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Quiz: Chemical Bonding and Lewis Structures

Test your understanding of ionic bonding, covalent bonding, Lewis structures, resonance, formal charge, and bond properties with these questions.

1. Which factor most directly determines whether two atoms will form an ionic bond rather than a covalent bond?

  1. A large electronegativity difference between the two atoms
  2. The total number of electrons in both atoms combined
  3. The identical atomic radii of both atoms
  4. The presence of d orbitals in both atoms
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The correct answer is A. Ionic bonds form when there is a large electronegativity difference (typically greater than 1.7) between the bonding atoms — usually a metal and a nonmetal — because one atom effectively strips electrons from the other. Total electron count (B) does not determine bond type. Ionic bonds typically form between atoms of very different sizes, not identical radii (C). d orbitals relate to transition metal chemistry and expanded octets, not the ionic vs. covalent distinction (D).

Concept Tested: Chemical Bonds and Bonding Type

2. Magnesium oxide (MgO) has a much higher melting point than sodium fluoride (NaF). What best explains this?

  1. NaF has a larger unit cell, so more energy is needed to melt it.
  2. MgO contains heavier atoms, and heavier atoms always produce higher melting points.
  3. MgO has covalent character that makes it harder to melt.
  4. The Mg²⁺ and O²⁻ ions carry higher charges than Na⁺ and F⁻, producing a much larger lattice energy.
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The correct answer is D. Lattice energy is proportional to the product of ion charges divided by interionic distance. MgO has ions with charges of +2 and −2, while NaF has charges of +1 and −1. The fourfold increase in the charge product dramatically increases MgO's lattice energy, giving it a melting point of ~2852°C vs. ~993°C for NaF. NaF has a smaller unit cell, not larger (A). Atomic mass alone does not determine melting point (B). MgO is fully ionic, not covalent (C).

Concept Tested: Lattice Energy and Ion Charges

3. What is the total number of valence electrons in the Lewis structure of sulfate ion (\(\ce{SO4^{2-}}\))?

  1. 24 electrons
  2. 30 electrons
  3. 32 electrons
  4. 34 electrons
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The correct answer is C. Sulfur contributes 6 valence electrons; four oxygen atoms contribute \(4 \times 6 = 24\) electrons; the 2− charge adds 2 more electrons. Total: \(6 + 24 + 2 = 32\) electrons. 24 electrons (A) accounts only for the four oxygens. 30 electrons (B) results from forgetting to add the charge contribution. 34 electrons (D) would result from adding 4 extra electrons instead of 2 for the charge.

Concept Tested: Lewis Structures and Valence Electron Counting

4. Which of the following molecules is an exception to the octet rule because its central atom has an incomplete octet?

  1. \(\ce{CO2}\)
  2. \(\ce{BF3}\)
  3. \(\ce{SF6}\)
  4. \(\ce{PCl5}\)
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The correct answer is B. Boron in \(\ce{BF3}\) has only three bonding pairs and no lone pairs, giving it just 6 electrons in its valence shell — an incomplete octet. This makes BF₃ a powerful Lewis acid. \(\ce{CO2}\) (A) satisfies the octet rule for all atoms (each O has 8 electrons and C has 8 electrons). \(\ce{SF6}\) (C) is an expanded octet exception, not an incomplete octet. \(\ce{PCl5}\) (D) is also an expanded octet exception, with 10 electrons on phosphorus.

Concept Tested: Exceptions to the Octet Rule — Incomplete Octets

5. What is the formal charge on the carbon atom in carbon dioxide (O=C=O), where carbon has no lone pairs and two double bonds?

  1. +2
  2. +1
  3. −1
  4. 0
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The correct answer is D. Using \(FC = V - N - B/2\): for carbon in CO₂, \(V = 4\) (valence electrons in neutral carbon), \(N = 0\) (no lone pairs on carbon), \(B = 8\) (two double bonds, contributing 4 bonding electrons each = 8 total). \(FC = 4 - 0 - 8/2 = 4 - 0 - 4 = 0\). +2 (A) would result from incorrectly counting only single bonds. +1 (B) results from an arithmetic error in the formula. −1 (C) would require more lone pair or bond electrons than the formula supports.

Concept Tested: Formal Charge Calculation

6. Ozone (O₃) is represented by two resonance structures connected by a double-headed arrow. Which statement best describes the actual ozone molecule?

  1. The molecule physically alternates between the two structures millions of times per second.
  2. Only one of the two resonance structures is the real structure; the other is a theoretical alternative.
  3. The real molecule is a resonance hybrid with both O–O bonds having a bond order of 1.5.
  4. Resonance structures can only be drawn for charged species, so ozone's structures are invalid.
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The correct answer is C. The actual ozone molecule is best described as a resonance hybrid — a weighted average of the two contributing structures — in which both O–O bonds are identical and have a bond order of 1.5, intermediate between a single and a double bond. Experimental bond length measurements confirm this. The molecule does not flip between structures (A) — it has one fixed real structure. Both resonance structures contribute equally to the hybrid; neither is the "true" one (B). Resonance is common in neutral molecules — ozone, benzene, and many others (D).

