`Ag_(2)S +underset("Excess")(KCN) to [Ag(CN)_(x)]^(-n)`
`[Ag (CN)_(x)]^(-n) +Zn to [Zn (CN)_(y)]^(-m)`
Find the sum of x and y

Options

A. 4
B. 6
C. 8
D. 10

Answer

To solve the question step by step, we will analyze the reactions involving silver sulfide (Ag₂S) and potassium cyanide (KCN), followed by the reaction of the resulting complex with zinc (Zn). We will find the values of x and y, which represent the number of cyanide (CN) ligands in the complexes formed, and then calculate their sum.

### Step 1: Reaction of Ag₂S with KCN
The first reaction involves silver sulfide (Ag₂S) reacting with excess potassium cyanide (KCN). The balanced equation for this reaction is:

\[ \text{Ag}_2\text{S} + 4 \text{KCN} \rightarrow 2 \text{AgCN} + \text{K}_2\text{S} \]

From this reaction, we can see that each silver ion (Ag⁺) forms a complex with two cyanide ions (CN⁻). Therefore, the complex formed can be represented as:

\[ \text{Ag(CN)}_2^- \]

Here, x (the number of CN ligands) is equal to 2, and the overall charge on the complex (n) is -1.

### Step 2: Reaction of [Ag(CN)₂]⁻ with Zn
Next, we consider the reaction of the complex [Ag(CN)₂]⁻ with zinc (Zn). The balanced equation for this reaction is:

\[ 2 \text{Ag(CN)}_2^- + \text{Zn} \rightarrow \text{Zn(CN)}_4^{2-} + 2 \text{Ag} \]

In this reaction, the zinc ion (Zn²⁺) forms a complex with four cyanide ions (CN⁻). Therefore, the complex formed can be represented as:

\[ \text{Zn(CN)}_4^{2-} \]

Here, y (the number of CN ligands) is equal to 4, and the overall charge on the complex (m) is -2.

### Step 3: Calculate the sum of x and y
Now that we have determined the values of x and y, we can calculate their sum:

\[ x + y = 2 + 4 = 6 \]

### Final Answer
The sum of x and y is 6.

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