Question

The correct order of strength of $-I$ effect is:

• A. $-\stackrel{+}{S}b{H}_3>-\stackrel{+}{A}s{H}_3>-\stackrel{+}{P}{H}_3>-\stackrel{+}{N}{H}_3$
• B. $-\stackrel{+}{S}b{H}_3>-\stackrel{+}{P}{H}_3>-\stackrel{+}{A}s{H}_3>-\stackrel{+}{N}{H}_3$
• C. $-\stackrel{+}{N}{H}_3>-\stackrel{+}{P}{H}_3>-\stackrel{+}{S}b{H}_3>-\stackrel{+}{A}s{H}_3$
• D. $-\stackrel{+}{N}{H}_3>-\stackrel{+}{P}{H}_3>-\stackrel{+}{A}s{H}_3>-\stackrel{+}{S}b{H}_3$

Answer 1

The correct option is D $-\stackrel{+}{N}{H}_3>-\stackrel{+}{P}{H}_3>-\stackrel{+}{A}s{H}_3>-\stackrel{+}{S}b{H}_3$

As we go down the group, the electronegativity of the elements decrease.
$N>P>As>Sb$
More the electronegativity of the element, higher the strength of its $-I$ effect.
Also as we move down the group, the size of the element increases thereby increasing the bond length between that element and the attached carbon chain. Therefore, the strength of $-I$ effect decreases down the group.
Therefore, the correct order of the $-I$ strength will be:
$-\stackrel{+}{N}{H}_3>-\stackrel{+}{P}{H}_3>-\stackrel{+}{A}s{H}_3>-\stackrel{+}{S}b{H}_3$