Question

Among ${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$, ${{{C}}_5}{{{H}}_5}{{N}}$ and ${{C}}{{{H}}_3}{{CN}}$, the electronegativity of nitrogen is in the order:
A. ${{C}}{{{H}}_3}{{CN > }}{{{C}}_5}{{{H}}_5}{{N}} > {\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$
B. ${{{C}}_5}{{{H}}_5}{{N}} > {\left( {{{C}}{{{H}}_3}} \right)_3}{{N}} > {{C}}{{{H}}_3}{{CN}}$
C. ${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}} > {{C}}{{{H}}_3}{{CN}} > {{{C}}_5}{{{H}}_5}{{N}}$
D. Electronegativity is the same in all of these.

Answer 1

Hint:Electronegativity can be measured using the steric number of the compound. Steric number is the sum of the number of $\sigma $ bonds and the number of lone pairs. Based on the number of lone pairs, the shape and geometry of the compound changes.

Complete answer:
Based on s character, we can find the electronegativity of nitrogen in each compound. s character is measured from the steric number.
Steric number $ = $ number of $\sigma $ bonds $ + $ number of lone pairs.
The structure of each compound is given below:
 $\left( {{{\left( {{{C}}{{{H}}_3}} \right)}_3}{{N}}} \right)$ $\left( {{{{C}}_5}{{{H}}_5}{{N}}} \right)$ $\left( {{{C}}{{{H}}_3}{{CN}}} \right)$
Now let’s calculate the steric number of each compound.
${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$ has one lone pair and three $\sigma $ bonds. Therefore its steric number is $3 + 1 = 4$, i.e. it has ${{s}}{{{p}}^3}$ hybridization. Its character is $25\% $.
${{{C}}_5}{{{H}}_5}{{N}}$ also has one lone pair and two $\sigma $ bonds. Thus its steric number is $2 + 1 = 3$, i.e. it has ${{s}}{{{p}}^2}$ hybridization. Its s character is $33\% $.
While ${{C}}{{{H}}_3}{{CN}}$ has one lone pair and only one $\sigma $ bond. Thus its steric number is $1 + 1 = 2$, i.e. it has ${{sp}}$ hybridization. Its s character is $50\% $.
${{C}}{{{H}}_3}{{CN}}$ has more s character. More the s character, more the electronegativity. Thus we can say that the nitrogen in ${{C}}{{{H}}_3}{{CN}}$ has more electronegativity than that in ${{{C}}_5}{{{H}}_5}{{N}}$ and ${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$. Between ${{{C}}_5}{{{H}}_5}{{N}}$ and ${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$, ${{{C}}_5}{{{H}}_5}{{N}}$ has more s character and ${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$ has least s character. Thus electronegativity is least in ${\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$.
Therefore the order is ${{C}}{{{H}}_3}{{CN > }}{{{C}}_5}{{{H}}_5}{{N}} > {\left( {{{C}}{{{H}}_3}} \right)_3}{{N}}$.

Hence, the correct option is A.

Note:
More s character indicates that the electrons are closer to the nucleus. Basicity is determined based on the electronegativity. Since the electrons are closer to the nucleus, it will be much more basic. This is because it is not ready to donate electrons to bond with a proton.