Among Me_{3}N,C_{5}H_{5}N and MeCN (Me = text | vedclass

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Question

(A) The electronegativity of an atom depends on the percentage of $s$-character in its hybrid orbitals. Higher $s$-character leads to higher electrone

Vedclass · Accepted Answer

$MeCN > C_{5}H_{5}N > Me_{3}N$

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(A) The electronegativity of an atom depends on the percentage of $s$-character in its hybrid orbitals. Higher $s$-character leads to higher electronegativity.In $Me_{3}N$,the nitrogen atom is $sp^{3}$-hybridized ($25\% \ s$-character).In $C_{5}H_{5}N$ (pyridine),the nitrogen atom is $sp^{2}$-hybridized ($33.3\% \ s$-character).In $MeCN$ (acetonitrile),the nitrogen atom is $sp$-hybridized ($50\% \ s$-character).Therefore,the order of electronegativity of the nitrogen atom is $MeCN > C_{5}H_{5}N > Me_{3}N$.

Among $Me_{3}N$,$C_{5}H_{5}N$ and $MeCN$ $(Me = \text{methyl group})$,the electronegativity of $N$ is in the order:

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