Question

Arrange the following atomic numbers in the decreasing order of ease of formation of ionic compounds with a metal ‘M’.

(A) 8

(B) 7

(C) 17

(D) 9

• A. DBAC
• B. CABD
• C. CDAB
• D. CBDA

Answer 1

The correct option is C

CDAB

Explanation for correct option:

• Electron affinity is the amount of energy released when an electron is added to the valence shell of an isolated gaseous atom in its ground state.
• To form an ionic bond the non-metal must be willing to accept electrons from the metal.
• So, a non-metal with have high electron affinity nothing but high tendency of accepting electrons and thus it can form a stronger ionic bond.
• So, the decreasing order of electron affinity among the given atomic numbers is Chlorine ($\mathrm{Cl};\mathrm{Z}=17$)$>$Fluorine ($\mathrm{F};\mathrm{Z}=9$)$>$Oxygen ($\mathrm{O};\mathrm{Z}=8$)$>$Nitrogen ($\mathrm{N};\mathrm{Z}=7$).
• The atomic number is represented by $\mathrm{Z}$.

So, the correct order is CDAB.

Hence, the correct option is (C).