Given below are two statements : Statement (I) : The correct...

The bond order of the given oxygen species can be calculated using molecular orbital theory:

For $O_2$ (16 electrons), the configuration 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.
Bond order = $\dfrac{10 - 6}{2} = 2$. Number of unpaired electrons = $2$.

For $O_2^+$ (15 electrons), one electron is removed from a $\pi^{*}$ orbital.
Bond order = $\dfrac{10 - 5}{2} = 2.5$. Number of unpaired electrons = $1$.

For $O_2^-$ (17 electrons), one electron is added to a $\pi^{*}$ orbital.
Bond order = $\dfrac{10 - 7}{2} = 1.5$. Number of unpaired electrons = $1$.

For $O_2^{2-}$ (18 electrons), two electrons are added to the $\pi^{*}$ orbitals.
Bond order = $\dfrac{10 - 8}{2} = 1$. Number of unpaired electrons = $0$.

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.

The number of unpaired electrons for the species are $2, 1, 1,$ and $0$ respectively. The correct sequence is $O_2 > O_2^+ = O_2^- > O_2^{2-}$. Thus, Statement II is false.

Answer: Statement I is true but Statement II is false