Concept Tested: Resonance Structures and the Resonance Hybrid

7. The bond data for carbon-carbon bonds are shown below. Which trend is correctly identified?

Bond Bond Order Bond Length (pm) Bond Energy (kJ/mol)
C–C 1 154 347
C=C 2 134 614
C≡C 3 120 839
  1. Bond order has no systematic relationship with bond length or bond energy.
  2. As bond order increases, bond length increases and bond energy decreases.
  3. As bond order increases, bond length decreases and bond energy increases.
  4. As bond order increases, both bond length and bond energy decrease.
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The correct answer is C. More shared electron pairs pull the nuclei closer (shorter bonds) and create a stronger attraction requiring more energy to break. The data confirm this: length decreases from 154 pm to 120 pm while energy increases from 347 to 839 kJ/mol as bond order rises from 1 to 3. Option (A) contradicts the clear, consistent pattern shown in the data. Option (B) reverses the bond length trend. Option (D) incorrectly states that bond energy also decreases.

Concept Tested: Bond Order, Bond Length, and Bond Energy

8. A student draws two Lewis structures for carbon monoxide (CO): Structure X has a triple bond with formal charges of −1 on C and +1 on O, and Structure Y has a double bond with formal charges of −2 on C and 0 on O. Which is preferred and why?

  1. Structure Y, because oxygen should always carry negative formal charge.
  2. Structure X, because smaller formal charge magnitudes make it the preferred representation.
  3. Structure Y, because double bonds are always preferred over triple bonds.
  4. Both are equally valid because the formal charges sum to zero in each structure.
Show Answer

The correct answer is B. Formal charge guidelines favor the structure with the smallest formal charge magnitudes. Structure X (triple bond) has formal charges of −1 and +1, with maximum magnitude of 1. Structure Y (double bond) has −2 and 0, with magnitude of 2. Structure X is preferred because its formal charges are smaller. Oxygen does not "always" carry negative formal charge — formal charge depends on the specific structure (A). Bond order preference is determined by formal charge analysis, not a blanket rule (C). The sum equaling zero (D) is a necessary condition for any valid Lewis structure, but it does not distinguish between two valid structures.

Concept Tested: Formal Charge and Choosing the Best Lewis Structure

9. Which of the following correctly explains why the electron sea model accounts for the malleability of metals?

  1. When metal cation layers slide past one another, the mobile electron sea adjusts instantly, maintaining bonding without fracture.
  2. Malleability results from the low density of metals, which allows atoms to rearrange freely.
  3. Metal atoms have very weak bonds that break easily when force is applied.
  4. When layers of metal atoms shift, like charges repel and the metal shatters along cleavage planes.
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The correct answer is A. In the electron sea model, metal cations sit in a delocalized electron sea. When a stress shifts one layer of cations relative to another, the mobile electrons instantly redistribute around the new arrangement, preserving the metallic bonding throughout. Unlike ionic crystals — where layer shifts bring like charges together and cause fracture — metals deform without breaking. Density is not the explanation for malleability (B). Metallic bonds are generally strong — metals have high melting points (C). The metal does not shatter (D); that describes ionic crystals.

Concept Tested: Metallic Bonds and the Electron Sea Model

10. In which compound does the central atom have an expanded octet, and what structural feature makes this possible?

  1. \(\ce{NH3}\), because nitrogen has five valence electrons available for bonding
  2. \(\ce{CO2}\), because carbon forms two double bonds, exceeding 8 electrons
  3. \(\ce{H2O}\), because oxygen has two lone pairs that expand the valence shell
  4. \(\ce{PCl5}\), because phosphorus is a third-period element with available d orbitals
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The correct answer is D. Phosphorus in \(\ce{PCl5}\) forms five bonds, placing 10 electrons in its valence shell — an expanded octet. This is possible because third-period and heavier elements have access to empty d orbitals. Second-period elements cannot expand their octets. Nitrogen in \(\ce{NH3}\) (A) has exactly 8 electrons (3 bonds + 1 lone pair) — a normal octet. Carbon in \(\ce{CO2}\) (B) has exactly 8 electrons (two double bonds, each 4 electrons) — a normal octet. Oxygen in \(\ce{H2O}\) (C) has 8 electrons (2 bonds + 2 lone pairs) — a normal octet; as a second-period element it cannot expand its octet.

Concept Tested: Expanded Octets and Exceptions to the Octet Rule