Advanced Bonding Theories

CHEM 50 — General Chemistry I · Dr. Karmach

build Jul 13 · 15:49 · CC BY-NC-SA 4.0 · OpenStax Chemistry 2e where noted
Dr. Karmach

By the end of this unit, you can…

  • Assign sp, sp², or sp³ hybridization to a central atom from its number of electron domains
  • Count sigma and pi bonds in a molecule from its Lewis structure
  • Build the molecular-orbital occupation of a second-period diatomic and compute its bond order
  • Predict paramagnetism or diamagnetism from a molecular-orbital diagram
Dr. Karmach

Today's route 🗺️

  1. Hybridization
  2. Sigma & Pi Bonds
  3. Molecular Orbitals & Bond Order
  4. MO Diagrams & Magnetism
Dr. Karmach

1 · Hybridization

Assign sp, sp², or sp³ hybridization to a central atom by counting its electron domains.

Dr. Karmach

Four identical bonds

Natural gas is methane. Its carbon makes four bonds of equal length and strength, spread evenly in space — yet carbon's electrons began in two different kinds of orbitals.

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Electron domains set the hybridization

2 domains → mix 2 orbitals · 3 domains → mix 3 · 4 domains → mix 4
the mixed orbitals are hybrids — equal in energy, each aimed at one domain

A central atom mixes its valence s and p orbitals into new, equivalent hybrid orbitals. The number it makes equals its number of electron domains, and each hybrid points at one domain.

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One s and three p become four alike

Carbon's one 2s and three 2p orbitals differ in shape and energy. Mixing them makes four identical sp³ orbitals, aimed at a tetrahedron's corners. Matched orbitals make matched bonds.

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Two, three, or four domains

Each domain count mixes that many orbitals into one matched set: two make sp, three make sp², four make sp³. The set points where VSEPR placed the domains: linear, trigonal planar, tetrahedral.

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A lone pair is a domain

H₂O: 2 bonds + 2 lone pairs = 4 domains → sp³
bonded atoms: 2 · lone pairs on O: 2 · total electron domains: 4

Every region of electron density around the central atom is a domain — a bond or a lone pair. Both take a hybrid orbital. Count bonds and lone pairs together.

Dr. Karmach

The method

  1. Draw the Lewis structure and find the central atom.
  2. Count the electron domains. Bonded groups and lone pairs each count once; a multiple bond counts once.
  3. Match the count: 2 → sp, 3 → sp², 4 → sp³.
Dr. Karmach

Worked example 1 — methane, CH₄

CH₄
central atom: C · bonded atoms: 4 H · lone pairs on C: 0

Methane's carbon bonds to four hydrogen atoms with no lone pairs left over. Assign the hybridization of the carbon: count its domains, then match.

Dr. Karmach

Worked example 1 — solution

CH₄
central atom: C · bonded atoms: 4 H · lone pairs on C: 0

Step 1 · Draw the Lewis structure and find the central atom

Carbon sits in the center, single-bonded to four hydrogens, with no lone pairs.

Dr. Karmach

Worked example 1 — solution

CH₄
central atom: C · bonded atoms: 4 H · lone pairs on C: 0
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains
4 bonded groups + 0 lone pairs = 4 domains
every bond is a single bond — four separate domains
Dr. Karmach

Worked example 1 — solution

CH₄
central atom: C · bonded atoms: 4 H · lone pairs on C: 0
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains
4 bonded groups + 0 lone pairs = 4 domains
every bond is a single bond — four separate domains
Step 3 · Match the count
4 domains → sp³
mix one 2s + three 2p → four equivalent sp³ hybrids, tetrahedral
Dr. Karmach

Worked example 1 — solution

CH₄
central atom: C · bonded atoms: 4 H · lone pairs on C: 0
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains
4 bonded groups + 0 lone pairs = 4 domains
every bond is a single bond — four separate domains
Step 3 · Match the count
4 domains → sp³
mix one 2s + three 2p → four equivalent sp³ hybrids, tetrahedral
Four matched hybrids point at a tetrahedron's corners, 109.5° apart — the shape methane actually has.
Dr. Karmach

Worked example 2 — ammonia, NH₃

NH₃
central atom: N · bonded atoms: 3 H · lone pairs on N: 1

Nitrogen bonds to three hydrogens and keeps one lone pair. Assign the hybridization of the nitrogen.

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Worked example 2 — solution

NH₃
central atom: N · bonded atoms: 3 H · lone pairs on N: 1

Step 1 · Draw the Lewis structure and find the central atom

Nitrogen is central, bonded to three hydrogens, with one lone pair on top.

Dr. Karmach

Worked example 2 — solution

NH₃
central atom: N · bonded atoms: 3 H · lone pairs on N: 1
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains
3 bonded groups + 1 lone pair = 4 domains
the lone pair is a domain — counting only the 3 bonded atoms would miss it
Dr. Karmach

Worked example 2 — solution

NH₃
central atom: N · bonded atoms: 3 H · lone pairs on N: 1
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains
3 bonded groups + 1 lone pair = 4 domains
the lone pair is a domain — counting only the 3 bonded atoms would miss it
Step 3 · Match the count
4 domains → sp³
three sp³ hybrids hold bonds · the fourth holds the lone pair
Dr. Karmach

Worked example 2 — solution

NH₃
central atom: N · bonded atoms: 3 H · lone pairs on N: 1
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains
3 bonded groups + 1 lone pair = 4 domains
the lone pair is a domain — counting only the 3 bonded atoms would miss it
Step 3 · Match the count
4 domains → sp³
three sp³ hybrids hold bonds · the fourth holds the lone pair
Same sp³ as methane, four domains, even though one hybrid holds a lone pair instead of a bond.
Dr. Karmach

Your turn — boron trifluoride, BF₃

BF₃
central atom: B · bonded atoms: 3 F · lone pairs on B: 0
step result
count the electron domains 3 bonded + 0 lone pairs = domains
match the count

Fill in the domain count, then the hybridization.

