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Question 63

Increasing order of electron withdrawing power of following functional groups is:
a. $$-\text{CN}$$, b. $$-\text{COOH}$$, c. $$-\text{NO}_2$$, d. $$-\text{I}$$

Electron-withdrawing power is determined mainly by two effects:
(i) Inductive effect ($$−I$$), which pulls electron density through σ-bonds, and
(ii) Resonance/Mesomeric effect ($$−M$$ when the group pulls electron density through $$π$$-bonds, $$+M$$ when it donates).

Step 1: Analyse each substituent.

1. $$-\text{I}$$ (iodo) : Iodine is less electronegative than $$N$$, $$O$$, or $$Cl$$, so its $$−I$$ effect is weak. Moreover, thanks to its lone pairs it can donate by $$+M$$ into an attached $$π$$-system, further reducing the net electron-withdrawing ability. 

Hence this is the weakest EW group in the list.

2. $$-\text{COOH}$$ (carboxyl) : Oxygen atoms are electronegative, so the group shows a moderate −I effect. Inside the group itself the C=O pulls electron density from the O−H bond, but relative to other strong −I groups, 

$$-\text{COOH}$$ is only moderately electron-withdrawing.

3. $$-\text{CN}$$ (cyano) : The carbon of the triple bond is sp-hybridised (50 % s-character) and directly attached to a highly electronegative nitrogen; therefore the −I effect is strong. 

There is no +M donation, so the net electron-withdrawing power is high.

4. $$-\text{NO}_2$$ (nitro) : The nitro group possesses two strongly electronegative oxygens and a positively charged nitrogen in its major resonance form $$\text{O}^{-}\!-\!\text{N}^{+}\!=\!\text{O}$$. It shows both a very powerful −I effect and a very powerful −M effect (withdraws electron density via resonance), 

making it the strongest electron-withdrawing group among common substituents.

Step 2: Arrange in increasing order of withdrawing power (weakest → strongest):
$$-\text{I} \lt -\text{COOH} \lt -\text{CN} \lt -\text{NO}_2$$

This sequence corresponds to Option C.

Final Answer: Option C $$\longrightarrow\ $$ $$d \lt b \lt a \lt c$$

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