The structure of `[BeF_4]^(-2)` is

Options

A. tetrahedral
B. octahedral
C. square planar
D. linear

Answer

To determine the structure of the ion \([BeF_4]^{2-}\), we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
- The central atom in this compound is beryllium (Be).
- Beryllium has 2 valence electrons.

### Step 2: Count the surrounding atoms and their contributions
- There are 4 fluorine (F) atoms surrounding the beryllium atom. Each fluorine contributes 1 electron.
- Therefore, the total contribution from the surrounding atoms is \(4 \times 1 = 4\) electrons.

### Step 3: Account for the charge on the ion
- The ion has a charge of \(2-\). This means we add 2 electrons to our total count.
- There are no positive charges to subtract in this case.

### Step 4: Calculate the total number of valence electrons
- Total valence electrons = Valence electrons from Be + Valence electrons from surrounding F + Charge
- Total = \(2 + 4 + 2 = 8\) electrons.

### Step 5: Determine the hybridization
- The hybridization index (H) can be calculated using the formula:
\[ H = \frac{(V + N - C)}{2} \]
where:
- \(V\) = number of valence electrons from the central atom (2 for Be)
- \(N\) = number of surrounding atoms (4 for F)
- \(C\) = number of positive charges (0 in this case)

Plugging in the values:
\[ H = \frac{(2 + 4 + 2)}{2} = \frac{8}{2} = 4 \]

### Step 6: Identify the hybridization type
- A hybridization index of 4 corresponds to \(sp^3\) hybridization.

### Step 7: Determine the molecular geometry
- For \(sp^3\) hybridization, the molecular geometry is tetrahedral.

### Conclusion
The structure of \([BeF_4]^{2-}\) is tetrahedral.

---