What is the molecular geometry of the BF_2^- ion?

• A. Trigonal planar
• B. Bent
• C. Linear
• D. Tetrahedral

Answer: (B)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons in the BF_2^- ion. Boron (B) has 3 valence electrons, each fluorine (F) has 7 valence electrons, and the negative charge adds 1 extra electron. So, total valence electrons = 3 + 2 \times 7 + 1.
2. Step 2: Draw the Lewis structure of BF_2^- by placing boron as the central atom and connecting it to two fluorine atoms with single bonds. Then, distribute the remaining electrons to satisfy the octet rule for fluorine atoms and place any leftover electrons on the central atom.
3. Step 3: Count the number of electron domains (regions of electron density) around the central boron atom. These include bonding pairs and lone pairs of electrons.
4. Step 4: Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the molecular geometry based on the number of bonding pairs and lone pairs around the central atom. Remember that lone pairs affect the shape by repelling bonding pairs.
5. Step 5: Conclude the molecular geometry by considering the arrangement that minimizes electron pair repulsions. For BF_2^-, with two bonding pairs and one lone pair on boron, the shape is bent (angular), not linear or trigonal planar.