Bond energy of sigma bond is greater than $$pi - {\text{bond}}$$  , thus their energies between $$C - {\text{atoms}}$$   are $$347\,kJ\,mo{l^{ - 1}}$$   and $$264\,kJ/mol$$   respectively.

$${H_2}S$$  have $$s{p^3}$$  hybridisation while remaining all have $$sp$$  hybridisation, so $${H_2}S$$  have bent structure and other have linear geometry.

The Lewis representation of $$NO_3^ - $$  is

Thus, $$N$$  atom has 4 bond pairs and 0 lone pairs of electrons.

For $$\pi - $$ overlap the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

Linus Pauling contributed to chemical bonding, so, he received the Nobel Prize for his work in chemical bonding.

$$\mathop N\limits^ + {O_2}$$  has linear shape due to $$sp$$  hybridisation of $$N$$  in $$\mathop N\limits^ + {O_2}$$

While $$S{O_2},NO_2^ - $$   and $${O_3}$$  have angular shape

In both the molecules the bond moments are not canceling with each other and hence the molecules has aresultant dipole and hence the molecule is polar.

Bond order $$ \propto $$  Bond dissociation energy.

$$\because S{F_4}\,{\text{have}}\,\mu > 0$$
$$\therefore $$  It has permanent dipole moment.

$${O_2} = $$  Oxygen $$\left( {Z = 8} \right)$$  has following molecular orbital configuration of $${O_2}.$$
$${O_2}\left( {16{e^ - }} \right) = \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},$$       $${\sigma ^ * }2{s^2},\sigma 2p_x^2,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,} \right.$$     $$\left\{ {{\pi ^ * }2p_y^1 = {\pi ^ * }2p_z^1} \right.$$    i.e., 2 unpaired and 14 paired electrons.