College Chemistry Quiz: Lewis Diagrams
20 questions · exam conditions
0:00
Lewis DiagramsQuestion 1 of 20

When drawing the Lewis structure for the nitrate ion (NO₃⁻), a student must consider resonance structures. How many equivalent resonance structures can be drawn for NO₃⁻, and what is the formal charge on the nitrogen atom in each structure?

2 resonance structures; nitrogen formal charge = 0
3 resonance structures; nitrogen formal charge = +1
3 resonance structures; nitrogen formal charge = 0
4 resonance structures; nitrogen formal charge = +1
2 resonance structures; nitrogen formal charge = +2
← Back to quizzes

College Chemistry Quiz

College Chemistry Quiz: Lewis Diagrams

Practice Lewis Diagrams in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Lewis Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

When drawing the Lewis structure for the nitrate ion (NO₃⁻), a student must consider resonance structures. How many equivalent resonance structures can be drawn for NO₃⁻, and what is the formal charge on the nitrogen atom in each structure?

  1. 2 resonance structures; nitrogen formal charge = 0
  2. 3 resonance structures; nitrogen formal charge = +1 (correct answer)
  3. 3 resonance structures; nitrogen formal charge = 0
  4. 4 resonance structures; nitrogen formal charge = +1
  5. 2 resonance structures; nitrogen formal charge = +2
Explanation: When you encounter questions about resonance structures and formal charges, you need to systematically draw all possible Lewis structures and calculate formal charges using the formula: Formal charge = (valence electrons) - (nonbonding electrons) - ½(bonding electrons). For the nitrate ion (NO₃⁻), start by counting total valence electrons: nitrogen has 5, each oxygen has 6, plus 1 extra from the negative charge, giving 24 total electrons. Place nitrogen in the center with three oxygens around it. To satisfy the octet rule, you'll need one double bond and two single bonds to the oxygens. The key insight is that the double bond can be placed between nitrogen and any of the three oxygen atoms, creating three equivalent resonance structures. In each structure, nitrogen forms one double bond (4 electrons) and two single bonds (4 electrons), using 8 bonding electrons total with no lone pairs. Using the formal charge formula: 5 - 0 - ½(8) = +1. The nitrogen formal charge is +1 in all structures. Choice A is wrong because there are three equivalent oxygens, not two, so three resonance structures exist. Choice C incorrectly calculates nitrogen's formal charge as 0, likely forgetting that nitrogen uses 8 bonding electrons. Choice D suggests four resonance structures, but there are only three equivalent positions for the double bond. Study tip: When drawing resonance structures for polyatomic ions, always count equivalent positions systematically. For formal charges, double-check your electron counting—bonding electrons are shared, so they contribute ½ to each atom's formal charge calculation.

Question 2

In the Lewis structure of phosphoric acid (H₃PO₄), the central phosphorus atom is bonded to four oxygen atoms. If one student draws all P-O bonds as single bonds while another draws one P=O double bond and three P-O single bonds, which approach gives the lower formal charges?

  1. All single bonds give lower formal charges on all atoms
  2. One double bond gives lower formal charges overall (correct answer)
  3. Both approaches give identical formal charge distributions
  4. The approach depends on which oxygen atom forms the double bond
  5. Neither approach satisfies the octet rule for phosphorus
Explanation: When drawing Lewis structures for molecules with atoms from the third period or below (like phosphorus), you need to consider formal charges to determine the most stable structure. Formal charge equals the number of valence electrons minus nonbonding electrons minus half the bonding electrons. Let's calculate formal charges for both approaches in H₃PO₄. Phosphorus has 5 valence electrons, and oxygen has 6. For all single P-O bonds: Each oxygen forms one bond to P and has three lone pairs (except the three bonded to hydrogen, which have two lone pairs). The central P has formal charge = 5 - 0 - 8/2 = +1. The double-bonded oxygen would have formal charge = 6 - 4 - 2/2 = +1, while other oxygens have formal charges of 0 or -1. For one P=O double bond with three P-O single bonds: The P atom has formal charge = 5 - 0 - 10/2 = 0. The double-bonded oxygen has formal charge = 6 - 4 - 4/2 = 0. The other oxygens maintain their charges of 0 or -1. The double bond structure minimizes formal charges, especially eliminating the +1 charge on phosphorus, making option B correct. Option A is wrong because single bonds create higher formal charges on phosphorus. Option C is incorrect since the formal charge distributions differ significantly between the two structures. Option D is wrong because regardless of which oxygen forms the double bond, the overall formal charge pattern remains the same. Remember: Lower formal charges generally indicate more stable structures, and third-period elements can expand their octets to achieve better formal charge distributions.

