Among the following ions which has the highest paramagntism ? .

Options

A. `[Cr(H_(2)O)_(6)]^(3+)`
B. `[Fe(H_(2)O)_(6)]^(2+)`
C. `[Cu(H_(2)O)_(6)]^(2+)`
D. `[Zn(H_(2)O)_(6)]^(2+)`

Explanation

To determine which ion has the highest paramagnetism among the given options, we need to find the number of unpaired electrons in each ion. The ion with the highest number of unpaired electrons will exhibit the highest paramagnetism. Let's analyze each ion step-by-step.

### Step 1: Analyze Cr(H2O)6^3+
1. **Determine the oxidation state of Chromium (Cr):**
- The overall charge is +3.
- Let the oxidation state of Cr be x.
- The equation becomes: \( x + 6(0) = +3 \) → \( x = +3 \).

2. **Write the electronic configuration:**
- Ground state configuration of Cr: [Ar] 4s^1 3d^5.
- For Cr^3+, we remove three electrons (two from 4s and one from 3d).
- Resulting configuration: 4s^0 3d^3.

3. **Count the unpaired electrons:**
- In 3d^3, there are 3 unpaired electrons.

### Step 2: Analyze Fe(H2O)6^2+
1. **Determine the oxidation state of Iron (Fe):**
- The overall charge is +2.
- Let the oxidation state of Fe be y.
- The equation becomes: \( y + 6(0) = +2 \) → \( y = +2 \).

2. **Write the electronic configuration:**
- Ground state configuration of Fe: [Ar] 4s^2 3d^6.
- For Fe^2+, we remove two electrons (both from 4s).
- Resulting configuration: 4s^0 3d^6.

3. **Count the unpaired electrons:**
- In 3d^6, there are 4 unpaired electrons.

### Step 3: Analyze Cu(H2O)6^2+
1. **Determine the oxidation state of Copper (Cu):**
- The overall charge is +2.
- Let the oxidation state of Cu be z.
- The equation becomes: \( z + 6(0) = +2 \) → \( z = +2 \).

2. **Write the electronic configuration:**
- Ground state configuration of Cu: [Ar] 4s^1 3d^10.
- For Cu^2+, we remove two electrons (one from 4s and one from 3d).
- Resulting configuration: 4s^0 3d^9.

3. **Count the unpaired electrons:**
- In 3d^9, there is 1 unpaired electron.

### Step 4: Analyze Zn(H2O)6^2+
1. **Determine the oxidation state of Zinc (Zn):**
- The overall charge is +2.
- Let the oxidation state of Zn be w.
- The equation becomes: \( w + 6(0) = +2 \) → \( w = +2 \).

2. **Write the electronic configuration:**
- Ground state configuration of Zn: [Ar] 4s^2 3d^10.
- For Zn^2+, we remove two electrons (both from 4s).
- Resulting configuration: 4s^0 3d^10.

3. **Count the unpaired electrons:**
- In 3d^10, there are 0 unpaired electrons.

### Conclusion
Now, let's summarize the number of unpaired electrons for each ion:
- Cr(H2O)6^3+: 3 unpaired electrons
- Fe(H2O)6^2+: 4 unpaired electrons
- Cu(H2O)6^2+: 1 unpaired electron
- Zn(H2O)6^2+: 0 unpaired electrons

The ion with the highest paramagnetism is **Fe(H2O)6^2+** with **4 unpaired electrons**.

### Final Answer
**The ion with the highest paramagnetism is Fe(H2O)6^2+.**