Given the hybridistaiton state of each carbon in the following compounds :
I. `CH_(2) = C = O` II. `CH_(3) - CH = CH_(2)`
III. `(CH_(3))_(2) CO` IV. `CH_(2) = CH = C -= N`
v. `C_(6) H_(6)`

Explanation

(i) `overset(1)CH_(2)=overset(2)C=O`
C–1 is `sp^(2)` hybridised.
C–2 is sp hybridised.
(ii) `overset(1)CH_(3)-overset(2)CH=overset(3)CH_(2)`
C–1 is `sp^(3)` hybridised.
C–2 is `sp^(2)` hybridised.
C–3 is `sp^(2)` hybridised.
(iii) `underset(1)CH_(3)-overset(O)overset(||)underset(2)C-underset(3)CH_(3)`
C–1 and C–3 are `sp^(3)` hybridised.
C–2 is `sp^(2)` hybridised.
(iv) `overset(1)CH_(2)=overset(2)CH-overset(3)CequivN`
C–1 is `sp^(2)` hybridised.
C–2 is `sp^(2)` hybridised.
C–3 is `sp^(2)` hybridised.
(v) `C_(6)H_(6)`
All the 6 carbon atoms in benzene are `sp^(2)` hybridised.