Multiple Choice
What is the molecular geometry of a molecule with four bonding domains and one nonbonding domain?
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12. Molecular Shapes & Valence Bond Theory
Molecular Geometry
Multiple Choice
Which of the following molecules would be expected to exhibit a trigonal planar geometry?
A
CH_4
B
H_2O
C
BF_3
D
NH_3
Verified step by step guidance1
Recall that molecular geometry depends on the number of regions of electron density (bonding and lone pairs) around the central atom, as described by VSEPR (Valence Shell Electron Pair Repulsion) theory.
Identify the central atom and count the number of bonded atoms and lone pairs for each molecule:
- CH_4: Carbon with 4 bonded hydrogens and 0 lone pairs.
- H_2O: Oxygen with 2 bonded hydrogens and 2 lone pairs.
- BF_3: Boron with 3 bonded fluorines and 0 lone pairs.
- NH_3: Nitrogen with 3 bonded hydrogens and 1 lone pair.
Use VSEPR theory to determine the geometry based on electron regions:
- 4 regions (all bonding) → tetrahedral geometry (e.g., CH_4).
- 4 regions (2 bonding + 2 lone pairs) → bent geometry (e.g., H_2O).
- 3 regions (all bonding) → trigonal planar geometry (e.g., BF_3).
- 4 regions (3 bonding + 1 lone pair) → trigonal pyramidal geometry (e.g., NH_3).
Therefore, the molecule with trigonal planar geometry is BF_3.
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