Question

Select the correct statement(s) about the compound $NO\left[B{F}_4\right]$.

• A. It has $5\sigma$ and $2\pi$ bonds.
• B. Nitrogen-oxygen bond length is higher than nitric oxide (NO).
• C. It is a diamagnetic species.
• D. $B-F$ bond length in this compound is lower than in $B{F}_3$.

Answer 1

Answer: (A), (B), (C)

It has $5\sigma$ and $2\pi$ bonds.
Nitrogen-oxygen bond length is higher than nitric oxide (NO).
It is a diamagnetic species.

$N{O}^{+}[B{F}_4{]}^{-}$
$\Rightarrow B.O.ofN{O}^{+}=3.0,$ i.e, one sigma bond and two $\pi$ bonds
$\therefore$ No. of $\pi$ bonds = 2
No. of $\sigma$ bond = 5
$\Rightarrow B.O.$ of $N{O}^{+}=3.0$
and $B.O.$ of $NO=2.5$
Bond order is inversely proportional to bond length.
$\Rightarrow N{O}^{+}$ is diamagnetic and $B{F}_4^{-}$ is also diamagnetic
$\Rightarrow B-F$ bonds are longer in $B{F}_4^{-}$ than in $B{F}_3$ due to absence of $p\pi -p\pi$ back bonding in $\left[B{F}_4^{-}\right]$