Question 3

A student attempts to draw the Lewis structure for ClF₃. After placing the chlorine atom in the center and bonding it to three fluorine atoms, the student finds that chlorine has 10 electrons around it. What is the molecular geometry around the central chlorine atom?

  1. Trigonal planar because there are three bonding pairs
  2. Tetrahedral because there are four electron pairs total
  3. T-shaped because there are three bonding pairs and two lone pairs (correct answer)
  4. Trigonal pyramidal because there are three bonding pairs and one lone pair
  5. Linear because the lone pairs cancel out the effect of bonding pairs
Explanation: When determining molecular geometry, you need to consider both bonding pairs and lone pairs around the central atom, then apply VSEPR theory to predict the three-dimensional arrangement. Let's work through ClF₃ systematically. Chlorine has 7 valence electrons, and each fluorine contributes 1 electron to the bonding, giving us 3 bonding pairs (6 electrons). The problem states chlorine has 10 electrons total around it, which means there are 4 additional electrons beyond the 6 bonding electrons. These 4 electrons form 2 lone pairs. So we have 5 electron pairs total around chlorine: 3 bonding pairs and 2 lone pairs. According to VSEPR theory, 5 electron pairs arrange themselves in a trigonal bipyramidal electron geometry. However, the lone pairs occupy the equatorial positions (where there's more space), leaving the three bonding pairs in a T-shaped molecular geometry. Answer A is wrong because it ignores the lone pairs entirely—molecular geometry must account for all electron pairs. Answer B incorrectly counts only 4 total electron pairs, missing one lone pair in the calculation. Answer D also miscounts, suggesting only 1 lone pair instead of 2. The key strategy here is to always count both bonding and lone pairs, then remember that molecular geometry describes only the arrangement of atoms (not lone pairs), while electron geometry includes everything. Practice identifying electron vs. molecular geometry—this distinction appears frequently on chemistry exams.

Question 4

When drawing the Lewis structure for the carbonate ion (CO₃²⁻), a student finds multiple valid structures. If the bond order between carbon and each oxygen atom is calculated using resonance, what is the average C-O bond order?

  1. 1.0
  2. 1.33 (correct answer)
  3. 1.5
  4. 2.0
  5. 2.33
Explanation: When you encounter Lewis structures with multiple valid forms, you're dealing with resonance—a key concept where the actual structure is a hybrid of all possible arrangements. This question tests your ability to calculate average bond order across resonance structures. To find the average C-O bond order in CO₃²⁻, start by drawing all valid resonance structures. The carbonate ion has three resonance forms: each places a double bond between carbon and one oxygen atom, while the other two C-O connections are single bonds. In each structure, one C-O bond has order 2, and two C-O bonds have order 1. Now calculate the average bond order for any C-O position. Since each oxygen atom participates in a double bond in exactly one of the three resonance structures, each C-O bond is double 13\frac{1}{3} of the time and single 23\frac{2}{3} of the time. The average bond order equals: 13(2)+23(1)=23+23=43=1.33\frac{1}{3}(2) + \frac{2}{3}(1) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} = 1.33 Choice A (1.0) assumes all bonds are single, ignoring the double bond character from resonance. Choice C (1.5) might result from incorrectly averaging a single and double bond without considering that there are three equivalent positions. Choice D (2.0) incorrectly assumes all bonds are double bonds. Remember this pattern: for resonance structures, calculate the fraction of time each bond type appears, then take the weighted average. This approach works for any molecule with delocalized bonding.

Question 5

When comparing Lewis structures for equivalent resonance forms, a student notices that formal charges help determine the most stable structure. For the azide ion (N₃⁻) with linear geometry, what is the formal charge on the central nitrogen atom in the most stable resonance structure?

  1. -1
  2. 0
  3. +1 (correct answer)
  4. +2
  5. -2
Explanation: When evaluating resonance structures, you need to determine which form minimizes formal charges while placing negative charges on the most electronegative atoms. For the azide ion (N₃⁻), start by calculating formal charges using the formula: formal charge = valence electrons - nonbonding electrons - (bonding electrons ÷ 2). The azide ion has 16 total valence electrons (5 from each nitrogen plus 1 from the negative charge). In the most stable linear resonance structure, the arrangement is N≡N⁺-N²⁻, where the central nitrogen forms a triple bond with one terminal nitrogen and a single bond with the other. For the central nitrogen: it has 5 valence electrons, 0 nonbonding electrons, and 8 bonding electrons (triple bond = 6, single bond = 2). Using the formula: 5 - 0 - (8 ÷ 2) = +1. This positive formal charge on the central nitrogen is correct answer C. Choice A (-1) would result from miscounting the bonding electrons or incorrectly assigning them. Choice B (0) represents a common error where students assume the central atom should be neutral, but this doesn't account for the actual electron distribution in the most stable form. Choice D (+2) occurs when students incorrectly place two double bonds on the central nitrogen, which violates the octet rule. Remember that the most stable resonance structure minimizes formal charges overall and places negative charges on the most electronegative atoms. Don't assume central atoms must be neutral—calculate formal charges systematically for each proposed structure.

