Question

$B{F}_3$ is planar and electron deficient compound. Hybridisation and number of electrons around the central atom, respectively are:

• A. $s{p}^3\text{and}4$
• B. $s{p}^3\text{and}6$
• C. $s{p}^2\text{and}6$
• D. $s{p}^2\text{and}8$

Answer 1

The correct option is C $s{p}^2\text{and}6$

The structure of $B{F}_3$ can be shown as follows,


Electronic configuration of Boron is $1s^{2}2s^{2}2p^1$
In excited state, its outer electronic configuration is $2s^{1}2p_x^{1}2p_y^1$
Three hybrid orbitals are required to form $B{F}_3$. Hence, the hydridisation of Boron is $s{p}^2$ and the number of electrons around the boron is 6 due to the presence of three $B-F$ bonds.
Hence, option (C) is the correct answer.