Use the molecular orbital energy level diagram to show that $\mathrm{N}_2$ would be expected to have a triple bond. $\mathrm{F}_2$, a single bond and $\mathrm{Ne}_2$, no bond.
Formation of $\mathrm{N}_2$ molecule Electronic configuration of N - atom ${ }_7 \mathrm{~N}=1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_z^1$
$N_2$ molecule $=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x{ }^2 \approx \pi 2 p_y{ }^2, \sigma 2 p_z^2$
Bond order $=\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-4)=3$.
Bond order value of 3 means that $\mathrm{N}_2$ contains a triple bond.
Formation of $\mathrm{F}_2$ molecule, ${ }_9 \mathrm{~F}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^2, 2 p_z^1$
$$\mathrm{F}_2 \text { molecule }=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^{\star} 2 p^2 x \approx \pi^{\star} 2 p^2 y$$
$$\text { Bond order }=\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-8)=1$$
Bond order value 1 means that $F_2$ contains single bond.
Formation of $\mathrm{Ne}_2$ molecule ${ }_{10} \mathrm{Ne}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^2, 2 p_z^2$
$$\mathrm{Ne}_2 \text { molecule }=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z{ }^2, \pi 2 p_x^2 \approx \pi 2 p_y{ }^2, \pi^{\star} 2 p^2 x \approx \pi^{\star} 2 p_y^2, \sigma^{\star} 2 p_z^2$$
Bond order $=\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-10)=0$
Bond order value zero means that there is no formation of bond between two Ne -atoms. Hence, $\mathrm{Ne}_2$ molecule does not exist.
Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Valence bond theory (VBT) was introduced by Heitler and London (1927) and developed further by Pauling and other. VBT is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridisation of atomic orbitals and the principles of variation and superposition.
Consider two hydrogen atoms $A$ and $B$ approaching each other having nuclei $N_A$ and $N_B$ and electrons present in them are represented by $e_A$ and $e_B$. When the two atoms are at large distance from each other, there is no interaction between them.
As these two atoms approach each other, new attractive and repulsive forces begin to operate.
Attractive forces arise between
(i) nucleus of one atom and its own electron
i.e., $$N_A-e_A \text { and } N_B-e_B$$
(ii) nucleus of one atom and electron of other atom
i.e., $$N_A-e_B, N_B-e_A$$
Similarly, repulsive forces arise between
(i) electrons of two atoms like $e_A-e_B$
(ii) nuclei of two atoms like $N_A-N_B$
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.
Experimentally, we have been found that the magnitude of new attractive force is more than the new repulsive forces. As a result two atoms approach each other and potential energy decreases.
Hence, a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage, two H -atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm .
Since, the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms.
The energy so released is called as bond enthalpy, which is corresponding to minimum in the curve depicted in the given figure. Conversely 435.8 kJ of energy is required to dissociate one mole of $\mathrm{H}_2$ molecule.
$$\mathrm{H}_2(g)+435.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)$$
The potential energy curve for the formation of $\mathrm{H}_2$ molecule as a function of internuclear distance of the H -atoms. The minimum in the curve corresponds to the most stable state or $\mathrm{H}_2$.
Describe hybridisation in the case of $\mathrm{PCl}_5$ and $\mathrm{SF}_6$. The axial bonds are longer as compared to equatorial bonds in $\mathrm{PCl}_5$ whereas in $\mathrm{SF}_6$ both axial bonds and equatorial bonds have the same bond length. Explain.
Formation of PCl$$_5$$
In $\mathrm{PCl}_5$, phosphorus is $s p^3 d$ hybridised to produce a set of five $s p^3 d$ hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal. These five $s p^3 d$ hybrid orbitals overlap with singly occupied $p$-orbitals of Cl -atoms to form five $\mathrm{P}-\mathrm{Cl}$ sigma bonds.
Three P-Cl bonds lie in one plane and make an angle of $120^{\circ}$ with each other. These bonds are called equatorial bonds. The remaining two $\mathrm{P}-\mathrm{Cl}$ bonds one lying above and other lying below the plane make an angle of $90^{\circ}$ with the equatorial plane. These bonds are called axial bonds. Axial bonds are slightly longer than equatorial bonds because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.
Formation of $\mathrm{SF}_6$
In $\mathrm{SF}_6$, sulphur is $s p^3 d^2$ hybridised to produce a set of $s i x ~ s p^3 d^2$ hybrid orbitals which are directed towards the six corners of a regular octahedron. These six $s p^3 d^2$ hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds. Thus, $\mathrm{SF}_6$ molecule has a regular octahedral geometry and all $\mathrm{S}-\mathrm{F}$ bonds have same bond length.
(a) Discuss the concept of hybridisation. What are its different types in a carbon atom?
(b) What is the type of hybridisation of carbon atoms marked with star?
Hybridisation It can be defined as the process of intermixing of the orbitals of slightly different energy or of same energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane. New orbitals formed are called hybrid orbitals.
Only the orbitals of an isolated single atom can undergo hybridisation. The hybrid orbitals generated are equal in number to that of the pure atomic orbitals which mix up.
Hybrid orbitals do not make $\pi$, pi-bonds. If there are $\pi$-bonds, equal number of atomic orbitals must be left unhybridised for $\pi$-bonding.Like atomic orbitals, hybrid orbitals cannot have more than two electrons of opposite spins. Types of hybridisation in carbon atoms
(a) (i) Diagonal or sp-hybridisation All compounds of carbon containing $\mathrm{C} \equiv \mathrm{C}$ triple bond like ethyne $\left(\mathrm{C}_2 \mathrm{H}_2\right)$.
(ii) Trigonal or $\mathrm{sp}^2$-hybridisation All compounds of carbon containing $\mathrm{C}=\mathrm{C}$ (double bond) like ethene $\left(\mathrm{C}_2 \mathrm{H}_4\right)$.
(iii) Tetrahedral or $\mathrm{sp}^3$-hybridisation All compounds of carbon containing $\mathrm{C}-\mathrm{C}$ single bonds only like ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti-bonding molecular orbital (ABMO). Energy of anti-bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals.
Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order
$\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x \approx \pi 2 p_y\right)<\sigma 2 p_z<\left(\pi^* 2 p_x \approx \pi^* 2 p_y\right)<\sigma^* 2 p_z$ and
For oxygen and fluorine order of energy of molecular orbitals is given below
$$\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\sigma p_z<\left(\pi 2 p_x \approx \pi 2 p_y\right)<\left(\pi^* 2 p_x \approx \pi^* 2 p_y\right)<\sigma^* 2 p_z$$
Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation.
Further, if the overlapping is head on, the molecular orbital is called 'sigma', ( $\sigma$ ) and if the overlap is lateral, the molecular orbital is called 'pi', $(\pi)$. The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals.
However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.
Which of the following statements is correct?