To determine the incorrect statement, we will analyze each option step-by-step.$\\$Option 1:$\\$The bond orders of $O_2^+, O_2, O_2^-$ and $O_2^{2-}$ are 2.5, 2, 1.5 and 1, respectively.$\\$• Bond order is calculated using the formula:$\\$Bond order$= \frac{1}{2} ($Number of bonding electrons$-$Number of antibonding electrons$) \\$• For $O_2: \\$Electronic configuration: $( \sigma_{1s})^2 ( \sigma^*_{1s})^2 ( \sigma_{2s})^2 ( \sigma^*_{2s})^2 ( \sigma_{2p_z})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 ( \pi^*_{2p_x})^1 ( \pi^*_{2p_y})^1 \\$• Bond order for $O_2 = \frac{1}{2} (10 - 6) = 2 \\$• For $O_2^+: \\$Remove one electron from $\pi^*_{2p_x}$ or $\pi^*_{2p_y}$ orbital.$\\$Bond order for $O_2^+ = \frac{1}{2} (10 - 5) = 2.5 \\$• For $O_2^-: \\$Add one electron to $\pi^*_{2p_x}$ or $\pi^*_{2p_y}$ orbital.$\\$Bond order for $O_2^- = \frac{1}{2} (10 - 7) = 1.5 \\$• For $O_2^{2-}: \\$Add two electrons to $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ orbitals.$\\$Bond order for $O_2^{2-} = \frac{1}{2} (10 - 8) = 1 \\$Thus, Option 1 is correct.$\\$Option 2:$\\$C$_2$ molecule has four electrons in its two degenerate $\pi$ molecular orbitals.$\\$• The electronic configuration of $C_2$ is $( \sigma_{1s})^2 ( \sigma^*_{1s})^2 ( \sigma_{2s})^2 ( \sigma^*_{2s})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 \\$• The $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals are degenerate and each contains 2 electrons.$\\$Thus, Option 2 is correct.$\\$Option 3:$\\$H$_2^+$ ion has one electron.$\\$• $H_2^+$ is formed by removing one electron from $H_2,$ leaving it with one electron.$\\$Thus, Option 3 is correct.$\\$Option 4:$\\ O_2^+$ ion is diamagnetic.$\\$• $O_2^+$ has an unpaired electron in the $\pi^*_{2p_x}$ or $\pi^*_{2p_y}$ orbital, making it paramagnetic.$\\$Thus, Option 4 is incorrect.$\\$Therefore, the incorrect statement is Option 4.