Central atom of $\mathrm{H}_2$ is S . There are 6 electrons in its valence shell $\left({ }_{16} \mathrm{~S}=2,8,6\right)$. Two electrons are shared with two H -atoms and the remaining four electrons are present as two lone pairs.

Hence, total pairs of electrons are four ( 2 bond pairs and 2 lone pairs). Due to the presence of 2 lone pairs the shape becomes distorted tetrahedral or angular or bent (non-linear).

$\mathrm{PCl}_3$-Central atom is phosphorus. There are 5 electrons in its valence shell $\left({ }_{15} \mathrm{P}=2,8,5\right)$. Three electrons are shared with three Cl -atoms and the remaining two electrons are present as one lone pair. Hence, total pairs of electrons are four ( 1 lone pair and 3 bond pairs). Due to the presence of one lone pair, the shape becomes pyramidal (non-planar).

According to molecular orbital theory electronic configurations of $\mathrm{O}_2^{+}$ and $\mathrm{O}_2^{-}$ species are as follows

$$\mathrm{O}_2^{+}:(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2, \pi 2 p_y^2\right)\left(\pi^* 2 p_x^1\right)$$

Bond order of $\mathrm{O}_2^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$

$$\mathrm{O}_2^{-}:(\sigma 1 s)^2\left(\sigma^{\star} 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^{\star} 2 s^2\right)\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x^2, \pi 2 p_y^2\right)\left(\pi^{\star} 2 p_x^2, \pi^* 2 p_y^1\right)$$

Bond order of $\mathrm{O}_2^{-}=\frac{10-7}{2}=\frac{3}{2}=1.5$

Higher bond order of $\mathrm{O}_2^{+}$shows that it is more stable than $\mathrm{O}_2^{-}$. Both the species have unpaired electrons. So, both are paramagnetic in nature.