Heteronuclear Diatomics & MO Diagrams - Physical Chemistry 2

1. The photoelectron spectrum of gaseous BF shows a peak at 11.0 eV corresponding to ionization from the HOMO. Molecular orbital calculations indicate that this orbital has 78% fluorine 2p character and 22% boron 2p character. If the isolated F atom has a 2p ionization energy of 17.4 eV and isolated B atom has a 2p ionization energy of 8.3 eV, what can be concluded about the bonding in BF?
   Answer: The HOMO is a bonding orbital, and its ionization energy is lowered relative to the F 2p value due to mixing with the lower-energy B 2p orbital The ionization energy (11.0 eV) is significantly lower than the F 2p value (17.4 eV) but higher than B 2p (8.3 eV). This indicates a bonding molecular orbital where the F 2p orbital is stabilized through bonding interaction with B 2p. The 78% F character explains why it's closer to the B value - the orbital energy reflects the stabilization of F 2p through bonding. A weighted average would give ~15.4 eV, much higher than observed. Choice A is incorrect about weighted averages. Choice C is wrong because this ionization energy indicates bonding stabilization, not non-bonding behavior. Choice D incorrectly identifies this as antibonding.
2. The molecular orbital diagram for carbon monoxide (CO) shows an interesting feature where the 4σ orbital energy lies above the 1π orbital energy, unlike in many homonuclear diatomics. What is the primary reason for this orbital energy ordering?
   Answer: The different 2s-2p mixing in C versus O creates asymmetric orbital contributions in the 4σ When analyzing molecular orbital diagrams for heteronuclear diatomics like CO, you need to consider how the different atomic orbital energies of each atom affect the resulting molecular orbitals. Unlike homonuclear molecules where both atoms contribute equally, heteronuclear systems create asymmetric mixing patterns. The unusual 4σ above 1π ordering in CO occurs because carbon and oxygen have significantly different 2s-2p energy gaps. Carbon's 2s and 2p orbitals are closer in energy than oxygen's, leading to different amounts of s-p mixing (hybridization) in each atom. When the molecular orbitals form, this asymmetric mixing creates unequal contributions from each atom in the 4σ orbital, destabilizing it relative to the 1π orbitals, which don't experience this same asymmetric s-p mixing effect. Choice A incorrectly focuses on electronegativity affecting π versus σ stabilization, but electronegativity primarily influences which atom's character dominates the orbital, not the relative σ/π ordering. Choice B suggests the energy gap prevents effective σ formation entirely, but σ orbitals do form successfully—they're just positioned higher than expected. Choice C reverses the mixing description and doesn't explain why π orbitals remain relatively unaffected. The correct answer is D because it captures the essence: different s-p mixing behavior in carbon versus oxygen creates asymmetric orbital contributions specifically in the σ system, leading to the unusual energy ordering. Study tip: For heteronuclear MO diagrams, always consider how different atomic orbital energy gaps between the two atoms will create asymmetric mixing patterns that can invert expected orbital orderings.
3. The molecular orbital correlation diagram for LiF shows a dramatic change in orbital character compared to molecules with smaller electronegativity differences. What is the most significant consequence of the large electronegativity difference (Li: 1.0, F: 4.0) on the molecular orbital description?
   Answer: The bonding orbitals become purely fluorine-based, approaching the ionic limit of Li⁺F⁻ When analyzing molecular orbital diagrams, the key factor driving orbital character is the electronegativity difference between atoms. Large electronegativity differences create highly polarized bonds that approach ionic character. In LiF, the massive electronegativity difference ($\Delta\chi = 3.0$) means fluorine attracts electrons much more strongly than lithium. This creates molecular orbitals that are heavily weighted toward the more electronegative atom. The bonding orbitals become predominantly fluorine-based because fluorine's atomic orbitals are much lower in energy than lithium's. As the electronegativity difference increases, the molecular orbital description approaches the ionic limit where electrons are essentially localized on the fluorine atom, giving us Li⁺F⁻. This makes A correct. B is wrong because large electronegativity differences don't create degeneracy—they actually split orbital energies further apart due to the energy mismatch between atomic orbitals. C incorrectly suggests an impossible scenario where antibonding orbitals drop below bonding orbitals, which would violate fundamental molecular orbital theory. D contradicts the actual effect—large electronegativity differences reduce orbital mixing because the atomic orbital energies become too mismatched to interact effectively. Study tip: Remember that electronegativity difference is the deciding factor in bond character. Small differences (< 0.5) give covalent bonds with balanced orbital mixing, moderate differences (0.5-1.7) give polar covalent bonds, and large differences (> 1.7) approach ionic character with heavily unequal electron distribution.
