Electronic configuration of four elements $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D are given below : (A)…
(A) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3$
(B) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4$
(C) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$
(D) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$
Which of the following is the correct order of increasing electronegativity (Pauling's scale)?
- A < D < B < C
- A < C < B < D
- A < B < C < D
- D < A < B < C
Solution
$\mathrm{O}:-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4$ (Electronegativity $=3.5$)
$\mathrm{F}:-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$ (Electronegativity $=4$)
$\mathrm{C}:-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$ (Electronegativity $=2.55$)
Correct order $=\mathrm{C} \gt \mathrm{B} \gt \mathrm{A} \gt \mathrm{D}$
Asked in: JEE Main 2025 (02 Apr Shift 2)
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