which from $\pi$ bond does not take part in hybridisation. Carbon has no lone pair so, according to VSEPR theory the $s p$ hybrid orbitals stay as far away as possible. In this case, the bond angle being $180^{\circ}$. $s p^{3}$ hybrid molecules are tetrahedral with bond angle $109^{\circ} 28', s p^{2}$ hybrid orbitals are triangular planar with bond angle $120^{\circ}$. $\overset{sp^2}{H_2C} = \overset{sp^2}{CH} - \overset{sp}{C} \equiv \overset{sp}{C} - \overset{sp^3}{CH_3}$