Chapter 4 · Chemical Bonding and Molecular Structure

4.4 Molecular Orbital Theory and Hydrogen Bonding

Molecular orbital theory spreads electrons over the whole molecule: atomic orbitals combine into bonding and antibonding molecular orbitals, filled in a fixed energy order. Bond order and magnetism of H₂ to Ne₂ and their ions follow at once — including the paramagnetism of O₂. Then hydrogen bonding, the strong attraction that makes water a liquid.

Two questions show the limits of everything so far. First, liquid oxygen is attracted by a magnet, so the $\mathrm{O_{2}}$ molecule has unpaired electrons — yet its Lewis structure $\mathrm{O{=}O}$ and its valence bond description pair every electron. Second, the ion $\mathrm{H_{2}^{+}}$, with only one electron, exists and holds its two protons together with a bond enthalpy of about $256\ \mathrm{kJ\,mol^{-1}}$; a bond with a single electron is impossible to draw as a shared pair.

The molecular orbital (MO) theory, developed by Friedrich Hund and Robert Mulliken from 1927 onwards, answers both. It treats a molecule the way Chapter 2 treated an atom: the electrons occupy orbitals that belong to the whole molecule, each with a definite energy, filled by the same Aufbau, Pauli and Hund rules. From the filling we read off the bond order, whether the molecule exists at all, and its magnetism. The last part of the section turns to a weaker but enormously important attraction between molecules — the hydrogen bond.

Molecular Orbitals from Atomic Orbitals

📘 Definition 4.12 — Molecular orbital theory
  1. The electrons of a molecule occupy molecular orbitals (MOs), which extend over the whole molecule and are influenced by all its nuclei (an atomic orbital is monocentric, a molecular orbital polycentric).
  2. Atomic orbitals of comparable energy and proper symmetry combine to form molecular orbitals; the number of MOs formed equals the number of atomic orbitals combined.
  3. When two atomic orbitals combine, two MOs form: a bonding MO of lower energy (and greater stability) than the atomic orbitals, and an antibonding MO of higher energy.
  4. Electrons fill the MOs in order of increasing energy (Aufbau principle), at most two per orbital with opposite spins (Pauli), and singly with parallel spins in degenerate orbitals before pairing (Hund's rule).

The approximate way to build molecular orbitals is the linear combination of atomic orbitals (LCAO). For two hydrogen atoms A and B, with $1s$ wave functions $\psi_{A}$ and $\psi_{B}$, the two combinations are $$\sigma=\psi_{A}+\psi_{B}\qquad\sigma^{*}=\psi_{A}-\psi_{B}$$ Adding the waves reinforces them between the nuclei (constructive interference): the electron density there is high, it attracts both nuclei and holds them together — a bonding MO. Subtracting them cancels the waves midway (destructive interference): there is a nodal plane between the nuclei, the electron density is pushed outside, and the nuclei repel each other — an antibonding MO, marked with an asterisk. The bonding MO lies as far below the atomic orbitals as the antibonding MO lies above them (to a first approximation).

📐 Rule 4.10 — Conditions for atomic orbitals to combine
  1. The atomic orbitals must have comparable energies: $1s$ combines with $1s$, $2s$ with $2s$, but not $1s$ with $2p$.
  2. They must have the same symmetry about the molecular axis (taken as $z$): $2p_{z}$ combines with $2p_{z}$, but $2p_{x}$ cannot combine with $2p_{z}$ — their overlap is zero.
  3. They must overlap to the maximum extent: the greater the overlap, the lower the bonding MO and the higher the antibonding MO.
Head-on combinations ($1s$, $2s$, $2p_{z}$) give σ molecular orbitals, symmetrical about the axis: $\sigma1s$, $\sigma^{*}1s$, $\sigma2s$, $\sigma^{*}2s$, $\sigma2p_{z}$, $\sigma^{*}2p_{z}$. Sideways combinations of $2p_{x}$ and of $2p_{y}$ give π molecular orbitals with electron density above and below the axis: the degenerate pairs $\pi2p_{x}$, $\pi2p_{y}$ (bonding) and $\pi^{*}2p_{x}$, $\pi^{*}2p_{y}$ (antibonding).

Energy-Level Diagrams, Bond Order and Magnetism

For the homonuclear diatomic molecules of the first two periods, the $1s$, $2s$ and $2p$ orbitals of the two atoms give ten molecular orbitals. Spectroscopy shows that their order of energy is not the same for every molecule.

📐 Rule 4.11 — Order of molecular orbital energies

(a) For $\mathrm{H_{2}}$ to $\mathrm{N_{2}}$ (atoms up to nitrogen):

$$\sigma1s\lt\sigma^{*}1s\lt\sigma2s\lt\sigma^{*}2s\lt(\pi2p_{x}=\pi2p_{y})\lt\sigma2p_{z}\lt(\pi^{*}2p_{x}=\pi^{*}2p_{y})\lt\sigma^{*}2p_{z}$$

(b) For $\mathrm{O_{2}}$, $\mathrm{F_{2}}$ and $\mathrm{Ne_{2}}$:

$$\sigma1s\lt\sigma^{*}1s\lt\sigma2s\lt\sigma^{*}2s\lt\sigma2p_{z}\lt(\pi2p_{x}=\pi2p_{y})\lt(\pi^{*}2p_{x}=\pi^{*}2p_{y})\lt\sigma^{*}2p_{z}$$

The difference is the position of $\sigma2p_{z}$. In B, C and N the $2s$ and $2p$ orbitals are close in energy, so $\sigma2s$ and $\sigma2p_{z}$ (which have the same symmetry) interact — s–p mixing — pushing $\sigma2p_{z}$ up above the $\pi2p$ pair. In O and F the $2s$–$2p$ gap is large and the mixing is negligible. The two inner orbitals, $\sigma1s^{2}\sigma^{*}1s^{2}$, are often abbreviated KK.

