Question

Number of electrons present in the antibonding axially overlapping orbital in $N{O}^{+}$, are:

• A. $4$
• B. $2$
• C. $6$
• D. $1$

Answer 1

The correct option is A $4$

Electronic configuration of $N{O}^{+}$:

$N{O}^{+}$: valence electrons=$5+6-1=10$

${\left(\sigma 1s\right)}^2{(\sigma \ast 1s)}^2{\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4$

2 each in ${\sigma }^{\ast }1s$ and ${\sigma }^{\ast }2s=4$