Of the interhalogen compounds, `CIF_3` is more reactive than `BrF_3` , but `BrF_3` has higher conductance in the liquid state. The reason is that

Options

A. `BrF_3` has higher molecular weight
B. `ClF_3` is volatile
C. `BrF_3` dissociates into `BeF_2^+` and `BrF_4^-` more easily
D. `CIF_3` is most reactive

Answer

To understand why `BrF3` has higher conductance in the liquid state compared to `ClF3`, despite `ClF3` being more reactive, we can break down the reasoning as follows:

### Step-by-Step Solution:

1. **Understanding Conductance**: Conductance in a liquid is primarily due to the presence of ions. The more ions present in a solution, the higher the conductance.

2. **Molecular Weight Consideration**: While molecular weight can influence physical properties, it is not the primary reason for the difference in conductance between `ClF3` and `BrF3`.

3. **Dissociation of `BrF3`**: In the liquid state, `BrF3` can dissociate into ions. Specifically, it can form `BrF2^+` (a cation) and `BrF4^-` (an anion). This dissociation increases the number of charge carriers in the liquid, which enhances its conductance.

4. **Reactivity of `ClF3`**: `ClF3` is indeed more reactive than `BrF3`, but this reactivity does not directly correlate with its ability to conduct electricity in the liquid state. `ClF3` does not dissociate into ions as readily as `BrF3`, which limits its conductance.

5. **Conclusion**: Therefore, the higher conductance of `BrF3` in the liquid state is due to its ability to dissociate into ions, while `ClF3` remains molecular and does not contribute to conductance in the same manner.

### Final Answer:
`BrF3` has higher conductance in the liquid state because it dissociates into `BrF2^+` and `BrF4^-` ions more easily, providing more charge carriers compared to `ClF3`.

---