📘 Definition 4.13 — Bond order (molecular orbital theory)
If $N_{b}$ is the number of electrons in bonding molecular orbitals and $N_{a}$ the number in antibonding molecular orbitals, $$\text{bond order}=\tfrac12\left(N_{b}-N_{a}\right)$$ (electrons in the inner KK orbitals may be counted or left out; they contribute $2-2=0$). Bonding electrons stabilise the molecule and antibonding electrons destabilise it.
📐 Rule 4.12 — What bond order and filling predict
  1. Stability: if $N_{b}\gt N_{a}$ (positive bond order) the molecule is stable; if $N_{b}\le N_{a}$ (bond order zero or negative) it does not exist. A fractional bond order ($\tfrac12$, $1\tfrac12$, $2\tfrac12$) is allowed.
  2. Bond strength and length: for related species, the higher the bond order, the greater the bond-dissociation enthalpy and the shorter the bond.
  3. Magnetism: a species with one or more unpaired electrons is paramagnetic (attracted into a magnetic field); one with all electrons paired is diamagnetic (weakly repelled).
Species$\mathrm{H_{2}}$$\mathrm{He_{2}}$$\mathrm{Li_{2}}$$\mathrm{Be_{2}}$$\mathrm{B_{2}}$$\mathrm{C_{2}}$$\mathrm{N_{2}}$$\mathrm{O_{2}}$$\mathrm{F_{2}}$$\mathrm{Ne_{2}}$
Electrons2468101214161820
$N_{b}$, $N_{a}$2, 02, 24, 24, 46, 48, 410, 410, 610, 810, 10
Bond order1010123210
Magnetismdia—dia—paradiadiaparadia—

Measured bond lengths and enthalpies follow the bond orders: $\mathrm{N_{2}}$ $110\ \mathrm{pm}$, $946\ \mathrm{kJ\,mol^{-1}}$; $\mathrm{O_{2}}$ $121\ \mathrm{pm}$, $498$; $\mathrm{F_{2}}$ $142\ \mathrm{pm}$, $155$. $\mathrm{He_{2}}$, $\mathrm{Be_{2}}$ and $\mathrm{Ne_{2}}$ (bond order 0) do not exist as stable molecules.

