Bond Order from MO Diagrams
Questions 1–4
Using the MO diagram above, calculate the bond order (BO) for each of the following oxygen species: O2, O2+, O2−, and O22−. Which species has the strongest bond?
Bond order = ½(bonding e− − antibonding e−). O2 has 16 electrons: (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)2, giving BO = ½(10 − 6) = 2.0. O2+ removes one π* electron: BO = ½(10 − 5) = 2.5. O2− adds one π* electron: BO = ½(10 − 7) = 1.5. O22− fills both π*: BO = ½(10 − 8) = 1.0. Higher bond order means a stronger bond, so O2+ is strongest at BO = 2.5.
Both N2 and CO are isoelectronic with bond order 3. Which molecule has the shorter bond length, and why?
N2 has a bond length of 109.8 pm while CO has a bond length of 112.8 pm — actually N2 is slightly shorter. However, the key teaching point is that CO (112.8 pm) has a very similar bond length to N2 (109.8 pm) despite being heteronuclear. The electronegativity difference in CO means the bonding MOs have more oxygen character, pulling electron density toward O. This uneven distribution creates a strong effective overlap. In practice the bond lengths are close, but CO's high bond enthalpy (1076 kJ/mol vs 945 kJ/mol for N2) reflects the electronegativity-enhanced bond strength from the polar covalent character.
Nitric oxide (NO) has a bond order of 2.5 with its electron configuration ending in (π*2p)1. NO readily loses one electron to form NO+. Why is this ionization so favorable?
NO has 15 electrons with one electron in an antibonding π* orbital, giving BO = ½(10 − 5) = 2.5. Removing this single antibonding electron to form NO+ (14 electrons) gives BO = ½(10 − 4) = 3.0. The resulting NO+ is isoelectronic with N2 and CO, both of which are exceptionally stable triple-bonded species. The loss of an antibonding electron simultaneously strengthens the bond and shortens it (from 115 pm in NO to 106 pm in NO+), making this ionization thermodynamically favorable.
F2 has a bond order of 1, yet its bond dissociation enthalpy (159 kJ/mol) is surprisingly low — even lower than that of Cl2 (242 kJ/mol), despite fluorine being smaller and having greater orbital overlap. What is the primary reason for the anomalously weak F–F bond?
Although F2 has BO = 1 and fluorine's small atomic radius should promote strong overlap, the extremely short bond length (141 pm) brings the three lone pairs on each fluorine atom into very close proximity. The resulting inter-electron (lone pair–lone pair) repulsion significantly weakens the bond. This is a well-known anomaly for period 2 elements: N–N, O–O, and F–F single bonds are all weaker than their period 3 counterparts (P–P, S–S, Cl–Cl) because the small atom size intensifies repulsion between non-bonding electron pairs on adjacent atoms.
Bond Length & Bond Enthalpy Correlations
Questions 5–8
The figure above illustrates the general relationships among bond order, bond length, and bond enthalpy. Which statement best summarizes these correlations?
The three bond properties are systematically related: as bond order increases, more electrons are shared between the two nuclei, pulling them closer together (shorter bond length) and requiring more energy to break apart (larger bond dissociation enthalpy). For example, in the carbon series: C–C (154 pm, 346 kJ/mol), C=C (134 pm, 614 kJ/mol), C≡C (120 pm, 839 kJ/mol). Each step up in bond order shortens the bond and increases the enthalpy. This is the fundamental triad relationship in chemical bonding.
Arrange the following carbon–carbon bonds in order of decreasing bond length (longest first): C–C (single), C=C (double), C≡C (triple).
Bond length decreases with increasing bond order. The values are: C–C = 154 pm (single, BO = 1), C=C = 134 pm (double, BO = 2), C≡C = 120 pm (triple, BO = 3). The additional shared electron pairs in multiple bonds pull the nuclei closer together. So from longest to shortest: C–C > C=C > C≡C. This trend holds generally for any element forming multiple bond types (e.g., N–N, N=N, N≡N follow the same pattern).
The N–N single bond enthalpy (159 kJ/mol) is unusually weak compared to the P–P single bond (200 kJ/mol), even though nitrogen is smaller and typically forms stronger bonds. What explains this anomaly?
