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Quiz: Molecular Geometry and Polarity

Test your understanding of VSEPR theory, orbital hybridization, sigma and pi bonds, bond polarity, and molecular polarity with these questions.

1. According to VSEPR theory, what is the electron geometry of a central atom surrounded by four electron groups — two bonding pairs and two lone pairs?

  1. Tetrahedral
  2. Bent
  3. Trigonal planar
  4. Linear
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The correct answer is A. VSEPR theory determines electron geometry from the total number of electron groups — bonding and lone pairs combined. Four electron groups always adopt a tetrahedral electron geometry to minimize mutual repulsion, regardless of how many are lone pairs. Bent (B) describes the molecular geometry of water, not its electron geometry. Trigonal planar (C) corresponds to three total electron groups. Linear (D) corresponds to two electron groups.

Concept Tested: VSEPR Theory and Electron Geometry

2. Water (H₂O) has a measured H–O–H bond angle of approximately 104.5°, which is less than the ideal tetrahedral angle of 109.5°. What is the best explanation?

  1. Oxygen is too small to accommodate angles larger than 104.5°.
  2. The two lone pairs on oxygen exert stronger repulsion than bonding pairs, compressing the H–O–H angle.
  3. The O–H bonds have partial double bond character that requires a smaller angle.
  4. The two hydrogen atoms repel each other and push their bonds toward the oxygen.
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The correct answer is B. In VSEPR theory, repulsion strength follows: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. The two lone pairs on oxygen occupy more space and push the O–H bonding pairs together, compressing the angle below 109.5°. Atomic size (A) does not directly limit bond angles. The O–H bonds are single bonds with no double bond character (C). Hydrogen–hydrogen repulsion (D) is not the VSEPR mechanism — it is the electron groups around the central atom that govern geometry.

Concept Tested: Lone Pair Repulsion and Bond Angle Compression

3. Which molecule has a trigonal pyramidal molecular geometry?

  1. \(\ce{BF3}\)
  2. \(\ce{CH4}\)
  3. \(\ce{H2O}\)
  4. \(\ce{NH3}\)
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The correct answer is D. Ammonia (\(\ce{NH3}\)) has four electron groups around nitrogen (three bonding pairs and one lone pair). The electron geometry is tetrahedral, but the molecular geometry — considering only the positions of the atoms — is trigonal pyramidal. \(\ce{BF3}\) (A) has three bonding pairs and no lone pairs → trigonal planar. \(\ce{CH4}\) (B) has four bonding pairs and no lone pairs → tetrahedral. \(\ce{H2O}\) (C) has two bonding pairs and two lone pairs → bent.

Concept Tested: Trigonal Pyramidal Molecular Geometry

4. A central atom has exactly three bonding pairs and zero lone pairs. Which combination of electron geometry, molecular geometry, and bond angles is correct?

  1. Tetrahedral electron geometry; trigonal pyramidal molecular geometry; ~107°
  2. Trigonal planar electron geometry; bent molecular geometry; ~120°
  3. Trigonal planar electron geometry; trigonal planar molecular geometry; 120°
  4. Trigonal bipyramidal electron geometry; trigonal planar molecular geometry; 90°/120°
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The correct answer is C. Three electron groups minimize repulsion by spreading to the corners of an equilateral triangle — trigonal planar electron geometry at 120°. When all groups are bonding pairs, the molecular geometry equals the electron geometry: trigonal planar with ideal 120° angles. Option (A) describes four electron groups with one lone pair. Option (B) would apply when one of three groups is a lone pair (e.g., \(\ce{SO2}\)). Option (D) describes five electron groups.

Concept Tested: Trigonal Planar Geometry and VSEPR

5. Carbon in ethylene (\(\ce{C2H4}\)) is sp² hybridized. Which statement correctly describes a consequence of this hybridization?

  1. Each carbon forms four equivalent sigma bonds and no pi bonds.
  2. Each carbon has one unhybridized p orbital that overlaps side-by-side with the adjacent carbon to form a pi bond.
  3. The sp² hybrid orbitals point toward the corners of a tetrahedron at 109.5°.
  4. The C=C double bond consists of two sigma bonds and no pi bonds.
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The correct answer is B. sp² hybridization mixes one s and two p orbitals into three sp² hybrid orbitals arranged in a plane at 120°, leaving one p orbital unhybridized on each carbon. These unhybridized p orbitals align parallel to each other and overlap side-by-side to form the pi bond that, together with the sigma bond between the carbons, constitutes the C=C double bond. Each carbon forms three sigma bonds — not four (A). sp² orbitals are in a flat plane at 120°, not tetrahedral at 109.5° (C). A double bond consists of one sigma plus one pi bond, not two sigma bonds (D).

