Question

Identify the element having general electronic configuration $ns^{2}np^{4}$ from following.

• A. Se
• B. Br
• C. Xe
• D. Kr

Hint:

Logic Tip: You can immediately associate the $p^4$ block with the chalcogens: O, S, Se, Te, Po. Since Br is a halogen ($p^5$) and Kr/Xe are noble gases ($p^6$), Se is the sole correct option.

Answer 1

The Correct Option is A

Concept:
The general outer electronic configuration $ns^2 np^4$ indicates that the element has $2 + 4 = 6$ valence electrons. Elements with 6 valence electrons belong to Group 16 (the oxygen family or chalcogens) of the periodic table.
Step 1: Identify the periodic groups of the given elements.
Let's evaluate the options based on their position in the periodic table:

• (A) Se (Selenium): Belongs to Group 16 (Oxygen family). Its configuration ends in $ns^2 np^4$ (specifically $4s^2 4p^4$).
• (B) Br (Bromine): Belongs to Group 17 (Halogens). Its configuration ends in $ns^2 np^5$.
• (C) Xe (Xenon): Belongs to Group 18 (Noble gases). Its configuration ends in $ns^2 np^6$.
• (D) Kr (Krypton): Belongs to Group 18 (Noble gases). Its configuration ends in $ns^2 np^6$.

Step 2: Match the configuration to the element.
Selenium (Se) is the only element among the choices that belongs to Group 16 and therefore possesses the $ns^2 np^4$ outer shell configuration.