Each of the compounmds `Pt(NH_(3))_(6)Cl_(4), Cr(NH_(3))_(6)Cl_(3) and K_(2)PtCl_(6)Co(NH_(3))_(4)Cl_(3)` were dissolved in water to make its 0.01 M solution. The correct order of their increasing conductivity in solution is

Options

A. `K_(2){T Cl_(6) lt Co(NH_(3))_(4).l_(3) lt Cr(NH_(3))_(6)Cl_(3) lt Pt (NH_(3))_(6)Cl_(4)`
B. `Cr(NH_(3))_(6)Clk_(3) lt Co(NH_(3))_(4).Cl_(3) lt K_(2)PtCl_(6) lt Pt (NH_(3))_(6)Cl_(4)`
C. `Co(NH_(3))_(4).Cl_(3)lt K_(2)PtCl_(6) lt Cr(NH_(3))_(6)Cl_(3) lt Pt (NH_(3))_(6)Cl_(4)`
D. `Pt(NH_(3))_(6)Cl_(4) lt Co(NH_(3))_(6).Cl_(3) lt Cr(NH_(3))_(6) Cl_(3)lt K_(2)PtCl_(6)`

Answer

To determine the correct order of increasing conductivity for the compounds \( Pt(NH_3)_6Cl_4 \), \( Cr(NH_3)_6Cl_3 \), and \( K_2PtCl_6 \), \( Co(NH_3)_4Cl_3 \) when dissolved in water to make a 0.01 M solution, we will analyze the dissociation of each compound into its constituent ions.

### Step-by-step solution:

1. **Identify the number of ions produced by each compound upon dissociation:**
- **For \( Pt(NH_3)_6Cl_4 \)**:
- This compound dissociates into 1 \( Pt^{2+} \) ion and 4 \( Cl^{-} \) ions.
- Total ions = \( 1 + 4 = 5 \) ions.

- **For \( Cr(NH_3)_6Cl_3 \)**:
- This compound dissociates into 1 \( Cr^{3+} \) ion and 3 \( Cl^{-} \) ions.
- Total ions = \( 1 + 3 = 4 \) ions.

- **For \( K_2PtCl_6 \)**:
- This compound dissociates into 2 \( K^{+} \) ions and 1 \( PtCl_6^{2-} \) ion.
- Total ions = \( 2 + 1 = 3 \) ions.

- **For \( Co(NH_3)_4Cl_3 \)**:
- This compound dissociates into 1 \( Co^{3+} \) ion and 3 \( Cl^{-} \) ions.
- Total ions = \( 1 + 3 = 4 \) ions.

2. **Count the total number of ions for each compound**:
- \( Pt(NH_3)_6Cl_4 \): 5 ions
- \( Cr(NH_3)_6Cl_3 \): 4 ions
- \( K_2PtCl_6 \): 3 ions
- \( Co(NH_3)_4Cl_3 \): 4 ions

3. **Determine the order of conductivity**:
- Conductivity is directly proportional to the number of ions in solution. Therefore, the more ions produced, the higher the conductivity.
- From the total ions calculated:
- \( K_2PtCl_6 \) (3 ions) < \( Cr(NH_3)_6Cl_3 \) (4 ions) = \( Co(NH_3)_4Cl_3 \) (4 ions) < \( Pt(NH_3)_6Cl_4 \) (5 ions)

4. **Final order of increasing conductivity**:
- The correct order of increasing conductivity is:
- \( K_2PtCl_6 < Cr(NH_3)_6Cl_3 = Co(NH_3)_4Cl_3 < Pt(NH_3)_6Cl_4 \)