4. The molecular orbital diagram for nitric oxide (NO) predicts it to be paramagnetic with one unpaired electron. However, solid NO exhibits temperature-dependent magnetic behavior. What molecular orbital consideration best explains this observation?
   Answer: NO molecules dimerize to form N₂O₂ with paired electrons in the solid state When you encounter questions about magnetic properties that change with temperature or phase, think about how molecular interactions can alter electronic structure from the isolated molecule prediction. Gaseous NO is indeed paramagnetic with one unpaired electron in its $2\pi^*$ orbital, as predicted by molecular orbital theory. However, in the solid state, NO molecules have a strong tendency to dimerize, forming N₂O₂ dimers. During dimerization, the unpaired electrons from two NO molecules pair up through orbital overlap, creating diamagnetic N₂O₂ units. This explains why solid NO shows temperature-dependent magnetic behavior - at lower temperatures, more dimers form, reducing the paramagnetic character. Option A incorrectly suggests crystal field effects cause spin-pairing within individual NO molecules. Crystal field theory applies to transition metal complexes, not simple diatomic molecules, and wouldn't cause intramolecular electron pairing. Option C proposes thermal population of the $2\pi^*$ orbital at low temperatures, which contradicts basic thermodynamics - higher energy orbitals are populated at higher temperatures, not lower ones. Option D vaguely mentions "different electronic configuration" but doesn't specify the mechanism. While intermolecular interactions are involved, this answer lacks the precision of explaining dimerization and electron pairing. Study tip: When molecular magnetic properties differ between gas and solid phases, always consider dimerization or polymerization as potential explanations. Many odd-electron molecules (NO, NO₂, ClO₂) form dimers in condensed phases to achieve electron pairing and greater stability.
5. The molecular orbital diagram for the hypothetical molecule MgO must account for the involvement of Mg 3s orbitals and the large electronegativity difference (Mg: 1.2, O: 3.5). Which factor most significantly determines the bonding character in this molecule?
   Answer: The electronegativity difference of 2.3 units places MgO firmly in the ionic bonding regime When analyzing bonding in compounds, you need to consider both orbital overlap potential and electronegativity differences to determine whether bonding is primarily covalent or ionic. The electronegativity difference serves as a crucial predictor of bonding character. With magnesium (electronegativity 1.2) and oxygen (3.5), the difference is $3.5 - 1.2 = 2.3$ units. This large difference means electrons are strongly pulled toward oxygen, creating significant charge separation. Generally, electronegativity differences greater than 1.7-2.0 indicate predominantly ionic bonding, where electron transfer rather than sharing dominates. MgO forms through Mg losing its 3s electrons to become Mg²⁺, while oxygen gains electrons to become O²⁻, resulting in electrostatic attraction between ions. Option A is incorrect because orbital size differences don't prevent molecular orbital formation entirely—they just affect the degree of overlap. Option C misrepresents the situation: while Mg 3s and O 2p orbitals can interact energetically, the huge electronegativity difference overwhelms any covalent character. Option D contains a fundamental error—Mg 3s orbitals aren't filled in bonding (Mg loses these electrons), and even filled orbitals can participate in bonding interactions. The electronegativity difference of 2.3 units in option B correctly identifies MgO as predominantly ionic, which matches experimental evidence showing MgO has a typical ionic crystal structure and properties. Study tip: For bonding character, use electronegativity differences as your primary guide: <0.5 = nonpolar covalent, 0.5-1.7 = polar covalent, >1.7 = predominantly ionic. This simple rule predicts bonding type more reliably than orbital considerations alone.
