`XCl_(2)("excess")+Ycl_(2) to XCl_(4)+Y darr` ,
`YO underset(gt400^(@))overset(Delta)to (1)/(2)O_(2)+Y`, Ore of Y would be :

Options

A. Siderite
B. Cinnabar
C. Malachite
D. Hornsilver

Explanation

To solve the given question step by step, we will analyze the reactions and identify the ore of element Y.

### Step 1: Analyze the Given Reaction
The first reaction provided is:
\[ \text{XCl}_2 + \text{YCl}_2 \rightarrow \text{XCl}_4 + \text{Y} \downarrow \]

Here, XCl₂ reacts with YCl₂ to form XCl₄ and precipitate Y.

### Step 2: Identify the Elements
From the reaction, we can infer that:
- X is an element that forms a chloride with a +2 oxidation state (XCl₂).
- Y is an element that forms a chloride with a +2 oxidation state (YCl₂) and is precipitated out.

### Step 3: Analyze the Decomposition Reaction
The second reaction is:
\[ \text{YO} \overset{\Delta}{\rightarrow} \frac{1}{2} \text{O}_2 + \text{Y} \]

This indicates that when the oxide of Y (YO) is heated above 400°C, it decomposes to produce oxygen gas and elemental Y.

### Step 4: Identify the Element Y
From the context of the reactions, we need to identify what Y could be. The oxide of Y (YO) decomposes to yield elemental Y and O₂.

### Step 5: Identify the Ore of Y
We need to find the ore from which Y is extracted. The video transcript mentions several ores:
- Citrate (FeCO₃)
- Cinnabar (HgS)
- Malachite (Cu(OH)₂·CuCO₃)
- Argentite (AgCl)

The key is to identify which ore corresponds to the element Y, which we suspect to be mercury (Hg) based on the mention of cinnabar.

### Step 6: Conclusion
Cinnabar (HgS) is known to be the primary ore of mercury. Therefore, the ore of Y is:
\[ \text{Cinnabar (HgS)} \]

### Final Answer
The ore of Y would be: **Cinnabar (HgS)**.

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