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

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Classification of Elements and Periodicity

Electronic configuration of four elements A, B, C and D are given below :
(A)

1 s^{2} 2 s^{2} 2 p^{3}

(B)

1 s^{2} 2 s^{2} 2 p^{4}

(C)

1 s^{2} 2 s^{2} 2 p^{5}

(D)

1 s^{2} 2 s^{2} 2 p^{2}

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✓ Correct

Answer: (D)

Solution

The configurations identify the elements as (A)

1 s^{2} 2 s^{2} 2 p^{3} = N

, (B)

2 p^{4} = O

, (C)

2 p^{5} = F

and (D)

2 p^{2} = C

, all in the second period. Electronegativity rises from left to right across a period as the nuclear charge grows, so carbon is the least and fluorine the most electronegative. The increasing order is

C < N < O < F

, that is D

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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