Which of the following atoms has the highest first ionization energy?

We begin by recalling the definition of the first ionization energy. It is the minimum energy required to remove the outermost electron from a neutral, gaseous atom, represented symbolically as $$X(g)\;\rightarrow\;X^{+}(g)+e^{-}$$. The magnitude of this energy depends mainly on three factors: the effective nuclear charge experienced by the outermost electron, the atomic radius, and the extent of shielding by inner electrons.

From periodic trends we know two key rules. First, as we move from left to right across a period, the effective nuclear charge $$Z_{\text{eff}}$$ increases because protons are added to the nucleus while the added electrons enter the same principal shell. The shielding does not increase appreciably, so the outer electrons are held more tightly, and the ionization energy rises. Second, as we travel down a group, a new principal shell is added. The atomic radius increases, shielding becomes stronger, the outer electron is held less tightly, and the ionization energy falls.

Now we examine each of the four elements individually and arrange them in order of their positions in the periodic table.

$$\text{Na}: \text{Group }1,\; \text{Period }3$$

$$\text{K}: \text{Group }1,\; \text{Period }4$$

$$\text{Rb}: \text{Group }1,\; \text{Period }5$$

$$\text{Sc}: \text{Group }3,\; \text{Period }4$$

Let us first compare the three alkali metals. All of them belong to Group 1, so each has a single valence electron in an $$ns^{1}$$ configuration. Moving downward from $$\text{Na}$$ to $$\text{K}$$ and then to $$\text{Rb}$$, the principal quantum number $$n$$ of the valence electron increases in the sequence $$3 \rightarrow 4 \rightarrow 5$$. The larger $$n$$ means a greater average distance from the nucleus, more shielding, and therefore a lower attraction toward the nucleus. Consequently, the ionization energy decreases down the group: $$I_{1}(\text{Na}) > I_{1}(\text{K}) > I_{1}(\text{Rb}).$$

Next we bring $$\text{Sc}$$ into the comparison. It is positioned in Period 4, to the right of $$\text{K}$$ and $$\text{Ca}$$. Because it lies three columns to the right of $$\text{K}$$, the nuclear charge has increased by three units, while the additional electrons are still in the same principal shell $$n = 4$$. The incomplete shielding means that the effective nuclear charge $$Z_{\text{eff}}$$ for the outer electrons in $$\text{Sc}$$ is appreciably higher than that in $$\text{K}$$.

To gauge the relative magnitudes quantitatively, we can look up the experimentally measured first ionization energies (all values in $$\text{kJ\,mol}^{-1}$$):

$$I_{1}(\text{Na}) \approx 496,$$

$$I_{1}(\text{K}) \approx 419,$$

$$I_{1}(\text{Rb}) \approx 403,$$

$$I_{1}(\text{Sc}) \approx 633.$$

Clearly, $$I_{1}(\text{Sc})$$ exceeds the other three values. Therefore $$\text{Sc}$$ possesses the highest first ionization energy among the four species listed.

Hence, the correct answer is Option B.