In aqueous solutions `Eu^(2+)` acts as

Options

A. an oxidising agent
B. a reducing agent
C. neither oxidant nor reductant
D. can act as redox agent

Answer

To determine how `Eu^(2+)` behaves in aqueous solutions, we need to analyze its oxidation states and electronic configuration. Here’s a step-by-step solution:

### Step 1: Identify the Element
- `Eu` stands for Europium, which is a lanthanide element.

### Step 2: Write the Electronic Configuration
- The electronic configuration of Europium (Eu) is:
\[ [Xe] 4f^7 5d^1 6s^2 \]

### Step 3: Determine the Electronic Configuration of `Eu^(2+)`
- When Europium loses two electrons to form `Eu^(2+)`, the two electrons are removed from the outermost shell (6s).
- Therefore, the electronic configuration of `Eu^(2+)` becomes:
\[ [Xe] 4f^7 5d^1 \]

### Step 4: Determine the Electronic Configuration of `Eu^(3+)`
- If Europium loses one more electron to form `Eu^(3+)`, the configuration will be:
\[ [Xe] 4f^7 \]
- This configuration is half-filled, which is a stable state.

### Step 5: Analyze the Stability of Oxidation States
- `Eu^(2+)` can oxidize to `Eu^(3+)`, which is more stable due to its half-filled 4f subshell.
- Oxidation involves the loss of electrons, and since `Eu^(2+)` readily loses an electron to become `Eu^(3+)`, it acts as a reducing agent.

### Step 6: Conclusion
- Therefore, in aqueous solutions, `Eu^(2+)` acts as a **reducing agent** because it oxidizes itself to form the more stable `Eu^(3+)`.

### Final Answer
- `Eu^(2+)` acts as a **reducing agent** in aqueous solutions.

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