Free VSEPR Theory MCQs with Answers

1. According to VSEPR theory, the shape of a molecule is decided by

• A. the mutual repulsion of the electron pairs around the central atom, which arrange themselves as far apart as possible
• B. the masses of the atoms attached
• C. the number of neutrons in the central atom
• D. the temperature at which the molecule is examined

Explanation: Bonding and lone pairs are regions of negative charge, so they repel and settle into the arrangement of minimum repulsion, which fixes the geometry. Repulsion is strongest between two lone pairs, weaker between a lone pair and a bond pair and weakest between two bond pairs, and this order explains why bond angles are squeezed when lone pairs are present. Atomic mass plays no part.

2. The shape of the ammonia molecule, NH3, is

• A. trigonal planar
• B. pyramidal, with a bond angle of about 107 degrees
• C. tetrahedral, with an angle of 109.5 degrees
• D. linear

Explanation: Nitrogen has three bond pairs and one lone pair, so the four electron regions point tetrahedrally but only the three bonded atoms are visible, giving a pyramid. The lone pair repels the bond pairs more strongly than they repel each other, so the angle is compressed from 109.5 to about 107 degrees. Describing a shape means naming the arrangement of the atoms, not of the electron pairs.

3. The bond angle in a water molecule is about 104.5 degrees rather than 109.5 degrees because

• A. oxygen is more electronegative than hydrogen
• B. water molecules are hydrogen bonded
• C. the two lone pairs on oxygen repel the bond pairs more strongly than bond pairs repel each other
• D. the molecule is linear

Explanation: Oxygen has four electron regions, two bonding and two lone pairs, so the basic arrangement is tetrahedral, but the two lone pairs squeeze the two O-H bonds closer together than in methane. Ammonia, with only one lone pair, is compressed less, which is why its angle of 107 degrees lies between the two. The progression from methane through ammonia to water is the standard illustration of the repulsion order.

4. The shape of the boron trifluoride molecule, BF3, is

• A. pyramidal
• B. bent
• C. tetrahedral
• D. trigonal planar with bond angles of 120 degrees

Explanation: Boron forms three bonds and has no lone pair, so the three electron regions lie flat at 120 degrees, and the molecule is an exception to the octet rule with only six electrons round the central atom. That electron deficiency is why boron trifluoride readily accepts a lone pair and acts as a Lewis acid. Ammonia looks similar on paper but is pyramidal because of its lone pair.

5. The shape of the methane molecule and the carbon dioxide molecule are respectively

• A. tetrahedral and bent
• B. trigonal planar and linear
• C. tetrahedral and linear
• D. pyramidal and bent

Explanation: Methane has four bond pairs and no lone pairs, so it is tetrahedral, while the carbon in carbon dioxide has only two regions of electron density, since each double bond counts as one region, giving a linear molecule at 180 degrees. Treating a double bond as a single region is the step candidates most often miss. This is also why carbon dioxide is non polar despite its polar bonds.

6. The bond angle in a molecule of methane, ammonia and water decreases in the order 109.5, 107 and 104.5 degrees because

• A. the central atoms become larger in that order
• B. the number of lone pairs on the central atom increases from zero to one to two
• C. the bonds become more ionic
• D. the molecules become heavier

Explanation: Each additional lone pair exerts stronger repulsion than a bond pair, squeezing the remaining bonds closer together, so the angle falls by roughly two and a half degrees for each one added. The underlying arrangement of four electron regions is tetrahedral in all three cases. This trio is the classic demonstration that lone pairs must be counted but are not seen in the shape.

7. According to VSEPR theory, a molecule with four bond pairs and no lone pairs on the central atom is

• A. linear
• B. trigonal planar
• C. tetrahedral
• D. bent

Explanation: Four regions of electron density spread as far apart as possible, which places them at the corners of a tetrahedron with angles of 109.5 degrees, as in methane. Adding a lone pair in place of a bond pair would give a pyramid, and two lone pairs a bent shape. The electron pair geometry and the molecular shape coincide only when there are no lone pairs.

8. The shape of the sulphur hexafluoride molecule, SF6, is

• A. tetrahedral
• B. octahedral
• C. trigonal bipyramidal
• D. square planar

Explanation: Six bond pairs around the central sulphur arrange themselves at 90 degrees to one another at the corners of an octahedron. This requires sulphur to expand its octet using d orbitals, which is possible from period 3 onwards but never for carbon, nitrogen or oxygen. Phosphorus pentachloride, with five pairs, is trigonal bipyramidal by the same reasoning.

9. The number of lone pairs on the central atom in a water molecule is

• A. zero
• B. one
• C. two
• D. three

Explanation: Oxygen has six valence electrons, two of which are used in bonding to the hydrogens, leaving two lone pairs. These compress the bond angle from the tetrahedral 109.5 degrees to about 104.5 and make the molecule bent and polar. They are also what allow water to accept two hydrogen bonds.

10. Which species is linear?

• A. H2O
• B. NH3
• C. BeCl2
• D. CH4

Explanation: Beryllium in beryllium chloride has only two bond pairs and no lone pairs, so they point in opposite directions at 180 degrees. Water is bent, ammonia pyramidal and methane tetrahedral, each shaped by the number of lone pairs present. Counting electron regions and lone pairs settles every VSEPR question of this kind.