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

Solution

Chemistry ma’am · 4 steps

Across period 3 the trend rises, but watch the two dips: Mg to Al, and P to S.

1. Identify the elements. All four are in period 3. 3s² is Mg, 3s² 3p¹ is Al, 3s² 3p³ is P and 3s² 3p⁴ is S.
2. General trend. Across a period the nuclear charge increases and the atom shrinks, so first ionisation enthalpy usually rises from left to right.
3. Two exceptions. Al loses a 3p electron, which is higher in energy and better shielded than Mg's 3s electron, so Al is lower than Mg. P has a half-filled 3p³ subshell, which is extra stable. S loses a paired 3p electron that feels pairing repulsion, so S is slightly lower than P.

   Label	Element	Electron removed	First IE (kJ mol⁻¹)
   (B)	Al	lone 3p electron	578
   (A)	Mg	paired 3s electron	738
   (D)	S	paired 3p electron	1000
   (C)	P	from half-filled 3p³	1012

4. So. The order is (B) < (A) < (D) < (C), which is option (b).

The idea to keep. In any period, group 13 dips below group 2 and group 16 dips below group 15. Remember that a filled s subshell and a half-filled p subshell are both extra stable.

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