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  • Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them

Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.

\mathbf{A}\) \quad \mathrm{O}_{2}^{-}<\mathrm{NO}<\mathrm{C}_{2}^{2-}<\mathrm{He}_{2}^{+}

\mathbf{B}\) \quad \mathrm{C}_{2}^{2-}<\mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{NO}

\mathbf{C}\) \quad \mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{N} \mathrm{O}<\mathrm{C}_{2}^{2-}

\mathbf{D}\) \quad \mathrm{N} \mathrm{O}<\mathrm{O}_{2}^{-}<\mathrm{C}_{2}^{2-}<\mathrm{H} \mathrm{e}_{2}^{+}

Answers (1)

\\\text{First we will calculate the bond order for all elements. } \\\\ \text{Bond order } =\frac{N_{b}-N_{a}}{2} \\ \mathrm{N}_{\mathrm{b}}= \text{ Number of electrons in bonding molecular orbital } \\ \mathrm{N}_{\mathrm{a}}=\text{ Number of electrons in antiobonding molecular orbital} \\

\begin{array}{l} \mathrm{NO}=15 \text { electrons } \\ \sigma_{1 \mathrm{s}^{2}}, \sigma_{1 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{s}^{2}}, \sigma_{2 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{p}_{z}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}=\pi_{2 \mathrm{p}_{y}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{4}}^{4} \\ \mathrm{B.O}=\frac{8-3}{2}=2.5 \end{array}

\begin{aligned} &\mathrm{O}_{2}^{-}=17 \text { electrons }\\ &\sigma_{1 \mathrm{s}^{2}}, \sigma_{1 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{s}^{2}}, \sigma_{2 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{p}_{z}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}=\pi_{2 \mathrm{p}_{\mathrm{y}}^{2}},, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}^{*}=\pi_{2 \mathrm{p}_{\mathrm{r}}^{2}}^{*}\\ &\mathrm{B} . \mathrm{O}=\frac{8-5}{2}=1.5 \end{aligned}

\begin{array}{l} \mathrm{C}_{2}^{2-}=14 \text { electrons } \\ \sigma_{1 \mathrm{s}^{2}}, \sigma_{1 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{s}^{2}}, \sigma_{2 \mathrm{s}^{2}}^{*}, \sigma_{2 \mathrm{p}_{z}^{2}}, \pi_{2 \mathrm{p}_{\mathrm{x}}^{2}}=\pi_{2 \mathrm{p}_{y}^{2}} \\ \mathrm{B.O}=\frac{8-2}{2}=3 \end{array}

\begin{aligned} &\mathrm{He}_{2}^{2+}=3 \text { electrons }\\ &\sigma_{1 s^{2}}, \sigma_{1 s^{1}}^{*}\\ &\mathrm{B.O}=\frac{2-1}{2}=0.5\\ &\text { Therefore order is } \mathrm{He}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{NO}<\mathrm{C}_{2}^{2-} \end{aligned}

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