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Pre AP Chemistry -2.2E Molecular Geometry (VSEPR Theory)- MCQ Exam Style Questions -New Syllabus 2025-2026

Pre AP Chemistry -2.2E Molecular Geometry (VSEPR Theory)- MCQ Exam Style Questions – New Syllabus 2025-2026

Pre AP Chemistry -2.2E Molecular Geometry (VSEPR Theory)- MCQ Exam Style Questions – Pre AP Chemistry – per latest Pre AP Chemistry Syllabus.

Pre AP Chemistry – MCQ Exam Style Questions- All Topics

Question

Which molecules have no overall dipole moment?

1 boron trifluoride
2 methane
3 phosphorus pentafluoride

▶️ Answer/Explanation
Solution

Step 1: Condition for zero dipole moment
A molecule has no overall dipole moment if:

  • bond dipoles are present but
  • the molecular shape is symmetrical so the dipoles cancel.

Step 2: Examine each molecule

  • BF₃ – trigonal planar and symmetrical → dipoles cancel → no dipole moment.
  • CH₄ – tetrahedral and symmetrical → dipoles cancel → no dipole moment.
  • PF₅ – trigonal bipyramidal and symmetrical → dipoles cancel → no dipole moment.

Step 3: Conclusion
All three molecules have no overall dipole moment.

Ans: A (1, 2 and 3)

Question

Which molecule does not have any 90° or 180° bond angles?

A \(C_{2}H_{6}\)     B \(CO_{2}\)     C \(PF_{5}\)     D \(SF_{6}\)

▶️ Answer/Explanation
Solution

Step 1: Determine molecular geometry

  • C₂H₆ – each carbon is tetrahedral, bond angles ≈ 109.5°. No 90° or 180° angles.
  • CO₂ – linear, bond angle = 180°.
  • PF₅ – trigonal bipyramidal, contains 90°, 120° and 180° angles.
  • SF₆ – octahedral, contains 90° and 180° angles.

Step 2: Identify the correct molecule

Only C₂H₆ has bond angles of about 109.5° and therefore does not contain 90° or 180° bond angles.

Ans: A

Question

The characteristic smell of garlic is due to alliin.

What are the approximate bond angles x, y and z in a molecule of alliin?

▶️ Answer/Explanation
Solution

Step 1: Identify hybridisation and geometry

  • x: The carbon is sp² hybridised in a C=C double bond → trigonal planar → bond angle ≈ 120°.
  • y: The second carbon in the double bond is also sp² hybridised → trigonal planar → bond angle ≈ 120°.
  • z: The carbon attached to NH₂, H, CO₂H and CH₂ is sp³ hybridised → tetrahedral → bond angle ≈ 109.5°.

Step 2: Match with the options

x ≈ 120°
y ≈ 120°
z ≈ 109.5°

This corresponds to option C.

Ans: C

Question

Which series shows molecules in order of increasing bond angle?

A CH4 → BF3 → NH3
B H2O → CO2 → BF3
C NH3 → CH4 → CO2
D NH3 → CH4 → H2O

▶️ Answer/Explanation
Solution

Step 1: Determine molecular shapes and bond angles

  • NH₃: trigonal pyramidal → bond angle ≈ 107°.
  • CH₄: tetrahedral → bond angle ≈ 109.5°.
  • CO₂: linear → bond angle = 180°.

Step 2: Arrange in increasing order

NH₃ (107°) < CH₄ (109.5°) < CO₂ (180°)

This sequence corresponds to option C.

Ans: C

Question

For which molecule is the dipole moment zero?

A \(CH_3Cl\)     B \(CH_2Cl_2\)     C \(CHCl_3\)     D \(CCl_4\)

▶️ Answer/Explanation
Solution

Step 1: Recall dipole moment concept

A molecule has a zero dipole moment when the bond dipoles cancel due to a symmetrical molecular geometry.

Step 2: Examine each molecule

  • CH₃Cl – tetrahedral but not symmetrical → dipoles do not cancel → polar.
  • CH₂Cl₂ – tetrahedral but two different substituents → dipoles do not cancel → polar.
  • CHCl₃ – tetrahedral but substituents are not identical → dipoles do not cancel → polar.
  • CCl₄ – perfectly symmetrical tetrahedral molecule → bond dipoles cancel.

Step 3: Conclusion

The molecule with zero dipole moment is CCl₄.

Ans: D

Question

When ammonia, \(NH_3\), is dissolved in water, a small concentration of ammonium ions, \(NH_4^+\), is formed.

