The minimum equivalent conductance in fused state is shown by-

Options

A. `MgCI_(2)`
B. `BeCI_(2)`
C. `CaCI_(2)`
D. `SrCI_(2)`

Answer

To solve the question regarding which compound shows the minimum equivalent conductance in the fused state among MgCl2, BeCl2, CaCl2, and SrCl2, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compounds**:
We have four compounds to consider:
- Magnesium chloride (MgCl2)
- Beryllium chloride (BeCl2)
- Calcium chloride (CaCl2)
- Strontium chloride (SrCl2)

2. **Determine the Group**:
All these compounds belong to Group 2 of the periodic table, which means they all have a +2 oxidation state for the cation (Mg²⁺, Be²⁺, Ca²⁺, Sr²⁺) and the same anion (Cl⁻).

3. **Analyze Ionic and Covalent Character**:
As we move down Group 2 from Be to Sr, the atomic size increases. This increase in size leads to a decrease in covalent character of the compounds. Covalent character is inversely proportional to ionic character.

4. **Evaluate Ionic Character**:
- Beryllium (Be) has the highest covalent character and therefore the lowest ionic character.
- Strontium (Sr) has the lowest covalent character and the highest ionic character.
- Thus, the order of ionic character is: SrCl2 > CaCl2 > MgCl2 > BeCl2.

5. **Relate Ionic Character to Conductance**:
More ionic character in a compound means that it will dissociate more easily in the fused state, leading to higher equivalent conductance. Conversely, a compound with lower ionic character will have lower equivalent conductance.

6. **Conclusion**:
Since BeCl2 has the lowest ionic character among the given compounds, it will exhibit the minimum equivalent conductance in the fused state. Therefore, the answer is BeCl2.

### Final Answer:
The compound that shows the minimum equivalent conductance in the fused state is **BeCl2**.