Consider the following complexes:
(I) `K_(2)Ptcl_(6)` (II) `PtCl_(4).2NH_(3)`
(III) `PtCl_(4).3NH_(3)` (IV) `PtCl_(4).5NH_(3)`
Their electrical conductances in an aqueous solution are:

Options

A. 256,0,97,404
B. 404,0,97,256
C. 256,97,0,404
D. 404,97,256,0

Explanation

To solve the question regarding the electrical conductance of the given complexes, we will analyze each complex and determine the number of ions produced when they dissociate in an aqueous solution.

### Step 1: Analyze the first complex `K2PtCl6`
- The formula indicates that it contains 2 potassium ions (K⁺) and one platinum hexachloride ion (PtCl₆²⁻).
- When it dissociates in water, it produces:
- 2 K⁺ ions
- 1 PtCl₆²⁻ ion
- Total ions produced = 2 + 1 = **3 ions**.

### Step 2: Analyze the second complex `PtCl4.2NH3`
- This complex is neutral and does not dissociate into ions in solution.
- Therefore, it produces:
- Total ions produced = **0 ions**.

### Step 3: Analyze the third complex `PtCl4.3NH3`
- This complex can be represented as Pt(NH₃)₃Cl₃, meaning it has 3 chloride ions outside the coordination sphere.
- When it dissociates, it produces:
- 1 Pt(NH₃)₃Cl³⁺ ion
- 3 Cl⁻ ions
- Total ions produced = 1 + 3 = **4 ions**.

### Step 4: Analyze the fourth complex `PtCl4.5NH3`
- This complex can be represented as Pt(NH₃)₅Cl²⁺, meaning it has 4 chloride ions outside the coordination sphere.
- When it dissociates, it produces:
- 1 Pt(NH₃)₅Cl²⁺ ion
- 4 Cl⁻ ions
- Total ions produced = 1 + 4 = **5 ions**.

### Summary of Ion Production:
- **Complex I (`K2PtCl6`)**: 3 ions
- **Complex II (`PtCl4.2NH3`)**: 0 ions
- **Complex III (`PtCl4.3NH3`)**: 4 ions
- **Complex IV (`PtCl4.5NH3`)**: 5 ions

### Conclusion:
The electrical conductance of a solution is directly proportional to the number of ions produced. Thus, the order of conductance from highest to lowest is:
1. `PtCl4.5NH3` (5 ions)
2. `PtCl4.3NH3` (4 ions)
3. `K2PtCl6` (3 ions)
4. `PtCl4.2NH3` (0 ions)

### Final Answer:
The complex with the maximum electrical conductance is `PtCl4.5NH3`.

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