$$L{i_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2},$$     ∴ Bond order $$\, = \frac{1}{2}\left( {4 - 2} \right) = 1$$
$$Li_2^ + = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^1},$$     $$B.O.$$  $$ = \frac{1}{2}\left( {3 - 2} \right) = 0.5$$
$$Li_2^ - = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^1},$$      $$B.O.$$  $$ = \frac{1}{2}\left( {4 - 3} \right) = 0.5$$
The bond order of $$Li_2^ + $$ and $$Li_2^ - $$ is same but $$Li_2^ + $$  is more stable than $$Li_2^ - $$  because $$Li_2^ + $$  is smaller in size and has $$2$$  electrons in Anti bonding orbital whereas $$Li_2^ - $$ has $$3$$ electrons in Anti bonding orbital. hence $$Li_2^ + $$  is more stable than $$Li_2^ - $$

$$C{O_2}$$  and $$C{H_4}$$  have zero dipole moment as these are symmetrical in nature. Between $$N{H_3}$$  and $$N{F_3}.$$  $$N{F_3}$$  has greater dipole moment though in $$N{H_3}$$  and $$N{F_3}$$  both, $$N$$  possesses one lone pair of electrons.

This is because in case of $$N{H_3},$$  the net $$N - H$$  bond dipole is in the same direction as the direction of dipole of lone pair but in case of $$N{F_3},$$  the direction of net bond dipole of three $$ - N - F$$   bonds is opposite than that of the dipole of the then lone pair.

$$\eqalign{ & \left( {\text{A}} \right)Br{F_5} = 12\,{\text{electrons}} \cr & \left( {\text{B}} \right)S{F_6} = 12\,{\text{electrons}} \cr & \left( {\text{C}} \right)I{F_7} = 14\,{\text{electrons}} \cr} $$
  $$\left( {\text{D}} \right)PCl_4^ + ;$$   $$ = 8\,{\text{electrons}}$$
Hence, except $$PCl_4^ + $$  all molecules violate octet rule.

In $$PO_4^{3 - }$$  ion, the formal charge on each $$O$$  atom of $$P-O$$  bond $$ = \frac{{{\text{Total charge}}}}{{{\text{No}}{\text{. of }}O{\text{ atoms}}}}$$   $$ = \frac{{ - 3}}{4} = - 0.75$$

In methanol, $$R$$  is $$ - C{H_3}$$  group.
Hydrogen bonding between methanol and water.

It has two nodal planes. It is $${\pi ^ * }2{p_y}$$

$$\mathop {C{H_2}}\limits^{s{p^2}} = \mathop {CH}\limits^{s{p^2}} - \mathop C\limits^{sp} \equiv N$$

Generally, $$A{B_5}$$  molecules have trigonal bipyramidal structure due to $$s{p^3}d$$  hybridisation but in some cases due to presence of lone pair of electrons, its geometry becomes distorted.

Molecule having zero bond order will not be a viable molecule.

NOTE THIS STEP: Write configuration ofall species. Half filled and full filled orbitals are more stable as compared to nearly half filled and nearly full filled orbitals.
$$L{i^ - } = 1{s^2},2{s^2};B{e^ - } = 1{s^2},2{s^2},2{p^1}$$
$${B^ - } = 1{s^2},2{s^2},2{p^2}:\,{C^ - } = 1{s^2},2{s^2},2{p^3}$$
∴ $$B{e^ - }$$  will be least stable. It has lowest I.E.