Let's analyze each statement:

The electronic configuration of $C_2$ (12 electrons) is $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2$. It has two degenerate $\pi$ molecular orbitals ($\pi 2p_x$ and $\pi 2p_y$), each containing two electrons, totaling four electrons. This statement is correct.

A neutral hydrogen atom (H) has one electron. $H_2$ molecule has two electrons. $H_2^{\oplus }$ ion is formed by removing one electron from $H_2$, so it has $2-1=1$ electron. This statement is correct.

The electronic configuration of $O_2$ (16 electrons) is $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{\ast }2p_x{)}^1({\pi }^{\ast }2p_y{)}^1$. It has two unpaired electrons in the antibonding ${\pi }^{\ast }$ orbitals, making it paramagnetic.

For $O_2^{\oplus }$ (15 electrons), one electron is removed from the antibonding ${\pi }^{\ast }$ orbital. The configuration becomes $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{\ast }2p_x{)}^1$. It still has one unpaired electron, making it paramagnetic. Therefore, the statement that $O_2^{\oplus }$ ion is diamagnetic is incorrect.

Let's calculate the bond order (BO) for each species using the formula: $BO=\frac{1}{2}(N_b-N_a)$, where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.

• $O_2^{\oplus }$ (15 electrons): $N_b=10$, $N_a=5$. $BO=\frac{1}{2}(10-5)=2.5$.
• $O_2$ (16 electrons): $N_b=10$, $N_a=6$. $BO=\frac{1}{2}(10-6)=2$.
• $O_2^{-}$ (17 electrons): $N_b=10$, $N_a=7$. $BO=\frac{1}{2}(10-7)=1.5$.
• $O_2^{2-}$ (18 electrons): $N_b=10$, $N_a=8$. $BO=\frac{1}{2}(10-8)=1$.

All bond orders are correctly stated. This statement is correct.

The incorrect statement is C.

Correct Answer: (C)