Physical Chemistry 2 Quiz: Mo Theory Bonding And Bond Order
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Mo Theory Bonding And Bond OrderQuestion 1 of 20

When constructing the molecular orbital diagram for NO+\text{NO}^+, which of the following statements correctly describes the relationship between bond order and orbital occupancy?

Bond order equals 2.5 because the unpaired electron occupies a bonding orbital, contributing 0.5 to the total
Bond order equals 3.0 because removal of an electron from antibonding π* increases bonding by 0.5 relative to NO
Bond order equals 2.0 because the 10 valence electrons fill exactly the bonding orbitals with no antibonding occupation
Bond order equals 2.5 because one antibonding π* orbital remains half-filled after electron removal from NO
Bond order equals 3.0 because the electron configuration matches that of N₂ with identical orbital filling
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Mo Theory Bonding And Bond Order

Practice Mo Theory Bonding And Bond Order in Physical Chemistry 2 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 Mo Theory Bonding And Bond Order, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

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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.

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Question 1

When constructing the molecular orbital diagram for NO+\text{NO}^+, which of the following statements correctly describes the relationship between bond order and orbital occupancy?

  1. Bond order equals 2.5 because the unpaired electron occupies a bonding orbital, contributing 0.5 to the total
  2. Bond order equals 3.0 because removal of an electron from antibonding π* increases bonding by 0.5 relative to NO (correct answer)
  3. Bond order equals 2.0 because the 10 valence electrons fill exactly the bonding orbitals with no antibonding occupation
  4. Bond order equals 2.5 because one antibonding π* orbital remains half-filled after electron removal from NO
  5. Bond order equals 3.0 because the electron configuration matches that of N₂ with identical orbital filling
Explanation: When tackling molecular orbital problems involving isoelectronic species and ion formation, you need to carefully track electron removal and its impact on bond order calculations. To solve this, start with NO, which has 11 valence electrons. The molecular orbital filling order places the "extra" electron in an antibonding π* orbital. When forming NO+\text{NO}^+, you remove this antibonding electron, leaving 10 electrons total. Using the bond order formula: Bond order=bonding electrons−antibonding electrons2\text{Bond order} = \frac{\text{bonding electrons} - \text{antibonding electrons}}{2}, you get 8−02=3.0\frac{8-0}{2} = 3.0. Since removing an electron from an antibonding orbital increases bond order by 0.5 compared to neutral NO (which has bond order 2.5), NO+\text{NO}^+ indeed has bond order 3.0. Choice A incorrectly suggests the removed electron was in a bonding orbital and uses faulty logic about unpaired electrons contributing fractional bond orders. Choice C correctly identifies 10 valence electrons filling bonding orbitals but miscalculates the bond order as 2.0 instead of 3.0. Choice D makes the fundamental error of claiming an antibonding orbital remains half-filled after electron removal, when actually all antibonding orbitals are empty in NO+\text{NO}^+. Remember this pattern: when an electron is removed from an antibonding orbital during ionization, bond order increases by 0.5. Always identify which orbital loses the electron first, then recalculate bond order from the new electron configuration rather than trying to adjust the neutral molecule's value.

Question 2

A hypothetical diatomic molecule XY\text{XY} has 12 valence electrons and exhibits a bond order of 2. If one electron is promoted from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO), what can be concluded about the new bond order and the nature of the HOMO and LUMO?

  1. New bond order is 1; HOMO was bonding and LUMO is antibonding, both contributing equally to bond order change (correct answer)
  2. New bond order is 1; HOMO was antibonding and LUMO is bonding, indicating an unusual electronic configuration
  3. New bond order is 3; HOMO was antibonding and LUMO is bonding, strengthening the bond upon promotion
  4. New bond order is 2; HOMO and LUMO are both nonbonding orbitals, so promotion doesn't affect bonding
  5. New bond order is 0; promotion creates equal numbers of bonding and antibonding electrons
Explanation: When you encounter molecular orbital problems involving electron promotion, you need to systematically analyze the initial configuration, then determine how moving an electron affects bond order. For a 12-electron diatomic molecule with bond order 2, you can deduce the molecular orbital filling. Using the general MO energy sequence, 12 electrons would fill through the σ2pz\sigma_{2p_z} orbital, with the HOMO being a bonding orbital and the LUMO being the π2p∗\pi_{2p}^* antibonding orbital. The initial bond order calculation gives: Bond order=8 bonding−4 antibonding2=2\text{Bond order} = \frac{8 \text{ bonding} - 4 \text{ antibonding}}{2} = 2 When you promote one electron from the HOMO (bonding) to the LUMO (antibonding), you lose one bonding electron and gain one antibonding electron. This changes the bond order to: Bond order=7 bonding−5 antibonding2=1\text{Bond order} = \frac{7 \text{ bonding} - 5 \text{ antibonding}}{2} = 1 Choice A correctly identifies that the new bond order is 1, with the HOMO being bonding and LUMO being antibonding, and both contribute equally to the bond order change (each changes by 1). Choice B incorrectly suggests the HOMO was antibonding, which contradicts the typical MO filling pattern for 12 electrons. Choice C incorrectly claims the bond order increases to 3, which would require the opposite effect of what actually happens. Choice D wrongly states that both orbitals are nonbonding, which would mean no bond order change. Remember: electron promotion from bonding to antibonding orbitals always decreases bond order by 1, since you simultaneously lose bonding character and gain antibonding character.

