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VSEPR theory and molecular shapes: NEET MCQs with answers

57 questions on this topic of Chemical Bonding and Molecular Structure (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. What is the molecular geometry of the chlorate ion (ClO₃⁻) predicted by VSEPR theory?

  1. Trigonal planar
  2. Tetrahedral
  3. T-shaped
  4. Trigonal pyramidal
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Answer: (D) Trigonal pyramidal

In ClO₃⁻, the central chlorine atom has 7 valence electrons plus 1 from the negative charge, giving 8 electrons. It forms three σ-bonds with oxygen atoms and retains one lone pair (total 4 electron pairs, steric number = 4). Hence, its molecular geometry is trigonal pyramidal.

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2. What is the molecular shape of the xenon difluoride (XeF₂) molecule according to VSEPR theory?

  1. Bent
  2. Linear
  3. V-shaped
  4. See-saw
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Answer: (B) Linear

Xenon has 8 valence electrons. In XeF₂, Xe forms 2 single bonds with fluorine and retains 3 lone pairs. The 5 electron pairs adopt a trigonal bipyramidal geometry with the 3 lone pairs at equatorial sites to minimize repulsions, producing a linear shape.

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3. In the carbonyl fluoride molecule (COF₂), the expected bond angle relationship according to VSEPR theory is:

  1. ∠F-C-F > ∠F-C-O
  2. ∠F-C-O = 90° and ∠F-C-F = 180°
  3. ∠F-C-F = ∠F-C-O = 120°
  4. ∠F-C-O > ∠F-C-F
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Answer: (D) ∠F-C-O > ∠F-C-F

Carbon forms a double bond with oxygen and single bonds with two fluorines (steric number = 3, trigonal planar electron geometry). Because the electron density in a C=O double bond is greater than in C-F single bonds, the double bond-single bond repulsion exceeds single bond-single bond repulsion. Hence, ∠F-C-O (> 120°, ~126°) is larger than ∠F-C-F (< 120°, ~108°).

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4. According to VSEPR theory, the shape of NH₃ is:

  1. T-shaped
  2. square planar
  3. trigonal pyramidal
  4. tetrahedral
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Answer: (C) trigonal pyramidal

The bond pairs and lone pairs around the central atom of NH₃ arrange to give a trigonal pyramidal shape.

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5. Which of the following molecules has an ideal square planar geometry according to VSEPR theory?

  1. XeF₄
  2. SF₄
  3. SiF₄
  4. BF₄⁻
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Answer: (A) XeF₄

XeF₄ has 8 valence electrons on Xe, forming 4 bond pairs and 2 lone pairs (sp³d² hybridization). To minimize repulsions, the two lone pairs occupy trans axial positions, giving a regular square planar geometry.

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6. How many 90° bond angles (F-Br-F) are present in the bromine pentafluoride (BrF₅) molecule according to VSEPR theory?

  1. 8
  2. 4
  3. 2
  4. 0
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Answer: (D) 0

BrF₅ has a square pyramidal geometry derived from an octahedral electron geometry with 5 bond pairs and 1 lone pair. The lone pair repels the four equatorial F atoms upward toward the axial F atom, distorting all F(axial)-Br-F(equatorial) angles to less than 90° (around 84.8°). Therefore, zero angles are strictly 90°.

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7. According to VSEPR theory, the shape of PCl₅ is:

  1. square planar
  2. tetrahedral
  3. trigonal bipyramidal
  4. square pyramidal
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Answer: (C) trigonal bipyramidal

The bond pairs and lone pairs around the central atom of PCl₅ arrange to give a trigonal bipyramidal shape.

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8. In which of the following molecules or ions does the central atom have exactly two lone pairs of electrons?

  1. ICl₂⁻
  2. ClF₄⁻
  3. NH₄⁺
  4. PCl₄⁺
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Answer: (B) ClF₄⁻

In ClF₄⁻, the central Cl atom has 7 + 1 = 8 valence electrons. It forms 4 bond pairs with fluorine atoms, leaving 4 non-bonding electrons, which make 2 lone pairs (AB₄E₂ type, square planar). In ICl₂⁻ there are 3 lone pairs, while NH₄⁺ and PCl₄⁺ have zero lone pairs.

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9. In the trigonal bipyramidal arrangement predicted by VSEPR theory, lone pairs preferentially occupy equatorial positions rather than axial positions because:

  1. Axial positions experience greater 180° repulsions than equatorial positions.
  2. Equatorial positions minimize repulsions by having only two 90° interactions.
  3. Equatorial positions have three 90° interactions with adjacent electron pairs.
  4. Bond angles between equatorial positions are 180°, which maximizes distance.
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Answer: (B) Equatorial positions minimize repulsions by having only two 90° interactions.

An electron pair at an equatorial position experiences 90° repulsive interactions with only two axial pairs. In contrast, an axial position experiences three 90° repulsive interactions with equatorial pairs, making the equatorial position energetically favorable for bulkier lone pairs.

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10. Consider the following two statements regarding VSEPR theory: Statement I: The lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which in turn is greater than bond pair-bond pair repulsion. Statement II: Lone pairs occupy more space around the central atom because they are under the attraction of only one nucleus. Choose the correct option:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
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Answer: (A) Both Statement I and Statement II are correct.

Both statements are fundamental NCERT postulates of VSEPR theory: lp-lp > lp-bp > bp-bp repulsion order arises directly because a lone pair is localized on a single nucleus and thus spreads out over a larger volume than a shared bond pair held between two nuclei.

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11. Arrange the following hydrides in the correct increasing order of their central atom bond angles: H₂O, H₂S, H₂Se, H₂Te.

  1. H₂O < H₂S < H₂Se < H₂Te
  2. H₂Te < H₂Se < H₂S < H₂O
  3. H₂S < H₂Se < H₂Te < H₂O
  4. H₂Te < H₂O < H₂S < H₂Se
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Answer: (B) H₂Te < H₂Se < H₂S < H₂O

As the electronegativity of the central atom decreases down the group (O > S > Se > Te), the bonded electron pairs are drawn less strongly toward the central atom. Consequently, bond pair-bond pair repulsion decreases, leading to narrower bond angles: H₂Te (90°) < H₂Se (91°) < H₂S (92.1°) < H₂O (104.5°).

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12. According to VSEPR theory, the shape of CH₄ is:

  1. tetrahedral
  2. T-shaped
  3. bent
  4. square planar
Show answer and explanation

Answer: (A) tetrahedral

The bond pairs and lone pairs around the central atom of CH₄ arrange to give a tetrahedral shape.

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