Physical Chemistry 2 Quiz: Heteronuclear Diatomics And Mo Diagrams
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Heteronuclear Diatomics And Mo DiagramsQuestion 1 of 20

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?

The 5.0 eV energy difference causes equal mixing of H 1s and F 2p in both bonding and antibonding orbitals
The energy difference causes the bonding MO to be primarily F 2p in character with minimal H 1s contribution
The energy difference causes the antibonding MO to be primarily H 1s in character with minimal F 2p contribution
The energy difference causes both MOs to have equal H 1s and F 2p character due to symmetry requirements
The energy difference prevents any significant orbital mixing, resulting in non-bonding molecular orbitals
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Heteronuclear Diatomics And Mo Diagrams

Practice Heteronuclear Diatomics And Mo Diagrams 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 Heteronuclear Diatomics And Mo Diagrams, 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

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?

  1. The 5.0 eV energy difference causes equal mixing of H 1s and F 2p in both bonding and antibonding orbitals
  2. The energy difference causes the bonding MO to be primarily F 2p in character with minimal H 1s contribution (correct answer)
  3. The energy difference causes the antibonding MO to be primarily H 1s in character with minimal F 2p contribution
  4. The energy difference causes both MOs to have equal H 1s and F 2p character due to symmetry requirements
  5. The energy difference prevents any significant orbital mixing, resulting in non-bonding molecular orbitals
Explanation: 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.

Question 2

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?

  1. Bond length increases in the order N₂⁺ < CO < NO⁺ due to decreasing nuclear charge
  2. Bond strength increases in the order NO⁺ < CO < N₂⁺ due to increasing covalent character (correct answer)
  3. Bond order remains constant at 3.0 for all three molecules due to identical electron configurations
  4. Vibrational frequency increases in the order CO < NO⁺ < N₂⁺ due to decreasing reduced mass
  5. Bond polarity increases in the order N₂⁺ < NO⁺ < CO due to increasing electronegativity difference
Explanation: 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.

Question 3

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?

  1. The electronegativity difference between C and O stabilizes π orbitals more than σ orbitals
  2. The 2s-2p energy gap difference between C and O prevents effective σ orbital formation
  3. Carbon's 2s orbital mixes with oxygen's 2p orbital, destabilizing the 4σ relative to 1π
  4. The different 2s-2p mixing in C versus O creates asymmetric orbital contributions in the 4σ (correct answer)
  5. Oxygen's higher nuclear charge preferentially stabilizes π-type interactions over σ-type
Explanation: 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.

Question 4

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?

  1. BF has stronger bonding due to better size matching between B 2p and F 2p orbitals
  2. AlF has stronger bonding due to the larger electronegativity difference between Al and F
  3. Both molecules have identical bonding due to the same electron configuration and group positions
  4. BF has weaker bonding due to greater 2s-2p energy separation in boron versus aluminum
  5. AlF has weaker bonding due to poor overlap between Al 3p and F 2p orbitals (correct answer)
Explanation: 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.

Question 5

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?

  1. The electronegativity difference prevents s-p mixing, making BN more like Ne₂ than C₂
  2. B and N atoms undergo different amounts of s-p mixing, creating asymmetric molecular orbitals (correct answer)
  3. The electronegativity difference enhances s-p mixing equally on both atoms, strengthening all bonds
  4. BN exhibits identical orbital mixing to C₂ because they are isoelectronic molecules
  5. The electronegativity difference causes only p orbital mixing, eliminating σ bonding components
Explanation: 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.

Question 6

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?

  1. The bonding orbitals become purely fluorine-based, approaching the ionic limit of Li⁺F⁻ (correct answer)
  2. The molecular orbitals become degenerate due to the large electronegativity difference
  3. The antibonding orbitals become lower in energy than the bonding orbitals
  4. All molecular orbitals become equally mixed between Li and F atomic orbitals
  5. The molecular orbital description becomes invalid, requiring a purely ionic model
Explanation: 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 (Δχ=3.0\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.

Question 7

The photoelectron spectrum of carbon monoxide shows distinct peaks corresponding to different molecular orbitals. If the 1π orbital appears at 16.9 eV and the 5σ orbital appears at 14.0 eV, what does this energy ordering reveal about the molecular orbital diagram of CO?