6. In heteronuclear diatomics, the mixing of s and p orbitals on each atom can lead to different hybridization effects. For the molecule BN, which has the same number of valence electrons as C₂, how does the electronegativity difference affect the orbital mixing compared to C₂?
   Answer: B and N atoms undergo different amounts of s-p mixing, creating asymmetric molecular orbitals When analyzing heteronuclear diatomics, you need to consider how electronegativity differences affect orbital mixing patterns. Unlike homonuclear molecules where atoms contribute equally to molecular orbitals, heteronuclear molecules create asymmetric bonding situations. In BN, boron (electronegativity ~2.0) and nitrogen (electronegativity ~3.0) have significantly different electron-attracting abilities. This difference means that when their atomic orbitals combine to form molecular orbitals, the s and p orbitals on each atom don't mix in the same way or to the same extent. Boron's orbitals will mix differently than nitrogen's orbitals because they experience different electronic environments. The result is asymmetric molecular orbitals where the electron density is not evenly distributed between the two atoms, creating polar character in the bonds. Option A is incorrect because electronegativity differences don't eliminate s-p mixing entirely - they just make it asymmetric. BN still forms covalent bonds, unlike the weak interactions in Ne₂. Option C incorrectly suggests the mixing is enhanced equally on both atoms, but electronegativity differences inherently create unequal environments. Option D misses the key point that being isoelectronic (same number of electrons) doesn't mean identical orbital mixing - the nuclear charges and electronegativities still matter significantly. Remember this pattern: isoelectronic doesn't mean identical behavior. When comparing hetero- and homonuclear diatomics with the same electron count, always consider how electronegativity differences create asymmetric orbital contributions and polar character in the heteronuclear case.
7. For the heteronuclear diatomic molecule lithium hydride (LiH), the molecular orbital diagram shows an unusual feature compared to typical heteronuclear diatomics. Given that Li has a much lower electronegativity (1.0) than H (2.1), what is the most likely explanation for the observed charge distribution?
   Answer: The bonding orbital is primarily hydrogen-based, but orbital size effects result in Li⁻-H⁺ polarity When analyzing heteronuclear diatomic molecules, you need to consider both electronegativity differences and orbital size effects, which can sometimes work in opposite directions. In LiH, hydrogen is more electronegative than lithium, so you might expect the electron density to shift toward hydrogen. However, the key insight is understanding how orbital overlap actually occurs. Lithium's valence orbital (2s) is much larger and more diffuse than hydrogen's 1s orbital. When these orbitals combine to form the bonding molecular orbital, the smaller, more compact hydrogen 1s orbital contributes more significantly to the electron density in the bonding region between the nuclei. This creates a counterintuitive situation: even though the bonding orbital is primarily hydrogen-based (meaning hydrogen contributes more to the orbital character), the actual charge distribution results in Li⁻-H⁺ polarity. This occurs because the lithium atom effectively "donates" its electron to the hydrogen-dominated bonding region, creating partial negative charge on lithium and partial positive charge on hydrogen. Answer A incorrectly assumes electronegativity alone determines polarity direction. Answer B correctly identifies the polarity but wrongly attributes it to lithium-based bonding character. Answer C is incorrect because LiH is definitely polar due to the significant electronegativity difference and orbital size mismatch. Remember that in heteronuclear molecules, orbital size effects can override simple electronegativity predictions. Always consider both factors when predicting molecular orbital character and resulting charge distributions, especially with atoms from different periods.
8. In constructing the MO diagram for the hypothetical molecule AlF, the energy levels are: Al 3s (-11.3 eV), Al 3p (-6.0 eV), F 2s (-40.1 eV), and F 2p (-18.6 eV). The large energy gaps prevent significant s-p mixing on each atom, but cross-atom interactions still occur. If this molecule has 8 valence electrons, which molecular orbitals are occupied and what is the predicted dipole moment direction?
