The f-block elements are those in which the differentiating electron enters the (n-2)f orbital. There are two series of f-block elements corresponding to filling of 4f and 5f-orbitals called lanthanides and actinides respectively. They show different oxidation states depending upon the stability of `f^0, f^7`and `f^14` configurations, though the principal oxidation state is +3. There is a regular decrease in size of lanthanide ions with increase in atomic number and it is known as lanthanide contraction. As a result of this, the basic character of oxides and hydroxides decreases from first element (La) to last element (Lu). All the actinides are radioactive and therefore, it is difficult to study their chemical nature.
Terbium has the electronic configuration : `[Xe]5f^(9) 6s^2 . It will show oxidation states of

Options

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

Explanation

To determine the oxidation states of Terbium (Tb), we will follow these steps:

### Step 1: Identify the Electronic Configuration
The electronic configuration of Terbium is given as:
\[ \text{[Xe]} 5f^9 6s^2 \]

### Step 2: Determine the Principal Oxidation State
For f-block elements, the principal oxidation state is typically +3. This is due to the loss of the three outermost electrons, which in the case of Terbium are the two 6s electrons and one from the 5f orbital.

### Step 3: Consider Other Possible Oxidation States
In addition to the principal oxidation state, we need to consider the stability of the f-orbitals:
- The 5f orbital can have configurations that are half-filled or fully filled, which are generally more stable.
- For Terbium, with the configuration \(5f^9\), it can also lose an additional electron from the 5f orbital, leading to a +4 oxidation state.

### Step 4: Conclusion on Oxidation States
Thus, Terbium can exhibit the following oxidation states:
- +3 (the principal oxidation state)
- +4 (due to the stability of the half-filled f-orbital)

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

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