Question 6

In the Lewis structure of the perchlorate ion (ClO₄⁻), a student can draw the structure with all single bonds or with some double bonds. If the student wants to minimize formal charges, how many Cl=O double bonds should be included?

  1. 0 double bonds
  2. 1 double bond
  3. 2 double bonds
  4. 3 double bonds (correct answer)
  5. 4 double bonds
Explanation: When drawing Lewis structures for polyatomic ions like perchlorate (ClO₄⁻), you need to consider formal charge to determine the most stable arrangement. Formal charge equals the number of valence electrons an atom should have minus the electrons it actually "owns" in the structure (bonding electrons count as half-owned). To minimize formal charges in ClO₄⁻, let's calculate what happens with different numbers of double bonds. Chlorine has 7 valence electrons, and oxygen has 6. With the extra electron from the negative charge, you have 32 total electrons to distribute. If you use all single bonds (choice A), chlorine would have a formal charge of +3 and each oxygen would be -1, giving large formal charges. Adding one double bond (choice B) reduces chlorine's formal charge to +2. With two double bonds (choice C), chlorine's formal charge becomes +1. However, with three Cl=O double bonds (choice D), chlorine achieves a formal charge of 0, while one oxygen remains at -1 (carrying the ion's negative charge) and the three double-bonded oxygens have formal charges of 0. Choice A gives unnecessarily high formal charges. Choice B still leaves chlorine at +2. Choice C improves things but doesn't achieve the minimum possible formal charges. Choice D minimizes the formal charges most effectively, with chlorine at 0 and most oxygens at 0. Remember that while chlorine can expand its octet (being in period 3), the goal is always to minimize formal charges for maximum stability. Look for structures where formal charges are closest to zero and distributed logically based on electronegativity.

Question 7

A student draws the Lewis structure for phosphorus pentachloride (PCl₅) and notes that phosphorus has 10 electrons in its valence shell. If the student were to calculate the formal charge on phosphorus, what value would be obtained?

  1. -3
  2. -1
  3. 0 (correct answer)
  4. +2
  5. +5
Explanation: When you encounter Lewis structures with expanded octets like PCl₅, formal charge calculations follow the same rules regardless of how many electrons surround the central atom. Formal charge reveals the electron "accounting" for each atom in a molecule. To calculate formal charge, use the formula: Formal charge=(valence electrons)−(nonbonding electrons)−12(bonding electrons)\text{Formal charge} = \text{(valence electrons)} - \text{(nonbonding electrons)} - \frac{1}{2}\text{(bonding electrons)} For phosphorus in PCl₅: Phosphorus normally has 5 valence electrons. In the Lewis structure, phosphorus forms five single bonds (no lone pairs), so it has 0 nonbonding electrons and 10 bonding electrons (5 bonds × 2 electrons each). Therefore: Formal charge=5−0−102=5−5=0\text{Formal charge} = 5 - 0 - \frac{10}{2} = 5 - 5 = 0 The answer is (C) 0. Let's examine why the other options are incorrect. Option (A) -3 would suggest phosphorus gained three electrons, which doesn't match the bonding pattern where phosphorus shares electrons equally in covalent bonds. Option (B) -1 would indicate phosphorus gained one electron, but the equal sharing in the P-Cl bonds doesn't support this. Option (D) +2 might seem reasonable since phosphorus is less electronegative than chlorine, but formal charge doesn't account for electronegativity differences—it assumes equal electron sharing. Remember that formal charge assumes perfect covalent bonding with equal electron sharing, regardless of actual electronegativity differences. This is why many molecules with polar bonds still show formal charges of zero on their atoms.

Question 8

In the Lewis structure for the hydroxylamine molecule (NH₂OH), a student must decide how to connect the atoms. Given that the most stable structure minimizes formal charges, what is the formal charge on the oxygen atom in the correct Lewis structure?

