Assume `100%` ionisation of the following aq. Solution of
(I) `[Pt(NH_(3))_(6)]Cl_(4)" (II) "[Cr(NH_(3))_(6)]Cl_(3)`
(III) `[Co(NH_(3))_(4)]Cl_(2)" (IV) "NaCl`
Increasing order of conductivity is :

Options

A. I lt II lt III lt IV
B. III lt IV lt II lt I
C. IV lt III lt II lt I
D. equal

Answer

To determine the increasing order of conductivity for the given complexes and NaCl, we need to analyze the number of ions produced when each compound dissociates in solution. The conductivity of a solution is directly proportional to the number of ions present; more ions result in higher conductivity.

Let's break down the dissociation of each compound step by step:

### Step 1: Analyze the first complex `[Pt(NH₃)₆]Cl₄`
- **Dissociation**:
\[ [Pt(NH₃)₆]Cl₄ \rightarrow [Pt(NH₃)₆]^{4+} + 4Cl^{-} \]
- **Total ions produced**:
\[ 1 + 4 = 5 \text{ ions} \]

### Step 2: Analyze the second complex `[Cr(NH₃)₆]Cl₃`
- **Dissociation**:
\[ [Cr(NH₃)₆]Cl₃ \rightarrow [Cr(NH₃)₆]^{3+} + 3Cl^{-} \]
- **Total ions produced**:
\[ 1 + 3 = 4 \text{ ions} \]

### Step 3: Analyze the third complex `[Co(NH₃)₄]Cl₂`
- **Dissociation**:
\[ [Co(NH₃)₄]Cl₂ \rightarrow [Co(NH₃)₄]^{2+} + 2Cl^{-} \]
- **Total ions produced**:
\[ 1 + 2 = 3 \text{ ions} \]

### Step 4: Analyze the fourth compound `NaCl`
- **Dissociation**:
\[ NaCl \rightarrow Na^{+} + Cl^{-} \]
- **Total ions produced**:
\[ 1 + 1 = 2 \text{ ions} \]

### Step 5: Summarize the total ions produced for each compound
- `[Pt(NH₃)₆]Cl₄`: 5 ions
- `[Cr(NH₃)₆]Cl₃`: 4 ions
- `[Co(NH₃)₄]Cl₂`: 3 ions
- `NaCl`: 2 ions

### Step 6: Determine the increasing order of conductivity
Based on the number of ions produced, we can arrange the compounds in increasing order of conductivity:
- `NaCl` (2 ions) < `[Co(NH₃)₄]Cl₂` (3 ions) < `[Cr(NH₃)₆]Cl₃` (4 ions) < `[Pt(NH₃)₆]Cl₄` (5 ions)

### Final Answer:
The increasing order of conductivity is:
\[ \text{NaCl} < [Co(NH₃)₄]Cl₂ < [Cr(NH₃)₆]Cl₃ < [Pt(NH₃)₆]Cl₄ \]