The third ionization enthalpy is minimum for :

Solution

Chemistry ma’am · 3 steps

Write the M²⁺ ion: if losing one electron makes it 3d⁵, that loss comes cheap.

1. Which electron goes. The third ionization enthalpy removes an electron from the M²⁺ ion. The two 4s electrons are already gone, so this third electron comes out of the 3d subshell.
2. Half-filled stability. A half-filled 3d⁵ subshell is extra stable because of exchange energy. Breaking a 3d⁵ costs a lot of energy, and reaching a 3d⁵ costs very little.

   Ion	Configuration	After losing one electron	Δᵢ H₃ (kJ mol⁻¹)
   Mn²⁺	3d⁵	breaks a half-filled d⁵	3248
   Fe²⁺	3d⁶	becomes half-filled 3d⁵	2957
   Co²⁺	3d⁷	3d⁶, no special stability	3232
   Ni²⁺	3d⁸	3d⁷, higher nuclear charge	3393

3. So. Fe²⁺ (3d⁶) loses one electron to become the stable Fe³⁺ (3d⁵), so iron has the lowest third ionization enthalpy. The answer is (d).

The idea to keep. For the nth ionization enthalpy, write the configuration of the ion with charge (n-1)+. If removing one more electron gives d⁵, that IE is low. If it breaks a d⁵, that IE is high.

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