`2Cu^(2+)+5I^(-) to 2CuI darr+[X]`
`[X]+2S_(2)O_(3)^(2-) to 3[Y]+S_(4)O_(6)^(2-),X and Y` are:

Options

A. `I_(3)^(-) and I^(-)`
B. `I_(2) and I_(3)^-`
C. `I_(2) and I^(-)`
D. `I_(3)^(-) and I_(2)`

Answer

To solve the problem step by step, we need to analyze the given chemical equations and identify the unknown compounds X and Y.

### Step 1: Analyze the first reaction
The first reaction is:
\[ 2Cu^{2+} + 5I^{-} \rightarrow 2CuI + [X] \]

In this reaction, copper ions (\(Cu^{2+}\)) react with iodide ions (\(I^{-}\)) to form copper iodide (\(CuI\)) and an unknown compound [X].

### Step 2: Determine the unknown compound X
To find [X], we need to balance the reaction.

- On the left side, we have:
- 2 copper ions (\(Cu^{2+}\)) contribute a total charge of +4.
- 5 iodide ions (\(I^{-}\)) contribute a total charge of -5.
- The total charge on the left is \(+4 - 5 = -1\).

- On the right side, we have:
- 2 \(CuI\) contributes 2 copper and 2 iodide ions, which gives a total charge of 0 (since \(CuI\) is neutral).
- Therefore, to balance the charges, we need to account for the charge of [X].

The only way to balance the charges is if [X] is the triiodide ion (\(I_3^{-}\)), which has a charge of -1. Thus, we can write:
\[ 2Cu^{2+} + 5I^{-} \rightarrow 2CuI + I_3^{-} \]
So, we conclude that:
\[ X = I_3^{-} \]

### Step 3: Analyze the second reaction
The second reaction is:
\[ [X] + 2S_2O_3^{2-} \rightarrow 3[Y] + S_4O_6^{2-} \]

Substituting [X] with \(I_3^{-}\):
\[ I_3^{-} + 2S_2O_3^{2-} \rightarrow 3[Y] + S_4O_6^{2-} \]

### Step 4: Determine the unknown compound Y
In this reaction, we need to determine what [Y] is.

- The triiodide ion (\(I_3^{-}\)) can react with thiosulfate ions (\(S_2O_3^{2-}\)) to produce iodide ions (\(I^{-}\)) and tetrathionate ions (\(S_4O_6^{2-}\)).
- The balanced reaction would be:
\[ I_3^{-} + 2S_2O_3^{2-} \rightarrow 3I^{-} + S_4O_6^{2-} \]

Thus, we conclude that:
\[ Y = I^{-} \]

### Final Answer
The values of X and Y are:
- \(X = I_3^{-}\) (triiodide ion)
- \(Y = I^{-}\) (iodide ion)