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(B) Total number of electrons in $OF = 8$ (from $O$) $+ 9$ (from $F$) $= 17$ electrons.The molecular orbital configuration for $17$ electrons is: $\si

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Total number of electrons in $OF = 8$ (from $O$) $+ 9$ (from $F$) $= 17$ electrons.The molecular orbital configuration for $17$ electrons is: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$.Number of bonding electrons $(N_b)$ $= 10$.Number of anti-bonding electrons $(N_a)$ $= 7$.Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(10 - 7) = \frac{3}{2} = 1.5$.

$A$ simplified application of $MO$ theory to the hypothetical molecule $OF$ would give its bond order as:

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