The correct order or radii of three species `Ca, Ca ^(+) and Ca ^(2+)` is :

Options

A. `Ca gt Ca ^(+) gt Ca ^(2+)`
B. `Ca ^(2+) gt Ca ^(+) gt Ca`
C. `Ca ^(+) gt Ca gt Ca ^(2+)`
D. `Ca ^(+) gt Ca ^(2+) gt Ca`

Answer

To determine the correct order of radii for the species \( \text{Ca} \), \( \text{Ca}^+ \), and \( \text{Ca}^{2+} \), we need to analyze the number of protons and electrons in each species and how they affect the atomic radius.

### Step-by-Step Solution:

1. **Identify the Atomic Number of Calcium**:
- Calcium (Ca) has an atomic number of 20, which means it has 20 protons and, in its neutral state, 20 electrons.

2. **Determine the Electron Configuration**:
- For neutral Calcium (Ca):
- Protons = 20
- Electrons = 20
- For Calcium ion \( \text{Ca}^+ \):
- Protons = 20
- Electrons = 19 (1 electron is removed)
- For Calcium ion \( \text{Ca}^{2+} \):
- Protons = 20
- Electrons = 18 (2 electrons are removed)

3. **Understand the Effect of Charge on Atomic Radius**:
- The atomic radius generally decreases with an increase in positive charge because the effective nuclear charge (the net positive charge experienced by electrons) increases as electrons are removed.
- In \( \text{Ca} \), there are 20 electrons being held by 20 protons.
- In \( \text{Ca}^+ \), there are 19 electrons being held by 20 protons, which means each electron experiences a stronger pull from the nucleus.
- In \( \text{Ca}^{2+} \), there are 18 electrons being held by 20 protons, leading to an even stronger pull on each electron.

4. **Compare the Sizes**:
- Since \( \text{Ca}^{2+} \) has the highest effective nuclear charge per electron, it will have the smallest radius.
- \( \text{Ca}^+ \) will be larger than \( \text{Ca}^{2+} \) but smaller than neutral \( \text{Ca} \).
- Therefore, the order of radii from largest to smallest is:
\[ \text{Ca} > \text{Ca}^+ > \text{Ca}^{2+} \]

5. **Final Answer**:
- The correct order of radii is:
\[ \text{Ca} > \text{Ca}^+ > \text{Ca}^{2+} \]