Question

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

• A. Magnetic moment of the species is zero
• B. $B-F$ bond energy is lower in this compound than in $B{F}_3$
• C. It has $5\sigma$ and $2\pi$ bond
• D. Nitrogen oxygen bond order is same as in $N_2$

Answer 1

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

Nitrogen oxygen bond order is same as in $N_2$

$N{O}^{+}[B{F}_4{]}^{-}$
Number of $\sigma$ bonds in $[B{F}_4{]}^{-}=4$
Bond order of $N{O}^{+}=3.0$, i.e., one sigma bond and two $\pi$ bonds.
$\therefore$ It has $5$ sigma bonds and two $\pi$ bonds.
Molecular orbital electronic configuration of $N{O}^{+}$ is:
$\sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi (2p_y{)}^2$

Bond Order of $N{O}^{+}=\frac{(N_b-N_a)}{2}=\frac{(10-4)}{2}=3$
Bond Order of $N_2$ is also $3$. So, the bond order of $N{O}^{+}$ is equal to bond order of $N_2$.
$N{O}^{+}$ has no unpaired electron. So it is diamagnetic.
$[B{F}_4{]}^{-}$ has no unpaired electron. So, it is also diamagnetic.
Hence the magnetic moment of $\left[NO\right]\left[B{F}_4\right]$ is zero and is diamagnetic.
$B-F$ bond energy is lower in $B{F}_4^{-}$ than in $B{F}_3$, due to presence of back bonding in $B{F}_3$.