As dipole moment = electric charge × bond length
$$D. M.$$  of $$AB$$  molecule
$$\eqalign{ & = 4.8 \times {10^{ - 10}} \times 2.82 \times {10^{ - 8}} = 13.53\,D \cr & D.M.\,{\text{of}}\,CD\,{\text{molecule}} \cr & = 4.8 \times {10^{ - 10}} \times 2.67 \times {10^{ - 8}} = 12.81\,D \cr & {\text{Now }}\% {\text{ ionic character}} \cr & {\text{ = }}\frac{{{\text{Actual dipole moment of the bond}}}}{{{\text{Dipole moment of pure ionic compound}}}} \times 100 \cr & {\text{then }}\% {\text{ ionic character in }}AB \cr & = \frac{{10.41}}{{13.53}} \times 100 \cr & = 76.94\% \cr & \% {\text{ ionic character in }}CD \cr & = \frac{{10.27}}{{12.81}} \times 100 \cr & = 80.17\% \cr} $$

Hybridisation of $$N = \frac{1}{2}\left[ {5 + 3 + 0 - 0} \right] = 4$$      hence $$s{p^3}$$

As bond angle decreases, hybridisation continuously changes, $$s$$ - characteristic increases in bonding hybridised orbital and thus bond length decreases.
\[\underset{\begin{smallmatrix} \text{no lone} \\ \text{pair} \end{smallmatrix}}{\mathop{C{{H}_{4}}}}\,\,\,\,\underset{\begin{smallmatrix} \text{one lone} \\ \text{pair} \end{smallmatrix}}{\mathop{N{{H}_{3}}}}\,\,\,\,\underset{\begin{smallmatrix} \text{two lone} \\ \text{pairs} \end{smallmatrix}}{\mathop{{{H}_{2}}O}}\,\]
$${\text{Since}}\,\,lp - lp > lp - bp > bp - bp.$$

$$2s \to \left( {\sigma 2s} \right)\left( {{\sigma ^ * }2s} \right)\left( 2 \right)$$
$$2p \to \left( {\sigma 2{p_z}} \right)\left( {{\sigma ^ * }2{p_z}} \right)\left( {\pi 2{p_x}} \right)$$      $$\left( {{\pi ^ * }2{p_x}} \right)\left( {\pi 2{p_y}} \right)\left( {{\pi ^ * }2{p_y}} \right)\left( 6 \right)$$

$$N{i^{2 + }}$$  with $$N{H_3}$$  shows $$CN = 6$$  forming $${\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}$$
(Octahedral)
$$P{t^{2 + }}$$  with $$N{H_3}$$  shows $$CN = 4$$  forming $$\,{\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$ 
$$\left( {5d\,\,{\text{series}}\,\,CMA,\,{\text{square}}\,{\text{planner}}} \right)$$
$$\,Z{n^{2 + }}$$  with $$N{H_3}$$  shows $$CN = 4$$  forming $${\left[ {Zn{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$
$$\left( {3{d^{10}}\,\,{\text{configuration,}}\,{\text{tetrahedral}}} \right)$$

$$\eqalign{ & {\text{Hybridisation in}}\,N{H_3} = s{p^3}, \cr & {\left[ {PtC{l_4}} \right]^{2 - }} = ds{p^2}{\text{(inner complex);}} \cr & PC{l_5} = s{p^3}d\,{\text{and}}\,BC{l_3}\,{\text{is}}\,\,s{p^2}. \cr} $$

Strength of hydrogen bonding depends on the size and electronegativity of the atom. Smaller the size of the atom, greater is the electronegativity and hence stronger is the $$H$$ - bonding. Thus, the order of strength of $$H$$ - bonding is $$H...F > H...O > H...N.$$      But each $$HF$$  molecule is linked only to two other $$HF$$  molecules while each $${H_2}O$$  molecule is linked to four other $${H_2}O$$  molecules through $$H$$ - bonding.
Hence, the decreasing order of boiling points is $${H_2}O > HF > N{H_3}.$$

In $${A_3}{\left( {B{C_4}} \right)_2}$$
3 × oxidation number of $$A$$  + 2 [ oxidation number of $$B$$  + 4 × [ oxidation number of $$C$$  ] = 0
3 × ( + 2 ) + 2 [ 5 + 4 × ( - 2 ) ] = 0
                              6 + 2 [ - 3 ] = 0

In regular tetrahedral structure, dipole moment of one bond is cancelled by opposite dipole moment of the other bonds.

Key Idea Molecules having same hybridisation have same number of hybrid orbitals,
$$H = \frac{1}{2}\left[ {V + X - C + A} \right]$$
where,
$$V = $$  number of valence electrons of central atom
$$X = $$  number of monovalent atoms
$$C = $$  charge on cation
$$A = $$  charge on anion
$$\eqalign{ & SbCl_5^{2 - } = s{p^3}{d^2},PC{l_5} = s{p^3}d \cr & S{F_4} = s{p^3}d,l_3^ - = s{p^3}d \cr} $$