✏️ Example 1 — Hydrogen and helium species
Write the MO configurations and find the bond orders of $\mathrm{H_{2}^{+}}$, $\mathrm{H_{2}}$, $\mathrm{He_{2}^{+}}$ and $\mathrm{He_{2}}$. Which exist? Which are paramagnetic? Arrange the existing ones in order of increasing stability.
Configurations $\mathrm{H_{2}^{+}}$ (1 e): $\sigma1s^{1}$. $\mathrm{H_{2}}$ (2): $\sigma1s^{2}$. $\mathrm{He_{2}^{+}}$ (3): $\sigma1s^{2}\sigma^{*}1s^{1}$. $\mathrm{He_{2}}$ (4): $\sigma1s^{2}\sigma^{*}1s^{2}$.
Bond orders $\tfrac12(1-0)=\tfrac12$; $\tfrac12(2-0)=1$; $\tfrac12(2-1)=\tfrac12$; $\tfrac12(2-2)=0$.
Existence, magnetism $\mathrm{H_{2}^{+}}$, $\mathrm{H_{2}}$ and $\mathrm{He_{2}^{+}}$ exist; $\mathrm{He_{2}}$ does not. $\mathrm{H_{2}^{+}}$ and $\mathrm{He_{2}^{+}}$ each have one unpaired electron — paramagnetic; $\mathrm{H_{2}}$ is diamagnetic.
Order $\mathrm{He_{2}^{+}}\lt\mathrm{H_{2}^{+}}\lt\mathrm{H_{2}}$. ($\mathrm{H_{2}^{+}}$ and $\mathrm{He_{2}^{+}}$ have the same bond order, but $\mathrm{He_{2}^{+}}$ is less stable: an antibonding electron raises the energy slightly more than a bonding electron lowers it, and $\mathrm{He_{2}^{+}}$ has one.)
✏️ Example 2 — Nitrogen and oxygen
(a) Write the full MO configurations of $\mathrm{N_{2}}$ and $\mathrm{O_{2}}$ and find their bond orders. (b) Explain why liquid oxygen is attracted by a magnet while liquid nitrogen is not. (c) Which has the stronger bond?
(a) N₂ 14 electrons, order (a): $\sigma1s^{2}\sigma^{*}1s^{2}\sigma2s^{2}\sigma^{*}2s^{2}(\pi2p_{x}^{2}=\pi2p_{y}^{2})\sigma2p_{z}^{2}$. $N_{b}=2+2+4+2=10$, $N_{a}=2+2=4$; bond order $\tfrac12(10-4)=3$.
(a) O₂ 16 electrons, order (b): $\mathrm{KK}\,\sigma2s^{2}\sigma^{*}2s^{2}\sigma2p_{z}^{2}(\pi2p_{x}^{2}=\pi2p_{y}^{2})(\pi^{*}2p_{x}^{1}=\pi^{*}2p_{y}^{1})$. $N_{b}=10$, $N_{a}=6$; bond order $2$.
(b) The last two electrons of $\mathrm{O_{2}}$ enter the two degenerate $\pi^{*}2p$ orbitals singly, with parallel spins (Hund's rule): two unpaired electrons, so $\mathrm{O_{2}}$ is paramagnetic. All the electrons of $\mathrm{N_{2}}$ are paired: diamagnetic.
(c) $\mathrm{N_{2}}$ (bond order 3 against 2; $946$ against $498\ \mathrm{kJ\,mol^{-1}}$).
✏️ Example 3 — The dioxygen family
Find the bond order and the number of unpaired electrons of $\mathrm{O_{2}^{+}}$, $\mathrm{O_{2}}$, $\mathrm{O_{2}^{-}}$ (superoxide) and $\mathrm{O_{2}^{2-}}$ (peroxide). Arrange them in order of increasing bond length, and say which is diamagnetic.
Electrons $15$, $16$, $17$, $18$. In each, $N_{b}=10$; the $\pi^{*}2p$ pair holds $1$, $2$, $3$, $4$ electrons, so $N_{a}=5$, $6$, $7$, $8$.
Bond orders $\tfrac12(10-5)=2.5$; $2$; $\tfrac12(10-7)=1.5$; $1$.
Unpaired $\pi^{*1}$: 1; $\pi^{*}$ $(1,1)$: 2; $\pi^{*}$ $(2,1)$: 1; $\pi^{*}$ $(2,2)$: 0. Only $\mathrm{O_{2}^{2-}}$ is diamagnetic.
Bond length Lower bond order, longer bond: $\mathrm{O_{2}^{+}}\lt\mathrm{O_{2}}\lt\mathrm{O_{2}^{-}}\lt\mathrm{O_{2}^{2-}}$ (about $112$, $121$, $128$ and $149\ \mathrm{pm}$).
✏️ Example 4 — Evidence for s–p mixing; removing an electron
(a) $\mathrm{B_{2}}$ is paramagnetic. Show that this agrees with order (a) of Rule 4.11 but not with order (b). (b) When $\mathrm{N_{2}}$ loses an electron to form $\mathrm{N_{2}^{+}}$ the bond becomes weaker ($946\rightarrow842\ \mathrm{kJ\,mol^{-1}}$), but when $\mathrm{O_{2}}$ forms $\mathrm{O_{2}^{+}}$ the bond becomes stronger ($498\rightarrow643\ \mathrm{kJ\,mol^{-1}}$). Explain.
(a) order (a) $\mathrm{B_{2}}$ has 10 electrons: $\mathrm{KK}\,\sigma2s^{2}\sigma^{*}2s^{2}$ then 2 electrons for the degenerate $\pi2p_{x}$, $\pi2p_{y}$ — one in each (Hund): 2 unpaired, paramagnetic ✓. Bond order $\tfrac12(6-4)=1$.
(a) order (b) The two electrons would pair in $\sigma2p_{z}$, giving a diamagnetic molecule — contrary to experiment. So in $\mathrm{B_{2}}$, $\pi2p$ lies below $\sigma2p_{z}$.
(b) N₂⁺ The highest occupied MO of $\mathrm{N_{2}}$ is the bonding $\sigma2p_{z}$: removing an electron lowers $N_{b}$ to 9, bond order $\tfrac12(9-4)=2.5$ — weaker bond.
(b) O₂⁺ The highest occupied MO of $\mathrm{O_{2}}$ is the antibonding $\pi^{*}2p$: removing an electron lowers $N_{a}$ to 5, bond order $2.5$ — stronger bond.
🚀 Beyond the board syllabus (enrichment only) — heteronuclear diatomic species

Not examined anywhere: CBSE, JEE Main, NEET and JEE Advanced all restrict molecular orbital theory to homonuclear diatomic species. Read this for interest only.

The same diagrams are used, approximately, for diatomic species made of two different period-2 atoms, by counting electrons. $\mathrm{CO}$, $\mathrm{CN^{-}}$ and $\mathrm{NO^{+}}$ have 14 electrons, like $\mathrm{N_{2}}$: bond order $3$, diamagnetic — matching the triple bonds of Section 4.1. $\mathrm{NO}$ has 15 electrons, one more than $\mathrm{N_{2}}$, in a $\pi^{*}$ orbital: bond order $2.5$ and one unpaired electron, so nitric oxide is paramagnetic — an odd-electron molecule that Lewis structures cannot describe properly. In reality the orbitals of the more electronegative atom lie lower, so the diagram is lopsided, but the simple electron count still gives the right bond order and magnetic behaviour for these species.
Interactive 1 — Molecular Orbital Diagram Filler

Choose a homonuclear diatomic species and its charge. Press ▶ Play to watch the electrons go into the molecular orbitals one at a time — lowest energy first, singly into degenerate π orbitals before pairing — or move the slider to add them yourself. The lab uses order (a) of Rule 4.11 for elements up to nitrogen and order (b) for oxygen to neon, and reports the configuration, $N_{b}$, $N_{a}$, bond order and magnetism.

14
species—
N_b, N_a—
bond order—
magnetism—

Hydrogen Bonding

A hydrogen atom bonded to fluorine, oxygen or nitrogen is in an unusual situation. The bond is very polar, and the tiny hydrogen atom, which has no inner electrons, is left almost as a bare proton on the surface of the molecule. It is strongly attracted by a lone pair on a small electronegative atom of a neighbouring molecule.

