Question

The electron structures of four elements are given below.

(i) $\text{A→2, 8, 1}$ (ii) $\text{B→2,8,8}$

(iii) $\text{C→2,8,7}$ (iv) $\text{D→2,8,18,18,1}$

Arrange the above elements in the order of elements with:

(A) Zero electron affinity

(B) Minimum ionization energy

(C) Good oxidizing agent

(D) Best reducing agent in the third period

• A. DBAC
• B. BDAC
• C. DBCA
• D. BDCA

Answer 1

The correct option is D

BDCA

Explanation for correct option: (d)

1. The electron affinity of an element is defined as the energy released when an electron is added to an isolated gaseous atom of that element in the ground state.
2. Here element ‘B’ has a stable octet configuration. In such a case addition of an electron is not favored and it has zero electron affinity.
3. It is very easy to remove the outermost electron of element ‘D’ which requires a small amount of energy. Hence it has low ionization energy.
4. Here element ‘C’ is Chlorine which has the ability to take electrons and it is a powerful oxidizing agent.
5. Here element ‘A’ is Sodium which can lose its outermost electron to attain the stable octet configuration and thereby undergoes oxidation. Hence it is a powerful reducing agent.

Explanation for the incorrect options:

a) Here the element with zero electron affinity is ‘B’, an element with minimum ionization energy is ‘D’, the good oxidizing agent is ‘C’ and the good reducing agent is ‘A’. Hence the given order DBAC is incorrect.

b) The given order DBCA is not correct. Here the element with zero electron affinity is ‘B’, an element with minimum ionization energy is ‘D’, the good oxidizing agent is ‘C’ and the good reducing agent is ‘A’.

c) The given order DBCA is not correct. The correct option is BDCA.

Hence option (d) is correct and the correct order is BDCA.