`underset(("element having atomic Number" = 53))"Ferric Ion+Mono valent anion" to x+y`
x and y are-

Options

A. `Fe^(0)` & Iodate ion
B. `Fe^(+2)` & Per iodate ion
C. `Fe^(2+)` & `I_(2)`
D. `Fe^(2+)` & `FeI_(3)`

Answer

To solve the question, we need to identify the products formed when a ferric ion (Fe³⁺) reacts with a monovalent anion. Let's break down the solution step by step.

### Step 1: Identify the Ferric Ion
The ferric ion is represented as Fe³⁺. This ion has a charge of +3.

### Step 2: Identify the Monovalent Anion
A monovalent anion typically has a charge of -1. Common examples include halides like fluoride (F⁻), chloride (Cl⁻), bromide (Br⁻), and iodide (I⁻). Given that the atomic number 53 corresponds to iodine (I), we will consider iodide (I⁻) as our monovalent anion.

### Step 3: Write the Reaction
The reaction between the ferric ion and the iodide ion can be written as:
\[ \text{Fe}^{3+} + \text{I}^- \rightarrow x + y \]

### Step 4: Determine the Products
In this reaction, the ferric ion (Fe³⁺) can be reduced to ferrous ion (Fe²⁺), and the iodide ion (I⁻) can be oxidized to iodine (I₂). The balanced reaction can be represented as:
\[ 2 \text{Fe}^{3+} + 2 \text{I}^- \rightarrow 2 \text{Fe}^{2+} + \text{I}_2 \]

From this reaction, we can identify:
- \( x = \text{Fe}^{2+} \) (ferrous ion)
- \( y = \text{I}_2 \) (iodine)

### Final Answer
Thus, the values of \( x \) and \( y \) are:
- \( x = \text{Fe}^{2+} \)
- \( y = \text{I}_2 \)