Dr. Karmach

Your turn — boron trifluoride, BF₃

BF₃
central atom: B · bonded atoms: 3 F · lone pairs on B: 0
step result
count the electron domains 3 bonded + 0 lone pairs = domains
match the count

Fill in the domain count, then the hybridization.

3 domains → sp²
mix one 2s + two 2p → three equivalent sp² hybrids, trigonal planar (120°)
Boron carries only six valence electrons here — three bonds, no lone pair, three domains, sp².
Dr. Karmach

Where this goes wrong

Counting only the bonded atoms. Water has two bonds, so its oxygen looks like sp. But the two lone pairs are domains too: 2 + 2 = 4, sp³. Every lone pair takes a hybrid orbital.
Reading the superscript as a domain count. Four domains is not sp⁴. The superscripts count the p orbitals mixed in: one s + three p = sp³. There is no fourth p orbital to add.
Skipping hybridization altogether. Pure s and p orbitals would bond at 90°. Real molecules bond at 109.5°, 120°, or 180°. Mixing the orbitals is what points them at the measured geometry.
Dr. Karmach

Practice 1

NI₃ — nitrogen triiodide
central atom: N · bonded atoms: 3 I · lone pairs on N: 1

What is the hybridization of the central nitrogen atom?

  1. sp² — three iodines bonded to the nitrogen
  2. sp³ — three bonds and one lone pair on the nitrogen
  3. sp⁴ — one s orbital plus three p orbitals, four in all
  4. Unhybridized — nitrogen bonds through pure s and p orbitals
Dr. Karmach

Practice 1 — answer: B

NI₃: 3 bonds + 1 lone pair = 4 domains → sp³ — answer B
central N · four electron domains · tetrahedral arrangement

A counted only the three bonded iodines and missed the lone pair: 3 domains → sp². C read the superscript as the domain count; the superscripts count orbitals mixed, one s + three p = sp³, and there is no sp⁴. D skipped hybridization; pure s and p would bond near 90°, not the wider angles nitrogen shows.

Nitrogen here matches ammonia — three bonds, one lone pair, four domains, sp³. The lone pair counts.
Dr. Karmach

Worked example 3 — carbon dioxide, CO₂

O=C=O
central atom: C · two double bonds · lone pairs on C: 0

Carbon sits between two oxygens, joined by a double bond on each side. A common first attempt: two double bonds, count two domains each, four total. Assign the hybridization of the carbon.

Dr. Karmach

Worked example 3 — solution

O=C=O
central atom: C · two double bonds · lone pairs on C: 0

A common first attempt

two double bonds, two domains each = 4 domains → sp³?
counting each line of a double bond separately ✗

Two double bonds look like four domains. But a domain is a region of electron density, and a double bond is one region.

Dr. Karmach

Worked example 3 — solution

O=C=O
central atom: C · two double bonds · lone pairs on C: 0
A common first attempt
two double bonds, two domains each = 4 domains → sp³?
counting each line of a double bond separately ✗
Step 1 · Draw the Lewis structure and find the central atom

Carbon is central, double-bonded to each oxygen, with no lone pairs.

Dr. Karmach

Worked example 3 — solution

O=C=O
central atom: C · two double bonds · lone pairs on C: 0
A common first attempt
two double bonds, two domains each = 4 domains → sp³?
counting each line of a double bond separately ✗
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains Step 3 · Match the count
1 double bond + 1 double bond = 2 domains → sp
each double bond is one domain · two sp hybrids, linear (180°)
Dr. Karmach

Worked example 3 — solution

O=C=O
central atom: C · two double bonds · lone pairs on C: 0
A common first attempt
two double bonds, two domains each = 4 domains → sp³?
counting each line of a double bond separately ✗
Step 1 · Draw the Lewis structure and find the central atom Step 2 · Count the electron domains Step 3 · Match the count
1 double bond + 1 double bond = 2 domains → sp
each double bond is one domain · two sp hybrids, linear (180°)
Two domains push apart to 180°. The double bonds hold extra electrons, but they still point in two directions.
Dr. Karmach

Take-home: a multiple bond is one domain

CO₂: O=C=O → 2 domains → sp
two double bonds, each one domain — not four ✓
counting each double bond as two → 4 domains → sp³
the classic error — a double or triple bond is one region of electrons ✗

A double or triple bond crowds more electrons between the same two atoms, but it stays one domain. Count regions of electron density, not lines.

Dr. Karmach

Practice 2

H₂C=O — formaldehyde
central atom: C · bonded groups: 2 H and 1 O · one C=O double bond · lone pairs on C: 0

What is the hybridization of the central carbon atom?

  1. sp — a carbon joined by a double bond, the same as in CO₂
  2. sp² — two C–H bonds and one C=O double bond, three domains
  3. sp³ — the C=O double bond counted as two domains, four in all
  4. Unhybridized — carbon bonds through pure s and p orbitals
Dr. Karmach

Practice 2 — answer: B

H₂C=O: 2 C–H + 1 C=O = 3 domains → sp² — answer B
the double bond counts once · three domains · trigonal planar

A over-generalized carbon dioxide; formaldehyde also has two C–H bonds, so three domains, not two. C counted the C=O double bond as two domains: 2 + 2 = 4 would give sp³, but a double bond is one domain. D skipped hybridization; pure orbitals cannot give the 120° angles formaldehyde shows.

Three groups around carbon — two hydrogens and one oxygen — spread to 120°. One double bond, still one domain.
Dr. Karmach

Check yourself

  1. Sulfur dioxide, SO₂, has a central sulfur with two bonding groups and one lone pair. How many domains, and what hybridization?
  2. In hydrogen cyanide, H–C≡N, the carbon has one single bond and one triple bond. What is its hybridization?

