Molecular orbital theory Questions in English

(C) Statement $I$: According to Molecular Orbital Theory $(MOT)$, the bond orders are: $O_2^+ (2.5)$, $O_2 (2.0)$, $O_2^- (1.5)$, and $O_2^{2-} (1.0)$.
Since bond length is inversely proportional to bond order, the correct sequence of bond lengths is $O_2^+ < O_2 < O_2^- < O_2^{2-}$. Thus, Statement $I$ is true.
Statement $II$: The number of unpaired electrons in these species are: $O_2$ $(2)$, $O_2^+$ $(1)$, $O_2^-$ $(1)$, and $O_2^{2-} (0)$.
The correct sequence is $O_2 > O_2^+ = O_2^- > O_2^{2-}$.
Therefore, the sequence given in Statement $II$ $(O_2 > O_2^+ > O_2^- > O_2^{2-})$ is incorrect because $O_2^+$ and $O_2^-$ have the same number of unpaired electrons. Thus, Statement $II$ is false.

(D) The total number of electrons in $Ne_2$ is $10 + 10 = 20$.
The molecular orbital configuration for $Ne_2$ is: $(\sigma_{1s})^2, (\sigma^*_{1s})^2, (\sigma_{2s})^2, (\sigma^*_{2s})^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, (\pi^*_{2p_x})^2, (\pi^*_{2p_y})^2, (\sigma^*_{2p_z})^2$.
Counting the electrons: $2+2+2+2+2+2+2+2+2+2 = 20$.
The highest occupied molecular orbital $(HOMO)$ is the last orbital filled, which is $\sigma^*_{2p_z}$ (often denoted as $\sigma^*_{2p}$).

(C) The atomic number of fluorine $(F)$ is $9$. Therefore, an $F_2$ molecule has $9 \times 2 = 18$ electrons.
The molecular orbital configuration for $F_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$.
Bonding electrons are those in orbitals without an asterisk $(*)$. Total bonding electrons = $2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_z) + 2 (\pi 2p_x) + 2 (\pi 2p_y) = 10$.
Antibonding electrons are those in orbitals with an asterisk $(*)$. Total antibonding electrons = $2 (\sigma^* 1s) + 2 (\sigma^* 2s) + 2 (\pi^* 2p_x) + 2 (\pi^* 2p_y) = 8$.
Thus, bonding electrons are $10$ and antibonding electrons are $8$.

(C) According to Molecular Orbital Theory $(MOT)$, the electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since $O_2$ has two unpaired electrons in its antibonding $\pi^*$ orbitals, it is paramagnetic.
$Li_2$, $N_2$, and $F_2$ have all paired electrons and are diamagnetic.

(A) According to Molecular Orbital Theory $(MOT)$, the electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since $O_2$ contains two unpaired electrons in the $\pi^*$ antibonding molecular orbitals, it exhibits paramagnetism.
$O_3$, $N_2$, and $F_2$ are diamagnetic because they do not contain any unpaired electrons.