Question

$BF _{3}$ easily reacts with a Lewis base such as $NH _{3}$ to complete octet around boron. This can be expressed as

• A. $F _{3} B +: NH _{3} \longrightarrow F _{3} B \leftarrow NH _{3}$
• B. $F _{3} B +: NH _{3} \longrightarrow F _{3} B \rightarrow NH _{3}$
• C. $: BF _{3}+ NH _{3} \longrightarrow BF _{3} \rightarrow NH _{3}$
• D. $: BF _{3}+ NH _{3} \longrightarrow BF _{3} \leftarrow NH _{3}$

Answer 1

Answer: (A)

The monomeric trihalides of elements of group $13$ , being electron deficient, are strong Lewis acids. Boron trifluoride $\left( BF _{3}\right)$ easily reacts with Lewis base such as $NH _{3}$ to complete its octet around boron. This can be expressed as shown below :
$F _{3} B +: NH _{3} \longrightarrow F _{3} B \longleftarrow NH _{3}$