Question 3

Consider the series C22−\text{C}_2^{2-}, C2−\text{C}_2^{-}, C2\text{C}_2, C2+\text{C}_2^{+}, and C22+\text{C}_2^{2+}. Which statement correctly describes the trend in bond order across this series?

  1. Bond order increases monotonically from 1 to 3 as electrons are removed from antibonding orbitals
  2. Bond order decreases from 3 to 1 because electrons are progressively removed from bonding π orbitals
  3. Bond order varies as 1, 1.5, 2, 1.5, 1 due to initial removal from antibonding orbitals, then from bonding orbitals (correct answer)
  4. Bond order remains constant at 2 because electron removal affects only nonbonding orbitals
  5. Bond order follows the pattern 3, 2.5, 2, 1.5, 1 as electrons are sequentially removed from bonding orbitals
Explanation: When analyzing bond order trends across isoelectronic series, you need to systematically track electron configuration changes using molecular orbital theory. For homonuclear diatomic molecules like C₂, electrons fill orbitals in order: σ₂s, σ₂s, σ₂pz, π₂px = π₂py, π₂px = π₂py, σ₂pz. Let's trace the electron configurations. C₂²⁻ has 14 electrons: σ₂s² σ₂s² σ₂pz² π₂px² π₂py² π₂px¹ π*₂py¹, giving bond order = (8-6)/2 = 1. C₂⁻ has 13 electrons (removes one antibonding electron): bond order = (8-5)/2 = 1.5. Neutral C₂ has 12 electrons (removes another antibonding electron): bond order = (8-4)/2 = 2. Here's the key insight: once antibonding orbitals are emptied, further electron removal comes from bonding orbitals. C₂⁺ has 11 electrons (removes from bonding π orbital): bond order = (7-4)/2 = 1.5. C₂²⁺ has 10 electrons: bond order = (6-4)/2 = 1. This gives the pattern 1 → 1.5 → 2 → 1.5 → 1, confirming answer C. Answer A incorrectly assumes electrons only come from antibonding orbitals and predicts monotonic increase. Answer B wrongly suggests electrons are removed from bonding orbitals throughout, predicting monotonic decrease. Answer D incorrectly claims bond order stays constant—there are no nonbonding orbitals in this system. Study tip: Always draw out the MO diagram and track which type of orbital (bonding vs. antibonding) loses electrons at each step. The transition from removing antibonding to bonding electrons creates the characteristic "peak" in bond order.

Question 4

The molecule BN\text{BN} is isoelectronic with C2\text{C}_2 but exhibits different bonding characteristics due to the polar nature of the B-N bond. If the nitrogen 2p2p orbitals are significantly lower in energy than the boron 2p2p orbitals, how does this affect the molecular orbital description compared to C2\text{C}_2?

  1. Bond order increases because the energy mismatch creates more effective orbital overlap between B and N
  2. Bond order decreases because poor orbital energy matching reduces the degree of covalent bonding character
  3. Bond order remains the same but orbital coefficients become unequal, with greater N character in bonding orbitals (correct answer)
  4. Bond order becomes non-integer because the energy mismatch prevents proper pairing of electrons in molecular orbitals
  5. Bond order increases because electron density shifts toward nitrogen, strengthening the overall molecular stability
Explanation: When analyzing heteronuclear diatomic molecules like BN, you need to consider how orbital energy mismatches affect molecular orbital formation compared to homonuclear molecules like C₂. Since BN is isoelectronic with C₂, both molecules have the same number of electrons (10 total) and will fill the same molecular orbitals in the same order. This means the bond order calculation remains identical: both molecules have a bond order of 2, with electrons filling σ₂s, σ₂s, π₂p, and σ₂p orbitals while leaving π₂p orbitals empty. However, the energy mismatch between nitrogen's lower-energy 2p orbitals and boron's higher-energy 2p orbitals creates unequal mixing. When atomic orbitals of different energies combine, the resulting molecular orbitals have unequal coefficients. The bonding orbitals will have greater character from the lower-energy nitrogen orbitals, while antibonding orbitals will have greater boron character. This creates the polar B-N bond while maintaining the same overall electron count and bond order. Option A incorrectly suggests bond order increases—energy mismatch actually reduces overlap efficiency, though not enough to change bond order. Option B wrongly claims bond order decreases; while covalent character is reduced, the electron configuration and formal bond order remain the same. Option D misunderstands molecular orbital theory—energy mismatches don't prevent electron pairing or create non-integer bond orders. Remember: for isoelectronic molecules, focus on how orbital energy differences affect electron distribution and polarity rather than changing the fundamental bonding framework.

Question 5

In the molecular ion O2+\text{O}_2^{+}, which has been observed in the upper atmosphere, the unpaired electron occupies a π∗\pi^* antibonding orbital. If this unpaired electron were to be excited to the σ∗(2pz)\sigma^*(2p_z) orbital, what would be the effect on bond order and bond length?

