Question

The correct decreasing order of basic strength in the given compound is:

• A. $a>b>c$
• B. $a>c>b$
• C. $c>b>a$
• D. $c>a>b$

Answer 1

The correct option is B $a>c>b$

Basic nature of nitrogen follows the order
$N\left(s{p}^3\right)>N\left(s{p}^2\right)>N\left(sp\right)$
This is due to electronegativity nature of hybridised orbital.
$sp>s{p}^2>s{p}^3$ [Electronegativity order]

The lone pair of N (b) is involved in resonance so it is least available for protonation. Hence, it is a least basic N atom.

Comparing lone pair of N of (a) and (c), (c) is less basic than (a) due to its $s{p}^2$ hybridisation.
Thus, the decreasing order of basicity of N is
$a>c>b$