Arrange the following solutions in decreasing order of `[Ag^(o+)]` ion:
I. `1M [Ag(CN)_(2)]^(Theta)`
II. Saturated `AgC1`
III. `1M [Ag(NH_(3))_(2)]^(o+) in 0.1M NH_(3)`
IV. Saturated `AgI`
`(K_(sp) of AgC1 = 10^(-10), K_(sp) of AgI = 8.3 xx 10^(-17) K_(f)` (formation constant) `[Ag(CN_(2))]^(Theta) = 10^(21), K_(f) [Ag(NH_(3))_(2)]^(o+) = 10^(8)`

Options

A. `I gt II gt III gt IV`
B. `II gt III gt I gt IV`
C. `IV gt II gt II gt I`
D. `I gt IV gt III gt II`

Explanation

To solve the problem of arranging the given solutions in decreasing order of `[Ag^(+) ]` ion concentration, we will analyze each solution step by step.

### Step 1: Analyze `[Ag(CN)2]^-` solution
- The complex ion `[Ag(CN)2]^-` has a formation constant \( K_f = 10^{21} \).
- The dissociation can be represented as:
\[ [Ag(CN)_{2}]^{-} \rightleftharpoons Ag^{+} + 2CN^{-} \]
- Let \( x \) be the concentration of \( Ag^{+} \) produced. Then, the concentration of \( CN^{-} \) will be \( 2x \).
- The equilibrium expression for the dissociation constant \( K_d \) is:
\[ K_d = \frac{[Ag^{+}][CN^{-}]^2}{[[Ag(CN)_{2}]^{-}]} = \frac{x(2x)^2}{1-x} \approx \frac{4x^3}{1} \quad (\text{since } x \text{ is small}) \]
- Given \( K_f = 10^{21} \), we find \( K_d \):
\[ K_d = \frac{1}{K_f} = 10^{-21} \]
- Setting the equation:
\[ 10^{-21} = 4x^3 \implies x^3 = \frac{10^{-21}}{4} \implies x = \sqrt[3]{\frac{10^{-21}}{4}} \approx 6.3 \times 10^{-8} \text{ M} \]

### Step 2: Analyze Saturated `AgCl` solution
- The dissociation of `AgCl` is:
\[ AgCl \rightleftharpoons Ag^{+} + Cl^{-} \]
- Let \( x \) be the concentration of \( Ag^{+} \):
\[ K_{sp} = [Ag^{+}][Cl^{-}] = x^2 \]
- Given \( K_{sp} = 10^{-10} \):
\[ 10^{-10} = x^2 \implies x = 10^{-5} \text{ M} \]

### Step 3: Analyze `1M [Ag(NH3)2]^{+}` solution
- The dissociation can be represented as:
\[ [Ag(NH_{3})_{2}]^{+} \rightleftharpoons Ag^{+} + 2NH_{3} \]
- The formation constant \( K_f = 10^{8} \) gives:
\[ K_d = \frac{1}{K_f} = 10^{-8} \]
- Let \( x \) be the concentration of \( Ag^{+} \):
\[ K_d = \frac{x(0.1 + 2x)^2}{1-x} \approx \frac{x(0.1)^2}{1} \quad (\text{since } x \text{ is small}) \]
- Setting the equation:
\[ 10^{-8} = x(0.1)^2 \implies x = \frac{10^{-8}}{0.01} = 10^{-6} \text{ M} \]

### Step 4: Analyze Saturated `AgI` solution
- The dissociation of `AgI` is:
\[ AgI \rightleftharpoons Ag^{+} + I^{-} \]
- Let \( y \) be the concentration of \( Ag^{+} \):
\[ K_{sp} = [Ag^{+}][I^{-}] = y^2 \]
- Given \( K_{sp} = 8.3 \times 10^{-17} \):
\[ 8.3 \times 10^{-17} = y^2 \implies y = \sqrt{8.3 \times 10^{-17}} \approx 2.8 \times 10^{-9} \text{ M} \]

### Step 5: Compare the concentrations
Now we have the following concentrations:
1. `[Ag(CN)2]^-`: \( 6.3 \times 10^{-8} \) M
2. `AgCl`: \( 10^{-5} \) M
3. `[Ag(NH3)2]^{+}`: \( 10^{-6} \) M
4. `AgI`: \( 2.8 \times 10^{-9} \) M

### Final Order
Arranging these in decreasing order:
1. `AgCl`: \( 10^{-5} \) M
2. `[Ag(NH3)2]^{+}`: \( 10^{-6} \) M
3. `[Ag(CN)2]^{-}`: \( 6.3 \times 10^{-8} \) M
4. `AgI`: \( 2.8 \times 10^{-9} \) M

The final order of decreasing concentration of `[Ag^(+) ]` is:
**II > III > I > IV**