Question

Which of the following has the smallest bond length?

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

Answer 1

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

NO has 15 electrons
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4,(2p_z{)}^2,({\pi }^{\ast }{)}^1$
So Bond Order$=\frac{1}{2}(a-b)$
a is the number of electrons in bonding molecular orbitals and
b is the number of electrons in antibondng molecular orbitals.
Bond Order$=\frac{1}{2}(10-5)$
$=2.5$
$N{O}^{+}$ has 14 electrons
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4,(2p_z{)}^2$
Bond Order$=\frac{1}{2}(10-4)$
$=3$
$N{O}^{-}$ has 16 electrons
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4,(2p_z{)}^2,({\pi }^{\ast }{)}^2$
Bond Order$=\frac{1}{2}(10-6)$
$=2$
Bond length is inversely proportional to bond order
$N{O}^{+}$ has highest bond order that means it has smallest bond length