Electronegativity and Electron affinity of an element A are X and Y resp. Hence Ionisation potential isa)(X+Y)/2b)2X-Yc)2Y-Xd)2X+YCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

E. N. = I. E. +E.A /2

Answer

According to Mulliken's scale for calculating electronegativity,
Electronegativity= (Ionisation energy+ Electron affinity)/2

So, here EN is X and EA is Y.

X= (IE+Y)/2
Therefore, IE= 2X-Y

Answer

Explanation:

Ionisation potential is the energy required to remove an electron from an atom or a molecule. Let us assume that an electron is removed from an element A to form a cation. The ionisation potential can be expressed as follows:

A → A+ + e-

The electronegativity of an element is the tendency of an atom to attract a shared pair of electrons towards itself. The electron affinity of an element is the energy change that occurs when an electron is added to a neutral atom to form a negative ion.

X = Electronegativity of element A
Y = Electron affinity of element A

To remove an electron from element A, the energy required is X. To add an electron to element A, the energy released is -Y.

Therefore, the energy required to remove an electron from element A and the energy released when an electron is added to element A can be expressed as:

A → A+ + e- (Energy required = X)
A + e- → A- (Energy released = -Y)

The total energy change for the above two reactions can be expressed as:

A → A+ + e- + A- (Energy change = X - Y)

The energy change for the above reaction is equal to the ionisation potential of element A.

Hence, Ionisation potential = X - Y

Now, let us simplify the above equation:

Ionisation potential = X - Y
= (-Y) + X
= 2X - Y

Therefore, the correct option is B) 2X - Y.

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