Among the following, VWF are maximum in :-

Options

A. HBr
B. LiBr
C. LiCl
D. AgBr

Explanation

To determine which compound has the maximum Van der Waals forces (VWF) among the given options (HBr, LiBr, LiCl, AgBr), we can follow these steps:

### Step 1: Understand Van der Waals Forces
Van der Waals forces are weak intermolecular forces that arise from temporary dipoles created when electrons move around the nucleus of an atom. The strength of these forces generally increases with the number of electrons present in a molecule or compound.

### Step 2: Analyze the Given Compounds
We need to analyze the compounds provided:
- **HBr (Hydrogen Bromide)**
- **LiBr (Lithium Bromide)**
- **LiCl (Lithium Chloride)**
- **AgBr (Silver Bromide)**

### Step 3: Determine the Number of Electrons
Next, we will determine the number of electrons in each compound:
- **HBr**: H has 1 electron, Br has 35 electrons → Total = 36 electrons
- **LiBr**: Li has 3 electrons, Br has 35 electrons → Total = 38 electrons
- **LiCl**: Li has 3 electrons, Cl has 17 electrons → Total = 20 electrons
- **AgBr**: Ag has 47 electrons, Br has 35 electrons → Total = 82 electrons

### Step 4: Compare the Total Electrons
Now, we compare the total number of electrons in each compound:
- HBr: 36 electrons
- LiBr: 38 electrons
- LiCl: 20 electrons
- AgBr: 82 electrons

### Step 5: Identify the Compound with Maximum VWF
From the comparison, **AgBr** has the highest number of electrons (82 electrons). Since Van der Waals forces increase with the number of electrons, AgBr will have the maximum Van der Waals forces among the given options.

### Conclusion
Therefore, the compound among the options that has the maximum Van der Waals forces is **AgBr**.

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