  1. Bond order increases by 0.5 and bond length decreases because σ* has less antibonding character than π*
  2. Bond order decreases by 0.5 and bond length increases because σ* is higher energy and more antibonding
  3. Bond order remains unchanged but bond length increases due to the stronger antibonding character of σ* (correct answer)
  4. Bond order remains unchanged and bond length stays constant because both orbitals contribute equally to bonding
  5. Bond order increases by 1.0 and bond length decreases significantly because σ* has partial bonding character
Explanation: When analyzing electron transitions in molecular orbitals, you need to consider both bond order changes and the relative antibonding character of different orbital types. Let's examine what happens when the unpaired electron in O2+\text{O}_2^{+} moves from π∗\pi^* to σ∗(2pz)\sigma^*(2p_z). Since both orbitals are antibonding and we're moving one electron from one antibonding orbital to another, the bond order calculation remains the same: Bond order=12(bonding electrons−antibonding electrons)\text{Bond order} = \frac{1}{2}(\text{bonding electrons} - \text{antibonding electrons}). The total number of antibonding electrons doesn't change, so bond order stays constant. However, bond length will increase because σ∗\sigma^* orbitals have stronger antibonding character than π∗\pi^* orbitals. This occurs because σ\sigma interactions involve direct head-to-head orbital overlap along the internuclear axis, creating more effective bonding (and correspondingly more effective antibonding when filled). The electron in σ∗(2pz)\sigma^*(2p_z) thus weakens the bond more than when it occupied π∗\pi^*. Answer A incorrectly suggests σ∗\sigma^* has less antibonding character—the opposite is true. Answer B wrongly claims bond order decreases; moving one antibonding electron to another antibonding orbital doesn't change the electron count difference. Answer D fails to recognize that different orbital types have different antibonding strengths. Study tip: Remember that σ\sigma interactions (bonding or antibonding) are always stronger than π\pi interactions due to better orbital overlap along the bond axis. This principle helps predict relative bond-weakening effects.

Question 6

A student claims that Be2\text{Be}_2 cannot exist because its molecular orbital configuration results in a bond order of zero. However, if s−ps-p mixing is considered, how might this conclusion change?

  1. Bond order becomes positive because s-p mixing creates new bonding orbitals that can accommodate the valence electrons
  2. Bond order remains zero because s-p mixing affects orbital energies but not the total number of bonding and antibonding electrons (correct answer)
  3. Bond order becomes negative because s-p mixing increases antibonding character in the occupied orbitals
  4. Bond order becomes positive because s-p mixing destabilizes the antibonding σ*(2s) orbital, making it unoccupied
  5. Bond order remains zero but the molecule becomes more stable due to orbital energy lowering from s-p mixing
Explanation: When analyzing diatomic molecules like Be2\text{Be}_2, you need to consider how s-p mixing affects molecular orbital energy levels and electron occupancy patterns. Without s-p mixing, Be2\text{Be}_2 has 4 valence electrons filling the σ(2s) and σ*(2s) orbitals equally, giving a bond order of zero. S-p mixing occurs when s and p atomic orbitals are close enough in energy to interact significantly. This mixing changes the relative energies of molecular orbitals but crucially doesn't change the total number of electrons or create entirely new orbitals. The 4 valence electrons in Be2\text{Be}_2 still occupy the same relative bonding and antibonding positions, just in orbitals with modified energies due to mixing. The bond order calculation (bonding electrons minus antibonding electrons, divided by 2) remains zero because the electron distribution between bonding and antibonding character is unchanged. Choice A incorrectly suggests s-p mixing creates new orbitals that can hold additional electrons - mixing only modifies existing orbital energies. Choice C misunderstands the effect, as s-p mixing doesn't make bond order negative. Choice D incorrectly claims the σ*(2s) orbital becomes unoccupied due to destabilization, but energy changes from mixing aren't dramatic enough to completely empty occupied orbitals. The correct answer is B because s-p mixing affects orbital shapes and energies but doesn't fundamentally alter the bonding/antibonding electron count that determines bond order. Remember: s-p mixing modifies orbital energies and shapes, but electron counting for bond order follows the same principles regardless of mixing effects.

Question 7

The CN−\text{CN}^- ion has the same number of valence electrons as CO\text{CO} but exhibits different chemical reactivity. Both have bond order 3, but CN−\text{CN}^- is a much stronger nucleophile than CO\text{CO}. In terms of molecular orbital theory, what best explains this difference in nucleophilic behavior?

  1. CN⁻ has higher energy occupied orbitals due to the negative charge, making electrons more readily available for donation (correct answer)
  2. The formal charge distribution in CN⁻ places more electron density on carbon, making it more nucleophilic than the carbon in CO
  3. CN⁻ has a different orbital energy ordering than CO due to the charge, creating more accessible lone pair electrons
  4. The extra electron in CN⁻ compared to a hypothetical CN species increases the overall electron density available for nucleophilic attack
  5. CN⁻ has less stable bonding orbitals than CO, making its electrons more reactive and readily donated to electrophiles
Explanation: When comparing molecules with similar bonding but different reactivity, molecular orbital theory helps explain how electron availability affects nucleophilic behavior. Both CN−\text{CN}^- and CO\text{CO} are isoelectronic with 14 valence electrons and bond order 3, but their charge differences create crucial distinctions in orbital energies. The key insight is that CN−\text{CN}^- carries a negative charge, which raises the energy levels of all its molecular orbitals compared to neutral CO\text{CO}. Higher energy occupied orbitals mean the electrons are less tightly bound and more readily available for donation to electrophiles. This makes CN−\text{CN}^- a much stronger nucleophile than CO\text{CO}, despite their similar bonding patterns. Answer A correctly identifies this orbital energy effect. The negative charge elevates the occupied orbital energies, making electron donation more favorable. Answer B incorrectly focuses on formal charge distribution rather than orbital energies. While electron density matters, it's the orbital energy that determines electron availability for nucleophilic attack. Answer C mentions orbital energy ordering changes, but this is misleading. The ordering doesn't fundamentally change; rather, all orbital energies are elevated by the negative charge. Answer D creates confusion by discussing an "extra electron compared to hypothetical CN." This comparison isn't relevant since we're comparing CN−\text{CN}^- to CO\text{CO}, not to CN. Remember: when analyzing nucleophilicity differences between isoelectronic species, focus on how charge affects orbital energies rather than just electron count or formal charges.