Each hybrid orbital overlaps end-to-end to form one sigma bond. The extra lines of a double or triple bond are pi bonds, built from the p orbitals left unmixed. Sorting every bond into sigma and pi builds straight on the hybridization counted here.

Dr. Karmach

2 · Sigma & Pi Bonds

Count the sigma and pi bonds in a molecule from its Lewis structure, using that every bond is one σ and each extra line of a double or triple bond is a π.

Dr. Karmach

Bonds that turn and bonds that don't

The ends of a single bond spin freely, like a wheel on an axle. A double bond is locked. That rigidity fixes a molecule's shape.

Dr. Karmach

Every bond is one sigma

Two kinds of bond form between atoms: sigma and pi. The first bond between any two atoms is always a σ. Every extra line of a double or triple bond is a π.

Dr. Karmach

Sigma and pi: two kinds of overlap

A sigma bond overlaps end-on, along the bond axis. A pi bond overlaps side-on, above and below it. The σ framework is built from hybrid orbitals.

Dr. Karmach

Why a double bond cannot twist

single bond → free rotation  ·  double bond → locked
the σ lies on the axis, so the ends turn · a π sits off the axis, and twisting would tear it

A σ bond's overlap lies on the axis, so the ends spin freely. A π bond's overlap sits off the axis, and twisting would tear it. So a double bond holds its shape.

Dr. Karmach

The method

  1. Read the structure. Mark every bond: single, double, or triple.
  2. Give each bond one σ. Every bonded pair has one σ.
  3. Count the extra lines as π, then total. A double adds 1 π; a triple adds 2 π.
Dr. Karmach

Worked example 1 — methane

Step 1 · Read the structure

CH₄ — four C–H single bonds
given: the Lewis structure of CH₄ · wanted: total σ and π

Methane has four single bonds and no double or triple bonds. Count the σ and the π.

Dr. Karmach

Worked example 1 — solution

CH₄ — four C–H single bonds
given: the Lewis structure of CH₄ · wanted: total σ and π

Step 2 · Give each bond one σ

Four single bonds, one σ apiece.

σ: 4 + 0 + 0 = 4
four C–H bonds · no multiple bonds to add
Dr. Karmach

Worked example 1 — solution

CH₄ — four C–H single bonds
given: the Lewis structure of CH₄ · wanted: total σ and π
Step 2 · Give each bond one σ
σ: 4 + 0 + 0 = 4
four C–H bonds · no multiple bonds to add
Step 3 · Count the extra lines as π

No double or triple bonds means no extra lines.

π: 0 + 2×0 = 0
π comes only from the extra lines of a double or triple
Dr. Karmach

Worked example 1 — solution

CH₄ — four C–H single bonds
given: the Lewis structure of CH₄ · wanted: total σ and π
Step 2 · Give each bond one σ
σ: 4 + 0 + 0 = 4
four C–H bonds · no multiple bonds to add
Step 3 · Count the extra lines as π
π: 0 + 2×0 = 0
π comes only from the extra lines of a double or triple
Four bonds, four σ, zero π. A molecule of only single bonds carries only σ bonds.
Dr. Karmach

Worked example 2 — ethylene

Step 1 · Read the structure

H₂C=CH₂ — four C–H bonds and one C=C double
given: the Lewis structure of C₂H₄ · wanted: total σ and π

Ethylene has one double bond. A common first attempt: count the double as two σ. Find the σ and the π.

Dr. Karmach

Worked example 2 — solution

H₂C=CH₂ — four C–H bonds and one C=C double
given: the Lewis structure of C₂H₄ · wanted: total σ and π

A common first attempt

count the C=C as two σ → 4 + 2 = 6 σ?
that reads both lines of the double as σ — only one σ fits along the axis ✗

Only one σ lies on the axis between the two carbons. The double's second line is not a second σ.

Dr. Karmach

Worked example 2 — solution

H₂C=CH₂ — four C–H bonds and one C=C double
given: the Lewis structure of C₂H₄ · wanted: total σ and π
A common first attempt
count the C=C as two σ → 4 + 2 = 6 σ?
that reads both lines of the double as σ — only one σ fits along the axis ✗
Step 2 · Give each bond one σ

Five bonded pairs — four C–H and one C=C — one σ each.

Dr. Karmach

Worked example 2 — solution

H₂C=CH₂ — four C–H bonds and one C=C double
given: the Lewis structure of C₂H₄ · wanted: total σ and π
A common first attempt
count the C=C as two σ → 4 + 2 = 6 σ?
that reads both lines of the double as σ — only one σ fits along the axis ✗
Step 2 · Give each bond one σ Step 3 · Count the extra lines as π
σ: 4 + 1 + 0 = 5  ·  π: 1 + 2×0 = 1
every bonded pair gives one σ · the double's second line is the one π
Dr. Karmach

Worked example 2 — solution

H₂C=CH₂ — four C–H bonds and one C=C double
given: the Lewis structure of C₂H₄ · wanted: total σ and π
A common first attempt
count the C=C as two σ → 4 + 2 = 6 σ?
that reads both lines of the double as σ — only one σ fits along the axis ✗
Step 2 · Give each bond one σ Step 3 · Count the extra lines as π
σ: 4 + 1 + 0 = 5  ·  π: 1 + 2×0 = 1
every bonded pair gives one σ · the double's second line is the one π
Five σ and one π. The double bond splits into one σ and one π, never two σ.
Dr. Karmach

Take-home: a double bond is not two σ

C=C → 1 σ + 1 π  (not 2 σ)
one σ lies on the axis · the second line is the π · a triple is 1 σ + 2 π

The two lines of a double bond are not two σ bonds. The first line is the σ; the second is a π.

