Question

$NO$ has $11$ valence electron. It is impossible for all to be paired, and hence this is an odd electron molecule and the gas is paramagnetic.

Which of the following species has the smallest $N-O$ bond length ?

• A. $NO$
• B. $N{O}^{-}$
• C. $N{O}^{+}$
• D. $N_{2}O$

Answer 1

The correct option is C $N{O}^{+}$

$NO$ has a total of $15$ electrons ($7$ from nitrogen and $8$ from oxygen). The electronic configuration is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2\ast 2p{x}^1$

$BondOrder=\frac{BondingOrbitalelectrons-AntibondingOrbitalelectrons}{2}$

Bond Order = $\frac{6-1}{2}$ = 2.5

$N{O}^{+}$ has a total of 14 electrons (7 from nitrogen and 8 from oxygen). The electronic configuration is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2$

Bond Order = $\frac{6-0}{2}$ = 3

$N{O}^{-}$ has a total of 16 electrons (7 from nitrogen and 8 from oxygen). The electronic configuration is $1s^2\ast 1s^{2}2s^2\ast 2s^{2}2p{z}^{2}2p{x}^{2}2p{y}^2\ast 2p{x}^1\ast 2p{y}^1$

$BondOrder=\frac{BondingOrbitalelectrons-AntibondingOrbitalelectrons}{2}$

Bond Order = $\frac{6-2}{2}$ = 2

The bond order decreases in the order $N{O}^{+}>NO>N{O}^{-}$.

The bond order decreases in the order $N{O}^{+}>NO>N{O}^{-}>N_{2}O$. The bond orders of $N{O}^{+},NO,N{O}^{-}and{N}_{2}O$ are 3, 2.5, 2 and 1.5 respectively. Higher is the bond order, shorter is the bond length. Thus $N{O}^{+}$ has the shortest N-O bond.