Using VSEPR theory, what is the molecular geometry for a molecule with the formula AX2E2 (two bonding pairs and two lone pairs on the central atom)?

Using VSEPR theory, what is the molecular geometry for a molecule with the formula AX2E2 (two bonding pairs and two lone pairs on the central atom)?

• A. Bent (angular), approximately 104.5°.
• B. Linear, 180°.
• C. Trigonal planar, 120°.
• D. Tetrahedral, 109.5°.

Answer: (A)

The main idea here is how lone pairs influence molecular shape in VSEPR theory. With four electron domains around the central atom, the electron-domain geometry is tetrahedral, but two of those domains are lone pairs. The two bonding pairs occupy positions that minimize repulsion, resulting in a bent (angular) molecular geometry. The presence of lone pairs pushes the bonds closer together, lowering the bond angle from the tetrahedral 109.5° to about 104.5°, a pattern seen in water. The other shapes correspond to different counts of bonding vs lone pairs: two bonds with no lone pairs would be linear (180°), three bonds with no lone pairs would be trigonal planar (120°), and four bonds with no lone pairs would be tetrahedral (109.5°).

The main idea here is how lone pairs influence molecular shape in VSEPR theory. With four electron domains around the central atom, the electron-domain geometry is tetrahedral, but two of those domains are lone pairs. The two bonding pairs occupy positions that minimize repulsion, resulting in a bent (angular) molecular geometry. The presence of lone pairs pushes the bonds closer together, lowering the bond angle from the tetrahedral 109.5° to about 104.5°, a pattern seen in water. The other shapes correspond to different counts of bonding vs lone pairs: two bonds with no lone pairs would be linear (180°), three bonds with no lone pairs would be trigonal planar (120°), and four bonds with no lone pairs would be tetrahedral (109.5°).