Q9 / 155 · WBJEE 2017 · Chemistry
Among $\mathrm{Me}_{3}\mathrm{N}, \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}$ and MeCN
(Me $=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order
- Correct answerA$\mathrm{MeCN}>\mathrm{C}_{\mathrm{g}} \mathrm{H}_{\mathrm{s}} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}$
- B$\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}>\mathrm{MeCN}$
- C$\mathrm{Me}_{3} \mathrm{N}>\mathrm{MeCN}>\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}} \mathrm{N}$
- DElectronegativity same in all
Solution
$\because$ In Me $_{3} \mathrm{N} \rightarrow \mathrm{N}$ is $s p^{3}$ -hybrid
and in $\mathrm{MeCN} \rightarrow \mathrm{N}$ is $s p$ -hybrid Also electronegativity of any element $\alpha \%$ $s$ -character in hybrid orbitals. Thus, order of electronegativity for nitrogen is
$$
\mathrm{MeCN}>\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} < \mathrm{Me}_{3} \mathrm{N}
$$
and in $\mathrm{MeCN} \rightarrow \mathrm{N}$ is $s p$ -hybrid Also electronegativity of any element $\alpha \%$ $s$ -character in hybrid orbitals. Thus, order of electronegativity for nitrogen is
$$
\mathrm{MeCN}>\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} < \mathrm{Me}_{3} \mathrm{N}
$$Hence, the correct answer is option: 'A'.