  1. The 5σ orbital is higher in energy than 1π, confirming the standard heteronuclear ordering
  2. The 1π orbital is higher in energy than 5σ, indicating unusual orbital mixing effects
  3. The energy difference confirms that 1π has more oxygen character than 5σ (correct answer)
  4. The photoelectron peaks confirm that 5σ is the HOMO in CO
  5. The similar energies indicate that 1π and 5σ orbitals are nearly degenerate
Explanation: When analyzing photoelectron spectroscopy data, remember that higher binding energies correspond to lower-energy (more stable) molecular orbitals. The energy required to remove an electron reflects how tightly it's bound in that orbital. In CO, the 1π orbital appears at 16.9 eV while the 5σ orbital appears at 14.0 eV. Since 16.9 eV > 14.0 eV, the 1π orbital is actually lower in energy than the 5σ orbital, meaning 1π electrons are more tightly bound. This unusual ordering occurs because of the significant electronegativity difference between carbon and oxygen. The correct answer is C because this energy difference reveals that the 1π orbital has greater oxygen character. Oxygen's higher electronegativity and lower-energy 2p orbitals contribute more to the 1π bonding orbital, making it more stable and requiring more energy to remove electrons from it. A is wrong because 5σ is actually higher in energy than 1π (lower binding energy), which is not the standard ordering. B incorrectly states that 1π is higher in energy when the binding energies show the opposite. D is incorrect because while 5σ has the lowest binding energy shown, making it the HOMO, this wasn't what the energy ordering specifically "reveals about the molecular orbital diagram." Study tip: In photoelectron spectroscopy, always convert binding energies to orbital energies by remembering that higher binding energy = lower orbital energy = more stable orbital. For heteronuclear molecules, expect the more electronegative atom to contribute more to lower-energy orbitals.

Question 8

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?

  1. The large size difference between Mg 3s and O 2p orbitals prevents effective molecular orbital formation
  2. The electronegativity difference of 2.3 units places MgO firmly in the ionic bonding regime (correct answer)
  3. The energy match between Mg 3s and O 2p orbitals enables significant covalent orbital mixing
  4. The filled Mg 3s and O 2p orbitals result in purely electrostatic interactions without orbital overlap
  5. The diffuse nature of Mg 3s orbitals creates weak van der Waals-type bonding with oxygen
Explanation: 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=2.33.5 - 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.

Question 9

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?

  1. The unpaired electron undergoes spin-pairing at low temperatures due to crystal field effects
  2. NO molecules dimerize to form N₂O₂ with paired electrons in the solid state (correct answer)
  3. The 2π* orbital becomes doubly occupied at low temperatures due to thermal population
  4. Intermolecular interactions stabilize a different electronic configuration in the solid
  5. The unpaired electron delocalizes across multiple NO molecules, reducing paramagnetism
Explanation: 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 2π∗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 2π∗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.

Question 10

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?

  1. The bonding orbital is primarily hydrogen-based, resulting in Li⁺-H⁻ polarity as expected from electronegativity
  2. The bonding orbital is primarily lithium-based, resulting in Li⁻-H⁺ polarity opposite to electronegativity predictions
  3. The bonding orbital has equal Li and H character, resulting in a nonpolar molecule despite electronegativity differences
  4. The bonding orbital is primarily hydrogen-based, but orbital size effects result in Li⁻-H⁺ polarity (correct answer)
  5. The bonding orbital energy is intermediate between Li 2s and H 1s, causing electron density to localize equally
Explanation: 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.

Question 11

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?

  1. The bonding orbital becomes purely Cl 3p in character with no H 1s contribution
  2. The antibonding orbital becomes purely H 1s in character with no Cl 3p contribution
  3. Both bonding and antibonding orbitals show significant mixing of H 1s and Cl 3p character (correct answer)
  4. The small energy gap prevents orbital mixing, resulting in nonbonding molecular orbitals
  5. The molecular orbitals become degenerate due to similar atomic orbital energies
Explanation: 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.

Question 12

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?

