- For the molecule BeH2, 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?
Answer: Delocalization of electron density across the Be-H-Be system reduces individual bond strength compared to localized bonds
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.
- The molecular orbital energy diagram for HF shows significant energy differences between H 1s and F 2p orbitals. If the bonding molecular orbital is 80% F 2p character and 20% H 1s character, while the antibonding orbital is 20% F 2p and 80% H 1s, what can be concluded about the electron density distribution and bond polarity?
Answer: Electron density in the bonding orbital is primarily on F, consistent with H-F bond polarity, while the antibonding orbital is largely H-centered
The 80% F character in the bonding MO means electron density is concentrated on the more electronegative F atom, consistent with bond polarity (δ⁻F—δ⁺H). The antibonding orbital being 80% H character means it's primarily localized on H. This explains both the polar nature and the energy levels. Choice A incorrectly states both orbitals concentrate density on F. Choice B misinterprets antibonding orbital character. Choice D incorrectly suggests equal sharing.
- The molecule B2 has been detected spectroscopically and found to be paramagnetic with a bond order of 1. However, simple molecular orbital theory without considering s−p mixing predicts a different magnetic behavior. What magnetic behavior does the simple theory predict, and why?
Answer: Simple theory predicts diamagnetism because the 6 valence electrons would completely fill the three lowest molecular orbitals
When analyzing diatomic molecules like B2, 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 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∗. Filling the 6 electrons according to this scheme gives: (σ2s)2(σ2s∗)2(σ2pz)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 involves s-p orbital mixing, which changes the energy ordering and places electrons in degenerate π orbitals.
Study tip: Always distinguish between simple MO theory predictions and experimental results - s-p mixing is crucial for light molecules like B2. - In the molecular orbital treatment of 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?
Answer: Bond order = 1; the unequal contributions indicate polar covalent bonding with significant charge separation
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.
- In heterodiatomic molecules like 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?
Answer: The σ* orbital has 70% carbon character and 30% oxygen character, exactly opposite to the bonding orbital
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.
- The molecular orbital diagram for O2 shows that the π2p∗ orbitals are singly occupied, giving O₂ its paramagnetic character. If one electron is added to form O2−, and then a second electron is added to form O22−, how do the bond orders and bond lengths change?
Answer: Bond orders: 2.0 → 1.5 → 1.0; bond lengths increase progressively due to increased antibonding character
O₂ has bond order 2.0 with configuration ending in π₂p*². Adding electrons to O₂⁻ (bond order 1.5) and O₂²⁻ (bond order 1.0) progressively fills the π₂p* antibonding orbitals, decreasing bond order and increasing bond length. Choice B incorrectly suggests bond order increases to 2.0 for O₂²⁻. Choice C reverses the trend entirely. Choice D correctly identifies bond orders but incorrectly suggests bond lengths remain constant.
- Consider a hypothetical molecule 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?
Answer: 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
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.
- A hypothetical diatomic molecule XY has the same number of valence electrons as CO but shows different bonding characteristics. If X has lower ionization energy than C and Y has higher electron affinity than O, how would this affect the molecular orbital energy levels and bond order compared to CO?
Answer: The bonding orbitals would be more Y-centered but bond order would remain the same since electron count is unchanged
Lower ionization energy for X means its orbitals are higher in energy than C's, while higher electron affinity for Y means its orbitals are lower than O's. This increases the energy gap between X and Y orbitals, making bonding MOs more Y-centered (like F in HF). However, since electron count remains 14, the bond order calculation (bonding - antibonding)/2 gives the same result as CO. Choice A incorrectly suggests bond order increases. Choice C focuses on wrong orbital type. Choice D incorrectly suggests both effects are equal.
- The molecular orbital diagram for F2 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?
Answer: Extensive occupation of antibonding π* orbitals nearly cancels the bonding contribution from filled π orbitals
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, the molecular orbital diagram shows that while the bond order is indeed 1 (calculated as 21(8 bonding−6 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. - Consider the isoelectronic series: N2, CO, and 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⁺?
Answer: Differences in effective nuclear charge cause varying degrees of orbital contraction and overlap efficiency
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+, 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 has moderate nuclear charge with two nitrogen atoms (Z=7 each), while 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+ 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. - Consider the molecular orbital configurations of NO and 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?
Answer: ΔBO = -0.5 with NO stronger; NO⁻ has an additional electron in an antibonding π* orbital compared to NO
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. 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∗ antibonding orbital: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2π2px2π2py2π2py∗1. This gives 8 bonding electrons and 4 antibonding electrons, so BO = (8-4)/2 = 2.0.