   Answer: Occupied: σ(F2s), σ(Al3s-F2p), non-bonding F2p; dipole points from Al to F due to bonding orbital polarization toward F and additional electron density in non-bonding F orbitals When approaching molecular orbital diagrams for heteronuclear diatomic molecules, you need to consider both the energy matching of atomic orbitals and how electrons fill the resulting molecular orbitals based on energy ordering. For AlF with 8 valence electrons, start by identifying which atomic orbitals can interact. The F 2s orbital (-40.1 eV) is too low in energy to mix significantly with Al orbitals, so it remains essentially non-bonding. The best energy match occurs between Al 3s (-11.3 eV) and F 2p (-18.6 eV), creating bonding and antibonding combinations. The Al 3p orbitals (-6.0 eV) are much higher than F 2p, so they participate minimally in bonding. Filling the 8 valence electrons in order of increasing energy gives: 2 electrons in the non-bonding F 2s, 2 electrons in the σ(Al3s-F2p) bonding orbital, and 4 electrons in non-bonding F 2p orbitals. This matches answer A. Answer B incorrectly suggests π bonding between Al 3p and F 2p, but the large energy gap prevents effective overlap. Answer C places electrons in non-bonding Al 3s, but this orbital actually hybridizes with F 2p to form the primary bonding interaction. Answer D omits the non-bonding F 2p orbitals, which must be filled before higher-energy antibonding orbitals. The dipole points from Al to F because the bonding orbital is polarized toward the more electronegative fluorine, and additional electron density resides in non-bonding F orbitals. Remember: energy matching determines orbital interactions, and electron density distribution on the more electronegative atom creates the dipole direction.
9. In the molecular orbital diagram for hydrogen chloride (HCl), the large energy difference between H 1s (-13.6 eV) and Cl 3p (-13.0 eV) orbitals is smaller than might be expected. What is the primary consequence of this relatively small energy gap on the molecular orbital formation?
   Answer: Both bonding and antibonding orbitals show significant mixing of H 1s and Cl 3p character When you encounter molecular orbital questions involving energy gaps between atomic orbitals, the key principle is that orbital mixing depends on how close the atomic orbitals are in energy. The closer they are, the more they can mix and interact. In HCl, the H 1s orbital (-13.6 eV) and Cl 3p orbital (-13.0 eV) have only a 0.6 eV energy difference, which is relatively small. This small energy gap allows for significant orbital overlap and mixing when the molecular orbitals form. When atomic orbitals are close in energy, both contribute substantially to both the resulting bonding and antibonding molecular orbitals. The bonding MO will have contributions from both H 1s and Cl 3p (though weighted toward the lower-energy Cl 3p), and the antibonding MO will also contain both characters (weighted toward the higher-energy H 1s). Option A is incorrect because even though Cl 3p is slightly lower in energy, the small gap means H 1s still contributes significantly to the bonding orbital. Option B fails because the antibonding orbital isn't purely H 1s—it retains substantial Cl 3p character due to the effective mixing. Option D contradicts the fundamental principle: small energy gaps actually promote orbital mixing, not prevent it. Large energy gaps would prevent mixing and potentially create nonbonding orbitals. Remember this pattern: small energy differences between atomic orbitals lead to strong mixing and hybrid molecular orbitals, while large energy differences lead to minimal mixing and orbitals that retain more of their original atomic character.
10. When comparing the molecular orbital diagrams of BF and AlF, both having the same number of valence electrons, which statement best describes the difference in their bonding characteristics?
