JEE Main 2025 · 2 April · evening · Single correct · +4 / −1

Electronic configuration of four elements A, B, C and D are given below :
(A) 1 s² 2 s² 2 p³
(B) 1 s² 2 s² 2 p⁴
(C) 1 s² 2 s² 2 p⁵
(D) 1 s² 2 s² 2 p²
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

  1. (a) D<A<B<C
  2. (b) A<B<C<D
  3. (c) A<C<B<D
  4. (d) A<D<B<C
Answer and solution

Answer (a) D<A<B<C

Solution

Chemistry ma’am · 3 steps

Name the elements first: all four sit in period 2, so electronegativity just climbs from left to right.

  1. Identify each element. Count the electrons in each configuration to get the atomic number. All four end in 2p, so they are all period 2 elements.
    LabelConfigurationElementPauling electronegativity
    D1s² 2s² 2p²C (Z = 6)2.5
    A1s² 2s² 2p³N (Z = 7)3.0
    B1s² 2s² 2p⁴O (Z = 8)3.5
    C1s² 2s² 2p⁵F (Z = 9)4.0
  2. Trend across a period. Going left to right, the nuclear charge rises and the atom gets smaller, so the nucleus pulls shared electrons more strongly. There is no half-filled exception for electronegativity, so it rises steadily: C < N < O < F.
  3. So. In terms of the labels, the order is D < A < B < C, which is option (a).

The idea to keep. Within one period, electronegativity rises smoothly from left to right with no anomalies. The half-filled dips you remember belong to ionisation enthalpy and electron gain enthalpy, not electronegativity.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Periodic Table & Periodicity

Asked about 120 times in JEE Main since 2020. This question is from 2025.