Dr. Karmach

Your turn — carbon dioxide

O=C=O — two C=O double bonds
each double bond splits into one σ and one π
bond σ π
first C=O
second C=O
total

Give each double bond one σ and one π, then total.

Dr. Karmach

Your turn — carbon dioxide

O=C=O — two C=O double bonds
each double bond splits into one σ and one π
bond σ π
first C=O
second C=O
total

Give each double bond one σ and one π, then total.

σ: 1 + 1 = 2  ·  π: 1 + 1 = 2
two doubles → 2 σ and 2 π · CO₂ has 2 σ and 2 π
Dr. Karmach

Where this goes wrong

Counting every line as a σ. A double or triple looks like two or three σ. Only one σ fits along the axis. Ethylene is 5 σ and 1 π, not 4 + 2 = 6 σ with no π.
Dropping the σ inside a multiple bond. Reading the double as pure π forgets the σ on the axis. Ethylene's double is 1 σ + 1 π, so 5 σ and 1 π — not 4 σ and 2 π.
Counting a double bond as 2 σ. A double bond is one σ plus one π, never two σ. Counting 2 σ turns ethylene's 5 σ into 6 σ, still with its 1 π.
Dr. Karmach

Practice 1

H₂C=O — two C–H bonds and one C=O double
formaldehyde · count every bond, single and double

Counting every bond, how many σ and how many π does formaldehyde contain?

  1. 4 σ and 0 π
  2. 3 σ and 1 π
  3. 2 σ and 2 π
  4. 4 σ and 1 π
Dr. Karmach

Practice 1 — answer: B

H₂C=O → 2 + 1 = 3 σ and 1 π — answer B
2 C–H + 1 σ in the C=O · the double's second line is the π

A counted every line as a σ: 2 + 2 = 4 σ, and no π. C dropped the σ inside the double, reading both of its lines as π: 2 σ and 2 π. D counted the double as 2 σ and then added its π: 2 + 2 = 4 σ and 1 π.

Three σ and one π. The double contributes one of each; the two C–H bonds add the other two σ.
Dr. Karmach

Worked example 3 — acetylene

Step 1 · Read the structure

H–C≡C–H — two C–H bonds and one C≡C triple
given: the Lewis structure of C₂H₂ · wanted: total σ and π

Acetylene has a triple bond. Count the σ and the π.

Dr. Karmach

Worked example 3 — solution

H–C≡C–H — two C–H bonds and one C≡C triple
given: the Lewis structure of C₂H₂ · wanted: total σ and π

Step 2 · Give each bond one σ

Three bonded pairs — two C–H and one C≡C — one σ each.

σ: 2 + 1 = 3
the triple counts as one σ, like every bonded pair
Dr. Karmach

Worked example 3 — solution

H–C≡C–H — two C–H bonds and one C≡C triple
given: the Lewis structure of C₂H₂ · wanted: total σ and π
Step 2 · Give each bond one σ
σ: 2 + 1 = 3
the triple counts as one σ, like every bonded pair
Step 3 · Count the extra lines as π

The triple bond has two extra lines beyond its σ.

π: 2×1 = 2
a triple adds two π · the σ is already counted
Dr. Karmach

Worked example 3 — solution

H–C≡C–H — two C–H bonds and one C≡C triple
given: the Lewis structure of C₂H₂ · wanted: total σ and π
Step 2 · Give each bond one σ
σ: 2 + 1 = 3
the triple counts as one σ, like every bonded pair
Step 3 · Count the extra lines as π
π: 2×1 = 2
a triple adds two π · the σ is already counted
Three σ and two π. The triple bond is one σ and two π; the two C–H bonds add the other two σ.
Dr. Karmach

Practice 2

H₃C–C≡N — three C–H bonds, one C–C, and one C≡N triple
acetonitrile · count every bond in the structure

Counting every bond, how many σ and how many π does acetonitrile contain?

  1. 7 σ and 0 π
  2. 4 σ and 3 π
  3. 5 σ and 2 π
  4. 7 σ and 2 π
Dr. Karmach

Practice 2 — answer: C

H₃C–C≡N → 3 + 1 + 1 = 5 σ and 2×1 = 2 π — answer C
4 single bonds + 1 σ in the triple · the triple adds 2 π

A counted every line as a σ: 4 + 3 = 7 σ, and no π. B dropped the σ inside the triple, reading its three lines as 3 π with only 4 σ. D counted every line as a σ and then added the triple's π: 7 σ and 2 π.

Five σ and two π. Four single bonds and the triple's σ make five; the triple's two extra lines are the π.
Dr. Karmach

Check yourself

  1. In HCN (H–C≡N): how many σ and how many π? Give each bond one σ, then count the extra lines.
  2. Why can a single bond rotate while a double bond cannot? Name the bond that would have to break.

A σ bond allows rotation; a π bond fixes a shape. Counting them reads a fixed Lewis structure. Molecular-orbital theory goes further: atomic orbitals combine into bonding and antibonding levels, and how they fill gives a bond order — the measure a Lewis structure cannot express.

Dr. Karmach

3 · Molecular Orbitals & Bond Order

Build the molecular-orbital picture of a second-period diatomic, count its bonding and antibonding electrons, and take the bond order as (bonding − antibonding) ÷ 2 — the number that says how strong the bond is and whether the molecule can exist at all.

Dr. Karmach

The air's unreactive gas

Nitrogen is most of the air, yet it barely reacts. Its two atoms are locked together by a triple bond that takes lightning or furnace heat to break.

Dr. Karmach

Why a molecule holds together

H₂: 2 bonding, 0 antibonding — bond order = (2 − 0)/2 = 1
bonding wins → the molecule exists
He₂: 2 bonding, 2 antibonding — bond order = (2 − 2)/2 = 0
bonding and antibonding cancel → no stable molecule

Electrons in bonding orbitals pull the two atoms together. Electrons in antibonding orbitals push them apart. A molecule is stable only when bonding wins. Bond order counts that surplus, in pairs.

