`BeF_(2)` has more dipole moment than `BeCl_(2)`

Answer

To determine why `BeF₂` has a greater dipole moment than `BeCl₂`, we can follow these steps:

### Step 1: Understanding Dipole Moment
The dipole moment (\( \mu \)) is a measure of the separation of positive and negative charges in a molecule. It is calculated as the product of the charge and the distance between the charges. The formula is:
\[ \mu = q \times r \]
where \( q \) is the charge and \( r \) is the distance between the charges.

### Step 2: Analyzing the Molecular Geometry
Both `BeF₂` and `BeCl₂` have a linear molecular geometry due to the beryllium atom being surrounded by two halogen atoms. This means that the dipole moments from each bond will be in opposite directions.

### Step 3: Electronegativity of Atoms
Electronegativity is the tendency of an atom to attract electrons. In this case:
- Fluorine (F) is more electronegative than chlorine (Cl).
- The electronegativity values are approximately:
- \( \text{F} = 4.0 \)
- \( \text{Cl} = 3.0 \)

### Step 4: Comparing Dipole Moments
Since the dipole moment depends on the electronegativity difference between the bonded atoms:
- In `BeF₂`, the dipole moment from beryllium to fluorine is larger because fluorine is more electronegative than chlorine.
- In `BeCl₂`, the dipole moment from beryllium to chlorine is smaller due to chlorine's lower electronegativity.

### Step 5: Resultant Dipole Moment
Although both molecules have their dipole moments canceling out due to their linear geometry, the individual dipole moments are stronger in `BeF₂` than in `BeCl₂`. Therefore, we can conclude:
\[ \text{Dipole moment of } BeF₂ > \text{Dipole moment of } BeCl₂ \]

### Conclusion
Thus, `BeF₂` has a greater dipole moment than `BeCl₂` because the electronegativity of fluorine is significantly higher than that of chlorine, leading to a stronger dipole in the `BeF₂` molecule.

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