The following is paramagnetic

Options

A. `CaCl_(2)`
B. `CuCl_(2)`
C. `ZnCl_(2)`
D. `NaCl`

Answer

To determine which of the given options is paramagnetic, we need to analyze the electronic configurations of the ions provided and check for the presence of unpaired electrons. Here's the step-by-step solution:

### Step 1: Identify the Ions
We have the following ions to analyze:
1. Ca²⁺
2. Cu²⁺
3. Zn²⁺
4. Na⁺

### Step 2: Write the Electronic Configurations
We will write the electronic configurations for each ion.

1. **Ca²⁺**: Calcium (Ca) has an atomic number of 20. The electronic configuration of neutral Ca is:
\[ \text{Ca}: [\text{Ar}] 4s^2 \]
For Ca²⁺, we remove 2 electrons from the 4s subshell:
\[ \text{Ca}^{2+}: [\text{Ar}] \]

2. **Cu²⁺**: Copper (Cu) has an atomic number of 29. The electronic configuration of neutral Cu is:
\[ \text{Cu}: [\text{Ar}] 3d^{10} 4s^1 \]
For Cu²⁺, we remove 2 electrons (1 from 4s and 1 from 3d):
\[ \text{Cu}^{2+}: [\text{Ar}] 3d^9 \]

3. **Zn²⁺**: Zinc (Zn) has an atomic number of 30. The electronic configuration of neutral Zn is:
\[ \text{Zn}: [\text{Ar}] 3d^{10} 4s^2 \]
For Zn²⁺, we remove 2 electrons from the 4s subshell:
\[ \text{Zn}^{2+}: [\text{Ar}] 3d^{10} \]

4. **Na⁺**: Sodium (Na) has an atomic number of 11. The electronic configuration of neutral Na is:
\[ \text{Na}: [\text{Ne}] 3s^1 \]
For Na⁺, we remove 1 electron from the 3s subshell:
\[ \text{Na}^{+}: [\text{Ne}] \]

### Step 3: Check for Unpaired Electrons
Now we will check each ion for unpaired electrons:

1. **Ca²⁺**:
- Configuration: [Ar]
- Unpaired Electrons: 0 (All orbitals are filled)

2. **Cu²⁺**:
- Configuration: [Ar] 3d⁹
- Unpaired Electrons: 1 (There is one unpaired electron in the 3d subshell)

3. **Zn²⁺**:
- Configuration: [Ar] 3d¹⁰
- Unpaired Electrons: 0 (All orbitals are filled)

4. **Na⁺**:
- Configuration: [Ne]
- Unpaired Electrons: 0 (All orbitals are filled)

### Step 4: Conclusion
From the analysis:
- **Ca²⁺**: Diamagnetic (0 unpaired electrons)
- **Cu²⁺**: Paramagnetic (1 unpaired electron)
- **Zn²⁺**: Diamagnetic (0 unpaired electrons)
- **Na⁺**: Diamagnetic (0 unpaired electrons)

Thus, the only paramagnetic ion among the options is **Cu²⁺**.

### Final Answer
**Cu²⁺ is paramagnetic.**