Dr. Karmach

How molecular orbitals form

Two atomic orbitals, one from each atom, combine into two molecular orbitals: a bonding MO at lower energy and an antibonding MO (starred) at higher energy. Electrons fill the lowest first.

Dr. Karmach

Bond order measures the bond

bond order = (bonding electrons − antibonding electrons) ÷ 2
the net number of bonding pairs

The formula counts the net bonding pairs. A higher bond order means a stronger, shorter bond. A bond order of zero means the atoms do not stay bonded.

Dr. Karmach

The method

  1. Fill the orbitals lowest-first: valence electrons into the MOs.
  2. Count bonding and antibonding electrons.
  3. Bond order = (bonding − antibonding) ÷ 2.
  4. Read the bond: higher order, stronger and shorter.
Dr. Karmach

Worked example 1 — the bond order of N₂

bond order = (bonding − antibonding) ÷ 2
given: N₂, 10 valence electrons · wanted: the bond order

The molecular orbitals of N₂ are filled with its 10 valence electrons. Count the bonding and antibonding electrons, then find the bond order.

Dr. Karmach

Worked example 1 — solution

bond order = (bonding − antibonding) ÷ 2
N₂: 10 valence electrons in the molecular orbitals

Step 1 · Fill the orbitals lowest-first

The 10 valence electrons fill σ2s, σ*2s, π2p, then σ2p — lowest energy first.

Dr. Karmach

Worked example 1 — solution

bond order = (bonding − antibonding) ÷ 2
N₂: 10 valence electrons in the molecular orbitals
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding

Bonding orbitals σ2s, π2p, σ2p hold 2 + 4 + 2 = 8 electrons. Antibonding σ*2s holds 2.

Dr. Karmach

Worked example 1 — solution

bond order = (bonding − antibonding) ÷ 2
N₂: 10 valence electrons in the molecular orbitals
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding Step 3 · Bond order = (bonding − antibonding) ÷ 2
bond order = 8 − 22 = 3
Dr. Karmach

Worked example 1 — solution

bond order = (bonding − antibonding) ÷ 2
N₂: 10 valence electrons in the molecular orbitals
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding Step 3 · Bond order = (bonding − antibonding) ÷ 2
bond order = 8 − 22 = 3
Step 4 · Read the bond
Bond order 3: a triple bond. That is the strong, short bond that makes nitrogen gas so hard to pull apart. ✓
Dr. Karmach

Worked example 2 — the bond order of O₂

bond order = (bonding − antibonding) ÷ 2
given: O₂, 12 valence electrons · wanted: the bond order

Oxygen has 12 valence electrons, two more than nitrogen. Fill the orbitals, count each kind, and find the bond order.

Dr. Karmach

Worked example 2 — solution

bond order = (bonding − antibonding) ÷ 2
O₂: 12 valence electrons in the molecular orbitals

Step 1 · Fill the orbitals lowest-first

Eight electrons fill the bonding orbitals (σ2s, σ2p, π2p). The last four fill antibonding orbitals (σ2s, π2p).

Dr. Karmach

Worked example 2 — solution

bond order = (bonding − antibonding) ÷ 2
O₂: 12 valence electrons in the molecular orbitals
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding

Bonding: 8 electrons. Antibonding: 4 electrons — two more than nitrogen carried.

Dr. Karmach

Worked example 2 — solution

bond order = (bonding − antibonding) ÷ 2
O₂: 12 valence electrons in the molecular orbitals
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding Step 3 · Bond order = (bonding − antibonding) ÷ 2
bond order = 8 − 42 = 2
Dr. Karmach

Worked example 2 — solution

bond order = (bonding − antibonding) ÷ 2
O₂: 12 valence electrons in the molecular orbitals
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding Step 3 · Bond order = (bonding − antibonding) ÷ 2
bond order = 8 − 42 = 2
Step 4 · Read the bond
Bond order 2: a double bond, weaker and longer than nitrogen's triple bond. Four antibonding electrons cancel four of the bonding ones. ✓
Dr. Karmach

Your turn — the bond order of F₂

bond order = (bonding − antibonding) ÷ 2
given: F₂ · bonding 8, antibonding 6 · wanted: the bond order

F₂ fills every bonding orbital and all but one antibonding orbital.

bond order = − 2 =

Fill in the bonding and antibonding counts, then compute.

Dr. Karmach

Your turn — the bond order of F₂

bond order = (bonding − antibonding) ÷ 2
given: F₂ · bonding 8, antibonding 6 · wanted: the bond order

F₂ fills every bonding orbital and all but one antibonding orbital.

bond order = − 2 =

Fill in the bonding and antibonding counts, then compute.

bond order = 8 − 62 = 1
Bond order 1: a single bond. Fluorine's six antibonding electrons cancel six of its eight bonding ones, leaving one net pair. ✓
Dr. Karmach

Where this goes wrong

bond order = (bonding − antibonding) ÷ 2
N₂: bonding 8, antibonding 2 · correct bond order = 3
Forgetting to divide by two. 8 − 2 = 6 is the surplus of bonding electrons, not the bond order. Bond order counts PAIRS, so halve it: (8 − 2)/2 = 3.
Adding the antibonding electrons. Antibonding electrons cancel bonding ones, so subtract. Adding gives (8 + 2)/2 = 5, a bond order larger than any real diatomic.
Subtracting in the wrong order. (2 − 8)/2 = −3. A negative bond order has no meaning. Bonding electrons come first: (8 − 2)/2 = 3.
Counting every electron. 8 + 2 = 10 is the total number of valence electrons, not the difference. Bond order uses bonding minus antibonding, then ÷ 2.
Dr. Karmach

Practice 1

bond order = (bonding − antibonding) ÷ 2
given: B₂ · bonding 4, antibonding 2 · wanted: the bond order

The molecular orbitals of B₂ hold 4 bonding electrons and 2 antibonding electrons. What is the bond order?

