Question

The one which is least basic is
(A) \[N{{H}_{3}}\]
(B) \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\]
(C) \[{{\left( {{C}_{6}}{{H}_{5}} \right)}_{3}}N\]
(D) \[{{\left( {{C}_{6}}{{H}_{5}} \right)}_{2}}NH\]

Answer 1

Hint: In this, we will use the Lewis concept of acid and base as in given options, nitrogen contains lone pairs of electrons. According to Lewis' concept, a base is a substance that donates its lone pair of electrons to an electron-deficient atom. So a good base in comparison will be that which can easily give its lone pair or make available a lone pair of electrons to electron-deficient species.

Complete Step by Step Answer:
In given options, nitrogen (valence of 3 and one lone pair) is the central metal and its basic character depends on other atoms with which it is bonded. In ammonia (\[N{{H}_{3}}\]), \[{{C}_{6}}{{H}_{5}}N{{H}_{2}}\], \[{{\left( {{C}_{6}}{{H}_{5}} \right)}_{3}}N\], and \[{{\left( {{C}_{6}}{{H}_{5}} \right)}_{2}}NH\], nitrogen has lone pair and it can donate its lone pair to electron-deficient species so, all can behave as lewis base. But which is least basic?

The basic character of nitrogen compounds depends on atoms bonded to nitrogen. Atoms bonded to nitrogen other than hydrogen is benzene in given options. In benzene, there is double bond conjugation therefore, it shows resonance through delocalization of pi electrons. Also, lone pairs can take part in resonance and so, lone pairs will be less available to electron-deficient species.

Thus, the greater the number of benzene compounds bonded with nitrogen, the less will lone pair of electron available to electron-deficient atom (as lone pair busy in resonance) and thus will be least basic as compared to those compounds in which nitrogen bonded with less number of benzenes. As in option C, nitrogen is bonded with the maximum number of benzene so it will be the least basic of all others.
Thus, the correct option is C.

Note: In all the above options central metal is nitrogen whose atomic number is 7. So its electronic configuration is \[1s{}^\text{2},\text{ }2s{}^\text{2},\text{ }2p{}^\text{3}\]. From its electronic configuration, it is clear that the valence (tendency to form a bond to complete its octet) of nitrogen is three and one lone pair in the 2s orbital. Thus, it behaves like a lewis base as it has a lone pair in its inner orbit.