Which row is correct?

▶️ Answer/Explanation
Solution

Step 1: Number of electrons in ammonium ion

Nitrogen contributes 5 valence electrons and each hydrogen contributes 1 electron.

Total electrons = \(5 + 4(1) = 9\)

The ion has a +1 charge, meaning one electron is removed:

\(9 – 1 = 8\) electrons around nitrogen in bonding pairs.

Thus the ammonium ion contains 4 bonding pairs (8 electrons) around nitrogen.

Step 2: Compare bond angles

  • NH₃: trigonal pyramidal with one lone pair → bond angle ≈ 107°.
  • NH₄⁺: tetrahedral with no lone pairs → bond angle ≈ 109.5°.

Therefore the H–N–H bond angle increases from ammonia to ammonium.

Step 3: Identify correct row

Number of electrons = 8
Bond angle change = increases

This corresponds to option D.

Ans: D

Question

Phosphorus forms two chlorides. Phosphorus(III) chloride, PCl3, is a covalent liquid.

Phosphorus(V) chloride is an ionic solid. One of the ions present is [PCl4]+.

What is the shape of the PCl3 molecule and the [PCl4]+ ion?

▶️ Answer/Explanation
Solution

Step 1: Determine the shape of PCl3

  • Phosphorus has 5 valence electrons.
  • Three P–Cl bonds are formed.
  • One lone pair remains on phosphorus.

This gives 4 electron regions (3 bonding pairs + 1 lone pair).

The electron geometry is tetrahedral but the molecular shape becomes trigonal pyramidal.

Step 2: Determine the shape of [PCl4]+

  • Phosphorus forms four P–Cl bonds.
  • There are no lone pairs on phosphorus.
  • Total electron regions = 4 bonding pairs.

This gives a tetrahedral shape.

Step 3: Match with the correct option

PCl3 → trigonal pyramidal
[PCl4]+ → tetrahedral

Ans: B

Question

Which molecules contain at least one bond angle of 120°?

1 C2H4
2 PF5
3 NCl3

▶️ Answer/Explanation
Solution

Step 1: Determine molecular shapes

  • C₂H₄ – each carbon is sp² hybridised, trigonal planar geometry → bond angles ≈ 120°.
  • PF₅ – trigonal bipyramidal structure → equatorial bond angles are 120°.
  • NCl₃ – trigonal pyramidal structure → bond angles ≈ 107°, not 120°.

Step 2: Identify molecules with 120° bond angles

C₂H₄ and PF₅ contain at least one 120° bond angle.

Ans: B (1 and 2)

Question

The eight species that follow all have covalent bonds.

In which pair do the species have different shapes from each other?

A BeCl2 and CO2
B CH4 and NH4+
C NH3 and BF3
D SCl2 and H2O

▶️ Answer/Explanation
Solution

Step 1: Determine molecular shapes

  • BeCl₂ → linear (180°)
  • CO₂ → linear (180°)

Option A: same shape.

  • CH₄ → tetrahedral
  • NH₄⁺ → tetrahedral

Option B: same shape.

  • NH₃ → trigonal pyramidal (lone pair on N)
  • BF₃ → trigonal planar (no lone pairs on B)

Option C: different shapes.

  • SCl₂ → bent
  • H₂O → bent

Option D: same shape.

Therefore the correct pair with different shapes is:

NH₃ and BF₃

Ans: C

Question

Histidine is an amino acid.

What are the approximate bond angles 1, 2 and 3?

▶️ Answer/Explanation
Solution

Step 1: Identify hybridisation at each labelled atom

  • Angle 1: Located in the aromatic imidazole ring where atoms are sp² hybridised. Trigonal planar geometry → bond angle ≈ 120°.
  • Angle 2: Around the nitrogen in the –NH2 group. Nitrogen has three bonds and one lone pair, giving a trigonal pyramidal arrangement → bond angle ≈ 107°.
  • Angle 3: Around the tetrahedral carbon in the side chain (sp³ hybridised). Bond angle ≈ 109.5°.

Step 2: Match with the given options

1 ≈ 120°
2 ≈ 107°
3 ≈ 109.5°

This corresponds to option B.

Ans: B

Question

Which molecule or ion contains the smallest bond angle?