📘 Definition 4.14 — Hydrogen bond
A hydrogen bond is the attractive force that binds a hydrogen atom, covalently bonded to a highly electronegative atom X (F, O or N), to a lone pair of another electronegative atom Y (F, O or N). It is written with a dotted line: $\mathrm{X{-}H\cdots Y}$. Hydrogen bonds are much weaker than covalent bonds — typically about $10$ to $40\ \mathrm{kJ\,mol^{-1}}$ against several hundred — but stronger than the other attractions between molecules. They are of two kinds:
  • intermolecular — between different molecules, of the same or of different substances: $\mathrm{HF}$, $\mathrm{H_{2}O}$, $\mathrm{NH_{3}}$, alcohols, and water with alcohols;
  • intramolecular — between an $\mathrm{X{-}H}$ group and an acceptor atom in the same molecule, when they are close enough to form a ring, usually of six atoms: 2-nitrophenol (o-nitrophenol), 2-hydroxybenzaldehyde (salicylaldehyde).
📐 Rule 4.13 — Consequences of hydrogen bonding
  1. High melting and boiling points. Extra energy is needed to break intermolecular hydrogen bonds, so $\mathrm{NH_{3}}$, $\mathrm{H_{2}O}$ and $\mathrm{HF}$ boil far higher than the trend of their groups predicts (Interactive 2).
  2. Solubility. Substances that can hydrogen-bond with water dissolve in it: ammonia, lower alcohols, carboxylic acids, sugars.
  3. Ice is less dense than water. In ice each $\mathrm{H_{2}O}$ forms four hydrogen bonds arranged tetrahedrally, giving an open cage-like structure; on melting, some hydrogen bonds break and the molecules pack more closely. So ice ($0.917\ \mathrm{g\,cm^{-3}}$) floats on water, and lakes freeze from the top down.
  4. Intramolecular hydrogen bonding ties up the $\mathrm{X{-}H}$ group inside the molecule, so the molecules hardly associate with each other or with water: 2-nitrophenol is more volatile (boiling point about $215\ ^{\circ}\mathrm{C}$; it can be steam-distilled) and less soluble in water than 4-nitrophenol (about $279\ ^{\circ}\mathrm{C}$).
✏️ Example 5 — How much does hydrogen bonding raise the boiling point of water?
The boiling points of the group-16 hydrides are $\mathrm{H_{2}O}$ $100$, $\mathrm{H_{2}S}$ $-60$, $\mathrm{H_{2}Se}$ $-41$ and $\mathrm{H_{2}Te}$ $-2\ ^{\circ}\mathrm{C}$. (a) Why do the boiling points rise from $\mathrm{H_{2}S}$ to $\mathrm{H_{2}Te}$? (b) Extend the straight line through $\mathrm{H_{2}Se}$ and $\mathrm{H_{2}S}$ by one period to estimate the boiling point water "should" have. By how much does hydrogen bonding raise it? (c) The $\mathrm{F{-}H\cdots F}$ hydrogen bond is stronger than $\mathrm{O{-}H\cdots O}$, yet water boils higher than HF ($20\ ^{\circ}\mathrm{C}$). Explain.
(a) The molecules get larger, with more electrons, so the ordinary attractions between them (which grow with molecular size) increase. S, Se and Te are not electronegative enough for hydrogen bonding.
(b) From $\mathrm{H_{2}Se}$ to $\mathrm{H_{2}S}$ the boiling point falls by $19\ ^{\circ}\mathrm{C}$; one more step gives $-60-19=-79\ ^{\circ}\mathrm{C}$. Water actually boils at $100\ ^{\circ}\mathrm{C}$: hydrogen bonding raises it by about $179\ ^{\circ}\mathrm{C}$. Without it, water would be a gas on Earth.
(c) Each $\mathrm{H_{2}O}$ has two H atoms and two lone pairs, so it can take part in up to four hydrogen bonds — a three-dimensional network. HF has three lone pairs but only one H per molecule, so on average it forms only one hydrogen bond per molecule as donor and makes chains. More hydrogen bonds per molecule outweigh the greater strength of each $\mathrm{F{-}H\cdots F}$ bond.
✏️ Example 6 — Recognising hydrogen bonds
(a) Which of these pure substances have intermolecular hydrogen bonds: $\mathrm{CH_{4}}$, $\mathrm{NH_{3}}$, $\mathrm{CH_{3}OH}$, $\mathrm{CH_{3}OCH_{3}}$, $\mathrm{HCl}$, $\mathrm{HF}$? (b) Ethanol, $\mathrm{CH_{3}CH_{2}OH}$, boils at $78\ ^{\circ}\mathrm{C}$; its isomer methoxymethane (dimethyl ether), $\mathrm{CH_{3}OCH_{3}}$, at $-24\ ^{\circ}\mathrm{C}$. Explain. (c) Why is 2-nitrophenol less soluble in water than 4-nitrophenol?
(a) $\mathrm{NH_{3}}$, $\mathrm{CH_{3}OH}$ and $\mathrm{HF}$ (H on N, O or F, plus a lone pair on N, O or F). Not $\mathrm{CH_{4}}$ (C–H is not polar enough), not $\mathrm{HCl}$ (Cl is too large and not electronegative enough), and not pure $\mathrm{CH_{3}OCH_{3}}$, which has no O–H; its O lone pairs can only accept hydrogen bonds, e.g. from water.
(b) Same molar mass ($46\ \mathrm{g\,mol^{-1}}$), but ethanol has an $\mathrm{O{-}H}$ group and its molecules are held together by $\mathrm{O{-}H\cdots O}$ hydrogen bonds; the ether's molecules cannot hydrogen-bond to each other.
(c) In 2-nitrophenol the O–H is used in an intramolecular hydrogen bond to the nitro group, so it is not available to hydrogen-bond with water. 4-nitrophenol's O–H and nitro group are both free to hydrogen-bond with water molecules.
Interactive 2 — Hydrogen Bonding and the Boiling Points of the Hydrides

