Incorrect order of radius is:

Options

A. `Sr^(2+) lt Rb^(+) lt Br^(-) lt Se^(2-)`
B. `Nb^(5+) lt Zr^(4+) lt Y^(3+)`
C. `Co gt Co^(2+) gt Co^(3+) gt Co^(4+)`
D. `Ba^(2+) lt Cs^(+) lt Se^(2-) lt As^(3-)`

Explanation

To determine the incorrect order of radii among the given elements, we will analyze the ionic forms of the elements in question and apply the principles of periodic trends in atomic and ionic radii. Here’s a step-by-step solution:

### Step 1: Identify the Elements and Their Ionic Forms
We need to identify the elements mentioned in the question and their respective ionic forms. The elements in question are:
- Strontium (Sr) as Sr²⁺
- Rubidium (Rb) as Rb⁺
- Bromine (Br) as Br⁻
- Cerium (Ce) as Ce³⁺

### Step 2: Understand the Periodic Trends
1. **Atomic Radius Trend**: Atomic radius increases down a group and decreases across a period.
2. **Ionic Radius Trend**: Cations (positively charged ions) are smaller than their neutral atoms because they lose electrons, while anions (negatively charged ions) are larger than their neutral atoms because they gain electrons.

### Step 3: Analyze the Ionic Radii of Each Element
1. **Strontium (Sr²⁺)**: When strontium loses two electrons, it becomes Sr²⁺. This ion has a configuration similar to krypton (Kr) but with a higher number of protons (38), resulting in a smaller ionic radius.
2. **Rubidium (Rb⁺)**: Rubidium loses one electron to become Rb⁺. It also has a configuration similar to krypton but has fewer protons (37) compared to strontium, resulting in a larger ionic radius than Sr²⁺.
3. **Bromine (Br⁻)**: When bromine gains an electron, it becomes Br⁻. This ion has a configuration similar to krypton but has fewer protons (35), resulting in a larger ionic radius than both Sr²⁺ and Rb⁺.
4. **Cerium (Ce³⁺)**: Cerium loses three electrons to become Ce³⁺. It has a higher atomic number (58) and thus a larger ionic radius compared to Sr²⁺, Rb⁺, and Br⁻.

### Step 4: Compare the Ionic Radii
Based on the analysis:
- Sr²⁺ < Rb⁺ < Br⁻ < Ce³⁺
This means that strontium ion is the smallest, followed by rubidium ion, then bromine ion, and finally cerium ion is the largest.

### Step 5: Identify the Incorrect Order
If we were given an order of radii in the options, we would check which one does not follow the trend established above. The incorrect order would be one that contradicts the established ionic radii.

### Conclusion
The incorrect order of radius among the given options is **Option D** (if it suggests a different order than Sr²⁺ < Rb⁺ < Br⁻ < Ce³⁺).

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