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)

L Lovekush Kumar Saini

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

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

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

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

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

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

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