Boiling points of the hydrides of groups 14 to 17 against the period of the heavier atom. Choose a group: its line is highlighted, and a dashed line extends the trend of periods 4 and 3 back to period 2 — what the boiling point would be without hydrogen bonding. Press ▶ Play to tour the four groups.

period-2 hydride—
actual boiling point—
trend estimate—
raised by—

Exercises

I
Molecular orbitals.
  • 1 Give two differences between an atomic orbital and a molecular orbital.
  • 2 What is meant by LCAO? Write the bonding and antibonding combinations of two $1s$ orbitals $\psi_{A}$ and $\psi_{B}$.
  • 3 State the three conditions for the combination of atomic orbitals.
  • 4 Compare a bonding and an antibonding MO with respect to energy and electron density between the nuclei.
  • 5 How many molecular orbitals are formed when the $2s$ and three $2p$ orbitals of each of two atoms combine?
  • 6 With $z$ as the molecular axis, state whether each pair forms σ MOs, π MOs or none: $2s+2s$; $2p_{z}+2p_{z}$; $2p_{x}+2p_{x}$; $2p_{x}+2p_{y}$.
  • 7 Why are $\pi2p_{x}$ and $\pi2p_{y}$ degenerate?
  • 8 How does the order of the $\sigma2p_{z}$ and $\pi2p$ levels differ between $\mathrm{N_{2}}$ and $\mathrm{O_{2}}$, and why?
1 An atomic orbital belongs to one nucleus (monocentric); a molecular orbital extends over two or more nuclei (polycentric). AOs have shapes fixed by the atom; MO shapes depend on the combining AOs. (Also: MOs are formed by combining AOs.)
2 Linear combination of atomic orbitals: MO wave functions are approximated as sums and differences of AO wave functions. $\sigma1s=\psi_{A}+\psi_{B}$ (bonding), $\sigma^{*}1s=\psi_{A}-\psi_{B}$ (antibonding).
3 Comparable energies; the same symmetry about the molecular axis; maximum overlap.
4 Bonding: lower energy than the AOs, high electron density between the nuclei. Antibonding: higher energy, a nodal plane between the nuclei with little electron density there.
5 $2\times4=8$ MOs: $\sigma2s$, $\sigma^{*}2s$, $\sigma2p_{z}$, $\sigma^{*}2p_{z}$, $\pi2p_{x}$, $\pi2p_{y}$, $\pi^{*}2p_{x}$, $\pi^{*}2p_{y}$.
6 $2s+2s$: σ. $2p_{z}+2p_{z}$: σ. $2p_{x}+2p_{x}$: π. $2p_{x}+2p_{y}$: none (zero overlap, different symmetry).
7 The $p_{x}$ and $p_{y}$ orbitals are identical except for a $90^{\circ}$ rotation about the molecular axis, so their sideways combinations have the same energy.
8 In $\mathrm{N_{2}}$, $\pi2p$ lies below $\sigma2p_{z}$; in $\mathrm{O_{2}}$, $\sigma2p_{z}$ lies below $\pi2p$. In N the $2s$–$2p$ energy gap is small, and mixing of $\sigma2s$ with $\sigma2p_{z}$ raises $\sigma2p_{z}$; in O the gap is large and the mixing negligible.
II
Configurations, bond order and magnetism of neutral molecules.
  • 1 Define bond order and find it for $\mathrm{H_{2}}$ and $\mathrm{He_{2}}$.
  • 2 Write the MO configuration of $\mathrm{Li_{2}}$ and find its bond order.
  • 3 Why does $\mathrm{Be_{2}}$ not exist?
  • 4 Write the configuration of $\mathrm{C_{2}}$ and give its bond order and magnetic behaviour.
  • 5 Write the configuration of $\mathrm{F_{2}}$. Why is the bond in $\mathrm{F_{2}}$ weaker than in $\mathrm{O_{2}}$?
  • 6 Find the bond order of $\mathrm{Ne_{2}}$. What does it predict?
  • 7 Arrange $\mathrm{N_{2}}$, $\mathrm{O_{2}}$ and $\mathrm{F_{2}}$ in order of increasing bond length and explain.
  • 8 Which of $\mathrm{H_{2}}$, $\mathrm{B_{2}}$, $\mathrm{C_{2}}$, $\mathrm{N_{2}}$, $\mathrm{O_{2}}$, $\mathrm{F_{2}}$ are paramagnetic?
1 Bond order $=\tfrac12(N_{b}-N_{a})$. $\mathrm{H_{2}}$: $\tfrac12(2-0)=1$. $\mathrm{He_{2}}$: $\tfrac12(2-2)=0$.
2 $\sigma1s^{2}\sigma^{*}1s^{2}\sigma2s^{2}$; $\tfrac12(4-2)=1$ (diamagnetic).
3 $\mathrm{KK}\,\sigma2s^{2}\sigma^{*}2s^{2}$: $N_{b}=N_{a}=4$, bond order $0$ — no net bonding.
4 $\mathrm{KK}\,\sigma2s^{2}\sigma^{*}2s^{2}(\pi2p_{x}^{2}=\pi2p_{y}^{2})$; $\tfrac12(8-4)=2$; all paired — diamagnetic.
5 $\mathrm{KK}\,\sigma2s^{2}\sigma^{*}2s^{2}\sigma2p_{z}^{2}(\pi2p^{4})(\pi^{*}2p^{4})$; bond order $\tfrac12(10-8)=1$, against $2$ for $\mathrm{O_{2}}$: two more antibonding electrons weaken the bond.
6 20 electrons: $N_{b}=N_{a}=10$, bond order $0$ — $\mathrm{Ne_{2}}$ does not exist.
7 Bond orders $3$, $2$, $1$, so $\mathrm{N_{2}}\ (110)\lt\mathrm{O_{2}}\ (121)\lt\mathrm{F_{2}}\ (142\ \mathrm{pm})$.
8 $\mathrm{B_{2}}$ (two unpaired electrons in $\pi2p$) and $\mathrm{O_{2}}$ (two in $\pi^{*}2p$). The others are diamagnetic.
III
Molecular ions.
  • 1 Find the bond orders and magnetic behaviour of $\mathrm{N_{2}^{+}}$ and $\mathrm{N_{2}^{-}}$.
  • 2 Which is more stable, $\mathrm{O_{2}^{+}}$ or $\mathrm{O_{2}^{-}}$? Explain.
