Animated Solution for Chemistry - Periodicity in Properties: The element having greatest difference between its first and second ionisation energy, is

Imagine you are a microscopic mechanic, tasked with plucking electrons out of an atom one by one. The energy you expend to remove the first electron is the first ionization energy ($I{E}_1$). The energy required for the second is the second ionization energy ($I{E}_2$).

The question asks us to find the element where the difference between these two efforts is the absolute greatest. In other words, where does removing the second electron suddenly feel like hitting an impenetrable brick wall?

To understand this, we must look at the blueprint of the atoms: their electronic configurations. Atoms are inherently lazy; they want to reach a state of maximum stability, which is the noble gas configuration ($n{s}^{2}n{p}^6$).

Once an atom achieves this perfect, symmetrical arrangement, it guards its electrons fiercely. Removing an electron from a noble gas core requires an astronomical amount of energy. This is the secret to finding our massive energy jump!

Let's line up our suspects and examine their outer shells: - Potassium (K): $[Ar]4s^1$ - Calcium (Ca): $[Ar]4s^2$ - Barium (Ba): $[Xe]6s^2$ - Scandium (Sc): $[Ar]3d^{1}4s^2$

Now, let's play out the scenario. For Calcium, Barium, and Scandium, removing the first electron leaves them with at least one more valence electron ($4s^1$, $6s^1$, and $3d^{1}4s^1$ respectively). Removing the second electron is certainly harder than the first, but it doesn't require breaking into a stable core.

But look at Potassium (K). It starts with a single valence electron: $[Ar]4s^1$. Plucking that first electron is relatively easy.

However, the moment it loses that electron, it becomes $K^{+}$, which has the exact same electron configuration as the noble gas Argon ($[Ar]$). It has achieved perfection! Now, trying to remove a second electron means you are trying to break that perfect Argon core. The energy required ($I{E}_2$) skyrockets.

Therefore, the difference between $I{E}_1$ and $I{E}_2$ is overwhelmingly the greatest for Potassium.