Integrated MO and Intermolecular Problems

Introduction

A molecular-orbital diagram explains electrons within a molecule, but a boiling point describes interactions among many molecules. Integrated questions may ask both. The main discipline is to change scale deliberately: first determine a diatomic's internal bond and spins, then discuss how entire molecules attract one another in the specified phase.

Core explanation

For oxygen, the standard MO filling of valence electrons yields eight in bonding orbitals and four in antibonding orbitals. The bond order is (8 − 4)/2 = 2. Two electrons occupy separate degenerate π orbitals according to Hund's rule, so O₂ is paramagnetic. This tells us about the internal O–O bond and magnetic response of isolated or weakly interacting oxygen molecules. It does not say that pure oxygen is an ionic solid or that O₂ molecules are joined to one another by covalent bonds under ordinary conditions.

In condensed molecular oxygen, separate O₂ molecules attract one another, principally through dispersion and other weak interactions appropriate to their electronic and magnetic properties. O₂ has no permanent electric dipole because it is homonuclear. Cooling enough can condense it without converting every O=O bond into a network. Melting or boiling molecular oxygen principally changes the spatial arrangement and interactions among molecules; chemical dissociation O₂ → 2O is a different process with a different energy requirement.

Compare nitrogen. N₂'s standard MO bond order is 3 and it is diamagnetic, with all electrons paired in the elementary diagram. It is also homonuclear and has no permanent dipole, so dispersion matters in its molecular condensed phases. It would be wrong to infer that N₂ must boil higher than O₂ just because its intramolecular bond order is larger. Their boiling behaviour depends on forces between whole molecules, not directly on the energy to break the N≡N versus O=O internal bond. Molecular size, polarisability and other effects enter the comparison; observed data should decide a close numerical ranking.

Ion formation changes the MO account. Removing an electron from O₂'s antibonding π orbital gives O₂⁺ with bond order 2.5 and one unpaired electron. Adding an electron gives O₂⁻ with bond order 1.5 and one unpaired electron. Neither ion should automatically be treated as if it were an ordinary neutral liquid at room conditions. Its charge strongly affects interactions with solvents, counterions and electric fields. An ionic species' surroundings must be specified before predicting a bulk phase or solubility.

Consider hydrogen. H₂ has simple MO bond order 1 and paired electrons. It is nonpolar, yet its molecules attract through dispersion, enough to condense at sufficiently low temperature. H₂⁺ has bond order 0.5 and is paramagnetic under the simple one-electron MO picture; its charge introduces interactions entirely different from neutral H₂. Confusing bond order with intermolecular force strength would miss that distinction.

The three questions must be separated: bond order estimates net within-molecule bonding; unpaired spins predict magnetic classification; intermolecular attractions contribute to phase behaviour. They can influence one another indirectly—for example, electronic structure affects polarisability—but there is no formula turning bond order alone into boiling point. State the structural unit and physical state before comparing.

Step-by-step reasoning

1. Count electrons and fill the appropriate MO ordering. 2. Compute bond order from bonding and antibonding occupancy. 3. Count unpaired electrons for magnetism independently. 4. Identify whether the bulk sample contains neutral molecules, ions or a network. 5. Explain phase behaviour through between-particle interactions, using measured data for close rankings.

Visual explanation

Draw one O₂ molecule with two unpaired π orbital arrows inside a solid border. Draw several separate O₂ molecules outside it linked by dotted dispersion attractions. Label the solid border “intramolecular MO account” and dotted lines “intermolecular cohesion.”

Real-world analogy

The strength of each bicycle's frame and the attraction that keeps bicycles parked close together are different questions. Strong frames do not imply that bicycles stick tightly to neighbours. Molecular bond order and liquid cohesion similarly concern different scales.

Real-world example

Liquid oxygen's attraction to a magnet demonstrates unpaired electrons predicted by MO theory, while its existence as a liquid at low temperature shows attraction among intact O₂ molecules. These two observations probe distinct aspects of one substance.

Why?

Why can a nonpolar gas condense? London dispersion arises from fluctuating electron clouds even without a permanent molecular dipole. At sufficiently low temperature, those attractions can hold molecules in a condensed phase.

Common misconception

“A triple bond guarantees a higher boiling point than a double bond.” Bond multiplicity describes intramolecular connectivity. Boiling requires overcoming between-molecule attractions, and size, shape and polarity also matter.

Worked example

Compare O₂ and O₂⁺ internally. O₂ has eight bonding and four antibonding valence electrons, giving order 2 and two unpaired electrons. O₂⁺ has one fewer antibonding electron, giving order (8 − 3)/2 = 2.5 and one unpaired electron. Both are paramagnetic in the simple diagram. Predicting their relative boiling points as pure substances would require specifying real phases and interactions, not just these bond orders.

Quick check

1. Does O₂'s paramagnetism imply that its molecules are held together by ionic bonds in liquid oxygen? Answer: No. Paramagnetism comes from unpaired molecular electrons; liquid cohesion is a separate issue.

Exam focus

Show MO filling, bond-order arithmetic and unpaired count. If a bulk property is asked, begin a new argument about interactions between the particles present. Never equate bond dissociation with boiling.

Advanced insight

Magnetic interactions between paramagnetic molecules can matter in specialised conditions, but a general introductory boiling comparison cannot be reduced to the yes/no magnetic label. Quantitative condensed-phase properties require thermodynamic and molecular-interaction data.

Summary

MO diagrams predict within-molecule bond order and spin pairing. Intermolecular forces explain cohesion and phase changes of molecular substances. O₂ illustrates both: bond order 2 and unpaired electrons internally, but separate intact molecules in a condensed sample.

Practice questions

1. What is N₂'s standard MO bond order? Answer: Three. 2. Why is O₂ paramagnetic? Answer: Two electrons remain unpaired in degenerate π orbitals. 3. Does boiling N₂ normally cleave each N≡N bond? Answer: No. It separates intact N₂ molecules from one another. 4. What changes in forming O₂⁻ from O₂? Answer: An antibonding π electron is added, reducing simple bond order from 2 to 1.5.