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

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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.

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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 ${BF}_{3}$ .