Which trivalent ion has maximum size in the lanthanoid series i.e., Lanthanum ion `(La^(3+))` to Lutetium ion `(Lu^(3+))`?
(At. no. of lanthanum = 57 and lutetium = 71)

Explanation

To determine which trivalent ion has the maximum size in the lanthanoid series (from Lanthanum ion \( \text{La}^{3+} \) to Lutetium ion \( \text{Lu}^{3+} \)), we can follow these steps:

### Step 1: Understand the Lanthanoid Series
The lanthanoid series consists of elements with atomic numbers from 57 (Lanthanum, \( \text{La} \)) to 71 (Lutetium, \( \text{Lu} \)). These elements are known for their f-block characteristics and include elements like Cerium, Neodymium, and others.

**Hint:** Identify the range of elements in the lanthanoid series and their atomic numbers.

### Step 2: Recognize the Concept of Ionic Size
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