    Answer: AlF has weaker bonding due to poor overlap between Al 3p and F 2p orbitals When analyzing molecular orbital diagrams and bonding in heteronuclear diatomics, you need to consider how orbital energy matching and size compatibility affect bond formation between different atoms. However, there appears to be an error in this question setup. The correct answer is listed as "E," but only four options (A-D) are provided. This suggests either a formatting issue or missing option E. Looking at the available choices: Option A correctly identifies that BF would have stronger bonding due to better orbital overlap. The B 2p and F 2p orbitals are much closer in size and energy compared to Al 3p and F 2p orbitals, leading to more effective molecular orbital formation and stronger covalent bonding. Option B incorrectly assumes that larger electronegativity differences always create stronger bonds. While Al-F has a larger electronegativity difference, this creates more ionic character but doesn't necessarily mean stronger overall bonding when considering the poor orbital overlap between the size-mismatched Al 3p and F 2p orbitals. Option C is wrong because identical electron counts don't guarantee identical bonding. The different principal quantum numbers (n=2 for B vs n=3 for Al) create significantly different orbital sizes and energies. Option D incorrectly focuses on s-p separation within atoms rather than the critical issue of orbital matching between different atoms in the molecule. Study tip: In heteronuclear diatomic molecules, effective bonding requires both energy similarity and size compatibility between atomic orbitals. Size mismatch often outweighs electronegativity differences in determining bond strength.
11. The molecular orbital diagram for hydrogen bromide (HBr) differs from that of HCl due to the involvement of Br 4p orbitals instead of Cl 3p orbitals. Which property is most significantly affected by this change in atomic orbitals?
    Answer: The overlap population decreases significantly due to the larger size of Br 4p orbitals When analyzing molecular orbital diagrams for hydrogen halides, the key factor to consider is how orbital size affects overlap efficiency. As you move down the periodic table from chlorine to bromine, the valence orbitals become significantly larger and more diffuse. The overlap population - a measure of how effectively two atomic orbitals combine to form molecular orbitals - depends critically on orbital size compatibility. The hydrogen 1s orbital is small and compact. When it interacts with Cl 3p orbitals in HCl, there's reasonable size matching that allows for effective overlap. However, when the same H 1s orbital attempts to overlap with the much larger Br 4p orbitals in HBr, the size mismatch becomes pronounced. The larger, more diffuse Br 4p orbitals have lower electron density at the bonding region, resulting in significantly weaker overlap and a decreased overlap population. Option A is incorrect because bond order remains 1.0 in both molecules - each forms a single covalent bond. Option B is wrong because increased orbital size actually makes the bonding orbital less polarized toward bromine, not more. The larger orbital spreads electron density over a greater volume. Option D incorrectly suggests antibonding orbitals become lower in energy with poor size matching - actually, poor overlap affects both bonding and antibonding orbitals similarly, and antibonding orbitals remain higher in energy. Remember this pattern: as atomic orbitals increase in size down a group, their overlap with smaller orbitals (like H 1s) becomes progressively less effective, weakening the resulting bonds.
12. In the molecular orbital treatment of hydrogen fluoride (HF), the overlap between H 1s and F 2p orbitals creates bonding and antibonding combinations. If the H 1s orbital energy is -13.6 eV and the F 2p orbital energy is -18.6 eV, which factor most significantly affects the relative contributions of atomic orbitals to the resulting molecular orbitals?
    Answer: The energy difference causes the bonding MO to be primarily F 2p in character with minimal H 1s contribution When atomic orbitals combine to form molecular orbitals, the energy difference between the participating orbitals determines how much each contributes to the resulting MOs. This is governed by the principle that orbitals of similar energy mix more effectively than those with large energy differences. In HF, the F 2p orbital at -18.6 eV is 5.0 eV lower (more stable) than the H 1s orbital at -13.6 eV. When orbitals with different energies combine, the bonding MO energy lies closer to the lower-energy atomic orbital, while the antibonding MO energy lies closer to the higher-energy atomic orbital. More importantly, the orbital contributions follow this energy pattern: the bonding MO has greater character from the lower-energy orbital (F 2p), while the antibonding MO has greater character from the higher-energy orbital (H 1s). Choice A incorrectly suggests equal mixing occurs despite the energy difference. Equal contributions only happen when atomic orbitals have identical or very similar energies. Choice C correctly identifies that the antibonding MO is primarily H 1s in character, but this alone doesn't address what the question asks about most significant factors. Choice D ignores the energy difference entirely, incorrectly invoking symmetry as the determining factor. Choice B correctly captures the key principle: the significant energy difference causes the bonding MO to be primarily F 2p in character. The lower-energy F 2p orbital dominates the bonding combination because it's energetically more favorable. Remember: in MO theory, energy differences drive orbital mixing—the closer in energy, the more equal the contributions. Large energy gaps create unequal mixing with the bonding MO resembling the lower-energy atomic orbital.
