Question

Match the species in List-I with their bond order given in List-II.

Answer 1

$BondOrder=\frac{No.of\overline{e}inbondingMO-Noof\overline{e}inantibondingMO}{2}$

(a) ${NO}^{+}=\frac{10-4}{2}=\frac{6}{2}=3$

(b) $O_2^{-}=\frac{10-7}{2}=\frac{3}{2}=1.5$

(c) $O_2^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$

(d) $NO=\frac{10-5}{2}=\frac{5}{2}=2.5$