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

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 arrow NH 3 (B) F 3 B +: NH 3 longrightarrow F 3 B arrow NH 3 (C) : BF 3+ NH 3 longrightarrow BF 3 arrow NH 3 (D) : BF 3+ NH 3 longrightarrow BF 3 arrow NH 3

Tardigrade Admin · Accepted Answer

The monomeric trihalides of elements of group

13

, being electron deficient, are strong Lewis acids. Boron trifluoride

(B F_{3})

easily reacts with Lewis base such as

N H_{3}

to complete its octet around boron. This can be expressed as shown below :

F_{3} B + : N H_{3} ⟶ F_{3} B ⟵ N H_{3}

Q. $B F_{3}$ easily reacts with a Lewis base such as $N H_{3}$ to complete octet around boron. This can be expressed as

$F_{3} B + : N H_{3} ⟶ F_{3} B \leftarrow N H_{3}$

$F_{3} B + : N H_{3} ⟶ F_{3} B \to N H_{3}$

$: B F_{3} + N H_{3} ⟶ B F_{3} \to N H_{3}$

$: B F_{3} + N H_{3} ⟶ B F_{3} \leftarrow N H_{3}$

Q. BF3​ easily reacts with a Lewis base such as NH3​ to complete octet around boron. This can be expressed as

F3​B+:NH3​⟶F3​B←NH3​

F3​B+:NH3​⟶F3​B→NH3​

:BF3​+NH3​⟶BF3​→NH3​

:BF3​+NH3​⟶BF3​←NH3​

The monomeric trihalides of elements of group 13 , being electron deficient, are strong Lewis acids. Boron trifluoride (BF3​) easily reacts with Lewis base such as NH3​ to complete its octet around boron. This can be expressed as shown below : F3​B+:NH3​⟶F3​B⟵NH3​

Q. $B F_{3}$ easily reacts with a Lewis base such as $N H_{3}$ to complete octet around boron. This can be expressed as

$F_{3} B + : N H_{3} ⟶ F_{3} B \leftarrow N H_{3}$

$F_{3} B + : N H_{3} ⟶ F_{3} B \to N H_{3}$

$: B F_{3} + N H_{3} ⟶ B F_{3} \to N H_{3}$

$: B F_{3} + N H_{3} ⟶ B F_{3} \leftarrow N H_{3}$