  1. 1
  2. 2
  3. 3
  4. 6
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Practice 1 — answer: A

bond order = (bonding − antibonding) ÷ 2
B₂: bonding 4, antibonding 2
bond order = 4 − 22 = 1 — answer A

B forgot to divide by two: 4 − 2 = 2 is the electron surplus, not the bond order. C added the antibonding electrons: (4 + 2)/2 = 3. D counted every electron: 4 + 2 = 6, the total, not the difference.

Bond order 1: B₂ is held by a single net bonding pair. ✓
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Worked example 3 — the bond order of O₂⁺

bond order = (bonding − antibonding) ÷ 2
given: O₂⁺ · bonding 8, antibonding 3 · wanted: the bond order

Remove one electron from O₂ and it leaves an antibonding orbital, giving O₂⁺: 8 bonding, 3 antibonding. A common first attempt: 8 − 3 = 5. Test it.

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Worked example 3 — solution

bond order = (bonding − antibonding) ÷ 2
given: O₂⁺ · bonding 8, antibonding 3 · wanted: the bond order
bonding − antibonding = 8 − 3 = 5 ✗

A common first attempt stops here. That is the surplus of bonding electrons, not the bond order. It skips the ÷ 2.

Dr. Karmach

Worked example 3 — solution

bond order = (bonding − antibonding) ÷ 2
given: O₂⁺ · bonding 8, antibonding 3 · wanted: the bond order
bonding − antibonding = 8 − 3 = 5 ✗
Step 1 · Fill the orbitals lowest-first

O₂ has 8 bonding and 4 antibonding electrons. Removing one empties an antibonding spot: 8 bonding, 3 antibonding.

Dr. Karmach

Worked example 3 — solution

bond order = (bonding − antibonding) ÷ 2
given: O₂⁺ · bonding 8, antibonding 3 · wanted: the bond order
bonding − antibonding = 8 − 3 = 5 ✗
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding

Bonding: 8. Antibonding: 3.

Dr. Karmach

Worked example 3 — solution

bond order = (bonding − antibonding) ÷ 2
given: O₂⁺ · bonding 8, antibonding 3 · wanted: the bond order
bonding − antibonding = 8 − 3 = 5 ✗
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding Step 3 · Bond order = (bonding − antibonding) ÷ 2
bond order = 8 − 32 = 2.5
Dr. Karmach

Worked example 3 — solution

bond order = (bonding − antibonding) ÷ 2
given: O₂⁺ · bonding 8, antibonding 3 · wanted: the bond order
bonding − antibonding = 8 − 3 = 5 ✗
Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding Step 3 · Bond order = (bonding − antibonding) ÷ 2
bond order = 8 − 32 = 2.5
Step 4 · Read the bond
Bond order 2.5: fractional, and real. Taking away an antibonding electron strengthens the bond, so O₂⁺ has a shorter, stronger bond than O₂. ✓
Dr. Karmach

Take-home: bond order counts pairs

bonding − antibonding = the surplus of bonding electrons
8 − 3 = 5 electrons ✗ — not a bond order
bond order = (bonding − antibonding) ÷ 2
(8 − 3)/2 = 2.5 ✓ — the surplus counted in pairs

A bond is a shared PAIR of electrons. The surplus of bonding electrons must be halved to count pairs. The ÷ 2 is never optional.

Dr. Karmach

Practice 2

bond order = (bonding − antibonding) ÷ 2
given: the superoxide ion O₂⁻ · bonding 8, antibonding 5 · wanted: the bond order

Add one electron to O₂ and it enters an antibonding orbital, giving the superoxide ion O₂⁻: 8 bonding, 5 antibonding. What is the bond order?

  1. 1.5
  2. 3
  3. 6.5
  4. -1.5
Dr. Karmach

Practice 2 — answer: A

bond order = (bonding − antibonding) ÷ 2
O₂⁻: bonding 8, antibonding 5
bond order = 8 − 52 = 1.5 — answer A

B forgot to divide by two: 8 − 5 = 3 is the electron surplus. C added the antibonding electrons: (8 + 5)/2 = 6.5. D subtracted backwards: (5 − 8)/2 = −1.5, a negative that has no meaning.

Bond order 1.5: the extra antibonding electron weakens the bond, so superoxide's bond is longer and weaker than O₂'s. ✓
Dr. Karmach

Check yourself

  1. C₂ has 8 valence electrons: 6 fill bonding orbitals and 2 fill antibonding. Find the bond order, and say whether C₂ is held more or less tightly than N₂.
  2. A molecule has equal numbers of bonding and antibonding electrons. What is its bond order, and can the molecule exist?

The same electron filling that sets the bond order also decides magnetism. A molecule with unpaired electrons in its orbitals is pulled toward a magnet. O₂'s two antibonding electrons are unpaired, which is why liquid oxygen clings to a magnet.

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4 · MO Diagrams & Magnetism

Fill the molecular-orbital diagram of a second-period diatomic, then read its bond order and whether it is paramagnetic or diamagnetic from the filled diagram.

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A liquid that clings to a magnet

Pour liquid oxygen between a magnet's poles and it bridges the gap, held there against gravity. The plain dot picture of oxygen gives no hint of this.

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Molecular orbitals fill like atomic ones

Molecular orbitals fill by the same rules as an atom's: lowest first, opposite spins per orbital, one electron in every equal-energy orbital before pairing. A leftover unpaired electron makes the molecule magnetic.

