Which is a good reducing agent '?

Options

A. `Yb^(2+)`
B. `Ce^(4+)`
C. `Tb^(4+)`
D. `La^(3+)`

Explanation

To determine which of the given options is a good reducing agent, we need to analyze the oxidation states and electronic configurations of each option. A reducing agent is a substance that can donate electrons and thereby reduce another substance while being oxidized itself.

### Step-by-Step Solution:

1. **Understanding Reducing Agents:**
- A reducing agent is a substance that can lose electrons (undergo oxidation) and, in doing so, reduce another substance. Therefore, we need to identify which of the given options can be oxidized.

2. **Analyzing Each Option:**
- **Option A: Ytterbium (Yb²⁺)**
- Atomic number: 70
- Electronic configuration: [Xe] 4f¹⁴ 5d⁰ 6s²
- Yb²⁺ loses 2 electrons: [Xe] 4f¹⁴ 5d⁰ 6s⁰
- Further oxidation to Yb³⁺: [Xe] 4f¹³ 5d⁰ 6s⁰
- Yb²⁺ can be oxidized to Yb³⁺, so it can act as a reducing agent.

- **Option B: Cerium (Ce⁴⁺)**
- Atomic number: 58
- Electronic configuration: [Xe] 4f² 5d⁰ 6s²
- Ce⁴⁺ loses 4 electrons: [Xe] 4f⁰ 5d⁰ 6s⁰
- Ce⁴⁺ has reached a stable noble gas configuration and cannot be oxidized further, so it does not act as a reducing agent.

- **Option C: Terbium (Tb⁴⁺)**
- Atomic number: 65
- Electronic configuration: [Xe] 4f⁹ 5d⁰ 6s²
- Tb⁴⁺ loses 4 electrons: [Xe] 4f⁷ 5d⁰ 6s⁰
- The 4f⁷ configuration is half-filled and stable, preventing further oxidation. Thus, Tb⁴⁺ does not act as a reducing agent.

- **Option D: Lanthanum (La³⁺)**
- Atomic number: 57
- Electronic configuration: [Xe] 4f⁰ 5d¹ 6s²
- La³⁺ loses 3 electrons: [Xe] 4f⁰ 5d⁰ 6s⁰
- Similar to Ce⁴⁺, La³⁺ has reached a stable state and cannot be oxidized further, so it does not act as a reducing agent.

3. **Conclusion:**
- Among the options, only Yb²⁺ can be oxidized to Yb³⁺, which means it can act as a good reducing agent. Therefore, the answer is:
- **Ytterbium (Yb²⁺) is a good reducing agent.**

### Final Answer:
**Yb²⁺ (Ytterbium 2+) is the good reducing agent.**