  1. -2
  2. -1
  3. 0 (correct answer)
  4. +1
  5. +2
Explanation: When drawing Lewis structures, you need to determine the most stable arrangement by minimizing formal charges across all atoms. Formal charge is calculated as: Formal charge=valence electrons−nonbonding electrons−12bonding electrons\text{Formal charge} = \text{valence electrons} - \text{nonbonding electrons} - \frac{1}{2}\text{bonding electrons} For hydroxylamine (NH₂OH), the most stable structure has nitrogen as the central atom bonded to two hydrogens and one oxygen, with the oxygen then bonded to the final hydrogen. In this arrangement, oxygen has 6 valence electrons, 4 nonbonding electrons (two lone pairs), and participates in 4 bonding electrons (two bonds: N-O and O-H). This gives oxygen a formal charge of 6−4−42=06 - 4 - \frac{4}{2} = 0. Choice A (-2) would require oxygen to have gained two electrons, which doesn't occur in any reasonable Lewis structure for this molecule. Choice B (-1) represents a common error where students might miscalculate the formal charge by incorrectly counting electrons or assuming oxygen carries a negative charge simply because it's electronegative. Choice D (+1) would mean oxygen has lost an electron, which contradicts oxygen's tendency to attract electrons. The correct answer is C (0) because this arrangement minimizes formal charges across the entire molecule, making it the most stable structure. Study tip: Always draw out the complete Lewis structure first, then systematically calculate formal charges using the formula. Remember that the most stable structure typically has formal charges closest to zero, especially on less electronegative atoms.

Question 9

When drawing the Lewis structure for xenon tetrafluoride (XeF₄), a student finds that xenon has 12 electrons around it. How many lone pairs are present on the central xenon atom?

  1. 0
  2. 1
  3. 2 (correct answer)
  4. 3
  5. 4
Explanation: When you encounter Lewis structure problems involving expanded octets, remember that atoms in the third period and beyond can accommodate more than 8 electrons by using d orbitals. To find the lone pairs on xenon in XeF₄, start with the total electrons around the central atom. The problem states xenon has 12 electrons surrounding it. Since XeF₄ has four Xe-F bonds, and each single bond contains 2 electrons, the bonding electrons account for 4×2=84 \times 2 = 8 electrons. The remaining electrons must exist as lone pairs: 12−8=412 - 8 = 4 electrons. Since each lone pair contains 2 electrons, xenon has 4÷2=24 ÷ 2 = 2 lone pairs. Looking at the wrong answers: Choice A (0 lone pairs) would mean all 12 electrons are in bonds, requiring 6 bonds, but XeF₄ only has 4 fluorine atoms. Choice B (1 lone pair) accounts for only 10 electrons total (8 bonding + 2 lone pair), leaving 2 electrons unaccounted for. Choice D (3 lone pairs) would require 14 total electrons (8 bonding + 6 lone pair), which exceeds the given 12 electrons. The correct answer is C: xenon has 2 lone pairs. For Lewis structures with expanded octets, always use this systematic approach: count total electrons around the central atom, subtract electrons in bonds, then divide the remainder by 2 to find lone pairs. This method prevents counting errors and ensures you account for all electrons properly.

Question 10

When drawing Lewis structures for molecules with multiple central atoms, a student encounters ethylene (C₂H₄). In the most stable Lewis structure, what is the formal charge on each carbon atom?

  1. -2
  2. -1
  3. 0 (correct answer)
  4. +1
  5. +2
Explanation: When you encounter Lewis structure problems involving formal charges, you're applying the concept that atoms "prefer" to have formal charges as close to zero as possible, which indicates the most stable electron distribution. To find the formal charge on carbon in ethylene (C₂H₄), you need to draw the Lewis structure and apply the formal charge formula: Formal charge=valence electrons−nonbonding electrons−12bonding electrons\text{Formal charge} = \text{valence electrons} - \text{nonbonding electrons} - \frac{1}{2}\text{bonding electrons} In ethylene's most stable structure, the two carbon atoms are connected by a double bond, with each carbon bonded to two hydrogen atoms. Each carbon has 4 valence electrons, participates in 8 bonding electrons (4 bonds × 2 electrons each), and has no nonbonding electrons. Using the formula: 4−0−82=04 - 0 - \frac{8}{2} = 0 Therefore, each carbon has a formal charge of 0, making C the correct answer. Option A (-2) would require carbon to have gained significant electron density, which doesn't match ethylene's bonding pattern. Option B (-1) suggests carbon has gained an electron, but the symmetrical double-bond structure doesn't support this charge distribution. Option D (+1) would indicate carbon has lost electron density, which would make the molecule less stable and doesn't reflect the actual electron sharing in the C=C double bond. Remember that formal charges are a bookkeeping tool to track electron distribution. The most stable Lewis structures typically minimize formal charges, with zero being ideal. When multiple resonance structures exist, favor those with formal charges closest to zero and negative formal charges on the most electronegative atoms.

Question 11

A student draws the Lewis structure for the dichromate ion (Cr₂O₇²⁻) and notices that each chromium atom is surrounded by more than 8 electrons. This occurs because chromium can form multiple bonds with oxygen atoms. What principle allows transition metals like chromium to exceed the octet rule?

