The group that has the species correctly listed in the order of decreasing radius is:

Options

A. `cu^(2+), cu^(+), Cu`
B. `V, V^(2+), V^(3+)`
C. `F^(-), Br^(-), I`
D. `B, Be, Li`

Answer

To solve the question regarding the order of decreasing atomic radius for the given species, we need to understand the trends in atomic and ionic radii in the periodic table.

### Step-by-Step Solution:

1. **Understand the Trends in Atomic Radius**:
- Atomic radius decreases across a period from left to right due to increasing nuclear charge, which pulls the electrons closer to the nucleus.
- Atomic radius increases down a group due to the addition of electron shells.

2. **Consider the Effect of Ionization**:
- Cations (positively charged ions) are smaller than their parent atoms because they lose one or more electrons, resulting in less electron-electron repulsion.
- Anions (negatively charged ions) are larger than their parent atoms because they gain electrons, increasing electron-electron repulsion.

3. **Analyze the Given Options**:
- **Option 1**: Cu²⁺, Cu⁺, Cu
- Cu²⁺ < Cu⁺ < Cu (increasing order, not decreasing).
- **Option 2**: V, V²⁺, V³⁺
- V > V²⁺ > V³⁺ (correct order of decreasing radius).
- **Option 3**: F⁻, Br⁻, I
- I > Br⁻ > F⁻ (Iodine has the largest radius, but the order is not correct as it should be I > Br⁻ > F⁻).
- **Option 4**: B, Be, Li
- Li > Be > B (this is incorrect as they are in the same period, and the order should be Be < B < Li).

4. **Conclusion**:
- The only option that correctly lists the species in the order of decreasing radius is **Option 2: V, V²⁺, V³⁺**.

### Final Answer:
The group that has the species correctly listed in the order of decreasing radius is **Option 2: V, V²⁺, V³⁺**.