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. `IIltIIIltIVltI`

Answer

To determine the increasing order of conductivity for the given compounds based on their ionization in aqueous solution, we need to analyze each compound to find out how many ions are produced when they dissolve in water. Conductivity is directly proportional to the number of ions in solution; more ions mean higher conductivity.

### Step-by-Step Solution:

1. **Analyze Compound I: `[Pt(NH3)6]Cl4`**
- When dissolved in water, this complex ion remains intact, while the chloride ions dissociate.
- The dissociation can be represented as:
\[ [Pt(NH3)6]Cl4 \rightarrow [Pt(NH3)6]^{4+} + 4Cl^{-} \]
- Total ions produced = 1 (complex ion) + 4 (chloride ions) = **5 ions**.

2. **Analyze Compound II: `[Cr(NH3)6]Cl3`**
- This complex will also remain intact, with the chloride ions dissociating.
- The dissociation can be represented as:
\[ [Cr(NH3)6]Cl3 \rightarrow [Cr(NH3)6]^{3+} + 3Cl^{-} \]
- Total ions produced = 1 (complex ion) + 3 (chloride ions) = **4 ions**.

3. **Analyze Compound III: `[Co(NH3)4]Cl2`**
- Similar to the previous complexes, this will dissociate as follows:
- The dissociation can be represented as:
\[ [Co(NH3)4]Cl2 \rightarrow [Co(NH3)4]^{2+} + 2Cl^{-} \]
- Total ions produced = 1 (complex ion) + 2 (chloride ions) = **3 ions**.

4. **Analyze Compound IV: `NaCl`**
- Sodium chloride dissociates completely into its ions:
- The dissociation can be represented as:
\[ NaCl \rightarrow Na^{+} + Cl^{-} \]
- Total ions produced = 1 (sodium ion) + 1 (chloride ion) = **2 ions**.

### Summary of Ion Production:
- `[Pt(NH3)6]Cl4` → 5 ions
- `[Cr(NH3)6]Cl3` → 4 ions
- `[Co(NH3)4]Cl2` → 3 ions
- `NaCl` → 2 ions

### Increasing Order of Conductivity:
Based on the number of ions produced, the increasing order of conductivity is:
\[ NaCl < [Co(NH3)4]Cl2 < [Cr(NH3)6]Cl3 < [Pt(NH3)6]Cl4 \]

### Final Answer:
**Increasing order of conductivity:**
\[ NaCl < [Co(NH3)4]Cl2 < [Cr(NH3)6]Cl3 < [Pt(NH3)6]Cl4 \]