A C2H4     B CH3COCH3     C NH4+     D NH3

▶️ Answer/Explanation
Solution

Step 1: Determine molecular geometries

  • C₂H₄ – trigonal planar around carbon → bond angle ≈ 120°.
  • CH₃COCH₃ – contains trigonal planar carbonyl carbon → bond angles ≈ 120°.
  • NH₄⁺ – tetrahedral → bond angle ≈ 109.5°.
  • NH₃ – trigonal pyramidal with one lone pair → bond angle ≈ 107°.

Step 2: Compare the bond angles

120° > 109.5° > 107°

The smallest bond angle occurs in NH₃ due to lone pair–bond pair repulsion compressing the bond angles.

Ans: D

Question

Which molecules and ions have a bond angle of 120°?

1 BF3
2 CH3–
3 NH3

▶️ Answer/Explanation
Solution

Step 1: Determine shapes of the species

  • BF₃: Boron has three bonding pairs and no lone pairs → trigonal planar → bond angle ≈ 120°.
  • CH₃⁻: Carbon has three bonding pairs and one lone pair → trigonal pyramidal → bond angle ≈ 107°, not 120°.
  • NH₃: Nitrogen has three bonding pairs and one lone pair → trigonal pyramidal → bond angle ≈ 107°, not 120°.

Step 2: Identify correct species

Only BF₃ has a bond angle of approximately 120°.

This corresponds to option D.

Ans: D

Question

What are the shapes of the molecules of water and boron trifluoride?

▶️ Answer/Explanation
Solution

Step 1: Determine the shape of H₂O

  • Oxygen has 6 valence electrons.
  • Two O–H bonds are formed.
  • Two lone pairs remain on oxygen.

This gives four electron regions (2 bonding pairs + 2 lone pairs).

The electron geometry is tetrahedral, but the molecular shape becomes bent (V-shaped) with a bond angle of about 104.5°.

Step 2: Determine the shape of BF₃

  • Boron has 3 valence electrons.
  • It forms three B–F bonds.
  • There are no lone pairs on boron.

This gives a trigonal planar geometry with bond angles of about 120°.

Step 3: Conclusion

H₂O → bent
BF₃ → trigonal planar

This corresponds to option D.

Ans: D

Question

Histamine is produced in the body to help fight infection. Its shape allows it to fit into receptors which expand blood vessels.

What are the bond angles x, y and z in histamine, from the smallest to the largest?

▶️ Answer/Explanation
Solution

Step 1: Identify the geometry at each labelled position

  • x is around the nitrogen in the –NH2 group. Nitrogen has 3 bonding pairs and 1 lone pair, giving a trigonal pyramidal shape with bond angle ≈ 107°.
  • y is around a carbon with four single bonds (sp³ hybridised). This gives a tetrahedral shape with bond angle ≈ 109.5°.
  • z is in the aromatic ring where atoms are sp² hybridised. This gives a trigonal planar geometry with bond angle ≈ 120°.

Step 2: Arrange from smallest to largest

x ≈ 107° < y ≈ 109.5° < z ≈ 120°

This corresponds to option A.

Ans: A

Question

Urea is a product of animal metabolism. It can also be used as a fertiliser.

The diagram shows angle x in this molecule.

Which statements about the structure of urea are correct?

1 Angle x is approximately 120°.
2 The molecule has two π bonds.
3 The molecule has only three lone pairs of electrons.

▶️ Answer/Explanation
Solution

Step 1: Geometry around the carbon atom

The carbon atom in urea is bonded to:

  • one oxygen by a C=O double bond
  • two nitrogen atoms by C–N single bonds

This gives three regions of electron density, meaning the carbon is sp² hybridised with a trigonal planar geometry.

Therefore the bond angle at x is approximately 120°.

Statement 1 → Correct

Step 2: Number of π bonds

The structure contains only one double bond (C=O). Each double bond contains one π bond.

So the molecule has one π bond, not two.

Statement 2 → Incorrect

Step 3: Count lone pairs

  • Oxygen has 2 lone pairs
  • Each nitrogen has 1 lone pair

Total lone pairs:

2 (O) + 1 (N) + 1 (N) = 4 lone pairs

So the molecule does not have only three lone pairs.

Statement 3 → Incorrect

Correct combination: Only statement 1 is correct.

According to the table of responses, this corresponds to Option D.

Ans: D

Question

\(AlCl_{3}\) vapour forms molecules with formula \(Al_{2}Cl_{6}\) as it is cooled.

What happens to the bond angles during the change from \(AlCl_{3}\) to \(Al_{2}Cl_{6}\)?

