₁₆S = 2,8,6 ∴ Lewis symbol =\(:\ddot { S } :\),S^2- ions = [\(:\ddot { S } :\)]^2-
₁₃AI = 2, 8, 3 ∴ Lewis symbol = ·\(\overset { \cdot }{ \cdot A } l\cdot \), AI³⁺ion = [AI]³⁺
₁H = 1 ∴ Lewis symbol = \(H\bullet \),\(H\bar { \bullet } \) ion =\(\dot { H } \cdot \)
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According to Kossel and Lewis, atoms combine together in order to complete their respective octets so as to acquire the stable inert gas configuration. This can occur in two ways; by transfer of one or more electrons from one atom to other or by sharing of electrons between two or more atoms.

In gaseous state at high temperature,  BeCl₂ exists as linear molecule Cl-Be-Cl, thus the hybridisation of the central atoms is sp  

In solid state,it has a polymeric structure with chlorine bridges as follows

Two Cl-atoms are listed to be atom by two coordination bonus and two by covalent bonds. For these bonds to be formed, Be in the excited state with the configuration undergoes Sp³ hybridization. Two half-filled hybrid orbitals will form normal covalent bonds with two Cl-atoms. The other two Cl-atoms are coordinated to Be-atom. The other two Cl-atoms are coordinated be Be-atom by donating electron pairs into the empty hybrid orbitals.

The central I_-atoms has the outer shell electronic configuration in the ground state as. It undergoes hybridisation.Out of the five hybrid orbitals, one is half-filled, one is empty and the remaining three are fully-filled.The half-filled orbital forms covalent bond with iodine atom.
    
The empty orbital accepts electron pair from I^-  ion to form a coordinate bond. The remaining fully-filled orbitals occupy equatorial position. Thus, the geometry of three lone pairs and two bond pair is trigonal bipyramidal and the shape of \({ I_{ 3 }^{ - } }\) is linear as shown in the figure.

Aluminium \(([Ne]^{ 3{ s }^{ 2 } } \ 3p'\) can achien the electronic configuration of the nearest noble gas (Ne) by losing only three electrons, Al³⁺ = 1s²2s²2p⁶ . Aluminium with not form the Al⁴⁺ ion because an extremely high amount of energy would be required to remove an electron from the stable noble gas configuration.

(a) \({ C }_{ 2 }{ H }_{ 2 }\left( { 180 }^{ 0 } \right) >{ CH }_{ 4 }\left( { 109 }^{ 0 }{ 28 }^{ ' } \right) >{ BF }_{ 3 }\left( { 120 }^{ 0 } \right) >{ NH }_{ 3 }\left( { 107 }^{ 0 } \right) >{ H }_{ 2 }O>\left( { 104.5 }^{ 0 } \right) \)This is because all of them involve sp³ hybridisation. The number of lone pair of electrons present on N-atom are 0,1 and 2 respectively. Greater the number of lone pairs, greater is the repulsion and lesser is the bond angle.
(b) \({ NH }_{ 4 }^{ + }>{ NH }_{ 3 }>{ NH }_{ 3 }^{ - }\)
This is because all of them involve sp3 hybridisation. The number of lone pair of electrons present on N-atom are 0,1 and 2 respectively. Greater the number of lone pairs, greater is the repulsion and lesser is the bond angle.

Greater is the electronegativity difference between the two bonded atoms, greater is the ionic character.

Therefore, increasing order of ionic character of the given bonds is as follows.
C-H.

Ionic character \(\propto \)  lattice energy
\(\propto \frac { 1 }{ size \ of \ ion } \propto charge \ on \ ion,\)
A non-polar molecule like N₂ has almost negligible ionic character.
\(\therefore \)The order of ionic character is
\( ({ N }_{ 2 }<{ SO }_{ 2 }<{ CIF }_{ 3 }<{ K }_{ 2 }O\)

This is due to greater repulsion on the axial bond pairs by the equatorial bond pairs of electrons.

Both have tetrahedral shape, i.e, four lobes of sp³ hybridised orbitals so called iso structural.

sp² , each C-atom is linked to two other C-atoms.