Question 8

Consider a hypothetical molecule AB\text{AB} where atom A has much lower ionization energy than atom B. In the molecular orbital diagram, how would this difference affect the relative energies and electron occupancy of the bonding and antibonding orbitals?

  1. Bonding orbitals will be closer in energy to atom A's levels and have more A character; antibonding orbitals will be closer to B's levels
  2. Bonding orbitals will be closer in energy to atom B's levels and have more B character; antibonding orbitals will be closer to A's levels (correct answer)
  3. Both bonding and antibonding orbitals will be closer to atom A's energy levels due to its lower ionization energy
  4. The energy difference will prevent molecular orbital formation, resulting in purely ionic bonding between A and B
  5. Orbital energies will be averaged between A and B levels, with equal contributions from both atoms in all molecular orbitals
Explanation: When analyzing molecular orbital formation between atoms with different ionization energies, you need to consider how electrons distribute based on atomic orbital energy differences. Lower ionization energy means an atom's valence electrons are less tightly bound and its atomic orbitals are higher in energy. In molecule AB, atom A has lower ionization energy than B, meaning A's atomic orbitals are higher in energy than B's. When these atomic orbitals combine to form molecular orbitals, electrons preferentially occupy the lowest available energy levels. The bonding molecular orbital forms at an energy closer to the lower-energy atomic orbitals (atom B's), and since electrons spend more time near the lower-energy atom, the bonding orbital has greater B character. Conversely, the antibonding orbital forms at higher energy, closer to atom A's atomic orbital levels. Answer B correctly describes this relationship: bonding orbitals are closer to B's energy levels with more B character, while antibonding orbitals are closer to A's levels. Answer A reverses this relationship incorrectly. Answer C incorrectly suggests both molecular orbitals would be closer to A's levels—this ignores that bonding orbitals must be lower in energy than the constituent atomic orbitals. Answer D incorrectly assumes that energy differences prevent molecular orbital formation; while large differences can lead to more ionic character, molecular orbitals can still form between atoms with different ionization energies. Remember: in molecular orbital theory, bonding orbitals always have greater character from the atom with lower-energy (more stable) atomic orbitals, regardless of ionization energy differences.

Question 9

In the molecular orbital treatment of HF\text{HF}, the large electronegativity difference between H and F results in molecular orbitals that are heavily polarized. If the bonding σ orbital has 85% F character and 15% H character, what is the most likely bond order and what does this suggest about the nature of bonding?

  1. Bond order = 1; the high F character indicates the bonding is primarily ionic with minimal covalent contribution
  2. Bond order = 1; the unequal contributions indicate polar covalent bonding with significant charge separation (correct answer)
  3. Bond order = 0.5; the extreme polarization reduces the effective overlap and weakens the covalent bond strength
  4. Bond order = 1.5; the strong electronegativity difference creates additional electrostatic bonding beyond the covalent component
  5. Bond order = 1; the molecular orbital description is equivalent to complete electron transfer from H to F
Explanation: When analyzing molecular orbitals in heteronuclear diatomic molecules like HF, you need to consider how electronegativity differences affect orbital character and bonding nature, while remembering that bond order depends on the number of bonding vs. antibonding electrons. The 85% F character and 15% H character in the bonding σ orbital reflects the large electronegativity difference (F = 4.0, H = 2.1). This creates a polar covalent bond where the bonding electrons spend more time near the fluorine atom, creating partial charges (δ⁻ on F, δ⁺ on H). Since HF has two electrons in the bonding orbital and none in antibonding orbitals, the bond order is 1. This polarization doesn't eliminate covalent character—it's still orbital overlap creating the bond, just with unequal electron sharing. Choice A incorrectly suggests the bonding is primarily ionic. While highly polarized, HF still forms through orbital overlap, making it polar covalent rather than ionic. Choice C misunderstands that polarization doesn't reduce bond order—you still have the same number of bonding electrons. The unequal orbital contributions don't change the fundamental electron count. Choice D incorrectly suggests bond order exceeds 1. Electronegativity differences create polarity but don't add extra bonding beyond what the orbital analysis predicts. Remember: bond order comes from electron counting (bonding minus antibonding electrons, divided by 2), while orbital character percentages tell you about polarity. High electronegativity differences create polar covalent bonds, not ionic bonds or altered bond orders.

Question 10

Consider the isoelectronic series C22−\text{C}_2^{2-}, BN\text{BN}, and BeO\text{BeO}, all having 12 valence electrons. Despite having the same electron count, these species have different bond characters. Which statement best describes how molecular orbital theory accounts for these differences?