13. Consider the molecular orbital energy level diagram for ClF, where the energy difference between Cl 3p (-13.0 eV) and F 2p (-18.6 eV) orbitals is moderate. The diagram shows both σ and π molecular orbitals formed from p orbital interactions. With 14 valence electrons total, what is the predicted magnetic behavior and the primary character of the LUMO (lowest unoccupied molecular orbital)?
    Answer: ClF is diamagnetic with all electrons paired; the LUMO has primarily Cl 3p character in a π* antibonding orbital When analyzing heteronuclear diatomic molecules like ClF, you need to construct molecular orbital diagrams considering the energy differences between atomic orbitals and count electrons to determine magnetic properties and orbital occupancy. ClF has 14 valence electrons (7 from Cl + 7 from F). The moderate energy difference between Cl 3p (-13.0 eV) and F 2p (-18.6 eV) means significant orbital mixing occurs, but the lower-energy F 2p orbitals contribute more to bonding MOs while higher-energy Cl 3p orbitals contribute more to antibonding MOs. Filling the molecular orbitals in order of increasing energy: the bonding σ, two bonding π, two antibonding π*, and finally the antibonding σ* orbitals. With 14 electrons, you fill through the π* orbitals completely, leaving the σ* orbital empty. Since all electrons are paired in this configuration, ClF is diamagnetic. The LUMO is the σ* antibonding orbital, which has primarily Cl 3p character due to the energy mismatch. Answer A incorrectly predicts paramagnetism - there are no unpaired electrons with 14 total electrons in this orbital arrangement. Answer B correctly identifies diamagnetic behavior but wrongly suggests mixed character in the LUMO; the energy difference ensures the σ* orbital is predominantly Cl-based. Answer D incorrectly predicts both paramagnetism and places the LUMO as a π* orbital, but the π* orbitals are filled before σ*. Remember: in heteronuclear molecules, antibonding orbitals have greater character from the higher-energy atomic orbital contributor, while bonding orbitals favor the lower-energy atom.
14. Consider the isoelectronic series CO, N₂⁺, and NO⁺, all with 10 valence electrons. Based on their molecular orbital diagrams, which statement correctly describes the trend in bond properties across this series?
    Answer: Bond strength increases in the order NO⁺ < CO < N₂⁺ due to increasing covalent character When analyzing isoelectronic molecules, you need to consider how nuclear charge affects electron distribution and bonding, even when electron count is identical. These three molecules all have 10 valence electrons, but their nuclear environments differ significantly. The correct answer is B because bond strength correlates with the effective nuclear charge and orbital overlap. N₂⁺ has the highest total nuclear charge (14), followed by CO (13), then NO⁺ (12). Higher nuclear charge pulls bonding electrons closer, creating stronger, more covalent bonds. The molecular orbital filling shows that while all have similar bonding patterns, the increased nuclear charge in N₂⁺ leads to greater orbital contraction and stronger bonds. Option A incorrectly relates bond length to nuclear charge in the wrong direction. Higher nuclear charge actually decreases bond length by pulling atoms closer together. Option C misses a crucial point: while these molecules have the same number of electrons, their molecular orbital energies differ due to varying nuclear charges, leading to different effective bond orders. The formal bond order calculation may be similar, but the actual bond strength varies. Option D confuses the relationship between vibrational frequency and reduced mass. While reduced mass does affect vibrational frequency, the primary factor here is bond strength - stronger bonds vibrate at higher frequencies. CO actually has one of the highest vibrational frequencies among diatomic molecules. Remember: isoelectronic doesn't mean identical properties. Always consider how nuclear charge affects electron density and bonding strength, even when electron count is the same.
