If the bond order in C_2 is 'x',then the bond | vedclass

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(C) The bond order $(B.O.)$ is calculated as $B.O. = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of an

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$\frac{1}{2} x, x$

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(C) The bond order $(B.O.)$ is calculated as $B.O. = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $C_2$ ($12$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. $B.O. = \frac{8 - 4}{2} = 2$. Thus,$x = 2$.For $B_2$ ($10$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$. $B.O. = \frac{6 - 4}{2} = 1$. Since $x = 2$,$B.O. = \frac{1}{2} x = 1$.For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. $B.O. = \frac{10 - 6}{2} = 2$. Since $x = 2$,$B.O. = x = 2$.Therefore,the bond orders for $B_2$ and $O_2$ are $\frac{1}{2} x$ and $x$,respectively.

If the bond order in $C_2$ is '$x$',then the bond order in $B_2$ and $O_2$,respectively,are

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