  • 3 Calcium carbide contains the ion $\mathrm{C_{2}^{2-}}$. Find its bond order and magnetic behaviour. Which neutral molecule is it isoelectronic with?
  • 4 Which has the higher bond-dissociation enthalpy: (i) $\mathrm{N_{2}}$ or $\mathrm{N_{2}^{+}}$; (ii) $\mathrm{O_{2}}$ or $\mathrm{O_{2}^{+}}$?
  • 5 Find the bond order of $\mathrm{H_{2}^{-}}$. Is it paramagnetic?
  • 6 How many unpaired electrons are there in $\mathrm{O_{2}^{-}}$, $\mathrm{O_{2}^{2-}}$ and $\mathrm{B_{2}}$?
  • 7 Find the bond order of $\mathrm{Li_{2}^{+}}$ and compare it with $\mathrm{Li_{2}}$.
  • 8 Applied. Potassium superoxide, $\mathrm{KO_{2}}$, is used in emergency breathing apparatus because it reacts with exhaled $\mathrm{CO_{2}}$ and moisture to release oxygen. Is solid $\mathrm{KO_{2}}$ paramagnetic? Find the bond order of its anion.
1 $\mathrm{N_{2}^{+}}$ (13 e): one electron removed from $\sigma2p_{z}$, $N_{b}=9$, $N_{a}=4$ → $2.5$, paramagnetic. $\mathrm{N_{2}^{-}}$ (15 e): one electron in $\pi^{*}2p$, $N_{b}=10$, $N_{a}=5$ → $2.5$, paramagnetic.
2 $\mathrm{O_{2}^{+}}$: bond order $2.5$ against $1.5$ for $\mathrm{O_{2}^{-}}$ — $\mathrm{O_{2}^{+}}$ is more stable.
3 $\mathrm{C_{2}^{2-}}$: $12+2=14$ electrons, configuration as $\mathrm{N_{2}}$: bond order $3$, diamagnetic; isoelectronic with $\mathrm{N_{2}}$ (the $\mathrm{C{\equiv}C}$ triple bond of the acetylide ion).
4 (i) $\mathrm{N_{2}}$ (3 vs 2.5: the electron lost from $\mathrm{N_{2}}$ was bonding). (ii) $\mathrm{O_{2}^{+}}$ (2.5 vs 2: the electron lost from $\mathrm{O_{2}}$ was antibonding).
5 3 electrons: $\sigma1s^{2}\sigma^{*}1s^{1}$; $\tfrac12(2-1)=0.5$; one unpaired electron — paramagnetic.
6 $\mathrm{O_{2}^{-}}$: 1. $\mathrm{O_{2}^{2-}}$: 0. $\mathrm{B_{2}}$: 2.
7 $\mathrm{Li_{2}^{+}}$ (5 e): $\sigma1s^{2}\sigma^{*}1s^{2}\sigma2s^{1}$, $\tfrac12(3-2)=0.5$ — half the bond order of $\mathrm{Li_{2}}$ (1), so a weaker, longer bond.
8 The superoxide ion $\mathrm{O_{2}^{-}}$ has 17 electrons: $(\pi^{*}2p)^{3}$, one unpaired electron, so $\mathrm{KO_{2}}$ is paramagnetic; bond order $\tfrac12(10-7)=1.5$.
IV
Hydrogen bonding.
  • 1 What is a hydrogen bond? Which atoms must be involved?
  • 2 Distinguish intermolecular and intramolecular hydrogen bonding, with one example of each.
  • 3 $\mathrm{NH_{3}}$ boils at $-33\ ^{\circ}\mathrm{C}$ and $\mathrm{PH_{3}}$ at $-88\ ^{\circ}\mathrm{C}$, although $\mathrm{PH_{3}}$ is heavier. Explain.
  • 4 Why is HF a liquid at $15\ ^{\circ}\mathrm{C}$ while HCl is a gas?
  • 5 Explain why ice floats on water.
  • 6 Which is more volatile, 2-nitrophenol or 4-nitrophenol? Why?
  • 7 Why do methanol and ethanol mix with water in all proportions, while hexane does not mix with water?
  • 8 Applied. Ethane-1,2-diol, $\mathrm{HOCH_{2}CH_{2}OH}$, is used as antifreeze in car radiators because it mixes completely with water. Explain this, and why it boils much higher ($197\ ^{\circ}\mathrm{C}$) than ethanol ($78\ ^{\circ}\mathrm{C}$).
1 The attraction between a hydrogen atom covalently bonded to F, O or N and a lone pair on another F, O or N atom: $\mathrm{X{-}H\cdots Y}$.
2 Intermolecular: between molecules, e.g. water, HF chains. Intramolecular: within one molecule, e.g. 2-nitrophenol (O–H···O of the nitro group).
3 $\mathrm{NH_{3}}$ molecules are linked by $\mathrm{N{-}H\cdots N}$ hydrogen bonds; P is not electronegative enough, so $\mathrm{PH_{3}}$ has only weaker attractions.
4 HF molecules form strong $\mathrm{F{-}H\cdots F}$ hydrogen bonds (boiling point $20\ ^{\circ}\mathrm{C}$); HCl cannot (boiling point $-85\ ^{\circ}\mathrm{C}$).
5 In ice every molecule forms four tetrahedral hydrogen bonds, giving an open structure with much empty space; liquid water is more closely packed, so ice is less dense ($0.917$ vs about $1.00\ \mathrm{g\,cm^{-3}}$) and floats.
6 2-nitrophenol: its O–H is tied up in an intramolecular hydrogen bond, so the molecules do not hydrogen-bond to each other; 4-nitrophenol molecules associate through intermolecular hydrogen bonds.
7 The O–H groups of the alcohols form hydrogen bonds with water molecules, replacing those broken in pure water. Hexane cannot hydrogen-bond, and mixing would break water's hydrogen bonds without replacing them.
8 Each molecule has two O–H groups (and four O lone pairs): it forms many hydrogen bonds with water, so it mixes in all proportions, and many hydrogen bonds between its own molecules, so much more energy is needed to boil it than ethanol, which has only one O–H.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET Which of the following molecules is paramagnetic? (A) $\mathrm{N_{2}}$ (B) $\mathrm{C_{2}}$ (C) $\mathrm{O_{2}}$ (D) $\mathrm{F_{2}}$