  1. Transition metals have lower electronegativity than main group elements
  2. Transition metals have d orbitals available for bonding in their valence shell (correct answer)
  3. Transition metals form only ionic bonds, which don't follow the octet rule
  4. Transition metals have more valence electrons than main group elements
  5. Transition metals can only form expanded octets when bonded to oxygen
Explanation: When you encounter questions about atoms exceeding the octet rule, you're dealing with expanded valence shells—a key concept that distinguishes different periods of the periodic table. Chromium can exceed the octet rule because it has d orbitals available for bonding in its valence shell. As a transition metal in the fourth period, chromium's valence shell includes both 4s and 3d orbitals. While the 3d orbitals are slightly lower in energy, they're still accessible for bonding and can accommodate additional electrons beyond the typical eight. In the dichromate ion, chromium forms multiple bonds with oxygen atoms, utilizing these d orbitals to hold more than eight electrons around each chromium center. Let's examine why the other options don't explain this phenomenon. Option A incorrectly suggests electronegativity differences cause octet rule violations—electronegativity affects bond polarity, not electron capacity around atoms. Option C is wrong because transition metals form both ionic and covalent bonds, and the octet rule applies to covalent bonding situations like those in polyatomic ions. Option D misses the mark because having more valence electrons doesn't automatically allow octet rule expansion—it's the availability of suitable orbitals that matters. Remember this pattern: elements in periods 1-2 strictly follow the octet rule because they only have s and p orbitals available, while elements in period 3 and beyond can exceed it due to accessible d orbitals. This orbital availability, not the number of electrons or electronegativity, determines expansion capability.

Question 12

In drawing Lewis structures, a student learns that formal charge calculations help determine the most reasonable structure. For the cyanide ion (CN⁻), what is the formal charge on the nitrogen atom in the most stable Lewis structure?

  1. -2
  2. -1 (correct answer)
  3. 0
  4. +1
  5. +2
Explanation: When you encounter Lewis structure problems involving formal charge, you're applying a key principle: the most stable structure minimizes formal charges and places negative charges on the most electronegative atoms. For the cyanide ion (CN⁻), start by drawing the Lewis structure. Carbon has 4 valence electrons, nitrogen has 5, and there's 1 additional electron from the negative charge, totaling 10 electrons. The most stable structure features a triple bond between carbon and nitrogen: C≡N−\text{C} \equiv \text{N}^- To calculate formal charge, use: Formal charge=valence electrons−nonbonding electrons−12(bonding electrons)\text{Formal charge} = \text{valence electrons} - \text{nonbonding electrons} - \frac{1}{2}(\text{bonding electrons}) For nitrogen: Formal charge=5−2−12(6)=5−2−3=0\text{Formal charge} = 5 - 2 - \frac{1}{2}(6) = 5 - 2 - 3 = 0 Wait—this gives 0, but we need to account for where the overall negative charge resides. In the most stable structure, the negative charge sits on nitrogen (the more electronegative atom), giving it a formal charge of -1. Choice A (-2) would result from incorrectly assigning too many electrons to nitrogen. Choice C (0) comes from forgetting to properly distribute the ion's negative charge. Choice D (+1) would place the negative charge on carbon, which violates the principle that negative charges prefer electronegative atoms. Remember this pattern: in polyatomic ions, negative charges typically reside on the most electronegative atoms, and the most stable Lewis structures minimize formal charges while respecting electronegativity trends.

Question 13

A student draws the Lewis structure for sulfur hexafluoride (SF₆) and counts 12 electrons around the central sulfur atom. The student is confused because this violates the octet rule. Which explanation best justifies this electron count?

  1. The structure is incorrect because no atom can have more than 8 electrons
  2. Sulfur can accommodate 12 electrons because it has d orbitals available for bonding (correct answer)
  3. The extra electrons are actually on the fluorine atoms, not sulfur
  4. The 12 electrons represent ionic bonding, not covalent bonding
  5. The student miscounted; sulfur actually has only 8 electrons around it
Explanation: When you encounter molecules that seem to violate the octet rule, you need to consider whether the central atom can expand its valence shell. This happens when atoms from the third period or below have access to empty d orbitals for bonding. Sulfur hexafluoride (SF₆) is a classic example of expanded valence. Sulfur, being in the third period, has empty 3d orbitals available beyond its 3s and 3p orbitals. When forming six bonds with fluorine atoms, sulfur uses these d orbitals to accommodate 12 electrons (6 bonding pairs) around itself. This expanded octet is perfectly legitimate and explains the molecule's stable, octahedral geometry. Looking at the incorrect options: Option A reflects a common misconception—while atoms in periods 1 and 2 are limited to 8 electrons, third-period and heavier atoms can exceed this limit. Option C is wrong because the 12 electrons are indeed around sulfur as bonding pairs; each S-F bond contributes 2 electrons to sulfur's count. Option D mischaracterizes the bonding—SF₆ involves covalent bonds where electrons are shared, not ionic transfer. The key study tip: Remember that only elements in period 3 and below can expand their octets using d orbitals. Common examples include SF₆, PCl₅, and ClF₃. When you see these molecules on exams, don't automatically assume octet rule violations are errors—check if the central atom can accommodate more than 8 electrons through orbital expansion.

