Question

Which of the following is the correct decreasing bond order for $NO,N{O}^{+}$, and $N{O}^{-}$ ?

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

Answer 1

The correct option is D $N{O}^{+}>NO>N{O}^{-}$

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 $N{O}^{+}>NO>N{O}^{-}$.