Idea $\mathrm{O_{2}}$ has $(\pi^{*}2p_{x}^{1}=\pi^{*}2p_{y}^{1})$: two unpaired electrons. $\mathrm{N_{2}}$, $\mathrm{C_{2}}$ and $\mathrm{F_{2}}$ have all electrons paired.
Answer (C).

E2. JEE Main The correct order of bond order is (A) $\mathrm{O_{2}^{2-}\lt O_{2}^{-}\lt O_{2}\lt O_{2}^{+}}$ (B) $\mathrm{O_{2}^{-}\lt O_{2}^{2-}\lt O_{2}\lt O_{2}^{+}}$ (C) $\mathrm{O_{2}^{+}\lt O_{2}\lt O_{2}^{-}\lt O_{2}^{2-}}$ (D) $\mathrm{O_{2}\lt O_{2}^{+}\lt O_{2}^{-}\lt O_{2}^{2-}}$

Bond orders $\mathrm{O_{2}^{2-}}$ $1$, $\mathrm{O_{2}^{-}}$ $1.5$, $\mathrm{O_{2}}$ $2$, $\mathrm{O_{2}^{+}}$ $2.5$ (each added electron goes into $\pi^{*}$).
Answer (A).

E3. Board · 3 marks Write the molecular orbital configurations of $\mathrm{N_{2}}$ and $\mathrm{O_{2}}$. Calculate their bond orders and predict their magnetic behaviour.

1 mark $\mathrm{N_{2}}$: $\sigma1s^{2}\sigma^{*}1s^{2}\sigma2s^{2}\sigma^{*}2s^{2}(\pi2p_{x}^{2}=\pi2p_{y}^{2})\sigma2p_{z}^{2}$. $\mathrm{O_{2}}$: $\sigma1s^{2}\sigma^{*}1s^{2}\sigma2s^{2}\sigma^{*}2s^{2}\sigma2p_{z}^{2}(\pi2p_{x}^{2}=\pi2p_{y}^{2})(\pi^{*}2p_{x}^{1}=\pi^{*}2p_{y}^{1})$.
1 mark $\mathrm{N_{2}}$: $\tfrac12(10-4)=3$. $\mathrm{O_{2}}$: $\tfrac12(10-6)=2$.
1 mark $\mathrm{N_{2}}$: all electrons paired — diamagnetic. $\mathrm{O_{2}}$: two unpaired electrons — paramagnetic.

E4. JEE Main Numerical answer. How many of the following species have a bond order of $2.5$? $\mathrm{O_{2}^{+}}$, $\mathrm{N_{2}^{+}}$, $\mathrm{O_{2}^{-}}$, $\mathrm{N_{2}^{-}}$, $\mathrm{C_{2}}$, $\mathrm{He_{2}^{+}}$

Bond orders $\mathrm{O_{2}^{+}}$ $\tfrac12(10-5)=2.5$; $\mathrm{N_{2}^{+}}$ $\tfrac12(9-4)=2.5$; $\mathrm{O_{2}^{-}}$ $1.5$; $\mathrm{N_{2}^{-}}$ $\tfrac12(10-5)=2.5$; $\mathrm{C_{2}}$ $2$; $\mathrm{He_{2}^{+}}$ $0.5$.
Answer 3.

