Which of the following ions has smallest radius?

Options

A. `Mn^(2+)`
B. `Ni^(2+)`
C. `Ti^(2+)`
D. `V^(2+)`

Explanation

To determine which of the given ions has the smallest radius, we need to analyze the ionic radii of the ions mentioned in the question. The ions in question are derived from the first transition series elements: manganese (Mn), nickel (Ni), titanium (Ti), and vanadium (V).

### Step-by-Step Solution:

1. **Identify the Ions**: The ions we are considering are Mn²⁺, Ni²⁺, Ti²⁺, and V²⁺.

2. **Understand Ionic Radius Trends**: In the first transition series, as the atomic number increases, the ionic radius generally decreases. This is due to the increase in effective nuclear charge, which pulls the electrons closer to the nucleus.

3. **Consider the Charge of the Ions**: All the ions listed are in the +2 oxidation state. The loss of electrons leads to a decrease in size compared to their neutral atoms.

4. **List the Ionic Radii**:
- Ti²⁺: Approximately 0.74 Å
- V²⁺: Approximately 0.67 Å
- Mn²⁺: Approximately 0.80 Å
- Ni²⁺: Approximately 0.69 Å

5. **Compare the Ionic Radii**: Now we compare the ionic radii:
- Ti²⁺ (0.74 Å) > Ni²⁺ (0.69 Å) > V²⁺ (0.67 Å) > Mn²⁺ (0.80 Å)

6. **Identify the Smallest Radius**: From the comparison, we see that V²⁺ has the smallest ionic radius of the ions listed.

### Final Answer:
The ion with the smallest radius among the given options is **V²⁺ (Vanadium ion)**.

---