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(D) To determine the number of electrons in antibonding molecular orbitals,we write the molecular orbital configuration for each molecule:$1$. $Li_2$

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$F_2$

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(D) To determine the number of electrons in antibonding molecular orbitals,we write the molecular orbital configuration for each molecule:$1$. $Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Antibonding electrons = $2$ (in $\sigma^* 1s$).$2$. $N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Antibonding electrons = $2+2 = 4$.$3$. $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Antibonding electrons = $2+2+2 = 6$.$4$. $F_2$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Antibonding electrons = $2+2+4 = 8$.Thus,$F_2$ contains the maximum number of electrons in antibonding molecular orbitals.

List-$I$	List-$II$
$A$. $\Psi_{MO} = \Psi_{A} - \Psi_{B}$	$I$. Dipole moment
$B$. $\mu = Q \times r$	$II$. Bonding molecular orbital
$C$. $\frac{N_{b} - N_{a}}{2}$	$III$. Anti-bonding molecular orbital
$D$. $\Psi_{MO} = \Psi_{A} + \Psi_{B}$	$IV$. Bond order

Which of the following molecules contains the maximum number of electrons in antibonding molecular orbitals?

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List-$I$ List-$II$

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