The sum of bond order values of C_2 and O_2^{ | vedclass

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(B) $1$. Calculate the bond order of $C_2$: The electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi

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$B_2, N_2, F_2$

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(B) $1$. Calculate the bond order of $C_2$: The electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(8-4)/2 = 2$. $2$. Calculate the bond order of $O_2^{2+}$: The total number of electrons is $16 - 2 = 14$. The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(10-4)/2 = 3$. $3$. The sum $x = 2 + 3 = 5$. $4$. Check the sum of bond orders for option $A$: $O_2^-$ $(1.5)$,$O_2^+$ $(2.5)$,$O_2$ $(2)$. Sum = $1.5 + 2.5 + 2 = 6$. $5$. Check the sum of bond orders for option $B$: $B_2$ $(1)$,$N_2$ $(3)$,$F_2$ $(1)$. Sum = $1 + 3 + 1 = 5$. $6$. Since the sum equals $5$,option $B$ is correct.

The sum of bond order values of $C_2$ and $O_2^{2+}$ is $x$,which is equal to the sum of bond order values of $a, b$ and $c$. What are $a, b$ and $c$?

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