What is the bond order of the F_2 molecule? | vedclass

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(B) The total number of electrons in an $F_2$ molecule is $18$. The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (

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$1$

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(B) The total number of electrons in an $F_2$ molecule is $18$. The molecular orbital configuration is: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2$. Number of bonding electrons $(N_b)$ $= 10$. Number of antibonding electrons $(N_a)$ $= 8$. Bond order $= \frac{N_b - N_a}{2} = \frac{10 - 8}{2} = 1$.

What is the bond order of the $F_2$ molecule?

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