Question

Electronic configuration of four elements A, B, C and D are given below:
(A) \( 1s^2 2s^2 2p^3 \) → N
(B) \( 1s^2 2s^2 2p^4 \) → O
(C) \( 1s^2 2s^2 2p^5 \) → F
(D) \( 1s^2 2s^2 2p^2 \) → C
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

• A. A < D < B < C
• B. A < C < B < D
• C. A < B < C < D
• D. D < A < B < C

Hint:

Just remember "FON" (Fluorine, Oxygen, Nitrogen). These are the three most electronegative elements in the entire periodic table.

Answer 1

The Correct Option is D

The question asks for the order of increasing electronegativity among four elements based on their electronic configurations. The elements provided are:

• (A) \(1s^2 2s^2 2p^3\) → Nitrogen (N)
• (B) \(1s^2 2s^2 2p^4\) → Oxygen (O)
• (C) \(1s^2 2s^2 2p^5\) → Fluorine (F)
• (D) \(1s^2 2s^2 2p^2\) → Carbon (C)

The concept of electronegativity refers to the ability of an atom to attract electrons towards itself in a chemical bond. Pauling's scale is a commonly used method for quantifying electronegativity.

Here's a brief overview of the electronegativities of these elements according to Pauling's scale:

• Carbon (C): 2.55
• Nitrogen (N): 3.04
• Oxygen (O): 3.44
• Fluorine (F): 3.98

Based on their electronegativities, the order of increasing electronegativity is:

• (D) < Carbon (C): 2.55
• (A) < Nitrogen (N): 3.04
• (B) < Oxygen (O): 3.44
• (C) < Fluorine (F): 3.98

Therefore, the correct order is: D < A < B < C.

Here is the logical reasoning:

• Fluorine is the most electronegative element; thus, it should be at the end of the order.
• Oxygen is more electronegative than nitrogen but less than fluorine.
• Nitrogen is more electronegative than carbon but less than oxygen.
• Carbon has the lowest electronegativity among the four.