The correct statement(s) about `Cr^(2+) and Mn^(3+)` is(are)
[Atomic number of `Cr=24 and Mn=25`]

Options

A. `Cr^(2+)` is a reducing agent
B. `Mn^(3+)` is an oxidizing agent
C. Both `Cr^(2+)` and `Mn^(3+)` exhibit `d^(4)` electronic configuration
D. When `Cr^(2+)` is used as a reducing agent, the chromium ion attains `d^(5)` electronic configuration

Explanation

To solve the question regarding the correct statements about `Cr^(2+)` and `Mn^(3+)`, we will analyze the oxidation states, electronic configurations, and their roles as reducing and oxidizing agents.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers and Electronic Configurations:**
- Chromium (Cr) has an atomic number of 24, and its ground state electronic configuration is:
\[ \text{Cr: } [Ar] 4s^1 3d^5 \]
- Manganese (Mn) has an atomic number of 25, and its ground state electronic configuration is:
\[ \text{Mn: } [Ar] 4s^2 3d^5 \]

2. **Determine the Electronic Configurations of `Cr^(2+)` and `Mn^(3+)`:**
- For `Cr^(2+)`, we remove two electrons from the electronic configuration of Cr:
\[ \text{Cr}^{2+}: [Ar] 3d^4 \]
- For `Mn^(3+)`, we remove three electrons from the electronic configuration of Mn:
\[ \text{Mn}^{3+}: [Ar] 3d^4 \]

3. **Analyze the Oxidizing and Reducing Properties:**
- `Cr^(2+)` acts as a reducing agent because it can lose an electron to become `Cr^(3+)`:
\[ \text{Cr}^{2+} \rightarrow \text{Cr}^{3+} + e^- \]
- `Mn^(3+)` acts as an oxidizing agent because it can gain an electron to become `Mn^(2+)`:
\[ \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \]

4. **Check the Electronic Configuration:**
- Both `Cr^(2+)` and `Mn^(3+)` have the same electronic configuration of `3d^4`.

5. **Summarize the Correct Statements:**
- **Statement 1:** `Cr^(2+)` is a reducing agent. **(True)**
- **Statement 2:** `Mn^(3+)` is an oxidizing agent. **(True)**
- **Statement 3:** Both `Cr^(2+)` and `Mn^(3+)` have a `3d^4` electronic configuration. **(True)**
- **Statement 4:** The electronic configuration of `Cr^(3+)` is `3d^3`, which is not the same as `3d^4`. **(False)**

### Conclusion:
The correct statements about `Cr^(2+)` and `Mn^(3+)` are:
1. `Cr^(2+)` is a reducing agent.
2. `Mn^(3+)` is an oxidizing agent.
3. Both have a `3d^4` electronic configuration.