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![VSEPR geometries](https://qallery.app/diagrams/seedneet-chembonding-5e47cba1/img-27.jpeg) In the molecular geometries shown for AB₂, AB₃, AB₄, AB₅ and AB₆ (with no lone pair on A), the central-atom bond angles are respectively:

A90°, 120°, 109.5°, 180°, 120°
B180°, 120°, 109.5°, 90/120°, 90°
C120°, 109.5°, 90°, 180°, 120°
D109.5°, 120°, 180°, 90°, 90°
Answer & Solution
Correct answer: B. 180°, 120°, 109.5°, 90/120°, 90°
Linear (AB₂) → 180°. Trigonal planar (AB₃) → 120°. Tetrahedral (AB₄) → 109.5°. Trigonal bipyramidal (AB₅) → 90° axial-equatorial / 120° equatorial-equatorial. Octahedral (AB₆) → 90°.
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