Ionization energy of atoms A and B are 360 and 260 ${{\text{ kcal mol }}^{\text{-}}}^{\text{1}}$ respectively. The electron affinities of these atoms are 70 ${{\text{ kcal mol }}^{\text{-}}}^{\text{1}}$ and 90 ${{\text{ kcal mol }}^{\text{-}}}^{\text{1}}$ respectively, then :

Given,

IE(A) = 360 kcal mol -1

IE(B) = 260 kcal mol -1

EA(A) = 70 kcal mol -1

EA(B) = 90 kcal mol -1

Electron cloud is more attracted to those atoms having more ionization energy. Hence, Electron cloud is more attracted by atom A.

The relationship between the electron affinity and ionization energy is

 A (IE)=−(EA). So that the atoms having the more ionization energy has high value of electron affinity.

$\begin{array}{l}\mathbf{=}\mathbf{E}\cdot {\mathbf{N}}_{\mathrm{A}}\propto 350+70\\ \mathbf{=}\mathbf{E}\cdot {\mathbf{N}}_{\mathrm{B}}\propto 250+90\\ \mathbf{=}\mathbf{E}\cdot {\mathbf{N}}_{\mathrm{A}}>\mathbf{E}\cdot {\mathbf{N}}_{\mathrm{B}}\end{array}$

Hence, option A and C are correct.