15. The molecular orbital diagram for CN⁻ shows that this ion is isoelectronic with N₂. However, the energy ordering of the molecular orbitals differs from N₂ due to the heteronuclear nature and formal charges. Given that CN⁻ has a calculated bond order of 3, but the C-N bond length (1.17 Å) is slightly longer than the N-N bond in N₂ (1.10 Å), what is the most likely explanation for this discrepancy?
    Answer: The difference in atomic sizes between C and N creates less effective orbital overlap compared to the identical atoms in N₂, weakening the bonds despite equal bond order Although CN⁻ and N₂ both have bond order 3, the heteronuclear nature of CN⁻ means the atomic orbitals have different sizes and energies, leading to less effective overlap compared to the perfectly matched orbitals in homonuclear N₂. This results in weaker bonds and longer bond length despite identical formal bond order. Choice A incorrectly invokes electrostatic repulsion. Choice B is wrong because both have similar MO filling. Choice C incorrectly attributes the effect to electron-electron repulsion from formal charges rather than orbital overlap efficiency.
16. For the heteronuclear diatomic molecule CO, the molecular orbital energy diagram shows that the 2s and 2p atomic orbitals mix significantly. If the electronegativity difference between C and O causes the oxygen 2s orbital to be 15.6 eV lower in energy than the carbon 2s orbital, and the oxygen 2p orbitals are 2.4 eV lower than the carbon 2p orbitals, which statement best describes the electron density distribution in the $\sigma_{2s}$ bonding molecular orbital?
    Answer: The electron density is primarily localized on the oxygen atom due to its lower energy 2s orbital In heteronuclear diatomics, the bonding molecular orbital has greater contribution from the lower energy atomic orbital. Since oxygen's 2s orbital is 15.6 eV lower than carbon's 2s orbital, the σ₂s bonding MO will have predominantly oxygen 2s character, leading to greater electron density on oxygen. Choice A is incorrect because equal distribution only occurs in homonuclear diatomics. Choice C incorrectly suggests carbon localization and misapplies LUMO concepts. Choice D incorrectly describes phase relationships as causing density alternation.
17. The MO diagram for HF shows that the fluorine 2s orbital is too low in energy to interact significantly with hydrogen's 1s orbital, remaining essentially non-bonding. However, the H 1s orbital does interact with F 2p orbitals. Given that HF has a bond order of 1 and contains 8 valence electrons, what is the primary reason that HF is more stable than would be predicted from a simple Lewis structure analysis?
    Answer: The energy lowering of the bonding σ orbital relative to the atomic orbitals is enhanced by the large energy gap between H 1s and F 2p orbitals The stability comes from the significant energy lowering when the bonding σ MO forms. Although there's an energy mismatch between H 1s and F 2p, the interaction still creates a bonding orbital that's substantially lower in energy than the starting atomic orbitals, providing extra stabilization beyond simple Lewis predictions. Choice A is wrong because π orbitals don't form in HF (no parallel p orbital overlap). Choice B confuses ionic character with MO stabilization. Choice C incorrectly attributes stability to non-bonding electrons rather than bonding orbital stabilization.
18. Consider the heteronuclear diatomic molecule LiF. The Li 2s orbital lies at -5.4 eV while the F 2p orbitals lie at -18.6 eV. When constructing the molecular orbital diagram, the large energy gap prevents significant orbital mixing. If we place the 4 valence electrons into the resulting molecular orbitals, which statement correctly describes the nature of the highest occupied molecular orbital (HOMO)?
