Animated Solution for Chemistry - Periodicity in Properties: Which of the following atoms has the highest first ionisation energy?

S-Block vs D-Block

When we talk about Ionization Energy (IE), we are essentially describing a microscopic tug-of-war. On one side, you have the positively charged nucleus pulling inward. On the other side, you have the outermost electron trying to break free. The energy required to finally rip that electron away is the first ionization energy.

In this problem, we are asked to find the champion of this tug-of-war among four contenders: Sodium ($\mathrm{Na}$), Potassium ($\mathrm{K}$), Rubidium ($\mathrm{Rb}$), and Scandium ($\mathrm{Sc}$).

Let's start by looking at the familiar faces in Group 1: Sodium, Potassium, and Rubidium. The periodic table is beautifully logical here. As we move down the group from $\mathrm{Na}$ to $\mathrm{Rb}$, we are adding entirely new electron shells.

Sodium has its outermost electron in the $3s$ orbital, Potassium in the $4s$, and Rubidium in the $5s$. Because the atomic radius is increasing, the outermost electron is getting further and further away from the nucleus. According to Coulomb's Law, greater distance means a weaker attractive force. Therefore, it becomes progressively easier to remove the electron.

This gives us a clear initial ranking: $I{E}_1(\mathrm{Na})>I{E}_1(\mathrm{K})>I{E}_1(\mathrm{Rb})$.

Now, we introduce Scandium into the mix. Scandium is a Period 4 transition metal with an atomic number of 21. Its electronic configuration is $[\mathrm{Ar}]3d^{1}4s^2$.

At first glance, you might think, "Wait, Scandium's outermost electrons are in the $4s$ shell, just like Potassium. And Sodium's outermost electron is in the $3s$ shell, which is closer to the nucleus. Shouldn't Sodium have a higher ionization energy than Scandium?"

This is a brilliant question, but it misses one crucial phenomenon: The Shielding Effect.

As we move across Period 4 from Potassium to Scandium, the extra electrons are not going into the outer shell; they are filling the inner $3d$ subshell.

Think of inner electrons as a frosted glass shield between the nucleus (the light bulb) and the outer electrons. S-orbitals and p-orbitals form a dense, effective shield. However, d-orbitals are highly diffused and spread out. They are like a very thin, patchy frosted glass. They offer poor shielding.

Because the $3d$ electrons fail to effectively screen the outer $4s$ electrons, the effective nuclear charge ($Z_{\text{eff}}$) felt by the $4s$ electrons in Scandium is exceptionally high. The nucleus, now armed with 21 protons, pulls those $4s$ electrons inward with a tremendous force.

This massive increase in $Z_{\text{eff}}$ completely overpowers the fact that Scandium's electrons are in the 4th shell. The nucleus holds onto Scandium's $4s$ electrons much more tightly than Sodium's nucleus holds onto its $3s$ electron.

Therefore, removing an electron from Scandium requires the most energy. The final correct order of first ionization energies is $\mathrm{Sc}>\mathrm{Na}>\mathrm{K}>\mathrm{Rb}$, making Scandium the undisputed winner of this tug-of-war.