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The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?
A
$O_2^+$, $O_2$ – Both paramagnetic
B
O, $O_2^{2-}$ – Both paramagnetic
C
$O_2^-$, $O_2^{2-}$ – Both diamagnetic
D
$O^+$, $O_2^{2-}$ – Both paramagnetic
Detailed Solution
$O_2$ (16 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \sigma2p_z^2\ \pi2p_x^2 = \pi2p_y^2\ \pi^*2p_x^1 = \pi^*2p_y^1$; 2 unpaired electrons, paramagnetic.
$O_2^+$ (15 electrons): ... $\pi^*2p_x^1\ \pi^*2p_y^0$; 1 unpaired electron, paramagnetic.
$O_2^-$ (17 electrons): ... $\pi^*2p_x^2\ \pi^*2p_y^1$; 1 unpaired electron, paramagnetic.
$O_2^{2-}$ (18 electrons): ... $\pi^*2p_x^2\ \pi^*2p_y^2$; no unpaired electron, diamagnetic.
O atom ($1s^22s^22p_x^22p_y^12p_z^1$) has 2 unpaired electrons and $O^+$ ($1s^22s^22p_x^12p_y^12p_z^1$) has 3 unpaired electrons; both are paramagnetic.
Every pair containing $O_2^{2-}$ described as paramagnetic is wrong, and $O_2^-$ is not diamagnetic.
The correct description is $O_2^+$, $O_2$ – both paramagnetic.
$O_2^+$ (15 electrons): ... $\pi^*2p_x^1\ \pi^*2p_y^0$; 1 unpaired electron, paramagnetic.
$O_2^-$ (17 electrons): ... $\pi^*2p_x^2\ \pi^*2p_y^1$; 1 unpaired electron, paramagnetic.
$O_2^{2-}$ (18 electrons): ... $\pi^*2p_x^2\ \pi^*2p_y^2$; no unpaired electron, diamagnetic.
O atom ($1s^22s^22p_x^22p_y^12p_z^1$) has 2 unpaired electrons and $O^+$ ($1s^22s^22p_x^12p_y^12p_z^1$) has 3 unpaired electrons; both are paramagnetic.
Every pair containing $O_2^{2-}$ described as paramagnetic is wrong, and $O_2^-$ is not diamagnetic.
The correct description is $O_2^+$, $O_2$ – both paramagnetic.