In    carbon number $$2$$  have five valency which is not possible, so it does not correctly represent the bonding capacities of $$C$$  atom.

$$\sigma $$ bond of $$O = O$$  is formed by axial overlapping of $$p{\text{ - }}p$$  orbital while $$\pi $$ bond is formed by sidewise overlapping of $$p - p$$  orbital.

$$S$$ can go into excited state with 6 unpaired electrons due to presence of vacant 3 $$d$$ - orbitals, which are overlapped by six $$F$$ electrons.
Ground state :
Excited state :

In $${H_2}S,$$  due to low electronegativity of sulphur the $$L.P.\,{\text{ - }}\,L.P.$$    repulsion is more than $$B.P.\,{\text{ - }}\,B.P.$$    repulsion and hence the bond angle is minimum.
\[\begin{matrix} {} \\ \text{Bond angle} \\ \end{matrix}\,\,\,\begin{matrix} S{{O}_{2}} \\ {{119.5}^{\circ }} \\ \end{matrix}\,\,\begin{matrix} {{H}_{2}}O \\ {{104.5}^{\circ }} \\ \end{matrix}\,\,\begin{matrix} {{H}_{2}}S \\ {{92.5}^{\circ }} \\ \end{matrix}\,\,\begin{matrix} N{{H}_{3}} \\ {{106.5}^{\circ }} \\ \end{matrix}\]

$$\eqalign{ & O_2^ - :B.O. = \frac{{10 - 7}}{2} = 1.5 \cr & {O_2}:B.O. = \frac{{10 - 6}}{2} = 2 \cr & O_2^ + :B.O. = \frac{{10 - 5}}{2} = 2.5 \cr & {\text{The order is}}\,O_2^ - < {O_2} < O_2^ + . \cr} $$

For atoms like nitrogen and below $$N$$ in atomic number have higher energy for $${p_z}$$ - orbital, hence the configuration for $${C_2}$$  molecule is $$\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {{\sigma ^ * }2{s^2}} \right)$$      $$\left( {\pi 2p_x^2 = \pi 2p_y^2} \right).$$

$$B{H_3}$$  have $$s{p^2}$$  hybridisation, so it have trigonal planar structure, not tetrahedral structure.

TIPS/Formulae : Compound having sp hybridisation will have linear shape.
$$\therefore C{O_2}$$   or $$\left( {O = C = O} \right)$$   which $$C$$ has in $$sp$$  hybrid state has linear shape.

Four $$s{p^3}$$ hybrid orbitals are formed when one $$s$$ and three $$p$$ - orbitals hybridise.

TIPS/Formulae: $$H = \frac{1}{2}\left( {V + M - C + A} \right)$$
$$\eqalign{ & \left( {\text{i}} \right)C{O_2};H = \frac{1}{2}\left( {4 + 0 - 0 + 0} \right) = 2 \cr & \therefore \,sp\,\,{\text{hybridisation}}{\text{.}} \cr & \left( {{\text{ii}}} \right)S{O_2};H = \frac{1}{2}\left( {6 + 0 - 0 + 0} \right) = 3 \cr & \therefore s{p^2}\,\,{\text{hybridisation}}{\text{.}} \cr & \left( {{\text{iii}}} \right)CO;H = \frac{1}{2}\left( {4 + 0 - 0 + 0} \right) = 2 \cr & \therefore sp\,\,{\text{hybridisation}}{\text{.}} \cr} $$