`SO_2+NO+H_2O to X+` a dibasic acid `X overset(NaNH_2)(rarr)Y + H_2O`
In the above sequence, X and Y are respectively

Options

A. `NOHSO_4, NaHSO_4`
B. `NO_2, NaNO_2`
C. `N_2O,NaN_3`
D. `N_2O,Na_2N`

Answer

To solve the problem step by step, we will analyze the given chemical reactions and identify the compounds X and Y.

### Step 1: Identify the reaction products
The initial reaction is:
\[ SO_2 + NO + H_2O \]

From the video transcript, we know that this reaction produces:
\[ N_2O + \text{dibasic acid} \]

### Step 2: Determine the dibasic acid
The dibasic acid mentioned is sulfuric acid, which can be represented as:
\[ H_2SO_4 \]

Thus, we can write the complete reaction as:
\[ SO_2 + NO + H_2O \rightarrow N_2O + H_2SO_4 \]

Here, we identify:
- \( X = N_2O \)
- The dibasic acid \( = H_2SO_4 \)

### Step 3: React X with NaNH2
Next, we look at the second part of the reaction:
\[ X + NaNH_2 \rightarrow Y + H_2O \]

Substituting \( X \) with \( N_2O \):
\[ N_2O + NaNH_2 \rightarrow Y + H_2O \]

### Step 4: Identify the product Y
When \( N_2O \) reacts with sodium amide (\( NaNH_2 \)), it produces sodium azide (\( NaN_3 \)) and water. Therefore, we can write:
\[ N_2O + NaNH_2 \rightarrow NaN_3 + H_2O \]

Thus, we identify:
- \( Y = NaN_3 \)

### Conclusion
From the above steps, we conclude that:
- \( X = N_2O \)
- \( Y = NaN_3 \)

### Final Answer
So, the answer to the question is:
- \( X = N_2O \)
- \( Y = NaN_3 \)