Question 14

A student draws Lewis structures for both carbon monoxide (CO) and nitrogen gas (N₂), both of which contain triple bonds. However, the student notices that CO has a formal charge separation while N₂ does not. What accounts for this difference?

  1. CO has different electronegativity between atoms while N₂ has identical atoms
  2. CO has an odd number of electrons while N₂ has an even number
  3. CO has different numbers of valence electrons on each atom while N₂ has identical atoms (correct answer)
  4. CO requires resonance structures while N₂ does not
  5. CO has a longer bond length than N₂ due to size differences
Explanation: When analyzing formal charges in Lewis structures, you need to understand how valence electrons are distributed among atoms in a molecule. Formal charge equals the number of valence electrons an atom normally has minus the electrons it "owns" in the molecule (lone pairs plus half the bonding electrons). In N₂, both nitrogen atoms start with 5 valence electrons each. When they form a triple bond, the electron distribution is perfectly symmetrical—each nitrogen ends up with the same formal charge of zero because they're identical atoms sharing electrons equally. CO presents a different situation. Carbon brings 4 valence electrons while oxygen brings 6. Even though they form a triple bond, this unequal starting point means you can't distribute the electrons without creating formal charges. The most stable Lewis structure for CO has carbon with a formal charge of -1 and oxygen with +1, giving the characteristic formal charge separation. Looking at the wrong answers: A) focuses on electronegativity differences, but formal charge calculations don't directly involve electronegativity—they're based on electron counting rules. B) is incorrect because both CO and N₂ have even numbers of total electrons (10 each). D) is wrong since neither molecule requires multiple resonance structures to describe their bonding. Remember this pattern: when identical atoms bond (like N₂), formal charges will always be zero due to symmetry. When different atoms bond, especially those with different numbers of valence electrons, formal charge separation often results from the unequal electron distribution needed to form stable bonds.

Question 15

A student draws two possible Lewis structures for the thiocyanate ion (SCN⁻). Structure I has the connectivity S-C-N, while Structure II has the connectivity C-S-N. If both structures satisfy the octet rule and minimize formal charges, which statement best describes these structures?

  1. Only Structure I is correct because sulfur must be the central atom
  2. Only Structure II is correct because carbon has the lowest electronegativity
  3. Structure I is preferred because carbon is more electronegative than sulfur
  4. Structure I is preferred because it places the negative formal charge on the most electronegative atom (correct answer)
  5. Both structures are equally valid resonance forms of the same molecule
Explanation: When evaluating multiple Lewis structures for polyatomic ions, you need to consider both formal charge minimization and electronegativity trends to determine the most stable arrangement. For thiocyanate ion (SCN⁻), let's examine both connectivities. In Structure I (S-C-N), when you calculate formal charges with optimal bonding, the negative charge ends up on nitrogen. In Structure II (C-S-N), the negative charge would be distributed differently. Since nitrogen is the most electronegative atom in this ion (electronegativity: N > C > S), the most stable structure places the negative formal charge on nitrogen, making Structure I preferred. Looking at the wrong answers: Choice A incorrectly assumes sulfur must be central based on position rather than stability - connectivity isn't determined by which atom "should" be central. Choice B misapplies electronegativity reasoning; while carbon does have lower electronegativity than nitrogen, this doesn't make Structure II correct. The key isn't which atom has the lowest electronegativity, but rather where formal charges end up. Choice C contains a factual error - carbon is actually more electronegative than sulfur, not less, and this supports Structure I for the wrong reason. Choice D correctly identifies that Structure I places the negative formal charge on nitrogen, the most electronegative atom, which maximizes stability through favorable charge distribution. Study tip: When comparing Lewis structures, always calculate formal charges and remember that the most stable structure places negative formal charges on the most electronegative atoms and positive charges on the least electronegative atoms.

Question 16

A student draws the Lewis structure for sulfur dioxide (SO₂) and finds that the central sulfur atom has 10 electrons around it. The student is concerned about violating the octet rule. Which of the following best describes the validity of this Lewis structure?

