There are two series of f-block elements as 4f and 5f, hence
Terbium has electronic configuration configuration `[Xe] 4f^(9) 6s^(2)`, Oxidation state will be:

Options

A. `+3, +4`
B. `+2, +3, +4`
C. `+3,+4, +5`
D. `+2, +3, +4,+5`

Answer

To determine the oxidation states of Terbium (Tb) based on its electronic configuration, we follow these steps:

### Step 1: Identify the Electronic Configuration
The electronic configuration of Terbium is given as:
\[ \text{[Xe]} 4f^9 6s^2 \]

### Step 2: Determine Common Oxidation States
Terbium is a member of the lanthanide series, which typically exhibits a common oxidation state of +3. This is because lanthanides generally lose three electrons from their outermost shells.

### Step 3: Calculate Possible Oxidation States
- **Common Oxidation State**:
- Terbium can lose three electrons to achieve the +3 oxidation state:
\[ \text{Tb} \rightarrow \text{Tb}^{3+} \quad \text{(losing 3 electrons)} \]
- **Higher Oxidation State**:
- Terbium can also lose four electrons, leading to a +4 oxidation state:
\[ \text{Tb} \rightarrow \text{Tb}^{4+} \quad \text{(losing 4 electrons)} \]
- The electronic configuration after losing four electrons becomes:
\[ \text{[Xe]} 4f^7 \]
- This configuration is half-filled, which provides extra stability.

### Step 4: Conclusion
Thus, the oxidation states that Terbium can exhibit are:
- +3 (common oxidation state for lanthanides)
- +4 (due to half-filled stability)

### Final Answer
The oxidation states of Terbium are +3 and +4.

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