A Some decrease, some remain the same.
B Some increase, some remain the same.
C They all decrease.
D They all increase.

▶️ Answer/Explanation
Solution

Step 1: Structure of AlCl₃

  • AlCl₃ is trigonal planar.
  • Bond angles are approximately 120°.

Step 2: Structure of Al₂Cl₆

When cooled, two AlCl₃ molecules dimerise to form \(Al_{2}Cl_{6}\).

  • Two chlorine atoms form bridges between the aluminium atoms.
  • The geometry around aluminium becomes closer to tetrahedral.
  • Bond angles become smaller than 120°.

Step 3: Compare bond angles

In AlCl₃ → 120°
In Al₂Cl₆ → smaller angles (around tetrahedral arrangements)

Therefore, the bond angles decrease during dimer formation.

Ans: C

Question

Valence shell electron pair repulsion (VSEPR) theory should be used to answer this question.

Which species are trigonal planar?

1 \(BH_3\)
2 \(CH_3^+\)
3 \(PH_3\)

The responses A to D should be selected on the basis of

▶️ Answer/Explanation
Solution

Step 1: Apply VSEPR theory

Molecular shape depends on the number of electron pairs around the central atom.

  • BH₃: Boron forms three bonds and has no lone pairs → trigonal planar → bond angles ≈ 120°.
  • CH₃⁺: Carbon forms three bonds and has no lone pairs → trigonal planar → bond angles ≈ 120°.
  • PH₃: Phosphorus forms three bonds and has one lone pair → trigonal pyramidal → bond angles ≈ 93°.

Step 2: Identify trigonal planar species

BH₃ and CH₃⁺ are trigonal planar.

PH₃ is not trigonal planar.

Correct combination: 1 and 2

Ans: B

Question

In which pair do the molecules have the same shape as each other?

  1. H2O and CO2
  2. H2O and SCl2
  3. NH3 and BH3
  4. SCl2 and BeCl2
▶️ Answer/Explanation
Solution

Step 1: Determine molecular shapes using VSEPR theory

  • H₂O: 2 bonding pairs + 2 lone pairs → bent (V-shaped).
  • CO₂: 2 bonding regions, no lone pairs → linear.

Option A: Shapes are different.

  • SCl₂: 2 bonding pairs + 2 lone pairs → bent.

Option B: Both H₂O and SCl₂ are bent → same shape.

  • NH₃: trigonal pyramidal.
  • BH₃: trigonal planar.

Option C: Shapes are different.

  • BeCl₂: linear.

Option D: Shapes are different.

Therefore, the correct pair with the same shape is:

H₂O and SCl₂

Ans: B

Question

Sodium borohydride, NaBH4, and boron trifluoride, BF3, are compounds of boron.

What are the shapes around boron in the borohydride ion and in boron trifluoride?

▶️ Answer/Explanation
Solution

Step 1: Shape of the borohydride ion, BH4−

  • Boron forms four B–H bonds.
  • There are no lone pairs on boron.
  • Total electron regions = 4.

According to VSEPR theory, this gives a tetrahedral geometry with bond angles ≈ 109.5°.

Step 2: Shape of boron trifluoride, BF3

  • Boron forms three B–F bonds.
  • There are no lone pairs on boron.
  • Total electron regions = 3.

This gives a trigonal planar geometry with bond angles ≈ 120°.

Step 3: Conclusion

BH4− → tetrahedral
BF3 → trigonal planar

Ans: D

Question

Two conversions are outlined below.

NH4+ → NH3

C2H4 → C2H6

What similar feature do these two conversions have?

  1. a lone pair of electrons in the product
  2. change in oxidation state of an element
  3. decrease in bond angle of the species involved
  4. disappearance of a π bond
▶️ Answer/Explanation
Solution

Step 1: Analyse the first conversion

NH4+ → NH3

  • NH4+ has a tetrahedral shape with bond angles ≈ 109.5°.
  • NH3 has a trigonal pyramidal shape due to one lone pair.
  • The lone pair–bond pair repulsion reduces the bond angle to about 107°.

Therefore, the bond angle decreases.

Step 2: Analyse the second conversion

C2H4 → C2H6

  • C2H4 (ethene) has sp² hybridised carbon with bond angles ≈ 120°.
  • C2H6 (ethane) has sp³ hybridised carbon with bond angles ≈ 109.5°.

Again, the bond angle decreases.

Step 3: Identify the common feature

In both reactions, the bond angles become smaller.

Ans: C

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