Concept Tested: sp² Hybridization and Pi Bonds

6. How many sigma bonds and pi bonds are present in one molecule of hydrogen cyanide (H–C≡N)?

  1. 3 sigma bonds and 0 pi bonds
  2. 1 sigma bond and 2 pi bonds
  3. 2 sigma bonds and 1 pi bond
  4. 2 sigma bonds and 2 pi bonds
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The correct answer is D. The H–C single bond is 1 sigma bond. The C≡N triple bond contains 1 sigma bond plus 2 pi bonds. Total: 2 sigma bonds and 2 pi bonds. Option (A) incorrectly counts all bonds as sigma bonds. Option (B) misses the H–C sigma bond. Option (C) counts only one pi bond in the triple bond, but a triple bond contains two pi bonds in addition to one sigma bond.

Concept Tested: Sigma Bonds and Pi Bonds in Multiple Bonds

7. Carbon dioxide (\(\ce{CO2}\)) has two polar C=O bonds yet is a nonpolar molecule overall. What best explains this?

  1. C and O have identical electronegativities, so C=O bonds are actually nonpolar.
  2. The linear geometry of CO₂ causes the two equal and opposite C=O bond dipoles to cancel exactly.
  3. The two lone pairs on each oxygen atom cancel the bond dipoles internally.
  4. Double bonds are always nonpolar regardless of the atoms involved.
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The correct answer is B. Each C=O bond is polar (oxygen is more electronegative than carbon, \(\Delta\chi \approx 0.9\)), but in the linear geometry of CO₂ the two bond dipoles point in exactly opposite directions (180° apart) and cancel completely, giving a net dipole moment of zero. Carbon and oxygen have different electronegativities — C=O bonds are polar (A). Lone pairs on terminal atoms are not bond dipoles and do not "cancel" the C=O dipoles (C). Bond polarity depends on electronegativity difference, not bond order (D).

Concept Tested: Molecular Polarity and Dipole Cancellation

8. Which of the following molecules is polar overall, even though all of its bonds are identical?

  1. \(\ce{CCl4}\) (tetrahedral geometry)
  2. \(\ce{BF3}\) (trigonal planar geometry)
  3. \(\ce{NH3}\) (trigonal pyramidal geometry)
  4. \(\ce{SF6}\) (octahedral geometry)
Show Answer

The correct answer is C. Ammonia (\(\ce{NH3}\)) has three identical N–H bonds, but its trigonal pyramidal geometry (caused by the lone pair on nitrogen) is asymmetric. The three N–H bond dipoles do not cancel; they add to give a net dipole pointing toward the lone pair — making NH₃ a polar molecule. \(\ce{CCl4}\) (A) has four identical polar bonds in a perfectly symmetric tetrahedral arrangement — dipoles cancel. \(\ce{BF3}\) (B) has three identical polar bonds in a symmetric trigonal planar arrangement — dipoles cancel. \(\ce{SF6}\) (D) has six identical polar bonds in a perfectly symmetric octahedral arrangement — dipoles cancel.

Concept Tested: Molecular Polarity and Molecular Geometry

9. What is the correct IUPAC name for the compound \(\ce{N2O5}\)?

  1. Dinitrogen pentoxide
  2. Nitrogen(V) oxide
  3. Nitrogen oxide
  4. Dinitrogen tetraoxide
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The correct answer is A. Binary covalent compounds are named using Greek prefixes: di- for 2 nitrogen atoms and penta- for 5 oxygen atoms, combined with the -ide suffix for oxygen. "Nitrogen(V) oxide" (B) uses the Roman numeral system, which is appropriate for variable-charge ionic compounds, not covalent compounds. "Nitrogen oxide" (C) is too vague — it specifies neither atom count nor subscripts. "Dinitrogen tetraoxide" (D) would be the name for \(\ce{N2O4}\), a different compound.

Concept Tested: Naming Covalent Compounds

10. Molecular orbital (MO) theory predicts that \(\ce{O2}\) is paramagnetic, but a Lewis structure of \(\ce{O2}\) shows all electrons paired. What does this reveal?

  1. MO theory is wrong about oxygen — Lewis structures correctly show all electrons paired.
  2. \(\ce{O2}\) is paramagnetic only in its excited state; the Lewis structure correctly represents the ground state.
  3. Both models agree that \(\ce{O2}\) is diamagnetic, but only under high-pressure conditions.
  4. MO theory correctly predicts two unpaired electrons in degenerate antibonding orbitals, demonstrating a case where Lewis structures fail.
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The correct answer is D. Liquid oxygen is experimentally observed to be attracted to magnets — it is paramagnetic, meaning it has unpaired electrons. MO theory correctly predicts this: the two highest-energy electrons of \(\ce{O2}\) occupy two degenerate antibonding \(\pi^*\) orbitals one each (Hund's rule), leaving them unpaired. The Lewis structure, which shows all electrons in paired bonding arrangements, cannot explain this behavior. This illustrates why evaluating competing bonding models is valuable — MO theory succeeds where Lewis structures fail. Options (A), (B), and (C) all contradict the experimental finding that ground-state \(\ce{O2}\) is paramagnetic.

Concept Tested: Molecular Orbital Theory and Paramagnetism