The common oxidation state shown by Europium in their compounds is

Options

A. `+1`
B. `+3`
C. `+5`
D. `+6`

Answer

To determine the common oxidation state shown by Europium in its compounds, we can follow these steps:

### Step 1: Identify the Element
Europium (Eu) is a member of the lanthanide series, which consists of elements with atomic numbers from 57 (Lanthanum) to 71 (Lutetium).

### Step 2: Determine the Electronic Configuration
The electronic configuration of Europium can be written as:
\[ \text{Eu: } [Xe] 4f^7 5d^1 6s^2 \]
This indicates that Europium has 7 electrons in the 4f subshell, 1 electron in the 5d subshell, and 2 electrons in the 6s subshell.

### Step 3: Identify the Outermost Electrons
The outermost electrons are those in the 5d and 6s orbitals. In this case, Europium has:
- 2 electrons in the 6s orbital
- 1 electron in the 5d orbital

### Step 4: Determine Possible Oxidation States
When forming compounds, Europium can lose electrons from its outermost orbitals:
1. The first two electrons (from the 6s orbital) can be lost, resulting in an oxidation state of +2.
2. Additionally, the single electron from the 5d orbital can also be lost, leading to a +3 oxidation state.

### Step 5: Consider Common Oxidation States in Lanthanides
While Europium can exhibit a +2 oxidation state, the most common oxidation state for lanthanides, including Europium, is +3. This is because:
- The +3 oxidation state is stable and prevalent among lanthanides.
- The energy required to remove an electron from the 4f subshell is significantly higher, making higher oxidation states (like +4, +5, etc.) less common.

### Conclusion
Based on the analysis, the common oxidation state shown by Europium in its compounds is:
\[ \text{+3} \]

### Answer
The common oxidation state shown by Europium in their compounds is **+3**.

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