Question

The atomic numbers of V, Cr, Mn and Fe are 23, 24, 25 and 26 respectively. Which one of these may be expected to have the highest second ionization enthalpy?

• A. Fe
• B. V
• C. Mn
• D. Cr

Answer 1

The correct option is D Cr

Let us look at the ground state electronic configuration of each element.
Vanadium has a $\left[Ar\right]3d^{3}4s^2$ electronic configuration.
Chromium has a $\left[Ar\right]3d^{5}4s^1$ electronic configuration. This configuration is much more stable than the otherwise expected configuration of $\left[Ar\right]3d^{4}4s^2$ because in the
$\left[Ar\right]3d^{5}4s^1$ arrangement, there is maximum stability due to more avenues of exchange energy. As a very crude generalization, exactly half-filled and fully-filled subshells (with respect to d orbitals) are considered more stable.
Manganese has a $\left[Ar\right]3d^{5}4s^2$ electronic configuration.
Iron has a $\left[Ar\right]3d^{6}4s^2$ electronic configuration.
It is known that exactly half-filled and completely filled d orbitals tend to be more stable than partially filled d orbitals. For instance, this is the reason why Cr and Cu have $\left[Ar\right]3d^{5}4s^{1}and\left[Ar\right]3d^{10}4s^1$ configurations respectively. As we can see, Cr has the most stable ground state electronic configuration. It also has the most stable M+ electronic configuration of $\left[Ar\right]3d^5$.
Hence, removing two electrons from Cr is going to require the maximum energy. So the answer is Cr.