Given below are two statements :
Statement (I) : The radii of isoelectronic species increases in the order.
Mg²⁺ < Na⁺ < F⁻ < O²⁻
Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order.
Cl > F > Br > I
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Same electron count, so nuclear charge decides size; and fluorine is too small to beat chlorine.

1. Statement I. Mg²⁺, Na⁺, F⁻ and O²⁻ all have 10 electrons, but their nuclear charges are 12, 11, 9 and 8. With fewer protons pulling on the same 10 electrons, the ion is larger. So Mg²⁺ < Na⁺ < F⁻ < O²⁻ is right, and Statement I is correct.
2. Statement II. Going down group 17, the electron is added farther from the nucleus, so less energy is released. Fluorine is the exception. Its 2p shell is so small and crowded that the incoming electron is strongly repelled, so F releases less energy than Cl.

   Halogen	Electron gain enthalpy (kJ mol⁻¹)	Magnitude rank
   Cl	-349	1 (largest)
   F	-328	2
   Br	-325	3
   I	-295	4 (smallest)

3. So. The magnitude order Cl > F > Br > I matches, so Statement II is also correct. Both statements are true, which is option (b).

The idea to keep. For isoelectronic ions, more protons means a smaller ion. For electron gain enthalpy, the period 2 member is the exception: Cl beats F and S beats O, because the tiny atom repels the extra electron.

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