Find the type of hybridization of each carbon in the following molecules.
(i) `CH_(3)F` , (ii) `CH_(3)CHO`
(iii) `CH_(3)CH_(2)CN` , (iv) `CH_(3)CONH_(2)`
(v) `CH_(3)C-=C CH_(3)` , (vi) `CH_(3)CH-=CH_(2)`
Strategy: Carbon atom forming four sigma bonds but no pi bond is always `sp^(3)` htbridized, carbon atom forming three sigma bonds and one pi bond is always `sp^(2)` hybridized, and carbon atom forming two sigma bonds and two pi bonds is always `sp` hybridized.

Explanation

(i) `H-underset(H)underset(|)overset(H)overset(|)(C)overset(sp^(3))-F`
(ii) `H - underset(H)underset(|)overset(H)overset(|)(C)overset(sp^(3))-overset(H)overset(|)(C)overset(sp^(2))=O`
(iii) `H-underset(H)underset(|)overset(H)overset(|)(C)overset(sp^(3))-underset(H)underset(|)overset(H)overset(|)(C)overset(sp^(3))-Coverset(sp)-=N`
(iv) `(iv)H-underset(H)underset(|)overset(H)overset(|)Coverset(sp^(3))-overset(O)overset(||)Coverset(sp^(2))-overset(H)overset(|)N-H`
(V) `Hoverset(sp^(3))-underset(H)underset(|)overset(H)overset(|)C-overset(sp)C-=overset(sp)C-underset(H)underset(|)overset(H)overset(|)Coverset(sp^(3))-H`
(vi) `Hoverset(sp^(3))-underset(H)underset(|)overset(H)overset(|)C-underset(H)underset(|)overset(sp^(2))C=overset(H)overset(|)Coverset(sp^(2))-H`