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The molecular-orbital ladder

Each set pairs a lower bonding orbital with a higher antibonding one. The two π orbitals share one energy — a degenerate pair. Only the σ2p and π2p order changes.

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Reading magnetism from the diagram

Count the unpaired electrons in the filled diagram. One or more unpaired: the molecule is paramagnetic, pulled toward a magnet. All paired: diamagnetic, not pulled. Bond order still measures the bond's strength.

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The method

  1. Count the valence electrons.
  2. Fill from the bottom up: opposite spins per orbital, one electron in each equal-energy orbital before pairing.
  3. Read the magnetism: any unpaired electron, paramagnetic; none, diamagnetic.
  4. Bond order = (bonding − antibonding) ÷ 2.
Dr. Karmach

Worked example 1 — nitrogen, N₂

Step 1 · Count the valence electrons

N₂ — each N contributes 5 valence electrons
2 × 5 = 10 valence electrons to place

Nitrogen sits early in the second period, so its π2p pair lies below σ2p. Fill the diagram, then read the magnetism and the bond order.

Dr. Karmach

Worked example 1 — filling the diagram

N₂ — 10 valence electrons
2 × 5 = 10 to place

Step 2 · Fill the MOs from the bottom up

Fill upward: σ2s, σ*2s, the degenerate π2p pair, then σ2p. Every filled orbital holds a pair.

Dr. Karmach

Worked example 1 — filling the diagram

N₂ — 10 valence electrons
2 × 5 = 10 to place

Step 2 · Fill the MOs from the bottom up

Fill upward: σ2s, σ*2s, the degenerate π2p pair, then σ2p. Every filled orbital holds a pair.

Every filled orbital holds a pair, so the N₂ diagram is ready to read for magnetism and bond order.
Dr. Karmach

Worked example 1 — reading it

N₂ — 10 valence electrons, diagram filled
σ2s, σ*2s, the π2p pair, then σ2p — every orbital paired

Step 3 · Read the magnetism

Not one orbital holds a lone electron, so nothing is left unpaired.

Dr. Karmach

Worked example 1 — reading it

N₂ — 10 valence electrons, diagram filled
σ2s, σ*2s, the π2p pair, then σ2p — every orbital paired
Step 3 · Read the magnetism Step 4 · Find the bond order

8 electrons fill bonding orbitals, 2 fill antibonding.

N₂ → diamagnetic · bond order = (8 − 2)/2 = 3
0 unpaired electrons · 8 bonding − 2 antibonding, halved · a triple bond
Both results come off the same diagram: N₂ carries a triple bond and no unpaired electrons, so it is diamagnetic.
Dr. Karmach

Worked example 2 — oxygen, O₂

Step 1 · Count the valence electrons

O₂ — each O contributes 6 valence electrons
2 × 6 = 12 valence electrons to place

Oxygen sits later in the period, so σ2p drops below π2p. The last two electrons reach the degenerate π*2p pair. A common first attempt pairs them in one orbital. Fill the diagram and read the magnetism.

Dr. Karmach

Worked example 2 — filling the diagram

O₂ — 12 valence electrons
2 × 6 = 12 to place

Step 2 · Fill the MOs from the bottom up

Hund's rule puts one electron in each π*2p orbital before either pairs.

Dr. Karmach

Worked example 2 — filling the diagram

O₂ — 12 valence electrons
2 × 6 = 12 to place

Step 2 · Fill the MOs from the bottom up

Hund's rule puts one electron in each π*2p orbital before either pairs.

Each π*2p orbital takes one electron before either pairs, so the O₂ diagram is ready to read for magnetism and bond order.
Dr. Karmach

Worked example 2 — reading it

O₂ — 12 valence electrons, diagram filled
σ2p below π2p · the last two electrons reach the π*2p pair

A common first attempt

Pairing the last two electrons in one π*2p orbital leaves none unpaired. That skips Hund's rule for a degenerate pair and wrongly predicts a diamagnetic molecule.

Dr. Karmach

Worked example 2 — reading it

O₂ — 12 valence electrons, diagram filled
σ2p below π2p · the last two electrons reach the π*2p pair
A common first attempt Step 3 · Read the magnetism

Each π*2p orbital holds a single electron: two electrons stay unpaired.

Dr. Karmach

Worked example 2 — reading it

O₂ — 12 valence electrons, diagram filled
σ2p below π2p · the last two electrons reach the π*2p pair
A common first attempt Step 3 · Read the magnetism Step 4 · Find the bond order

8 electrons fill bonding orbitals, 4 fill antibonding.

O₂ → paramagnetic · bond order = (8 − 4)/2 = 2
2 unpaired electrons · 8 bonding − 4 antibonding, halved · a double bond
The dot structure pairs every electron and predicts no magnetism. The diagram keeps the last two apart, so O₂ is paramagnetic — the liquid clings to a magnet.
Dr. Karmach

Take-home: fill a degenerate pair one at a time

✓ Hund: one electron in each π*2p orbital
2 unpaired · O₂ is paramagnetic
✗ paired early: both crammed into one π*2p orbital
0 unpaired · wrongly predicts diamagnetic

Two electrons entering an equal-energy pair take one orbital each before either pairs. That single choice is why O₂ is paramagnetic — the result a dot structure misses.

