For `Br_2(l)` Enthalpy of atomisation = x kJ/mol, Bond dissociation enthalpy of bromine = y kJ/mole, then

Options

A. `x gt y`
B. `x lt y`
C. `x = y`
D. Relation does not exist

Explanation

To solve the problem regarding the enthalpy of atomization and bond dissociation enthalpy of bromine, we can follow these steps:

### Step 1: Understand the Definitions
- **Enthalpy of Atomization (x)**: This is the energy required to convert one mole of a substance in its standard state (liquid bromine, Br₂(l)) into gaseous atoms (Br(g)). This process involves breaking all the bonds in the molecule to produce individual atoms.

- **Bond Dissociation Enthalpy (y)**: This is the energy required to break a bond in a gaseous molecule to form gaseous atoms. For bromine, this refers to the energy needed to break the Br-Br bond in Br₂(g) to produce two Br(g) atoms.

### Step 2: Analyze the Process
1. **Convert Liquid to Gas**: To convert liquid bromine (Br₂(l)) to gaseous bromine (Br₂(g)), we need to provide energy, which is the **enthalpy of vaporization** (let's denote it as z).

2. **Dissociate Gaseous Bromine**: Once we have Br₂(g), we need to break the Br-Br bond to get two Br(g) atoms. This requires the bond dissociation enthalpy (y).

### Step 3: Write the Energy Relationship
From the above steps, we can express the enthalpy of atomization (x) as:
\[ x = z + y \]
where:
- \( z \) is the enthalpy of vaporization,
- \( y \) is the bond dissociation enthalpy.

### Step 4: Compare x and y
Since \( z \) (enthalpy of vaporization) is always a positive value (energy is required to vaporize a liquid), it follows that:
\[ x = z + y > y \]
This implies that the enthalpy of atomization (x) is greater than the bond dissociation enthalpy (y).

### Conclusion
Thus, we conclude that:
\[ x > y \]

### Final Answer
The correct option is **A: x is greater than y**.

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