  1. The structure is invalid because the octet rule must never be violated for any atom
  2. The structure is invalid because sulfur can only form two bonds maximum
  3. The structure is valid because sulfur is in period 3 and can have an expanded octet (correct answer)
  4. The structure is valid only if the sulfur atom carries a negative formal charge
  5. The structure is invalid because oxygen atoms must always have exactly 8 electrons
Explanation: When you encounter Lewis structures that seem to violate the octet rule, you need to consider which elements can actually accommodate more than eight electrons. The key is understanding periodic trends and electron capacity. Sulfur is in period 3 of the periodic table, which means it has access to 3d orbitals in addition to its 3s and 3p orbitals. This allows sulfur to accommodate more than eight electrons in its valence shell - a phenomenon called an expanded octet. In SO₂, sulfur commonly forms double bonds with each oxygen atom, resulting in 10 electrons around the central sulfur atom (4 bonding pairs + 1 lone pair). This is perfectly valid and actually represents the most stable structure. Looking at why the other options fail: Option A incorrectly suggests the octet rule is absolute - while it's a useful guideline for period 2 elements, it's frequently exceeded by period 3 and higher elements. Option B is factually wrong since sulfur routinely forms more than two bonds (think SF₆ with six bonds). Option D misunderstands formal charge - the validity of expanded octets doesn't depend on the formal charge being negative. The correct answer is C because sulfur's position in period 3 gives it the orbital capacity to exceed eight electrons. Study tip: Remember that only period 2 elements (B, C, N, O, F) are strictly limited by the octet rule. Elements in period 3 and beyond can have expanded octets due to available d orbitals. When drawing Lewis structures, don't automatically reject structures that exceed eight electrons for these larger atoms.

Question 17

A student draws resonance structures for the formate ion (CHO₂⁻) and finds two equivalent forms. In these resonance structures, what is the average formal charge on each oxygen atom?

  1. -1.0
  2. -0.5 (correct answer)
  3. 0
  4. +0.5
  5. +1.0
Explanation: When you encounter resonance structures, remember that the actual molecule is a hybrid of all possible forms, and properties like formal charge are averaged across all structures. To solve this, you need to draw the two resonance structures for formate ion (CHO₂⁻) and calculate formal charges. The formula for formal charge is: Formal charge=valence electrons−nonbonding electrons−12bonding electrons\text{Formal charge} = \text{valence electrons} - \text{nonbonding electrons} - \frac{1}{2}\text{bonding electrons} In formate's two resonance structures, carbon is bonded to one hydrogen and two oxygens. The key difference is which oxygen carries the double bond. In structure 1, oxygen A has a double bond (formal charge = 0) while oxygen B has a single bond plus an extra electron (formal charge = -1). In structure 2, these roles reverse: oxygen B has formal charge 0, and oxygen A has formal charge -1. Since both structures contribute equally to the resonance hybrid, you average the formal charges: for each oxygen, 0+(−1)2=−0.5\frac{0 + (-1)}{2} = -0.5. This makes (B) -0.5 correct. (A) -1.0 represents the formal charge on each oxygen in individual resonance structures, but ignores the averaging effect. (C) 0 would be the formal charge on the doubly-bonded oxygen in each individual structure, not the average. (D) +0.5 has the wrong sign—oxygen is more electronegative than carbon, so it should carry negative formal charge. Study tip: Always remember that resonance structures are hypothetical—the real molecule has averaged properties. When calculating formal charges in resonance hybrids, draw all structures first, then average the values across equivalent atoms.

Question 18

A student draws the Lewis structure for the phosphate ion (PO₄³⁻) and considers whether to include P=O double bonds. When comparing a structure with all single bonds versus one with one P=O double bond, which statement about formal charges is correct?

  1. All single bonds give the lowest formal charges on all atoms
  2. One double bond reduces the formal charge on phosphorus from +1 to 0
  3. One double bond increases the formal charge on the double-bonded oxygen from -1 to 0 (correct answer)
  4. Both structures have identical total formal charges but different distributions
  5. The double bond structure violates the octet rule for phosphorus
Explanation: When analyzing Lewis structures with different bonding patterns, you need to calculate formal charges to determine which structure is most reasonable. Formal charge equals valence electrons minus nonbonding electrons minus half the bonding electrons. For phosphate ion (PO₄³⁻), let's compare the all-single-bond structure versus one with a P=O double bond. In the all-single-bond structure, phosphorus has formal charge +1 and each oxygen has formal charge -1. When you create one P=O double bond, the double-bonded oxygen gains an additional bond, which changes its formal charge calculation. The oxygen goes from having 3 lone pairs and 1 bond (formal charge = 6 - 6 - 1 = -1) to having 2 lone pairs and 2 bonds (formal charge = 6 - 4 - 2 = 0). This confirms answer C is correct. Answer A is wrong because the all-single-bond structure actually gives phosphorus a +1 formal charge, which isn't the lowest possible. Answer B incorrectly states the phosphorus formal charge change - phosphorus goes from +1 to +2 when forming the double bond, not +1 to 0. Answer D is incorrect because while both structures have the same total formal charge (-3, matching the ion charge), the question asks specifically about individual atom formal charges, not totals. Remember that formal charges help you evaluate Lewis structure quality - structures with formal charges closest to zero on individual atoms are generally preferred, though the total must always equal the overall charge on the species.

