`Al^(3+)` is smaller than `Ca^(2+)`.

Answer

To determine why \( \text{Al}^{3+} \) is smaller than \( \text{Ca}^{2+} \), we can analyze their electronic configurations and the effect of nuclear charge on ionic size.

### Step 1: Determine the electronic configuration of both ions.

- **Aluminum (Al)** has an atomic number of 13. Its electronic configuration is:
\[ \text{Al: } 2, 8, 3 \]
When it loses 3 electrons to form \( \text{Al}^{3+} \), its electronic configuration becomes:
\[ \text{Al}^{3+}: 2, 8 \]

- **Calcium (Ca)** has an atomic number of 20. Its electronic configuration is:
\[ \text{Ca: } 2, 8, 8, 2 \]
When it loses 2 electrons to form \( \text{Ca}^{2+} \), its electronic configuration becomes:
\[ \text{Ca}^{2+}: 2, 8, 8 \]

### Step 2: Analyze the number of electron shells.

- \( \text{Al}^{3+} \) has 2 electron shells (the 1st shell has 2 electrons, and the 2nd shell has 8 electrons).
- \( \text{Ca}^{2+} \) has 3 electron shells (the 1st shell has 2 electrons, the 2nd shell has 8 electrons, and the 3rd shell has 8 electrons).

### Step 3: Consider the effect of nuclear charge.

- The nuclear charge for \( \text{Al}^{3+} \) is +13 (13 protons in the nucleus).
- The nuclear charge for \( \text{Ca}^{2+} \) is +20 (20 protons in the nucleus).

### Step 4: Compare the ionic sizes.

- \( \text{Al}^{3+} \) has a higher effective nuclear charge relative to the number of electrons it has (which is 10 electrons).
- \( \text{Ca}^{2+} \) also has a high nuclear charge but has more electron shells (3 shells) and 18 electrons.

### Conclusion:

Because \( \text{Al}^{3+} \) has fewer electron shells and a higher effective nuclear charge acting on fewer electrons, it pulls its electrons closer to the nucleus, resulting in a smaller ionic radius compared to \( \text{Ca}^{2+} \).

Thus, we conclude that:
\[ \text{Al}^{3+} \text{ is smaller than } \text{Ca}^{2+}. \]

---