    Answer: The HOMO consists of non-bonding F 2p orbitals (π-type) that retain essentially pure fluorine character When you encounter molecular orbital problems with heteronuclear diatomics having large energy gaps between atomic orbitals, the key insight is that orbitals of very different energies don't mix effectively. This dramatically affects the resulting molecular orbital diagram. In LiF, the enormous 13.2 eV energy difference between Li 2s (-5.4 eV) and F 2p (-18.6 eV) orbitals prevents significant mixing. This means the F 2p orbitals remain essentially unchanged when forming molecular orbitals. With 4 valence electrons (1 from Li, 7 from F), electron filling follows energy order: the lowest energy F 2p orbital becomes bonding and fills first, while the Li 2s becomes a higher-energy orbital. The remaining F 2p orbitals that cannot interact with Li 2s (the π-type orbitals perpendicular to the bond axis) remain as non-bonding orbitals with pure fluorine character, and these end up as the HOMO. Answer A is wrong because the large energy gap prevents significant mixing between Li 2s and F 2p orbitals. Answer B incorrectly suggests the HOMO has Li character and is antibonding, when actually the non-bonding F orbitals are higher in energy than any Li-based orbitals. Answer D is impossible because s orbitals are spherically symmetric and cannot form π bonds through sideways overlap. Remember: when atomic orbital energy differences exceed ~10-12 eV, treat them as non-interacting. The more electronegative atom's orbitals often become non-bonding rather than strongly bonding, especially for lone pairs.
19. In the molecular orbital treatment of NO, which has 11 valence electrons, the unpaired electron occupies a π* antibonding orbital. If NO⁺ is formed by removing this unpaired electron, and NO⁻ is formed by adding an electron to the same orbital, which statement correctly compares the bond lengths and vibrational frequencies of these three species?
    Answer: Bond length order: NO⁺ < NO < NO⁻; vibrational frequency order: NO⁺ > NO > NO⁻ Removing an electron from π* (antibonding) increases bond order: NO⁺ (bond order 3) > NO (bond order 2.5) > NO⁻ (bond order 2). Higher bond order means shorter bond length and higher vibrational frequency. Therefore: bond lengths NO⁺ < NO < NO⁻ and frequencies NO⁺ > NO > NO⁻. Choice B reverses the bond length order. Choice C correctly orders bond lengths but incorrectly reverses vibrational frequencies. Choice D places neutral NO as having the shortest bond, which contradicts the effect of removing the antibonding electron.
20. When constructing the molecular orbital diagram for the CF⁺ ion, which has 12 valence electrons, the large electronegativity difference between C and F affects the orbital mixing patterns. What is the most likely description of the highest occupied molecular orbital (HOMO)?
    Answer: A π orbital with primarily fluorine character due to electronegativity stabilization When analyzing molecular orbital diagrams for heteronuclear diatomic molecules, you need to consider how electronegativity differences affect orbital energy levels and electron distribution. The large electronegativity difference between carbon and fluorine ($\Delta\chi \approx 1.5$) significantly polarizes the molecular orbitals toward the more electronegative fluorine atom. For CF⁺ with 12 valence electrons, you fill the molecular orbitals in order of increasing energy. The electronegativity difference means fluorine's atomic orbitals are much lower in energy than carbon's, so bonding molecular orbitals will have predominantly fluorine character while antibonding orbitals will have more carbon character. After filling the lower-energy σ bonding orbitals, the remaining electrons occupy π orbitals. The HOMO ends up being a π orbital that, due to fluorine's electronegativity stabilization, has primarily fluorine character. Option A is incorrect because while carbon's larger size might suggest some orbital mixing advantages, electronegativity effects dominate in determining orbital character. Option B incorrectly identifies the orbital type as σ rather than π, and while it correctly recognizes fluorine's influence, the HOMO in this electron configuration is a π orbital. Option D is wrong because the large electronegativity difference prevents equal character distribution—symmetry requirements don't override the energetic preference for electrons to localize on the more electronegative atom. Remember: in heteronuclear molecules with large electronegativity differences, bonding orbitals favor the more electronegative atom, while antibonding orbitals favor the less electronegative atom.

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