Consider the carbonyl compounds I $p\text{-}\mathrm{O}{}_2\mathrm{N}-\mathrm{C}{}_6\mathrm{H}{}_4-\mathrm{CHO}$, II $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CHO}$, III $p\text{-}\mathrm{CH}{}_3\mathrm{O}-\mathrm{C}{}_6\mathrm{H}{}_4-\mathrm{CHO}$ and IV $p\text{-}\mathrm{CH}{}_3\mathrm{O}-\mathrm{C}{}_6\mathrm{H}{}_4-\mathrm{CO}-\mathrm{CH}{}_3$. The correct order of decreasing rate of addition of $\mathrm{CN}{}^{-}$ is

The nitro group (strong -M, -I) makes the carbonyl carbon most electrophilic. Benzaldehyde has no donor. The methoxy group donates by +M and lowers reactivity in III. IV combines this +M donation with the extra methyl group, which reduces electrophilicity and increases hindrance, so it is the slowest.