`Be_2C+4H_2Orarr2X+CH_4X+2HClrarrY`
'X' and 'Y' formed in the above two reactions are -

Options

A. `BeCO_3 and Be(OH)_2` , respectively
B. `Be(OH)_2and BeCl_2` respectively
C. `Be(OH)_2and [Be(OH)_4]Cl_2` respectively
D. `[Be(OH)_4]^(2) and BeCl_2` , respectively

Answer

To solve the question, we need to analyze the given reactions step by step.

### Step 1: Identify the first reaction
The first reaction given is:
\[ \text{Be}_2\text{C} + 4\text{H}_2\text{O} \rightarrow 2X + \text{CH}_4 \]

In this reaction, we have beryllium carbide (\(\text{Be}_2\text{C}\)) reacting with water (\(\text{H}_2\text{O}\)). The products are \(2X\) and methane (\(\text{CH}_4\)).

### Step 2: Determine the products (X)
Beryllium carbide reacts with water to produce beryllium hydroxide and methane. The balanced reaction can be written as:
\[ \text{Be}_2\text{C} + 4\text{H}_2\text{O} \rightarrow 2\text{Be(OH)}_2 + \text{CH}_4 \]

From this, we can see that:
\[ X = \text{Be(OH)}_2 \]

### Step 3: Identify the second reaction
The second reaction given is:
\[ X + 2\text{HCl} \rightarrow Y \]

Substituting \(X\) from the previous step:
\[ \text{Be(OH)}_2 + 2\text{HCl} \rightarrow Y \]

### Step 4: Determine the product (Y)
When beryllium hydroxide reacts with hydrochloric acid, it forms beryllium chloride and water:
\[ \text{Be(OH)}_2 + 2\text{HCl} \rightarrow \text{BeCl}_2 + 2\text{H}_2\text{O} \]

Thus, we find:
\[ Y = \text{BeCl}_2 \]

### Final Answer
The substances \(X\) and \(Y\) formed in the reactions are:
- \(X = \text{Be(OH)}_2\) (Beryllium hydroxide)
- \(Y = \text{BeCl}_2\) (Beryllium chloride)

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