If the bond order in $\mathrm{C}_2$ is ' $\mathrm{x}$ ' then the bond order in $\mathrm{B}_2$ and…
If the bond order in $\mathrm{C}_2$ is ' $\mathrm{x}$ ' then the bond order in $\mathrm{B}_2$ and $\mathrm{O}_2$, respectively are
- $\frac{1}{2} x, 2 x$
- $x, x$
- $\frac{1}{2} \mathrm{x}, \mathrm{x}$
- $x, 2 x$
Solution
$\begin{aligned} & \text { B.O. }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2} \\ & \mathrm{C}_2: \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \\ & \Rightarrow \text { B.O. }=\frac{8-4}{2}=2, \mathrm{sox}=2 \\ & \mathrm{~B}_2: \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^1 \pi 2 \mathrm{p}_{\mathrm{y}}{ }^1 \\ & \Rightarrow \text { B .O. }=\frac{6-4}{2}=1, \text { so B.O. }=\frac{1}{2} \mathrm{x} \\ & \mathrm{O}_2: \sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 2 \mathrm{p}^1 \mathrm{x} \pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1 \\ & \Rightarrow \text { B.O. }=\frac{10-6}{2}=2, \text { so B.O. }=\mathrm{x} .\end{aligned}$
Asked in: AP EAMCET 2023 (15 May Shift 1)
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