  1. All three have identical molecular orbital diagrams and bond orders, with differences arising only from nuclear charges affecting bond lengths
  2. Bond orders remain constant at 2, but increasing electronegativity differences create progressively more ionic character in the series
  3. Molecular orbital energies and mixing change across the series due to different nuclear charges, but electron configurations remain similar (correct answer)
  4. The increasing polarity causes different orbital occupations, leading to different bond orders: 1, 2, and 3 respectively
  5. All have bond order 2, but the molecular orbitals become increasingly localized on the more electronegative atom across the series
Explanation: When you encounter isoelectronic species with different bonding characteristics, molecular orbital theory reveals how nuclear charge differences create distinct electronic environments despite identical electron counts. For C22−\text{C}_2^{2-}, BN\text{BN}, and BeO\text{BeO}, the key insight is that increasing nuclear charges across this series (C < N < O and Be < B < C) dramatically affect molecular orbital energies and s-p mixing. In C22−\text{C}_2^{2-}, the similar atomic orbitals create well-mixed molecular orbitals with minimal energy gaps. As you move to BN\text{BN} and then BeO\text{BeO}, the increasing electronegativity differences cause the atomic orbitals to have increasingly different energies, reducing orbital mixing and creating more polarized molecular orbitals. However, the 12 valence electrons still occupy analogous molecular orbitals in similar patterns. Option A is wrong because while bond orders might be similar, the molecular orbital diagrams definitely change due to different atomic orbital energies. Option B incorrectly assumes constant bond orders of 2 and oversimplifies the orbital changes to just ionic character. Option D makes a fundamental error by claiming different electron configurations lead to bond orders of 1, 2, and 3 - this contradicts the isoelectronic nature and basic MO filling principles. Option C correctly identifies that nuclear charge differences alter orbital energies and mixing while maintaining similar (not identical) electron configurations in analogous molecular orbitals. Remember: isoelectronic doesn't mean identical bonding - nuclear charges create the crucial differences in orbital interactions.

Question 11

In the molecular orbital treatment of CO\text{CO}, the large electronegativity difference between C and O causes significant mixing of the 2s2s and 2pz2p_z orbitals. How does this s−ps-p mixing affect the bond order calculation compared to a homonuclear diatomic with the same number of electrons?

  1. Bond order increases because s-p mixing stabilizes bonding orbitals more than antibonding orbitals
  2. Bond order decreases because s-p mixing creates additional antibonding character in nominally bonding orbitals
  3. Bond order remains unchanged because s-p mixing affects bonding and antibonding orbitals equally (correct answer)
  4. Bond order increases because s-p mixing reduces the antibonding character of the σ* orbital
  5. Bond order becomes fractional because s-p mixing creates partial bonding character in all orbitals
Explanation: When analyzing molecular orbital theory for heteronuclear diatomics like CO, you need to understand how electronegativity differences affect orbital mixing and energy levels, not just count electrons in orbitals. In CO, the large electronegativity difference between carbon and oxygen causes the atomic orbitals to have significantly different energies before bonding. This creates extensive s−ps-p mixing, where the 2s2s and 2pz2p_z orbitals hybridize more than they would in a homonuclear molecule. However, this mixing affects the overall bonding picture in a balanced way. The key insight is that s−ps-p mixing redistributes electron density and orbital character, but it doesn't fundamentally change the number of bonding versus antibonding electrons. The mixing stabilizes some orbitals while destabilizing others in a compensatory manner. Since bond order is calculated as 12(bonding electrons−antibonding electrons)\frac{1}{2}(\text{bonding electrons} - \text{antibonding electrons}), and the total electron count remains the same, the bond order stays essentially unchanged compared to an isoelectronic homonuclear molecule. Choice A is wrong because while mixing does affect orbital energies, it doesn't preferentially stabilize bonding over antibonding orbitals. Choice B incorrectly suggests that mixing creates net antibonding character - the effects are balanced. Choice D misrepresents the specific effect on the σ∗\sigma^* orbital and oversimplifies the overall impact. Remember: in MO theory, focus on the net effect of orbital interactions. Electronegativity differences change orbital shapes and energies, but bond order calculations depend on the balance of bonding versus antibonding electrons, which mixing preserves.

Question 12

The molecule B2\text{B}_2 has been detected spectroscopically and found to be paramagnetic with a bond order of 1. However, simple molecular orbital theory without considering s−ps-p mixing predicts a different magnetic behavior. What magnetic behavior does the simple theory predict, and why?

  1. Simple theory predicts diamagnetism because the 6 valence electrons would completely fill the three lowest molecular orbitals (correct answer)
  2. Simple theory predicts stronger paramagnetism with 4 unpaired electrons because all orbitals remain singly occupied
  3. Simple theory predicts diamagnetism because electrons pair in bonding orbitals before occupying antibonding orbitals
  4. Simple theory predicts no magnetic moment because B₂ would have bond order zero and not exist as a stable molecule
  5. Simple theory predicts the same paramagnetism but with different orbital occupancy involving π* rather than π orbitals
Explanation: When analyzing diatomic molecules like B2\text{B}_2, you need to understand how molecular orbital theory predicts electron configuration and magnetic properties. The key is comparing simple MO theory (without s-p mixing) to experimental observations. Simple molecular orbital theory for B2\text{B}_2 starts with 6 valence electrons (3 from each boron atom). Without considering s-p mixing, the energy order of molecular orbitals is: σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗<σ2pz∗\sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*. Filling the 6 electrons according to this scheme gives: (σ2s)2(σ2s∗)2(σ2pz)2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2. All electrons are paired, predicting diamagnetic behavior with bond order 1. This matches answer choice A perfectly. However, answer choice B incorrectly suggests 4 unpaired electrons, which would require electrons to remain unpaired across multiple orbitals - this violates proper electron filling rules. Answer choice C mentions the correct pairing principle but incorrectly implies this leads to the wrong prediction when simple theory actually does predict diamagnetism. Answer choice D is wrong because simple MO theory does predict a stable molecule with bond order 1, not zero. The discrepancy between theory and experiment (paramagnetic with 2 unpaired electrons) arises because real B2\text{B}_2 involves s-p orbital mixing, which changes the energy ordering and places electrons in degenerate π\pi orbitals. Study tip: Always distinguish between simple MO theory predictions and experimental results - s-p mixing is crucial for light molecules like B2\text{B}_2.

