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

Options

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

Answer

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

### Step 1: Identify the central atom and its valency
- The central atom in this anion is Beryllium (Be).
- Beryllium has an atomic number of 4, with an electron configuration of \(1s^2 2s^2\). This means it has 2 valence electrons.

### Step 2: Count the number of surrounding atoms and their valencies
- The surrounding atoms are Fluorine (F). Each Fluorine atom has a valency of 1.
- In \([BeF_4]^{-2}\), there are 4 Fluorine atoms.

### Step 3: Use the hybridization formula
The formula for determining hybridization is:
\[ \text{Hybridization} = \frac{1}{2} [B + M - C + A] \]
Where:
- \(B\) = number of valence electrons of the central atom (Be = 2)
- \(M\) = number of monovalent atoms surrounding the central atom (F = 4)
- \(C\) = number of cations (in this case, it is 0 since there are no positive charges)
- \(A\) = number of anions (here, it is 2 due to the -2 charge)

Substituting the values:
\[ \text{Hybridization} = \frac{1}{2} [2 + 4 - 0 + 2] = \frac{1}{2} [8] = 4 \]

### Step 4: Determine the hybridization type
- A result of 4 corresponds to \(sp^3\) hybridization.

### Step 5: Determine the molecular geometry
- The \(sp^3\) hybridization indicates a tetrahedral geometry. Therefore, the structure of \([BeF_4]^{-2}\) is tetrahedral.

### Final Structure
The structure of \([BeF_4]^{-2}\) is tetrahedral, with the Beryllium atom at the center and the four Fluorine atoms located at the corners of a tetrahedron.

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