Which of the following electrolytes will have maximum coagulating value for `AgI//Ag^(+)` sol?
a) Na_2S` b) `Na_3PO_4` c) `Na_2SO_4` d) `NaCl`

Options

A. `Na_2S`
B. `Na_3PO_4`
C. `Na_2SO_4`
D. `NaCl`

Explanation

To determine which electrolyte will have the maximum coagulating value for the `AgI//Ag^(+)` sol, we can follow these steps:

### Step 1: Understand Coagulation
Coagulation is the process where colloidal particles aggregate to form a precipitate. The coagulating power of an electrolyte is influenced by the valency of the ions it dissociates into. According to Hardy-Schulze rule, the coagulating power is directly proportional to the valency of the ions that carry opposite charges to the colloidal particles.

### Step 2: Identify the Colloidal System
In this case, we have a colloidal solution of `AgI` which has a positive charge (`Ag^+`). To coagulate this sol, we need to add an electrolyte that provides negatively charged ions.

### Step 3: List the Electrolytes and Their Ions
The given electrolytes and their dissociated ions are:
- a) `Na2S` → `2Na^+` + `S^2-`
- b) `Na3PO4` → `3Na^+` + `PO4^3-`
- c) `Na2SO4` → `2Na^+` + `SO4^2-`
- d) `NaCl` → `Na^+` + `Cl^-`

### Step 4: Determine the Valency of the Anions
Now we will look at the valency of the negatively charged ions (anions):
- `S^2-` from `Na2S` has a valency of 2.
- `PO4^3-` from `Na3PO4` has a valency of 3.
- `SO4^2-` from `Na2SO4` has a valency of 2.
- `Cl^-` from `NaCl` has a valency of 1.

### Step 5: Compare the Valencies
From the above analysis, we can see:
- `S^2-` has a valency of 2.
- `PO4^3-` has a valency of 3 (highest).
- `SO4^2-` has a valency of 2.
- `Cl^-` has a valency of 1.

### Step 6: Conclusion
Since the coagulating power is directly proportional to the valency of the ions, the electrolyte with the highest valency of the negatively charged ion will have the maximum coagulating value. Therefore, `Na3PO4` with the `PO4^3-` ion has the maximum coagulating power.

### Final Answer
The electrolyte with the maximum coagulating value for `AgI//Ag^(+)` sol is **b) `Na3PO4`**.

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