Question 13

Consider the molecular orbital configurations of NO\text{NO} and NO−\text{NO}^-. Both species have unpaired electrons, but in different types of orbitals. What is the difference in bond orders, and what does this suggest about the relative bond strengths?

  1. Δ\DeltaBO = 0.5 with NO⁻ stronger; the unpaired electron in NO⁻ occupies a bonding orbital while NO has unpaired electrons in antibonding orbitals
  2. Δ\DeltaBO = -0.5 with NO stronger; NO⁻ has an additional electron in an antibonding π* orbital compared to NO (correct answer)
  3. Δ\DeltaBO = 1.0 with NO⁻ much stronger; electron addition converts an antibonding electron pair to a bonding configuration
  4. Δ\DeltaBO = 0 with equal bond strengths; both molecules have the same number of net bonding electrons despite different configurations
  5. Δ\DeltaBO = -1.0 with NO much stronger; NO⁻ has additional antibonding character that significantly weakens the bond
Explanation: When analyzing molecular orbital configurations and bond orders, you need to systematically count bonding and antibonding electrons, then apply the bond order formula: BO = (bonding electrons - antibonding electrons)/2. For NO (11 total electrons), the MO configuration is: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2π2px2π2py1\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1. This gives 8 bonding electrons and 3 antibonding electrons, so BO = (8-3)/2 = 2.5. For NO⁻ (12 total electrons), you add one electron to the next available orbital, which is the π2py∗\pi_{2p_y}^* antibonding orbital: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2π2px2π2py2π2py∗1\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_y}^{*1}. This gives 8 bonding electrons and 4 antibonding electrons, so BO = (8-4)/2 = 2.0. The difference is Δ\DeltaBO = 2.0 - 2.5 = -0.5, making NO stronger than NO⁻. Answer B correctly identifies this relationship. Answer A is wrong because it claims NO⁻ is stronger and misidentifies which orbitals contain unpaired electrons. Answer C incorrectly calculates Δ\DeltaBO = 1.0 and the wrong strength relationship. Answer D falsely claims equal bond orders and strengths. Remember: when electrons are added to molecules, they occupy the lowest available energy orbital. If that's an antibonding orbital, the bond order decreases, weakening the bond. Always work through MO diagrams systematically rather than making assumptions about electron placement.

Question 14

A diatomic molecule has a bond order of 2.5 and exhibits paramagnetic behavior. If the molecule undergoes a one-electron oxidation, the resulting cation has a bond order of 3.0 and is diamagnetic. Which molecular orbital was the electron removed from, and what type of orbital is it?

  1. The electron was removed from a σ bonding orbital, which must have been singly occupied to account for paramagnetism
  2. The electron was removed from a π antibonding orbital, which was singly occupied in the neutral molecule (correct answer)
  3. The electron was removed from a σ antibonding orbital, which had unpaired electrons causing the original paramagnetism
  4. The electron was removed from a π bonding orbital, and the increase in bond order indicates this orbital had antibonding character
  5. The electron was removed from a nonbonding orbital, which explains why bond order increased despite losing an electron
Explanation: When you encounter molecular orbital questions involving bond order changes and magnetic properties, focus on how electron removal affects both bonding and unpaired electrons. Let's analyze what happens here. The neutral molecule has a bond order of 2.5 and is paramagnetic (has unpaired electrons). After losing one electron, the cation has a bond order of 3.0 and becomes diamagnetic (all electrons paired). The key insight is that removing an electron increased the bond order by 0.5, which only happens when you remove an electron from an antibonding orbital. The electron must have been removed from a π* antibonding orbital that was singly occupied. Removing this electron eliminates the unpaired electron (explaining the shift from paramagnetic to diamagnetic) while simultaneously reducing antibonding character, which increases the net bond order from 2.5 to 3.0. Option A is wrong because removing an electron from a bonding orbital would decrease bond order, not increase it. Option C incorrectly identifies the orbital type as σ* rather than π*, and while it correctly recognizes an antibonding orbital, σ* orbitals are typically higher in energy than π* orbitals in this context. Option D makes no sense—it claims a bonding orbital has antibonding character, which is contradictory. Remember this pattern: when electron removal increases bond order, the electron came from an antibonding orbital. When magnetic behavior changes from paramagnetic to diamagnetic upon oxidation, you've removed an unpaired electron. These clues together point directly to a singly-occupied antibonding orbital.

Question 15

In heterodiatomic molecules like CO\text{CO}, the molecular orbitals have unequal contributions from the two atomic orbitals. If the bonding σ orbital has 70% oxygen character and 30% carbon character, what does this suggest about the antibonding σ* orbital composition?

