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Outermost electronic configurations of four elements A, B, C, D are given below : (A) 3s² (B) 3s² 3p¹ (C) 3s² 3p³ (D) 3s² 3p⁴ The correct order of fist ionization enthalpy for them is:

1. (A) < (B) < (C) < (D)
2. (B) < (A) < (D) < (C)
3. (B) < (D) < (A) < (C)
4. (B) < (A) < (C) < (D)

Correct answer: (B) < (A) < (C) < (D)

Solution

The first ionization enthalpy generally increases with increasing nuclear charge and decreasing atomic radius. In this case, element B has the least electrons in the outer shell, making it easier to remove an electron, while element D has the most, making it harder to remove an electron, thus establishing the order of ionization enthalpy as (B) < (A) < (C) < (D).

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