`Ag_(2)S+NaCNrarrAoverset(Zn)rarrB`, Hence A and B are -

Options

A. `Na_(2)[Zn(CN)_(4], Zn`
B. `Na[Ag(CN)_(2)], Ag`
C. `Na_(2)[Ag(CN)_(4)], Ag`
D. `Na_(2)[Ag(CN)_(4)], Ag`

Answer

To solve the question step by step, we will analyze the reactions involved and identify the compounds A and B.

### Step 1: Identify the Reaction of Ag₂S with NaCN
When silver sulfide (Ag₂S) reacts with sodium cyanide (NaCN), it forms a complex compound. The reaction can be represented as follows:

\[ \text{Ag}_2\text{S} + 4 \text{NaCN} \rightarrow 2 \text{NaAg(CN)}_2 + \text{Na}_2\text{S} \]

### Step 2: Identify Compound A
From the reaction above, we can see that the product formed from the reaction of Ag₂S and NaCN is a complex salt, which we denote as A:

\[ A = \text{NaAg(CN)}_2 \]

### Step 3: Reaction of A with Zn
Next, we consider the reaction of compound A with zinc (Zn). When zinc is passed through the solution containing A, it reduces the silver ions present in the complex. The reaction can be represented as follows:

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

### Step 4: Identify Compound B
From the reaction above, we can see that the product formed after passing Zn through A is:

\[ B = 2 \text{Ag} \]

### Summary of Compounds
Thus, we have identified:
- A = NaAg(CN)₂ (the complex formed from Ag₂S and NaCN)
- B = 2Ag (the silver metal produced after reduction by Zn)

### Final Answer
Hence, A and B are:
- A = NaAg(CN)₂
- B = 2Ag