JEE Main 2026 — Chemistry Inorganic Chemistry

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

For $O_2$O2​ (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$210−6​=2. Number of unpaired electrons = $2$2.

For $O_2^+$O2+​ (15 electrons), one electron is removed from a $\pi^{*}$π∗ orbital.

Bond order = $\dfrac{10 - 5}{2} = 2.5$210−5​=2.5. Number of unpaired electrons = $1$1.

For $O_2^-$O2−​ (17 electrons), one electron is added to a $\pi^{*}$π∗ orbital.

Bond order = $\dfrac{10 - 7}{2} = 1.5$210−7​=1.5. Number of unpaired electrons = $1$1.

For $O_2^{2-}$O22−​ (18 electrons), two electrons are added to the $\pi^{*}$π∗ orbitals.

Bond order = $\dfrac{10 - 8}{2} = 1$210−8​=1. Number of unpaired electrons = $0$0.

Since bond length is inversely proportional to bond order, the correct sequence of bond lengths is $O_2^+ < O_2 < O_2^- < O_2^{2-}$O2+​<O2​<O2−​<O22−​. Thus, Statement I is true.

The number of unpaired electrons for the species are $2, 1, 1,$2,1,1, and $0$0 respectively. The correct sequence is $O_2 > O_2^+ = O_2^- > O_2^{2-}$O2​>O2+​=O2−​>O22−​. Thus, Statement II is false.

Answer: Statement I is true but Statement II is false