Question

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

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

Answer 1

Answer: (C)

D < A < B < C

Answer 2

1. Identify the elements from their electronic configurations:

   • 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)

2. Using electronegativity values (Pauling scale):

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

3. Increasing order (from lowest to highest): Carbon < Nitrogen < Oxygen < Fluorine → D < A < B < C.