Element Cu has two oxidation states `Cu^(+1)` & `Cu^(+2)` the right order of radii of these ions.

Options

A. `Cu^(+1)gtCu^(+2)`
B. `Cu^(+2)gtCu^(+1)`
C. `Cu^(+1)=Cu^(+2)`
D. `Cu^(+2)geCu^(+1)`

Answer

To determine the correct order of radii for the oxidation states of copper, \( Cu^{+1} \) and \( Cu^{+2} \), we need to consider the effect of the positive charge on the ionic radius.

### Step-by-Step Solution:

1. **Understanding Oxidation States**:
- Copper can exist in two oxidation states: \( Cu^{+1} \) (cuprous) and \( Cu^{+2} \) (cupric). The oxidation state indicates the charge of the ion.

2. **Effect of Charge on Ionic Size**:
- When an atom loses electrons to form a positive ion, the number of electrons decreases. This leads to a reduction in electron-electron repulsion within the ion.
- As the positive charge increases (from \( Cu^{+1} \) to \( Cu^{+2} \)), the effective nuclear charge experienced by the remaining electrons increases, pulling them closer to the nucleus.

3. **Comparing the Two Ions**:
- For \( Cu^{+1} \): This ion has lost one electron. The remaining electrons experience less nuclear charge compared to \( Cu^{+2} \).
- For \( Cu^{+2} \): This ion has lost two electrons. The remaining electrons are pulled closer to the nucleus due to the higher positive charge.

4. **Conclusion on Ionic Radii**:
- Since \( Cu^{+2} \) has a greater positive charge than \( Cu^{+1} \), it will have a smaller ionic radius.
- Therefore, the correct order of ionic radii is:
\[ Cu^{+1} > Cu^{+2} \]

### Final Answer:
The right order of radii of these ions is:
\[ Cu^{+1} > Cu^{+2} \]