In the reaction ${NH}_{3} + {BF}_{3} \to {NH}_{3} \cdot {BF}_{3}$ (Adduct).
Choose the correct statement.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Hybridization of $' N'$ an adduct is ${sp}^{3}$

Hybridization of $' B'$ in ${BF}_{3}$ changes in the adduct.

Here, ${NH}_{3}$ is acts as a Lewis base while ${BF}_{3}$ is a Lewis acid.

Hybridization of $' N'$ in ${NH}_{3}$ changes in adduct

answer is B, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Despite not having a proton, ${BF}_{3}$ functions as a Lewis acid because it is an electron-deficient substance. It completes its octet when it reacts with ${NH}_{3}$ by taking a single pair of electrons from ${NH}_{3}$ . The equation for the reaction is ${BF}_{3}$ + ${NH}_{3} \to$ ${BF}_{3}$ $\leftarrow {NH}_{3}$ . It can be explained using the Lewis electrical theory of acids and bases. While N is ${sp}^{3}$ hybridized in ${NH}_{3}$ , boron is ${sp}^{2}$ hybridized in ${BF}_{3}$ .