Question

Arrange the following in order of ${\mathrm{K}}_{\mathrm{b}}$value:

• A. P > Q > R
• B. Q > P > R
• C. R > P > Q
• D. R > Q > P

Answer 1

The correct option is C

R > P > Q

Explanation for the correct options:

Case 1: Pyridine

(P) $S{P}^2$ hybridization (Because L.P is not conjugate, not considered for hybridization.)

Case 2: Pyrrole

(Q) Nitrogen has three $\mathrm{\sigma }$ bonds and is $\left({\mathrm{SP}}^2\right)$ hybridized. The lone pair inside the ring is delocalized, thus it doesn't participate in hybridization, making this molecule aromatic in nature.

$\left({\mathrm{SP}}^2\right)$ hybridization as ($1\mathrm{S}+2\mathrm{P}$) orbitals taking part in the hybridization.

Case 3: Piperidine

(R) There is no lone pair delocalization, hence its $\left({\mathrm{SP}}^3\right)$ hybridization.

It is evident from the above 3 cases that the greater the $"\mathrm{p}"$ character, the more basic it (compound) is.

Therefore, the correct order is R>P>Q

Hence Option (C) is correct.