Select the most appropriate statemen. In `BF_(3)`

Options

A. all the bonds are completely ionic
B. the B-F bond is partially ionic
C. B-F bond has partial double bond character
D. all the bonds are covalent

Explanation

To solve the question regarding the most appropriate statement about \( BF_3 \), we can break down the information systematically:

### Step 1: Determine the Structure of \( BF_3 \)
- **Boron (B)** has an atomic number of 5, which means it has 3 valence electrons (since it is in group 13).
- **Fluorine (F)** has an atomic number of 9, which means it has 7 valence electrons. In \( BF_3 \), there are three fluorine atoms contributing a total of 21 valence electrons.

### Step 2: Hybridization of Boron in \( BF_3 \)
- Boron uses its 3 valence electrons to form bonds with three fluorine atoms.
- The hybridization of boron in \( BF_3 \) is \( sp^2 \) because it forms three equivalent bonds with fluorine atoms.

### Step 3: Geometry of \( BF_3 \)
- The geometry of \( BF_3 \) is trigonal planar due to the \( sp^2 \) hybridization. The bond angles are approximately 120 degrees.

### Step 4: Electron Deficiency of Boron
- Boron is electron-deficient in \( BF_3 \) because it only has 6 electrons in its valence shell after forming bonds with three fluorine atoms, while it requires 8 electrons to achieve a stable octet.

### Step 5: Nature of Bonds in \( BF_3 \)
- The bonds between boron and fluorine are primarily covalent due to the sharing of electrons. However, there is some degree of ionic character due to the electronegativity difference between boron and fluorine.
- Additionally, \( BF_3 \) can exhibit some back-bonding from the filled \( p \) orbitals of fluorine to the empty \( p \) orbital of boron, which gives it some double bond character.

### Conclusion
Based on the analysis, the most appropriate statement about \( BF_3 \) would be that it has a trigonal planar geometry, is electron-deficient, and exhibits covalent bonding with some ionic character.