  1. The σ* orbital has 70% carbon character and 30% oxygen character, exactly opposite to the bonding orbital (correct answer)
  2. The σ* orbital has 30% carbon character and 70% oxygen character, maintaining the same ratio as the bonding orbital
  3. The σ* orbital has equal 50% contributions from both atoms because antibonding orbitals are always symmetric
  4. The σ* orbital composition depends on the energy difference between the atomic orbitals and cannot be determined from bonding orbital data
  5. The σ* orbital has 70% carbon character and 30% oxygen character, with the percentages representing antibonding contributions
Explanation: When analyzing molecular orbitals in heterodiatomic molecules, you need to understand how bonding and antibonding orbitals relate through the principle of orbital conservation and symmetry properties. In molecular orbital theory, when two atomic orbitals combine, they must form two molecular orbitals: one bonding and one antibonding. The key insight is that these orbitals have inverted compositions - if the bonding orbital has greater character from one atom, the antibonding orbital will have greater character from the other atom. This occurs because the antibonding orbital must maintain orthogonality to the bonding orbital while conserving the total orbital contributions. Since the bonding σ orbital has 70% oxygen character and 30% carbon character, the antibonding σ* orbital will have the inverted composition: 70% carbon character and 30% oxygen character. This inversion is a fundamental consequence of how atomic orbitals mix when they have different energies. Answer B incorrectly suggests the same ratio is maintained - this would violate orbital orthogonality requirements. Answer C wrongly assumes antibonding orbitals are always symmetric; this only occurs in homodiatomic molecules where atoms are identical. Answer D is incorrect because the antibonding orbital composition is directly related to the bonding orbital composition through the inversion principle. Study tip: Remember that in heterodiatomic molecules, bonding and antibonding orbital compositions are mirror images of each other. The atom with less character in the bonding orbital will dominate the antibonding orbital, and vice versa.

Question 16

Consider the molecular orbital diagram for O22−\text{O}_2^{2-}. If the 2px2p_x and 2py2p_y orbitals are degenerate and lower in energy than the 2pz2p_z orbital, what is the bond order and magnetic behavior of this species?

  1. Bond order = 1, diamagnetic with all electrons paired in bonding orbitals
  2. Bond order = 2, paramagnetic with two unpaired electrons in antibonding orbitals
  3. Bond order = 1, paramagnetic with unpaired electrons in degenerate antibonding orbitals
  4. Bond order = 2, diamagnetic with complete filling of bonding orbitals only
  5. Bond order = 1, diamagnetic with complete pairing in both bonding and antibonding orbitals (correct answer)
Explanation: When analyzing molecular orbital diagrams, you need to systematically fill orbitals according to energy levels and apply Hund's rule, then calculate bond order using the formula: (bonding electrons - antibonding electrons)/2. For O22−\text{O}_2^{2-}, you start with 16 electrons (8 from each oxygen plus 2 extra from the 2- charge). Given that 2px2p_x and 2py2p_y orbitals are lower in energy than 2pz2p_z, the molecular orbital filling order is: σ2s<σ2s∗<π2px=π2py<σ2pz<π2px∗=π2py∗<σ2pz∗\sigma_{2s} < \sigma_{2s}^* < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^* The electron configuration becomes: σ2s2σ2s∗2π2px2π2py2σ2pz2π2px∗2π2py∗2\sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*2} This gives 10 bonding electrons and 6 antibonding electrons, yielding a bond order of (10-6)/2 = 2. Since all orbitals are completely filled with paired electrons, the species is diamagnetic. Answer A is incorrect because the bond order is 2, not 1. Answer B wrongly predicts paramagnetism and unpaired electrons in antibonding orbitals. Answer C incorrectly states both the bond order (1 instead of 2) and magnetic behavior (paramagnetic instead of diamagnetic). Answer D correctly identifies the bond order and magnetic behavior but incorrectly suggests only bonding orbitals are filled—the antibonding π∗\pi^* orbitals are also completely filled. Remember: always count total electrons carefully (including charge), fill orbitals systematically, and check for unpaired electrons to determine magnetic properties.

Question 17

The molecular orbital diagram for F2\text{F}_2 shows that this molecule has a relatively weak bond despite having a bond order of 1. Which aspect of the molecular orbital configuration best explains this apparent contradiction?

  1. The high electronegativity of fluorine creates strong electrostatic repulsion that opposes the covalent bonding
  2. Extensive occupation of antibonding π* orbitals nearly cancels the bonding contribution from filled π orbitals (correct answer)
  3. The small size of fluorine atoms leads to poor orbital overlap and ineffective molecular orbital formation
  4. Strong lone pair-lone pair repulsion between fluorine atoms destabilizes the molecule despite the covalent bond
  5. The σ bonding orbital has significant antibonding character due to s-p mixing in fluorine's molecular orbitals
Explanation: When analyzing molecular orbital diagrams, you need to consider not just bond order, but also the relative energies and occupations of all orbitals to understand bond strength. For F2\text{F}_2, the molecular orbital diagram shows that while the bond order is indeed 1 (calculated as 12(8 bonding−6 antibonding)=1\frac{1}{2}(8 \text{ bonding} - 6 \text{ antibonding}) = 1), the molecule has extensive occupation of antibonding π* orbitals. These high-energy antibonding orbitals contain four electrons that actively oppose the bonding interactions from the filled π orbitals. The antibonding electrons nearly cancel out the stabilizing effect of the bonding electrons, resulting in a weak overall bond despite the positive bond order. Option A incorrectly focuses on electronegativity effects. While fluorine is highly electronegative, this doesn't create the type of electrostatic repulsion described that would weaken covalent bonding. Option C is wrong because fluorine's small size actually promotes good orbital overlap—the issue isn't poor orbital formation but rather the population of antibonding orbitals. Option D misidentifies the problem as lone pair repulsion, but the weakness comes from the electronic structure within the molecular orbitals themselves, not from nonbonding electron interactions. Remember that bond order alone doesn't tell the full story about bond strength. Always examine the energy levels and populations of both bonding and antibonding orbitals. When antibonding orbitals are significantly populated (especially π* orbitals), they can dramatically weaken bonds even when bond order suggests moderate strength.

