Question

Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy?
A. Vanadium ($Z = 23$)
B. Chromium ($Z = 24$)
C. Iron ($Z = 26$)
D. Manganese ($Z = 25$)

Answer 1

The ionization enthalpy is defined as the amount of energy required to release an electron from a neutral isolated gaseous atom to convert it into a positive charged ion. The ionization enthalpy of half filled and fully filled orbitals is more due to their higher stability.

Complete step by step answer: Let us discuss each option one by one.
A. Vanadium ($Z = 23$): The electronic configuration of Vanadium in its ground state can be written as:
$V = [Ar]4{s^2}3{d^3}$
In order to extract the electrons from the $4s$ and $3d$ orbitals, less amount of energy is required. When two electrons are extracted from the vanadium atom, the electronic configuration comes to:
${V^{2 + }} = [Ar]4{s^0}3{d^3}$
Now, as we can see that for extracting the third electron, less amount of ionization enthalpy will be vested in it.
B. Chromium ($Z = 24$): The electronic configuration of Chromium in its ground state can be written as:
$Cr = [Ar]4{s^2}3{d^4}$
When two electrons are extracted from the chromium atom, the electronic configuration comes to:
$C{r^{2 + }} = [Ar]4{s^0}3{d^4}$
Now, as we can see that for extracting the third electron, less amount of ionization enthalpy will be vested in it since the valence electron is not stabilized by any kind of effect.
C. Iron ($Z = 26$): The electronic configuration of Iron in its ground state can be written as:
$Fe = [Ar]4{s^2}3{d^6}$
When two electrons are extracted from the iron atom, the electronic configuration comes to:
$F{e^{2 + }} = [Ar]4{s^0}3{d^6}$
Now, as we can see that for extracting the third electron, less amount of ionization enthalpy will be vested in it since the valence electron is not stabilized by any kind of effect. Also, the third electron will get released immediately as the $C{r^{2 + }}$ ion will try to attain the half-filled stable electronic configuration.
D. Manganese ($Z = 25$): The electronic configuration of Manganese in its ground state can be written as:
$Mn = [Ar]4{s^2}3{d^5}$
When two electrons are extracted from the iron atom, the electronic configuration comes to:
$M{n^{2 + }} = [Ar]4{s^0}3{d^5}$
The value of the third ionization enthalpy will be the largest in the case of manganese as the electrons in the orbitals are in half-filled stable electronic configuration and a very large amount of energy is required to overcome this effect.
Thus, the correct option is D. Manganese ($Z = 25$).

Note: For ionizing a neutral isolated gaseous atom, energy is required to extract an electron from the valence shell of the atom. This energy is the ionization enthalpy and the more stable the electronic