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.

The central atom Br has seven electrons in the valence shell. Five of these will form bonds with five fluorine atoms and the remaining two electrons are present as one lone pair.

Hence, total pairs of electrons are six ( 5 bond pairs and 1 lone pair). To minimize repulsion between lone pairs and bond pairs, the shape becomes square pyramidal.