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

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$F_{3} B + : {NH}_{3} ⟶ F_{3} B \to {NH}_{3}$

$: {BF}_{3} + {NH}_{3} ⟶ {BF}_{3} \leftarrow {NH}_{3}$

$F_{3} B + : {NH}_{3} ⟶ F_{3} B \leftarrow {NH}_{3}$

$: {BF}_{3} + {NH}_{3} ⟶ {BF}_{3} \to {NH}_{3}$

answer is A.

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Detailed Solution

The monomeric trihalides of elements of group 13, being electron deficient, are strong Iewis acids. Boron trifluoride easily reacts with Lewis bases such as ${NH}_{3}$ to complete its octet around boron.

$F_{3} B + : {NH}_{3} ⟶ F_{3} B \leftarrow {NH}_{3}$

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