The difference is ΔBO = 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 ΔBO = 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. - A student claims that Be2 cannot exist because its molecular orbital configuration results in a bond order of zero. However, if s−p mixing is considered, how might this conclusion change?
Answer: Bond order remains zero because s-p mixing affects orbital energies but not the total number of bonding and antibonding electrons
When analyzing diatomic molecules like Be2, you need to consider how s-p mixing affects molecular orbital energy levels and electron occupancy patterns. Without s-p mixing, Be2 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 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. - 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?
Answer: The electron was removed from a π antibonding orbital, which was singly occupied in the neutral molecule
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.
- Consider the molecular orbital diagram for O22−. If the 2px and 2py orbitals are degenerate and lower in energy than the 2pz orbital, what is the bond order and magnetic behavior of this species?
Answer: Bond order = 1, diamagnetic with complete pairing in both bonding and antibonding orbitals
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−, you start with 16 electrons (8 from each oxygen plus 2 extra from the 2- charge). Given that 2px and 2py orbitals are lower in energy than 2pz, the molecular orbital filling order is: σ2s<σ2s∗<π2px=π2py<σ2pz<π2px∗=π2py∗<σ2pz∗
The electron configuration becomes: σ2s2σ2s∗2π2px2π2py2σ2pz2π2px∗2π2py∗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∗ 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. - Consider the isoelectronic series C22−, BN, and 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?
Answer: Molecular orbital energies and mixing change across the series due to different nuclear charges, but electron configurations remain similar
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−, BN, and 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−, the similar atomic orbitals create well-mixed molecular orbitals with minimal energy gaps. As you move to BN and then 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. - In the molecular orbital treatment of CO, the large electronegativity difference between C and O causes significant mixing of the 2s and 2pz orbitals. How does this s−p mixing affect the bond order calculation compared to a homonuclear diatomic with the same number of electrons?
Answer: Bond order remains unchanged because s-p mixing affects bonding and antibonding orbitals equally
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−p mixing, where the 2s and 2pz 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−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 21(bonding electrons−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∗ 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. - The CN− ion has the same number of valence electrons as CO but exhibits different chemical reactivity. Both have bond order 3, but CN− is a much stronger nucleophile than CO. In terms of molecular orbital theory, what best explains this difference in nucleophilic behavior?
Answer: CN⁻ has higher energy occupied orbitals due to the negative charge, making electrons more readily available for donation
When comparing molecules with similar bonding but different reactivity, molecular orbital theory helps explain how electron availability affects nucleophilic behavior. Both CN− and 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− carries a negative charge, which raises the energy levels of all its molecular orbitals compared to neutral CO. Higher energy occupied orbitals mean the electrons are less tightly bound and more readily available for donation to electrophiles. This makes CN− a much stronger nucleophile than 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− to 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. - When constructing the molecular orbital diagram for NO+, which of the following statements correctly describes the relationship between bond order and orbital occupancy?
Answer: Bond order equals 3.0 because removal of an electron from antibonding π* increases bonding by 0.5 relative to NO
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+, you remove this antibonding electron, leaving 10 electrons total. Using the bond order formula: Bond order=2bonding electrons−antibonding electrons, you get 28−0=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+ 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+.
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. - In the molecular orbital treatment of NO, which has 11 valence electrons, the unpaired electron occupies a π* antibonding orbital. If NO forms NO− by gaining an electron, and then dimerizes to form N2O2, how does the bond order change from NO to the N-O bonds in the dimer?
Answer: Bond order decreases from 2.5 to 2.0 because electron pairing in antibonding orbitals weakens individual N-O bonds
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.
- Consider the molecular orbital diagram for N2+. If the 2pz atomic orbitals combine to form σ2p and σ2p∗ molecular orbitals, and the 2px and 2py orbitals form π2p and π2p∗ molecular orbitals, what is the bond order and magnetic behavior of this ion?
Answer: Bond order = 2.5, paramagnetic with one unpaired electron
N₂⁺ has 13 total electrons. The MO configuration is: σ₁s² σ₁s*² σ₂s² σ₂s*² π₂p⁴ σ₂p¹. Bond order = (8 bonding - 3 antibonding)/2 = 2.5. The unpaired electron in σ₂p makes it paramagnetic. Choice B incorrectly calculates bond order as 2.0. Choice C uses the bond order of neutral N₂ (3.0). Choice D incorrectly assumes diamagnetic behavior and wrong bond order.