  1. The bond order decreases from 1.0 in HCl to 0.5 in HBr due to poorer orbital overlap
  2. The bonding orbital becomes more polarized toward bromine due to increased orbital size
  3. The overlap population decreases significantly due to the larger size of Br 4p orbitals (correct answer)
  4. The antibonding orbital becomes lower in energy due to better size matching with H 1s
  5. The molecular orbital energies become more widely separated due to increased nuclear charge
Explanation: 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.

Question 13

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)?

  1. A π orbital with primarily carbon character due to size matching effects
  2. A σ orbital with primarily fluorine character due to electronegativity effects
  3. A π orbital with primarily fluorine character due to electronegativity stabilization (correct answer)
  4. A σ orbital with equal C and F character due to orbital symmetry requirements
  5. An antibonding π orbital with primarily carbon character due to energy matching
Explanation: 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 (Δχ≈1.5\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.

Question 14

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?

  1. The formation of π-type molecular orbitals from F 2p orbitals creates additional bonding character beyond the single σ bond
  2. The large electronegativity difference creates ionic character that strengthens the bond beyond pure covalent predictions
  3. The non-bonding F 2s electrons provide additional electrostatic stabilization through core-valence electron correlation effects
  4. 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 (correct answer)
Explanation: 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.

Question 15

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?

  1. The HOMO is a bonding orbital that has been stabilized relative to a weighted average of the atomic ionization energies due to constructive orbital overlap
  2. 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 (correct answer)
  3. The HOMO is a non-bonding orbital with primarily F character, explaining why its ionization energy lies between the two atomic values
  4. The HOMO is an antibonding orbital that has been destabilized relative to the B 2p value but stabilized relative to the F 2p value
Explanation: 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.

Question 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 σ2s\sigma_{2s} bonding molecular orbital?

  1. The electron density is equally distributed between carbon and oxygen atoms due to orbital mixing effects
  2. The electron density is primarily localized on the oxygen atom due to its lower energy 2s orbital (correct answer)
  3. The electron density is primarily localized on the carbon atom due to better energy matching with the LUMO
  4. The electron density alternates between atoms depending on the phase relationships of the constituent orbitals
Explanation: 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.

Question 17

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)?

  1. The HOMO is a bonding σ orbital with mixed Li 2s and F 2p character, containing paired electrons
  2. The HOMO is an antibonding σ* orbital with predominantly Li 2s character due to energy mismatch effects
  3. The HOMO consists of non-bonding F 2p orbitals (π-type) that retain essentially pure fluorine character (correct answer)
  4. The HOMO is a bonding π orbital formed from sideways overlap of Li 2s and F 2p orbitals
Explanation: 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.

Question 18

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?

  1. The ionic character in CN⁻ creates electrostatic repulsion that counteracts the covalent bonding, resulting in bond lengthening despite equivalent bond order
  2. The molecular orbitals in CN⁻ have greater antibonding character due to poor energy matching between C and N atomic orbitals compared to the homonuclear N₂ case
  3. The formal charges in CN⁻ cause electron-electron repulsion in the bonding molecular orbitals, reducing their effectiveness compared to N₂
  4. 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 (correct answer)
Explanation: 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.

Question 19

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?

  1. 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 (correct answer)
  2. Occupied: non-bonding F2s, σ(Al3s-F2p), π(Al3p-F2p); dipole points from F to Al due to formal charge distribution effects in the bonding molecular orbitals
  3. Occupied: non-bonding F2s, non-bonding Al3s, σ(Al3p-F2p); dipole points from Al to F due to electronegativity difference between aluminum and fluorine atoms
  4. Occupied: σ(Al3s-F2p) bonding orbital, non-bonding F2s orbital, and non-bonding F2p orbitals; dipole points from Al to F due to greater electron density accumulation on the more electronegative fluorine atom
Explanation: 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.

Question 20

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?

  1. Bond length order: NO⁺ < NO < NO⁻; vibrational frequency order: NO⁺ > NO > NO⁻ (correct answer)
  2. Bond length order: NO⁻ < NO < NO⁺; vibrational frequency order: NO⁻ > NO > NO⁺
  3. Bond length order: NO⁺ < NO < NO⁻; vibrational frequency order: NO⁻ > NO > NO⁺
  4. Bond length order: NO < NO⁺ < NO⁻; vibrational frequency order: NO > NO⁺ > NO⁻
Explanation: 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.