Dr. Karmach

Your turn — fluorine, F₂

F₂ has 14 valence electrons, two more than O₂. Those two complete the π*2p pair. Fill the blanks, then name the magnetism.

quantity value
electrons in the π*2p pair 4 — both orbitals full
unpaired electrons
bonding − antibonding 8 −
bond order
magnetism
Dr. Karmach

Your turn — fluorine, F₂

F₂ has 14 valence electrons, two more than O₂. Those two complete the π*2p pair. Fill the blanks, then name the magnetism.

quantity value
electrons in the π*2p pair 4 — both orbitals full
unpaired electrons
bonding − antibonding 8 −
bond order
magnetism
F₂ → diamagnetic · bond order = (8 − 6)/2 = 1
π*2p full: 0 unpaired · 8 bonding − 6 antibonding, halved
Dr. Karmach

Where this goes wrong

Pairing the degenerate electrons. Forcing O₂'s last two electrons into one π*2p orbital leaves none unpaired and wrongly calls O₂ diamagnetic. Hund's rule puts one in each orbital first: 2 unpaired, paramagnetic.
Reading magnetism from the dot structure. The O=O dot structure pairs every electron and predicts diamagnetic. Magnetism is read off the filled MO diagram, where the two π*2p electrons stay unpaired.
Forgetting to halve the bond order. For O₂, bonding minus antibonding is 8 − 4 = 4. Bond order counts electron pairs: (8 − 4)/2 = 2, not 4.
Adding the antibonding electrons. Antibonding electrons weaken the bond. Bond order is (8 − 4)/2 = 2, not (8 + 4)/2 = 6. They subtract, never add.
Dr. Karmach

Practice 1

peroxide ion, O₂²⁻ — 14 valence electrons
O₂ plus 2 more electrons, both into π*2p

Give the bond order and the magnetic behavior of the peroxide ion.

  1. Bond order 1; diamagnetic
  2. Bond order 1; paramagnetic
  3. Bond order 2; diamagnetic
  4. Bond order 7; diamagnetic
Dr. Karmach

Practice 1 — answer: A

O₂²⁻ → bond order 1, diamagnetic — answer A
π*2p full: 0 unpaired · (8 − 6)/2 = 1

B keeps the right bond order but calls it paramagnetic; the two extra electrons complete the π*2p pair, so 0 unpaired — diamagnetic. C forgets to halve: 8 − 6 = 2 counts single electrons, but bond order counts pairs, (8 − 6)/2 = 1. D adds the antibonding electrons, (8 + 6)/2 = 7, instead of subtracting them.

Two electrons past O₂ fill both π*2p orbitals. A full degenerate set leaves every electron paired: diamagnetic.
Dr. Karmach

Worked example 3 — the O₂⁺ ion

Step 1 · Count the valence electrons

O₂⁺ — O₂ with one electron removed
12 − 1 = 11 valence electrons to place

An electron leaves from the highest occupied orbital, a π*2p. Fill the diagram, then read the magnetism and the bond order.

Dr. Karmach

Worked example 3 — solution

O₂⁺ — 11 valence electrons
12 − 1 = 11 to place

Step 2 · Fill the MOs from the bottom up

Same order as O₂, one electron short. The π*2p pair now holds a single electron between its two orbitals.

Dr. Karmach

Worked example 3 — solution

O₂⁺ — 11 valence electrons
12 − 1 = 11 to place
Step 2 · Fill the MOs from the bottom up Step 3 · Read the magnetism

One π*2p orbital holds a lone electron; the other is empty.

O₂⁺ → paramagnetic
1 unpaired electron · still pulled toward a magnet
Dr. Karmach

Worked example 3 — solution

O₂⁺ — 11 valence electrons
12 − 1 = 11 to place
Step 2 · Fill the MOs from the bottom up Step 3 · Read the magnetism
O₂⁺ → paramagnetic
1 unpaired electron · still pulled toward a magnet
Step 4 · Find the bond order

Removing an antibonding electron leaves 8 bonding and 3 antibonding.

bond order = (8 − 3)/2 = 2.5
8 bonding − 3 antibonding, halved · stronger than O₂
Pulling an electron from an antibonding orbital raises the bond order from 2 to 2.5. One electron stays unpaired, so O₂⁺ is paramagnetic.
Dr. Karmach

Practice 2

superoxide ion, O₂⁻ — 13 valence electrons
O₂ plus 1 more electron, into π*2p

Give the bond order and the magnetic behavior of the superoxide ion.

  1. Bond order 1.5; paramagnetic
  2. Bond order 1.5; diamagnetic
  3. Bond order 3; paramagnetic
  4. Bond order 6.5; paramagnetic
Dr. Karmach

Practice 2 — answer: A

O₂⁻ → bond order 1.5, paramagnetic — answer A
π*2p holds 3 electrons: 1 unpaired · (8 − 5)/2 = 1.5

B reads the right bond order but calls it diamagnetic; three electrons cannot all pair in a two-orbital set, so one is left over — paramagnetic. C forgets to halve: 8 − 5 = 3 counts single electrons, but bond order counts pairs, (8 − 5)/2 = 1.5. D adds the antibonding electrons, (8 + 5)/2 = 6.5, instead of subtracting them.

An odd electron count guarantees at least one unpaired electron. The superoxide ion is paramagnetic, and its bond order sits between O₂'s 2 and peroxide's 1.
Dr. Karmach

Check yourself

  1. Fill the MO diagram of B₂ (6 valence electrons, π2p below σ2p). How many electrons are unpaired, and is B₂ paramagnetic or diamagnetic?
  2. C₂ has 8 valence electrons. Give its bond order, and state whether it is paramagnetic or diamagnetic.

Magnetism and bond order both come from how electrons pack into orbitals. The next unit turns to how whole molecules pack against one another: the intermolecular forces that set boiling points, and the gas laws that describe those particles once the attractions between them become negligible.

Dr. Karmach

Can you…?

  • ☐ assign sp, sp², or sp³ hybridization to a central atom from its number of electron domains?
  • ☐ count sigma and pi bonds in a molecule from its Lewis structure?
  • ☐ build the molecular-orbital occupation of a second-period diatomic and compute its bond order?
  • ☐ predict paramagnetism or diamagnetism from a molecular-orbital diagram?

If any box stays empty, the practice site has a drill for it. 🧪

Dr. Karmach