This is the same lone pair repulsion effect seen in F2. Nitrogen's small covalent radius means that when two N atoms form a single bond, the lone pairs on each atom are forced into close proximity, generating strong inter-electron repulsion that weakens the bond. Phosphorus, being larger (period 3), has its lone pairs further apart and suffers less repulsion. This is a general trend for period 2 elements: N–N (159 kJ/mol), O–O (144 kJ/mol), and F–F (159 kJ/mol) single bonds are all anomalously weak compared to their period 3 analogs P–P (200 kJ/mol), S–S (266 kJ/mol), and Cl–Cl (242 kJ/mol).
Compare the bond dissociation enthalpies of the hydrogen halides: H–F (568 kJ/mol), H–Cl (431 kJ/mol), H–Br (366 kJ/mol), H–I (298 kJ/mol). What trend do you observe, and what is the underlying cause?
The dominant factor is atomic size and orbital overlap. As you descend group 17, the halogen atom becomes larger and its valence orbital (2p for F, 3p for Cl, 4p for Br, 5p for I) becomes more diffuse. The overlap with hydrogen's compact 1s orbital decreases progressively, leading to longer and weaker bonds: H–F (92 pm, 568 kJ/mol), H–Cl (127 pm, 431 kJ/mol), H–Br (141 pm, 366 kJ/mol), H–I (161 pm, 298 kJ/mol). While decreasing electronegativity difference also plays a role (reducing ionic stabilization), the primary cause is the increasingly poor orbital overlap.
Bond Character — Ionic, Covalent, Metallic
Questions 9–12
The Ketelaar triangle (also called the van Arkel–Ketelaar triangle) shown above classifies bonding into three limiting types: ionic, covalent, and metallic. What two quantities define the axes of this triangle?
The Ketelaar triangle uses two electronegativity-based parameters. The vertical axis is the electronegativity difference (Δχ) between the two bonded atoms, while the horizontal axis is the average electronegativity (χavg). High Δχ (top) corresponds to ionic bonding (e.g., NaCl). High χavg with low Δχ (bottom right) corresponds to covalent bonding (e.g., F2). Low χavg with low Δχ (bottom left) corresponds to metallic bonding (e.g., Na metal). Most real compounds fall somewhere within the triangle, reflecting mixed bonding character.
According to Fajans' rules, which combination of cation properties promotes covalent character in an ionic compound?
Fajans' rules predict when an "ionic" compound will have significant covalent character. Covalent character is promoted by: (1) a small, highly charged cation (high charge density, strong polarizing power), (2) a large, highly charged anion (easily polarizable), and (3) cations with non-noble-gas electron configurations (e.g., d10 like Cu+ or Ag+). A small cation with high charge creates an intense electric field that distorts the anion's electron cloud toward the cation, effectively sharing electron density — i.e., creating covalent character. Conversely, a large cation with low charge has weak polarizing power and the bonding remains predominantly ionic.
Using the Ketelaar triangle above, where would you place NaCl, diamond (C–C), and sodium metal (Na)?
The three corners of the Ketelaar triangle represent the three idealized bond types. NaCl: Na (χ = 0.93) and Cl (χ = 3.16) give Δχ = 2.23 (large) and χavg = 2.05. The large Δχ places NaCl near the ionic vertex (top of the triangle). Diamond: both atoms are carbon (χ = 2.55), so Δχ = 0 and χavg = 2.55 (high). This places it at the covalent vertex (bottom-right). Sodium metal: both atoms are Na (χ = 0.93), so Δχ = 0 and χavg = 0.93 (low). This places it at the metallic vertex (bottom-left).
AlCl3 exhibits significantly more covalent character than NaCl, despite both being "metal + non-metal" compounds. Using Fajans' rules, which explanation best accounts for this difference?
This is a classic application of Fajans' rules. Al3+ has an ionic radius of only ~54 pm and a charge of +3, giving it an exceptionally high charge density (charge/volume ratio). This creates an intense electric field that strongly distorts (polarizes) the electron cloud of the Cl− anion, pulling electron density back toward the aluminium. The result is significant sharing of electron density — covalent character. This is why AlCl3 has a low melting point (192 °C), exists as a dimer Al2Cl6, and is soluble in non-polar solvents — all properties of covalent compounds. Na+, being much larger (102 pm) with only +1 charge, has weak polarizing power, so NaCl remains predominantly ionic (m.p. 801 °C).