Question

Among ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$, ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ and ${\text{MeCN}}$ (${\text{Me}} =$methyl group) the electronegativity of ${\text{N}}$ is in the order of:
A) ${\text{MeCN}} > {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}} > {\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$
B) ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}} > {\text{M}}{{\text{e}}_{\text{3}}}{\text{N}} > {\text{MeCN}}$
C) ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}} > {\text{MeCN}} > {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$
D) Electronegativity is the same in all.

Answer 1

Electronegativity of an atom depends on its state of hybridisation. Determine the hybridisation of nitrogen i.e. ${\text{N}}$ in each of the given compounds. From the hybridisation, calculate the percent s-character and then decide the most electronegative carbon atom.

Complete step-by-step answer:
We are given three compounds ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$, ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ and ${\text{MeCN}}$ (${\text{Me}} =$methyl group). Let us calculate the percent s-character for nitrogen atom in each of the given compounds.
Consider the compound ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$:
The structure of the compound ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$ is as follows:

The nitrogen atom in ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$ forms three bond pairs and also there is one lone pair of electron. Thus, ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$ has a total four bond pairs. Thus, the nitrogen atom in ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$ is $s{p^3}$ hybridised.
Percentage s-character in $s{p^3}$ hybridised orbital $= \dfrac{1}{4} = 0.25 \times 100\% = 25\%$.
Thus, the percentage s-character of the nitrogen atom in ${\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$ is $25\%$.
Consider the compound ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$:
The structure of the compound ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ is as follows:

The nitrogen atom in ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ forms two bond pairs and also there is one lone pair of electron. Thus, ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ has total three bond pairs. Thus, the nitrogen atom in ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ is $s{p^2}$ hybridised.
Percentage s-character in $s{p^2}$ hybridised orbital $= \dfrac{1}{3} = 0.33 \times 100\% = 33\%$.
Thus, the percentage s-character of the nitrogen atom in ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ is $33\%$.

Consider the compound ${\text{MeCN}}$:
The structure of the compound ${\text{MeCN}}$ is as follows:

The nitrogen atom in ${\text{MeCN}}$ forms one bond pair and also there is one lone pair of electrons. Thus, ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$ has total two bond pairs. Thus, the nitrogen atom in ${\text{MeCN}}$ is sp hybridised.
Percentage s-character in sp hybridised orbital $= \dfrac{1}{2} = 0.5 \times 100\% = 50\%$.
Thus, the percentage s-character of the nitrogen atom in ${\text{MeCN}}$ is $50\%$.
We know that more the s-character, more is the electronegativity of the nitrogen atom.
Thus, the electronegativity of ${\text{N}}$ is in the order of ${\text{MeCN}} > {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}} > {\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$.
Thus, the correct option is (A) ${\text{MeCN}} > {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}} > {\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}$.

Note: Remember that more the s-character, more is the electronegativity of the nitrogen atom. Thus, the sp hybridised nitrogen atom is most electronegative than the $s{p^2}$ hybridised nitrogen atom which is more electronegative than the $s{p^3}$ hybridised nitrogen atom