Question 18

Consider the isoelectronic series: N2\text{N}_2, CO\text{CO}, and NO+\text{NO}^+. Although all have 14 electrons, their bond lengths differ significantly. Using molecular orbital theory, which factor primarily accounts for the observed bond length order: CO > N₂ > NO⁺?

  1. Electronegativity differences in CO cause polarization that weakens the effective bonding orbital overlap
  2. Asymmetric charge distribution in heteronuclear molecules reduces orbital overlap compared to homonuclear species
  3. The formal charges in NO⁺ create stronger electrostatic attraction between nuclei than in neutral molecules
  4. Differences in effective nuclear charge cause varying degrees of orbital contraction and overlap efficiency (correct answer)
Explanation: When you encounter isoelectronic species with different bond lengths, the key is understanding how nuclear charge affects orbital overlap and bonding strength. All three molecules have 14 electrons in identical molecular orbital configurations, but their nuclei differ significantly. In NO+\text{NO}^+, you have a nitrogen (Z=7) and oxygen (Z=8) nucleus with one electron removed, creating high effective nuclear charge. This strongly contracts the atomic orbitals, leading to better overlap and shorter, stronger bonds. N2\text{N}_2 has moderate nuclear charge with two nitrogen atoms (Z=7 each), while CO\text{CO} has the lowest effective nuclear charge affecting bonding orbitals due to the formal charge distribution (C carries partial positive charge, O carries partial negative charge), resulting in less contracted orbitals and longer bonds. This explains the bond length order: CO > N₂ > NO⁺. Option A incorrectly focuses on electronegativity causing bond weakening through polarization, but electronegativity differences don't directly weaken orbital overlap. Option B suggests heteronuclear molecules have inherently weaker bonding, which isn't necessarily true - the nuclear charges matter more than symmetry. Option C misidentifies the mechanism by focusing on electrostatic attraction between nuclei rather than orbital contraction effects, and incorrectly suggests formal charges create stronger attraction in NO+\text{NO}^+ compared to neutral molecules. Remember: for isoelectronic series, effective nuclear charge determines orbital size and overlap efficiency. Higher nuclear charge means more contracted orbitals, better overlap, and shorter bonds.

Question 19

For the molecule BeH2\text{BeH}_2, molecular orbital theory predicts linear geometry with Be 2s and 2p orbitals mixing with H 1s orbitals. If the bond order between Be and each H is calculated to be 1.0, but experimental evidence shows Be-H bonds are weaker than expected for single bonds, which MO theory concept best explains this discrepancy?

  1. The Be 2s orbital is too low in energy to effectively overlap with H 1s orbitals, reducing actual bonding efficiency
  2. Significant ionic character reduces the covalent bond strength despite the calculated bond order of 1.0
  3. The linear geometry forces unfavorable orbital overlap angles that weaken the Be-H interactions
  4. Delocalization of electron density across the Be-H-Be system reduces individual bond strength compared to localized bonds (correct answer)
Explanation: In BeH₂, the molecular orbitals are delocalized across the entire Be-H-H-Be system, not localized between individual Be-H pairs. This delocalization means the electron density contributing to bonding is spread over the entire molecule, making individual Be-H bonds weaker than typical localized single bonds despite the calculated bond order. Choice A incorrectly focuses on energy mismatch. Choice B overemphasizes ionic character. Choice C incorrectly blames geometry for orbital overlap issues.

Question 20

In the molecular orbital treatment of NO\text{NO}, which has 11 valence electrons, the unpaired electron occupies a π* antibonding orbital. If NO forms NO−\text{NO}^- by gaining an electron, and then dimerizes to form N2O2\text{N}_2\text{O}_2, how does the bond order change from NO to the N-O bonds in the dimer?

  1. Bond order increases from 2.5 to 3.0 due to elimination of antibonding character upon dimerization
  2. Bond order decreases from 2.5 to 2.0 because electron pairing in antibonding orbitals weakens individual N-O bonds (correct answer)
  3. Bond order remains 2.5 since dimerization doesn't affect individual N-O molecular orbital configurations
  4. Bond order decreases from 2.5 to 1.5 due to formation of N-N bonding that withdraws electron density
Explanation: NO has bond order 2.5 with one unpaired electron in π*. When NO⁻ forms (bond order 2.0), the additional electron pairs the π* orbital, reducing bond order. In N₂O₂ dimer, each N-O unit essentially maintains the NO⁻ character with paired electrons in antibonding orbitals, giving bond order ≈ 2.0. Choice A incorrectly suggests bond order increases. Choice C ignores the effect of electron addition. Choice D overcorrects the bond order reduction.