The dissociation energy will be more when the bond order will be greater and bond order $$ \propto $$  dissociation energy
$${\text{Molecular orbital configuration of}}$$
$${N_2}\left( {14} \right) = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 1{s^2},$$      $$\mathop \sigma \limits^* 2{s^2},\pi 2p_y^2 \approx \pi 2p_z^2,\sigma 2p_x^2$$
$$\eqalign{ & {\text{So, bond order of}} \cr & {N_2} = \frac{{{N_b} - {N_a}}}{2} \cr & \,\,\,\,\,\,\,\, = \frac{{10 - 4}}{2} \cr & \,\,\,\,\,\,\,\, = 3 \cr & {\text{and bond order of}} \cr & N_2^ + = \frac{{9 - 4}}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 2.5 \cr} $$
As the bond order of $${N_2}$$  is greater than $$N_2^ + $$  so, the dissociation energy of $${N_2}$$  will be greater than $$N_2^ + $$ .

Hybridisation of $$S\,\,in\,\,{H_2}S = \frac{1}{2}\left( {6 + 2 + 0 - 0} \right) = 4$$
∴ $$S$$   has $$s{p^3}$$  hybridisation and $$2$$  lone pair of electrons in $${H_2}S$$
∴ It has angular geometry and so it has non-zero value of dipole moment.

$$KI$$  is ionic compound, so it is not soluble in non-polar solvent ( i.e. dipole moment $$\left( \mu \right)$$  for benzene $$ = 0$$ ).

$$\eqalign{ & {\rm{Partial}}\,{\rm{charge}} = {{{\rm{Dipole}}\,{\rm{moment}}} \over {{\rm{Bond}}\,{\rm{distance}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{1.2 \times {{10}^{ - 18}}\,esu\,cm} \over {1 \times {{10}^{ - 8}}\,cm}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.2 \times {10^{ - 10}}\,esu \cr & {\rm{Fractional}}\,{\rm{charge}} = {{1.2 \times {{10}^{ - 10}}} \over {4.8 \times {{10}^{ - 10}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.25 \cr & \left( {4.8 \times {{10}^{ - 10}}\,{\rm{is}}\,{\rm{a}}\,{\rm{theoretical}}\,{\rm{value}}\,{\rm{of}}\,\mu .} \right) \cr} $$

In (i), the $$lp$$  are at equatorial position so there are less $$lp{\text{ - }}bp$$   repulsions as compared to other positions. Hence, $$T$$ - shape is most stable.

In between $$C{H_3}OH$$  molecules intermolecular $$H - {\text{bonding}}$$    exist.

Hence, it is the intermolecular $$H - {\text{bonding}}$$    that must be overcome in converting liquid $$C{H_3}OH$$   to gas.

Greater the electronegativity and smaller the size of the atom, stronger is the $$H$$ - bond.

Both $$N{O_2}\,\,{\text{and}}\,\,\,{O_3}$$   have angular shape and hence will have net dipole moment.

As the number of lone pair of electrons on central element increases, repulsion between those lone pair of electrons increases and therefore, bond angle decreases.
Molecules Bond angle
$$C{H_4}$$   ( no lone pair of electrons ) $${109.5^ \circ }$$
$$N{H_3}$$   ( one lone pair of electrons ) $${107.5^ \circ }$$
$${H_2}O$$   ( two lone pair of electrons ) $${104.45^ \circ }$$