Question
Atomic orbitals of carbon in carbon dioxide are
A.
$$s{p^2}$$ - hybridised
B.
$$s{p^3}d$$ - hybridised
C.
$$sp$$ - hybridised
D.
$$s{p^3}$$ - hybridised
Answer :
$$sp$$ - hybridised
Solution :
$$\eqalign{
& {C_{{\text{excited}}\,\,{\text{state}}}} = 2{s^1}2p_x^1p_y^1p_z^1 \cr
& {O_{{\text{ground}}\,\,{\text{state}}}} = 2{s^2}2p_x^2p_y^1p_z^1 \cr} $$
In carbon dioxide $$C$$ is $$sp$$ hybridised and $$O$$ is $$s{p^2}$$ hybridised. $$sp$$ hybridised orbitals of carbon atom overlap with two $$p$$ orbitals of two oxygen atoms to form 2 $$sigma$$ bond and remaining $$p$$ - electrons ( unhybridised $$p$$ - orbital ) will be used to form $$pi$$ bonds.