E5. JEE Advanced One or more options correct. (A) $\mathrm{B_{2}}$ is paramagnetic because $\pi2p$ lies below $\sigma2p_{z}$ in boron. (B) $\mathrm{He_{2}^{+}}$ has bond order $0.5$ and can exist. (C) $\mathrm{N_{2}^{+}}$ has a larger bond-dissociation enthalpy than $\mathrm{N_{2}}$. (D) Removing an electron from $\mathrm{O_{2}}$ shortens the oxygen–oxygen bond.

(A) True: with s–p mixing the last two electrons enter the degenerate $\pi2p$ singly.
(B) True: $\tfrac12(2-1)=0.5\gt0$.
(C) False: the electron removed from $\mathrm{N_{2}}$ was bonding; bond order falls from 3 to 2.5.
(D) True: the electron leaves $\pi^{*}$; bond order rises from 2 to 2.5 ($121\rightarrow112\ \mathrm{pm}$).
Answer (A), (B), (D).

E6. Board · Case study The boiling points of $\mathrm{H_{2}O}$, $\mathrm{H_{2}S}$, $\mathrm{H_{2}Se}$ and $\mathrm{H_{2}Te}$ are $100$, $-60$, $-41$ and $-2\ ^{\circ}\mathrm{C}$. (i) Explain why water does not follow the trend shown by the other three. (2 marks) (ii) Why does ice float on water? (1 mark) (iii) Which of 2-nitrophenol and 4-nitrophenol is steam-volatile, and why? (1 mark)

(i) From $\mathrm{H_{2}S}$ to $\mathrm{H_{2}Te}$ the boiling point rises as the molecules get larger. Oxygen is small and highly electronegative, so water molecules are linked by extensive $\mathrm{O{-}H\cdots O}$ hydrogen bonds; extra energy is needed to break them, and water boils about $180\ ^{\circ}\mathrm{C}$ above the trend.
(ii) Hydrogen bonding gives ice an open tetrahedral structure that is less dense than liquid water.
(iii) 2-nitrophenol: intramolecular hydrogen bonding prevents association with other molecules (and with water), so it is volatile in steam.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. The two strands of DNA are held together by hydrogen bonds between base pairs. Why does a PCR test heat a DNA sample to about $95\ ^{\circ}\mathrm{C}$, and why do DNA samples rich in G–C pairs need higher temperatures to separate?

Why it matters. The polymerase chain reaction (PCR) copies DNA millions of times; it was the standard test for SARS-CoV-2 during the COVID-19 pandemic and is used in forensic science, food safety and the diagnosis of inherited diseases. Every cycle begins by "melting" the double helix — breaking the hydrogen bonds between the strands without breaking any covalent bond.

What to do.

  1. Adenine pairs with thymine through two hydrogen bonds and guanine with cytosine through three ($\mathrm{N{-}H\cdots N}$ and $\mathrm{N{-}H\cdots O}$). Which atoms in these hydrogen bonds satisfy Definition 4.14?
  2. A PCR primer is a short strand of 20 bases. Count the hydrogen bonds holding it to its target if it contains 9 G–C pairs and 11 A–T pairs; repeat for 13 G–C and 7 A–T.
  3. An empirical formula used for long DNA in a standard salt solution is $T_{m}=69.3+0.41\times(\%\mathrm{GC})$, where $T_{m}$ in $^{\circ}\mathrm{C}$ is the temperature at which half the DNA has separated. Find $T_{m}$ for DNA with 40% and with 60% G–C content.
  4. Explain, using bond enthalpies from Section 4.1, why heating to $95\ ^{\circ}\mathrm{C}$ separates the strands but leaves the covalent backbone intact.

Tools: Definition 4.14, Rule 4.13, Section 4.1 bond enthalpies, and a molecular-biology reference showing the base-pair structures.

Hand in: A labelled sketch of an A–T and a G–C pair with the hydrogen bonds marked, your counts and $T_{m}$ values, and a paragraph on why PCR cycles between about $95$, $55$ and $72\ ^{\circ}\mathrm{C}$.

Going further. Proteins fold into helices and sheets held by $\mathrm{N{-}H\cdots O{=}C}$ hydrogen bonds. Why does cooking an egg (heating its proteins) change it irreversibly, while melted DNA can re-form its double helix on cooling?

Instructor note: donors are N–H groups, acceptors N or O lone pairs — all N/O, as required. Counts: $9\times3+11\times2=49$; $13\times3+7\times2=53$. $T_{m}$: $69.3+0.41\times40=85.7\ ^{\circ}\mathrm{C}$; $69.3+0.41\times60=93.9\ ^{\circ}\mathrm{C}$. Hydrogen bonds are about $10$–$40\ \mathrm{kJ\,mol^{-1}}$ against several hundred for covalent bonds, so thermal energy at $95\ ^{\circ}\mathrm{C}$ (RT ≈ $3\ \mathrm{kJ\,mol^{-1}}$, but many cooperative bonds) unzips the strands without breaking the backbone. Egg proteins unfold and then tangle and aggregate irreversibly; DNA's complementary strands can find each other again (annealing).

🧪 Self-check quiz

0/5 correct

Q1. The bond order of the superoxide ion, $\mathrm{O_{2}^{-}}$, is

Q2. Which species is diamagnetic?

Q3. $\mathrm{He_{2}}$ does not exist because

Q4. Intramolecular hydrogen bonding is present in

Q5. In which molecule does the $\sigma2p_{z}$ orbital lie below the $\pi2p$ orbitals?

← 4.3 Valence Bond Theory and HybridisationChapter 5 — Thermodynamics →