The steady decreases in the ionic radius from `La^(3+)` to `Lu^(3+)` is termed as

Options

A. lanthanoid contraction
B. actionids contraction
C. Both of these
D. None of these

Answer

To solve the question regarding the steady decrease in the ionic radius from `La^(3+)` to `Lu^(3+)`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Elements**:
- We are dealing with the lanthanide series, specifically the ions `La^(3+)` (Lanthanum) and `Lu^(3+)` (Lutetium).

2. **Understand the Electron Configuration**:
- Lanthanum (`La`) has the electron configuration of `[Xe] 6s² 5d¹`. When it loses three electrons to form `La^(3+)`, it loses the two 6s electrons and one 5d electron, resulting in the configuration `[Xe]`.
- Lutetium (`Lu`) has the electron configuration of `[Xe] 6s² 4f¹⁴ 5d¹`. When it loses three electrons to form `Lu^(3+)`, it loses the two 6s electrons and one 5d electron, also resulting in the configuration `[Xe] 4f¹⁴`.

3. **Consider the 4f Electrons**:
- As we move from `La` to `Lu`, the number of electrons in the 4f subshell increases from 0 (in La) to 14 (in Lu).

4. **Understand Shielding Effect**:
- The 4f electrons have poor shielding effect. This means that they do not effectively shield the outer electrons from the nucleus's positive charge.

5. **Effective Nuclear Charge (Z_eff)**:
- Due to the poor shielding by the 4f electrons, the effective nuclear charge experienced by the outer electrons increases as we move from `La` to `Lu`. Higher effective nuclear charge pulls the electrons closer to the nucleus.

6. **Resulting Ionic Radius**:
- As a result of the increased effective nuclear charge and poor shielding, the ionic radius decreases from `La^(3+)` to `Lu^(3+)`.

7. **Conclusion**:
- The steady decrease in ionic radius from `La^(3+)` to `Lu^(3+)` is known as **lanthanide contraction**.

### Final Answer:
The steady decrease in the ionic radius from `La^(3+)` to `Lu^(3+)` is termed as **lanthanide contraction**.

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