Statement I: Fluorine molecule has bond order one
Statement II : The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

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Statement I and Statement II are true.

Statement I is false and Statement II is true.

Statement I is true and statement II is false.

Statement I and Statement II are false.

answer is A.

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Detailed Solution

$F_{2} {KK}^{1} {(σ_{2 s})}^{2} {(σ_{2 s}^{*})}^{2} {(σ_{2 p_{z}})}^{2} {(π_{2 p_{x}})}^{2} = {(π_{2 p_{y}})}^{2} {(π_{2 p_{x}}^{*})}^{2} {(π_{2 p_{y}}^{*})}^{2}$ $Bond order = \frac{NumberofelectronsinBMO - NumberofelectronsinABMO}{2}$

$= \frac{10 - 8}{2}$

= 1

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