Question

Bond orders of $NO$ and $N{O}^{+}$ are respectively $x\text{and}y$. Calculate the value of $x+y$

Answer 1

$N{O}^{+}$ and $NO$ are derivative of $N_2.$
For $NO$, we have total 7 + 8 = 15 electrons
$NO$ has 10 bonding electrons and 5 antibonding elelctrons.
We know, Bond order = $\frac{1}{2}$(Bonding electron - non bonding electron)
Bond order for $NO$ = $\frac{1}{2}$(10- 5) = 2.5
For $N{O}^{+}$, we have 7 + 8 - 1 = 14 electrons
i.e 10 bonding electrons and 4 antibonding electrons
So, Bond order = $\frac{1}{2}$(10 -4) = 3


$N{O}^{+}$ bond order = 3
$NO$ bond order = 2.5
$\Rightarrow 3+2.5=5.5$