Question

Outermost electronic configurations of four elements A, B, C, D are given below :

(A) 3s²

(B) 3s²3p¹

(C) 3s²3p³

(D) 3s²3p⁴

The correct order of fist ionization enthalpy for them is:
A (A)<(B)<(C)<(D)
B (B)<(D)<(A)<(C)
C (B)<(A)<(D)<(C)
D (B)<(A)<(C)<(D)

Answer 1

The correct order is:(A)<(B)<(D)<(C). Therefore, the correct answer is option D: (B)<(A)<(C)<(D).

The first ionization enthalpy generally increases across a period from left to right due to increasing effective nuclear charge. Let's analyze the given configurations:

(A) 3s²

(B) 3s²3p¹

(C) 3s²3p³

(D) 3s²3p⁴

The correct order of increasing ionization enthalpy is determined by the number of electrons and their distribution. More protons in the nucleus generally mean a higher ionization enthalpy.

So, the correct order is:(A)<(B)<(D)<(C)

Therefore, the correct answer is option D: (B)<(A)<(C)<(D).