Question

Correct relation among \[{X_A},{X_B}~and~\Delta \], where \[{X_A}~and~{X_B}\] are the electronegativities of element A and B
(A) \[{X_A} + {X_B} = 0.208\sqrt \Delta \]
(B) \[\sqrt {{X_A} + {X_B}} = 0.208 \times \Delta \]
(C) \[{X_A} - {X_B} = 0.208\sqrt \Delta \]
(D) \[{X_A} + {X_B} = \sqrt {0.208 \times \Delta } \]

Answer 1

Hint: In order to determine the relation among these three \[{X_A},{X_B}~and~\Delta \], we should be knowing about Pauling’s scale. Pauling’s scale is used to determine the relation between the bond energies of the atoms and their electronegativities.

Complete Solution :
Before moving to the relation between \[{X_A},{X_B}~and~\Delta \], we must know what electronegativity is. Electronegativity is the measure of tendency of a gaseous atom to attract the bonded pair of electrons towards itself. Earlier there was no unit of measurement for electronegativity. Later, Pauling was the one who came up with a scale for electronegativity. This scale is called Pauling’s scale. This scale will compare the electronegativity of various elements and rank them with respect to each other.
- Pauling’s scale was based on the relation between the electronegativities of the bonded atom and the energy of those bonds.

Let us consider two dissimilar atoms A and B, these are having a bond A-B between them. We can represent the bond energies of A-A, B-B and A-B as \[{{\text{E}}_{{\text{A - A}}}}{\text{,}}{{\text{E}}_{{\text{B - B}}}}{\text{ and }}{{\text{E}}_{{\text{A - B}}}}\] respectively.
It has been found that bond dissociation energies of A-B were found to be higher than the geometric mean of bond dissociation energies of A-A and B-B.
\[{\text{ }}{{\text{E}}_{{\text{A - B}}}} > \sqrt {{{\text{E}}_{{\text{A - A}}}} \times {{\text{E}}_{{\text{B - B}}}}} \]

- The difference between these two were given in
i.e. \[{\text{ }}\Delta {\text{ }} = {{\text{E}}_{{\text{A - B}}}} - \sqrt {{{\text{E}}_{{\text{A - A}}}} \times {{\text{E}}_{{\text{B - B}}}}} \]

The difference in bond energies is related to the difference in the electronegativities of A and B
\[\Delta {\text{ }} = {{\text{E}}_{{\text{A - B}}}} - \sqrt {{{\text{E}}_{{\text{A - A}}}} \times {{\text{E}}_{{\text{B - B}}}}} = {({X_A} - {X_B})^2}\]
Therefore,
\[{\text{ 0}}{\text{.208}}\sqrt \Delta = {X_A} - {X_B}\]
The value 0.208 came as we converted Kcals to electron volt.

Therefore, relation among \[{X_A}, {X_B}~ and~ \Delta \] are given as
\[{\text{ 0}}{\text{.208}}\sqrt \Delta = {X_A} - {X_B}\]
So, the correct answer is “Option C”.

Additional Information.
- Pauling’s scale and Mulliken’s scale are the measure of electron attraction power in an atom.
- The Mulliken scale tells us the arithmetic mean of first ionization energy and electron affinity should be the measure of the tendency of an atom to attract electrons, i.e. the electronegativity.
\[{X_m} = \dfrac{{I.{E_V} + E.{A_V}}}{2}\]

Note: Other methods of calculation of the electronegativities are:
- Sanderson electronegativity equalization
- Mulliken’s electronegativity
- Allen electronegativity
- Allred-Rochow electronegativity