Question

Answer the following questions with respect to the compound $NO\left[B{F}_4\right]$
(P) Bond order of the part underline.
(Q) Total number of $\sigma$ bonds in the compound.
(R) Total number of $\pi$ bonds in the compound.
(S) Number of hybrid orbitals involved in the hybridisation of boron.

Answer 1

$\left(NO{)}^{+}\right(B{F}_4{)}^{-}$

(Refer to Image)

(P) Bond order= $3$

(Q) Total $\sigma$ bonds= $5$

(R) Total $\Pi$ bonds= $2$

(S) Total hybrid orbitals of involved in hybridization of B= $s{p}^3$

$=3(p-orbitals)+1(s-orbitals)$

$=4$

P	Q	R	S
$3$	$5$	$2$	$4$