TIPS/Formulae :
i) Dipole moment is vector quantity. When vector sum of all dipoles in molecule will be zero, then molecule will not have net dipole moment.
(ii) NOTE: For net dipole moment to be equal to zero, all the atoms attached to central atom must be identical and geometry must be regular.

∴ Carbon tetrachloride having regular geometry and identical atoms attached to bonds has zero dipole moment.

$${\text{Lattice energy}} = \frac{{{\text{Product of charges}}}}{{{\text{interionic distance}}}}$$
In $$NaCl$$  the product of charges $$ = 1 \times 1;$$   In $$CaO$$  product of charges $$ = 2 \times 2 = 4$$   while the inter ionic distance is almost same in both. Thus lattice energy of $$CaO$$  is almost four times the lattice energy of $$NaCl.$$

$$N_2^ + \left( {13} \right) = \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},$$      $${\sigma ^ * }2{s^2},\pi 2p_x^2 = \pi 2p_y^2,\sigma 2p_z^1$$
$${O_2}\left( {16} \right) = \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},$$      $${\sigma ^ * }2{s^2},\sigma 2p_z^2,\pi 2p_x^2 = \pi 2p_y^2,$$     $${\pi ^ * }2p_x^1, = {\pi ^ * }2p_y^1$$
$$O_2^{2 - }\left( {18} \right) = \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},$$      $${\sigma ^ * }2{s^2},\sigma 2p_z^2,\pi 2p_x^2 = \pi 2p_y^2,$$     $${\pi ^ * }2p_x^2, = {\pi ^ * }2p_y^2$$
$${B_2}\left( {10} \right) = \sigma 1{s^2},{\sigma ^ * }1{s^2},\sigma 2{s^2},$$      $${\sigma ^ * }2{s^2},\pi 2p_x^1 = \pi 2p_y^1$$
Thus, $$O_2^{2 - }$$  does not contain unpaired electrons.

$${H_2}O$$   molecule can form four hydrogen bonds per molecule, two via lone pairs and two via hydrogen atoms.

The molecular orbital configuration of $$NO$$  is
$$KK{\left( {\sigma 2s} \right)^2}{\left( {\mathop \sigma \limits^* 2s} \right)^2}{\left( {\sigma 2{p_x}} \right)^2}$$     $${\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {\mathop \pi \limits^* 2{p_y}} \right)^1}$$
So, $$NO$$  is paramagnetic because it contains one unpaired electron.

The angle corresponds to $$s{p^2}$$  hybridisation, triangular planar is $${120^ \circ }.$$

In the formation of $${d^2}s{p^3}$$  hybrid orbitals, two $$\left( {n - 1} \right)d$$   - orbitals of $${{\text{e}}_g}$$  set, i.e. $$\left( {n - 1} \right){d_{{z^2}}}$$   and $$\left( {n - 1} \right){d_{{x^2} - {y^2}}}$$    orbitals, one $$ns$$  and three $$np\left( {n{p_x},n{p_y}\,{\text{and}}\,n{p_z}.} \right)$$     orbitals combine together.

$$H\mathop - \limits_\sigma C\mathop \equiv \limits_\pi ^{\pi ,\sigma } N$$
i.e $$2\pi $$  and $$2\sigma $$  bonds.

The distribution of electrons in $$MOs$$  is as follows:

$$\eqalign{ & N_2^ + \,\left( {{\text{electrons}}\,13} \right)\,{\sigma ^2}{\sigma ^{*2}}{\sigma ^2}{\sigma ^{*2}}{}_{{\pi ^2}}^{{\pi ^2}}{\sigma ^1}_{{\pi ^*}}^{{\pi ^*}}{\sigma ^*} \cr & {O_2}\,\left( {{\text{electrons}}\,16} \right){\sigma ^2}{\sigma ^{*2}}{\sigma ^2}{\sigma ^{*2}}{\sigma ^2}{}_{{\pi ^2}}^{{\pi ^2}}\,_{{\pi ^{*1}}}^{{\pi ^{*1}}}{\sigma ^*} \cr & O_2^{2 - }\,\left( {{\text{electrons}}\,\,18} \right)\,{\sigma ^2}{\sigma ^{*2}}{\sigma ^2}{\sigma ^{*2}}{\sigma ^2}{}_{{\pi ^2}}^{{\pi ^2}}\,_{{\pi ^*}}^{{\pi ^*}}{\sigma ^*} \cr & {B_2}\,\,\left( {{\text{electrons}}\,\,10} \right){\sigma ^2}{\sigma ^{*2}}{\sigma ^2}{\sigma ^{*2}}_{{\pi ^1}}^{{\pi ^1}} }$$

Only $$O_2^{2 - }$$  does not contain any unpaired electron.

$$B{e_2}:\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)$$     $$\left( {{\sigma ^ * }2{s^2}} \right)$$
$$\eqalign{ & B.O. = \frac{1}{2}\left( {{N_b} - {N_a}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times \left( {4 - 4} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0 \cr} $$
Since $$B.O.$$  is zero, the molecule does not exist.