The orbital electronic configuration of fluorine (Z = 9)
\(={ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }_{ x }^{ 2 }{ 2p }_{ y }^{ 2 }{ 2p }_{ z }^{ 1 }\)
M.O.E.C. of fluorine =[σ2s]² [σ*2s]² [σ2p_z]² [\(\pi\)2p_x]² [\(\pi\)2p_y]² [\(\pi\)*2p_x]² [\(\pi\)*2p_y]²
Due to presence of all filled orbitals, F2 is diamagnetic

The molecular orbital fonned by the subtractive effect of the electron waves of the combining atomic orbitals, is called antibonding molecular orbital.

Bond length is inversely proportional to bond order.

The molecular orbital electronic configuration of both B₂ and C₂ are
B₂: [σ1s]² [σ*1s]² [σ2s]² [σ*2s]² [\(\pi\)2px]¹[\(\pi\)*2py]¹
C₂:  [σ1s]² [σ*1s]² [σ2s]² [σ*2s]² [\(\pi\)2px]²[\(\pi\)*2py]²
Since B₂ has two unpaired electrons, B₂ is paramagnetic. C₂ has no unpaired electron. Thus, C2 is diamagnetic.

(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
\( ({ N }_{ 2 }<{ SO }_{ 2 }<{ CIF }_{ 3 }<{ K }_{ 2 }O\)

σ bond is stronger. This is because  σ-bond is formed by head-on overlapping of atomic orbitals and therefore, the overlapping is large.  π -bond is formed by sideway overlapping which is small.

A covalent bond is formed by the overlap of half-filled atomic orbitals.

Electronic configuration of Be₂ molecule (4 + 4 = 8) 
σ1s²,σ1s²,σ1s²,σ1s²
Bond order=1/2[Nb−Na]=1/2(4−4)=0
Since, bond order in zero, so Be₂ does not exist.

No, because bond order becomes zero, e.g in case of He₂, Be₂, Ne₂ etc. Note that in H₂ ,σ1s molecular orbitals is full but  σ1s is empty.

Atomic orbitals  with comparable energies and proper orientation overlap to form molecular orbitals

Electronic configuration of carbon atom: C: σ1s² 2s² 2p².
In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp³ hybridization in CH₄ molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp³. However, an atom of carbon does not have d-orbitals to undergo dsp3 hybridization. Hence, the structure of CH₄ is tetrahedral.

If the molecule where 100% ionic, then
\(\mu \) calculated for 100% ionic = \(q\times d\)
\(=1.602\times { 10 }^{ -19 } \ (C) \ 246\times { 10 }^{ -10 } \ (m) \ =3.94\times { 10 }^{ -27 } \ cm\)
% ionic character \(=\frac { { \mu }_{ observed } }{ { \mu }_{ calculated } } \times 100\)

\({ \mu }_{ observed }\)\(=\frac { { ionic \ character\times { \mu }_{ calculated } } }{ 100 } \)
\(=\frac { 54.76\times 3.94\times { 10 }^{ -27 } }{ 100 } \)
\(=2.15\times { 10 }^{ -27 } \ Cm=\frac { 2.15\times { 10 }^{ -27 } }{ 3.35\times { 10 }^{ -30 } } \ D=644.0D\)

The Lewis structure for BeH₂ is as follows:
H: Be: H
There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH₂ is of the type AB₂. It has a linear structure.
\(\nrightarrow \nleftarrow \)
H------Be-----H
Dipole moments of each H - Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, BeH₂ has a zero dipole moment.

Let us consider the combination between atoms of hydrogen H_A and H_B and e_A and e_B be their respective electrons.
As they tend to come closer, two different forces operate between the nucleus and the electron of the other and vice versa. The nuclei of the atoms as well as their electrons repel each other. Energy is needed to overcome the force of repulsion. Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more. Thus, when two hydrogen atoms approach each other, the overall potential energy of the system decreases. Thus, a stable molecule of hydrogen is formed.

(a) \(\underset { Tetrahedral }{ { CCl }_{ 4 }({ 109.5 }^{ 0 } } ),\underset { Pyramidal }{ { NH }_{ 3 }({ 107 }^{ 0 }) } ,\underset { Angular }{ { H }_{ 2 }O{ (104.5 }^{ 0 } } ),\underset { Octahedral }{ { SF }_{ 6 }{ (90 }^{ 0 }) } \)
(b) \(\underset { Planar }{ { BF }_{ 3 }({ 120 }^{ 0 } } ),\underset { Tetrahedral }{ { CH }_{ 4 }({ 109.5 }^{ 0 }) } ,\underset { Pyramidal }{ { NH }_{ 3 }{ (107 }^{ 0 } } ),\underset { Angular }{ { H }_{ 2 }{ O(104.5 }^{ 0 }) } \)