Question 19

A student draws the Lewis structure for the hypochlorite ion (ClO⁻) and wants to verify the formal charges. Given that chlorine and oxygen are bonded with a single bond and the remaining electrons are distributed as lone pairs, what are the formal charges on Cl and O respectively?

  1. Cl: 0, O: -1 (correct answer)
  2. Cl: -1, O: 0
  3. Cl: 0, O: 0
  4. Cl: +1, O: -2
  5. Cl: -1, O: -1
Explanation: When you encounter Lewis structure problems involving formal charges, you're working with a systematic way to track electron ownership in molecules and ions. Formal charge helps you determine the most stable electron arrangement. To calculate formal charge, use the formula: Formal charge=Valence electrons−Nonbonding electrons−12Bonding electrons\text{Formal charge} = \text{Valence electrons} - \text{Nonbonding electrons} - \frac{1}{2}\text{Bonding electrons} For the hypochlorite ion (ClO⁻), start by drawing the Lewis structure. Chlorine has 7 valence electrons, oxygen has 6, and the negative charge adds 1 more electron, totaling 14 electrons. With a single Cl-O bond (2 electrons) and the remaining 12 electrons as lone pairs, you get chlorine with 6 lone pair electrons and oxygen with 6 lone pair electrons. For chlorine: 7−6−12(2)=7−6−1=07 - 6 - \frac{1}{2}(2) = 7 - 6 - 1 = 0 For oxygen: 6−6−12(2)=6−6−1=−16 - 6 - \frac{1}{2}(2) = 6 - 6 - 1 = -1 This confirms answer A is correct: Cl has a formal charge of 0, O has -1. Answer B incorrectly assigns the negative charge to chlorine instead of the more electronegative oxygen. Answer C suggests both atoms are neutral, ignoring that the ion's negative charge must reside somewhere. Answer D gives unrealistic charges that don't match the electron distribution in the Lewis structure. Remember: the sum of formal charges must equal the overall charge of the species. Here, 0 + (-1) = -1, matching the ion's charge. Always verify your formal charges add up correctly as a final check.

Question 20

In drawing Lewis structures, a student learns that some molecules have an odd number of valence electrons. For the nitrogen dioxide molecule (NO₂), which has 17 valence electrons total, what is the most reasonable Lewis structure characteristic?

  1. All atoms satisfy the octet rule with no unpaired electrons
  2. The nitrogen atom has one unpaired electron and an incomplete octet
  3. The nitrogen atom has one unpaired electron but still satisfies the octet rule (correct answer)
  4. One oxygen atom must have an unpaired electron to balance the structure
  5. The molecule cannot form a stable Lewis structure due to the odd electron count
Explanation: When you encounter molecules with odd numbers of valence electrons, you're dealing with free radicals - species that must have at least one unpaired electron. The key insight is understanding how electrons can be distributed while still maximizing octet satisfaction. For NO₂ with 17 total valence electrons, let's work through the structure systematically. Nitrogen contributes 5 electrons and each oxygen contributes 6, giving us 5 + 6 + 6 = 17 electrons total. Since this is odd, at least one electron must remain unpaired. The most stable arrangement places the unpaired electron on nitrogen while still giving nitrogen an octet. This occurs when nitrogen forms one double bond and one single bond with the two oxygens. The double-bonded oxygen gets 8 electrons (complete octet), the single-bonded oxygen gets 8 electrons (complete octet with three lone pairs), and nitrogen gets 8 electrons total - but one is unpaired. This makes option C correct. Option A is impossible because 17 electrons cannot all be paired. Option B incorrectly suggests nitrogen has an incomplete octet, but nitrogen actually achieves 8 electrons around it in the optimal structure. Option D places the unpaired electron on oxygen, which creates a less stable structure since nitrogen can better accommodate the unpaired electron due to its central position and ability to form multiple bonds. Remember: when dealing with odd-electron molecules, the unpaired electron typically resides on the least electronegative atom that can still maintain reasonable formal charges and octet satisfaction.