Which of the following is correct statement ?
(a) `AlCl_(3)` is conducting in fused state
(b) Mobility of `Li^(+)`ion in water is greater than `Cs^(+)`ion
( c) `MCl_(2)` is more volatile than ` MCl_(4)`
(d) `BeSO_(4)` is more soluble in water than `BaSO_(4)`

Options

A. a,b
B. b,c,d
C. b,d
D. Only d

Answer

To solve the question, we need to evaluate each statement one by one to determine which one is correct.

### Step 1: Evaluate Statement (a)
**Statement:** `AlCl_(3)` is conducting in fused state.

**Analysis:**
- Aluminum chloride (AlCl3) in its fused state does not conduct electricity. This is because, in the fused state, AlCl3 tends to dimerize to form Al2Cl6, which does not produce free ions necessary for conduction. Therefore, this statement is **incorrect**.

### Step 2: Evaluate Statement (b)
**Statement:** Mobility of `Li^(+)` ion in water is greater than `Cs^(+)` ion.

**Analysis:**
- The mobility of ions in water is influenced by their size and hydration enthalpy. Lithium ion (Li⁺) is smaller than cesium ion (Cs⁺), which means it has a higher hydration enthalpy. However, due to its smaller size, it is surrounded by more water molecules, which can hinder its movement. Therefore, the mobility of Li⁺ is actually less than that of Cs⁺. This statement is also **incorrect**.

### Step 3: Evaluate Statement (c)
**Statement:** `MCl_(2)` is more volatile than `MCl_(4)`.

**Analysis:**
- MCl2 typically has a higher ionic character due to the +2 oxidation state of the metal, while MCl4 has a higher covalent character due to the +4 oxidation state. Ionic compounds are generally more volatile than covalent compounds. Therefore, this statement is **incorrect**.

### Step 4: Evaluate Statement (d)
**Statement:** `BeSO_(4)` is more soluble in water than `BaSO_(4)`.

**Analysis:**
- Beryllium sulfate (BeSO4) is more soluble in water compared to barium sulfate (BaSO4). This is due to the smaller size of the Be²⁺ ion, which has a higher hydration enthalpy, making it more soluble. In contrast, the larger Ba²⁺ ion has a lower hydration enthalpy and is less soluble. Therefore, this statement is **correct**.

### Conclusion
After evaluating all statements, the correct answer is:
**(d) `BeSO_(4)` is more soluble in water than `BaSO_(4)`**.

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