The paramagnetic species among the following is- `Na^(+), Zn^(2+), Cu^(+), Fe^(3+)`

Options

A. `Na^(+)`
B. `Zn^(2+)`
C. `Cu^(+)`
D. `Fe^(3+)`

Explanation

To determine the paramagnetic species among the given ions: `Na^(+), Zn^(2+), Cu^(+), Fe^(3+)`, we need to analyze the electronic configurations of each ion and check for unpaired electrons. A species is considered paramagnetic if it has at least one unpaired electron.

### Step-by-Step Solution:

1. **Determine the electronic configuration of each species:**
- **Na^(+):**
- The atomic number of Na (Sodium) is 11. The ground state electronic configuration is:
\[ \text{Na: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^1 \]
- For Na^(+), it loses one electron from the outermost shell (3s):
\[ \text{Na}^{+}: 1s^2 \, 2s^2 \, 2p^6 \quad (\text{No unpaired electrons}) \]

- **Zn^(2+):**
- The atomic number of Zn (Zinc) is 30. The ground state electronic configuration is:
\[ \text{Zn: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \]
- For Zn^(2+), it loses two electrons (from 4s):
\[ \text{Zn}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \quad (\text{No unpaired electrons}) \]

- **Cu^(+):**
- The atomic number of Cu (Copper) is 29. The ground state electronic configuration is:
\[ \text{Cu: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^1 \]
- For Cu^(+), it loses one electron (from 4s):
\[ \text{Cu}^{+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \quad (\text{No unpaired electrons}) \]

- **Fe^(3+):**
- The atomic number of Fe (Iron) is 26. The ground state electronic configuration is:
\[ \text{Fe: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \, 4s^2 \]
- For Fe^(3+), it loses three electrons (two from 4s and one from 3d):
\[ \text{Fe}^{3+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \quad (\text{5 unpaired electrons}) \]

2. **Identify the paramagnetic species:**
- **Na^(+):** Diamagnetic (no unpaired electrons)
- **Zn^(2+):** Diamagnetic (no unpaired electrons)
- **Cu^(+):** Diamagnetic (no unpaired electrons)
- **Fe^(3+):** Paramagnetic (5 unpaired electrons)

### Conclusion:
The paramagnetic species among the given options is **Fe^(3+)**.