Question

Select the correct statement(s) about the compound $NO\left[BF4\right]$ :

• A. It has $5\sigma$ and $2\pi$ bonds
• B. Nitrogen-Oxygen bond length in given compound is higher than in nitric oxide $\left(NO\right)$
• C. It is a diamagnetic species
• D. $B{F}_4^{-}$ and $B{F}_3$ both are lewis acids

Answer 1

Answer: (A), (C)

The correct options are
A It has $5\sigma$ and $2\pi$ bonds
C It is a diamagnetic species

Given compound is : $\left[NO{]}^{+}\right[B{F}_4{]}^{-}$
We know that number of $\sigma$ bonds in $[B{F}_4{]}^{-}$ = $4$.
As per MOT, Bond order of $[NO{]}^{+}$ is $3$, which means there is triple bond between $N$ $\&$ $O$ i.e., one $\sigma$ and two $\pi$ bonds.
Hence, overall count of $\sigma$ bonds are $5$ and $\pi$ bonds are $2$.
$\therefore$ option $\left(A\right)$ is correct.

Also, as per MOT, Bond order of $\left[NO\right]$ is $2.5$ and we know that bond order is inversely proportional to bond length. Hence nitrogen-oxygen bond length in $NO$ is higher than $[NO{]}^{+}$.
$\therefore$ option $\left(B\right)$ is incorrect.

Both $[NO{]}^{+}$ and $[B{F}_4{]}^{-}$ are diamagnetic individually hence overall compound is also diamagnetic.
$\therefore$ option $\left(C\right)$ is correct.

$[B{F}_4{]}^{-}$ is not a lewis acid because its octet is complete whereas in $B{F}_3$ although its octet is not complete but due to presence of $p\pi -p\pi$ back bonding there is partial completion of octet hence it is a weak lewis acid.
$\therefore$ option $\left(D\right)$ is incorrect.