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

 
(ii) AB₄- Tetrahedral

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

Covalent bond can be formed by the overlap of the orbitals belonging to the two atoms having opposite spins of electrons

Bond-length: It is the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond-lengths are measured by spectroscopic methods.

Therefore, only B_eCl₂ is linear and rest of the molecules are non-linear.

sp = orbital has greater directional character than p = orbital. This is because p-orbital has equal sized lobes, with equal electron density in both the lobes where as sp-hybrid orbital has greater electron density on one side.

This is because in NaCl, there is strong ionic bond between Na⁺ and Cl^-1 whereas in Na metal, there is weak metallic bond.

Hybridisation: It is the phenomenon of intermixing of atomic orbitals of slightly different energies to form new hybrid orbitals of equivalent energy. Formation of water. In water (H₂0), the atomic number of oxygen is 8 and its orbitals electronic configuration is 1s² 2s² \({ 2 }_{ x }^{ 2 }{ \ 2p }_{ y }^{ 1 }{ \ 2p }_{ z }^{ 1 }\) . The oxygen atom is also Sp³ hybridised. However, in this case, the two orbitals with one electron each (half filled) are involved in overlap with the hydrogen orbitals.

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.

In ethyne molecule, both the carbon atoms are sp hybridised having two unhybridised orbitals, i.e. 2P_x and 2P_y. The two sp hybrid orbitals of both the carbon atoms are oriented in opposite direction forming an angle of 180^o.
\(H-C\overset { \sigma }{ \underset { 2\pi -bond }{ \equiv } C } -H\)
That's why ethyne molecule is linear.

The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are

Phosphorus atom is sp³ d hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as:
PCI₅

The five Sp³ d hybrid orbitals are directed towards the five corners of the trigonal bipyramidal. Hence, the geometry of PCI₅ can be represented as:

There are five P-Cl sigma bonds in PCl₅ Three P-Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds. The remaining two P-CI bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

When two atoms with different electronegativity are linked to each other by covalent bond, the shared electron pair will not in the centre because of the difference in electronegativity. For example, in hydrogen flouride molecule, flouride haspeater electronegativity than hydrogen. Thus, the shared electron pair is displaced more towards flourine atom, the later will acquire a partial negative charge (\(\delta ^{ - }\)).At the same time hydrogen atom will have a partial positive charge (\(\delta ^{ + }\)). Such a covalent bond is known as polar covalent bond or simply polar bond. It is represented as
\(\overset { { \delta }^{ + } }{ \underset { 2.1 }{ H } - } \overset { { \delta }^{ - } }{ \underset { 4.0 }{ F } } \)

Teacher told her that the former two compounds dissolve because of their ionic nature whereas the later two because of H-bonding with the water molecules. 

\({ (CH }_{ 3 }{ ) }_{ 3 }\ddot { N } \) is trimethyl amine.It has pyramidal geometry as shown below

[(CH₃)₃Si]₃ N has triangular planar geometry as shown in diagram below.

Due to the presence of empty d-orbital on Si, there is an effective back bonding.
Thus, the two species are not isostructural.In (CH₃)₃N, the N-atom assumes sp³ hybrid state where as in [(CH₃)₃Si]₃N, the N-atom assumes sp² hybridtate.

The Lewis of  O₃ may be draw as

The oxygen atoms have been numbered as 1, 2 and 3. The formal charge on
(i) The central O-atom marked as 1 = V - L - \(\frac { 1 }{ 2 } \)S
= 6 - 2 - \(\frac { 1 }{ 2 } \)(6) = +1
(ii) The end O-atom marked as 2 = V - L - \(\frac { 1 }{ 2 } \)S
= 6 - 4 - \(\frac { 1 }{ 2 } \)(4) = 0
(iii) The end O-atom marked as 3 = V - L - \(\frac { 1 }{ 2 } \)S
= 6 - 6 - \(\frac { 1 }{ 2 } \)(2) = -1
Hence, the O₃ molecule along with the formal charges can be represented as follows