Advanced Bonding Theories

General Chemistry · Dr. Karmach

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, sp³, sp³d, or sp³d² 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.

Dr. Karmach

Domain counts from VSEPR

2 domains: linear, 180° · 3: trigonal planar, 120° · 4: tetrahedral, 109.5°
5: trigonal bipyramidal, 120° and 90° · 6: octahedral, 90° · domains = bonded atoms + lone pairs, a multiple bond counted once

VSEPR counts the electron domains on a central atom and spreads them apart. The three p orbitals point 90° apart, so they cannot aim along 109.5° or 120°. Hybridization starts from the same count.

Dr. Karmach

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 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

Pure s and p orbitals would give 90° bonds; real methane is 109.5°. So carbon mixes its one 2s and three 2p into four identical sp³ orbitals, aimed at a tetrahedron's corners.

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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.

memory hook: count the domains, not the bonds
a lone pair takes a hybrid just as a bond does · a double or triple bond is still one domain
Dr. Karmach

The method

  1. Draw the Lewis structure and find the central atom.
  2. Count the electron domains. A lone pair counts; a multiple bond counts once.
  3. Match the count: 2 → sp, 3 → sp², 4 → sp³, 5 → sp³d, 6 → sp³d².

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: methane's real shape.

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

Guided example: hydrogen selenide, H₂Se

H₂Se
central atom: Se · bonded atoms: 2 H

Only the formula is given. Assign the hybridization of the selenium, naming each method step as it runs: the lone pairs on Se come from the valence electrons.

Dr. Karmach

Guided example: solution

H₂Se
central atom: Se · bonded atoms: 2 H

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

6 + 2(1) = 8 valence e⁻ · 8 − 2 bonds × 2 = 4 left for Se
Se brings 6, each H brings 1 · H holds no lone pairs · 4 e⁻ = 2 lone pairs on Se

Selenium sits in the center, single-bonded to each hydrogen. The 4 electrons left over pair up on Se.

Dr. Karmach

Guided example: solution

H₂Se
central atom: Se · bonded atoms: 2 H
Step 1 · Draw the Lewis structure and find the central atom
6 + 2(1) = 8 valence e⁻ · 8 − 2 bonds × 2 = 4 left for Se
Se brings 6, each H brings 1 · H holds no lone pairs · 4 e⁻ = 2 lone pairs on Se
Step 2 · Count the electron domains Step 3 · Match the count
2 bonded atoms + 2 lone pairs = 4 domains → sp³
both lone pairs take a hybrid · one 4s + three 4p → four sp³ hybrids

Counting only the 2 H would give 2 domains and sp. The lone pairs count too.

Dr. Karmach

Guided example: solution

H₂Se
central atom: Se · bonded atoms: 2 H
Step 1 · Draw the Lewis structure and find the central atom
6 + 2(1) = 8 valence e⁻ · 8 − 2 bonds × 2 = 4 left for Se
Se brings 6, each H brings 1 · H holds no lone pairs · 4 e⁻ = 2 lone pairs on Se
Step 2 · Count the electron domains Step 3 · Match the count
2 bonded atoms + 2 lone pairs = 4 domains → sp³
both lone pairs take a hybrid · one 4s + three 4p → four sp³ hybrids
Same count as the oxygen in water: two bonds, two lone pairs, sp³. The formula shows neither lone pair; the valence count finds both.

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°. Methane, BF₃, and CO₂ bond at 109.5°, 120°, and 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

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
5 + 3(7) = 26 valence e⁻ · 3 bonds use 6 and 9 I lone pairs use 18 · 2 left = 1 lone pair on N

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.

Dr. Karmach

Practice 1 answer: B

NI₃: 3 bonds + 1 lone pair = 4 domains → sp³ → answer B
5 + 3(7) = 26 valence e⁻ · 3 bonds use 6 and 9 I lone pairs use 18 · 2 left = 1 lone pair on N
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 sit 180° apart. A double bond adds electrons, not a direction.

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

CH₃–CH=O: acetaldehyde
a classmate's labels: CH₃ carbon sp³ · carbonyl carbon sp · O sp² · the O keeps two lone pairs

Three atoms in acetaldehyde carry labels. Which label is wrong, and what should it be?

  1. The CH₃ carbon: sp², its three hydrogens make three domains
  2. The oxygen: sp³, the double bond counted twice plus two lone pairs make four domains
  3. The carbonyl carbon: sp², one H, one C, and one O make three domains
  4. None of them: a carbon joined by a double bond is sp, as in CO₂
Dr. Karmach

Practice 2 answer: C

CH₃ carbon: 3 H + 1 C = 4 → sp³ ✓ · carbonyl carbon: H + C + O = 3 domains → sp², not sp · O: 1 double bond + 2 lone pairs = 3 → sp² ✓
a double bond is one domain · the wrong label is the carbonyl carbon → answer C

A counted only the three hydrogens and skipped the bond to the carbonyl carbon: 3 + 1 = 4 domains, sp³. B counted the C=O double bond as two domains, 2 + 2 = 4, sp³; one bond is one domain, so 1 + 2 = 3, sp². D over-generalized CO₂; that carbon has two domains, this one has three, so sp² and 120°.

Dr. Karmach

Practice 2 answer: C

CH₃ carbon: 3 H + 1 C = 4 → sp³ ✓ · carbonyl carbon: H + C + O = 3 domains → sp², not sp · O: 1 double bond + 2 lone pairs = 3 → sp² ✓
a double bond is one domain · the wrong label is the carbonyl carbon → answer C
Count regions, not lines. A carbon with three neighbors is sp² whatever the bond orders; only two neighbors and no lone pair make sp.

Dr. Karmach

Five and six domains: sp³d and sp³d²

5 domains → sp³d: one s + three p + one d · 6 domains → sp³d²: one s + three p + two d
sp³d aims at a trigonal bipyramid (120° and 90°) · sp³d² aims at an octahedron (90°) · lone pairs still count as domains

Four hybrids use up one s and three p. A fifth or sixth domain needs d orbitals. The n = 2 shell has none: C, N, O, and F stop at sp³.

5 or 6 domains: central atoms from period 3 and below only
the atoms that expand their octets · the VSEPR frames · single bonds and lone pairs only
Dr. Karmach

Worked example 4: dichlorobromate ion, BrCl₂⁻

BrCl₂⁻
central atom: Br (period 4) · bonded atoms: 2 Cl · valence electrons: 7 + 2(7) + 1 = 22

Bromine bonds to two chlorine atoms, and the ion is linear. Assign the hybridization of the bromine: find its lone pairs, count its domains, then match.

Dr. Karmach

Worked example 4: solution

BrCl₂⁻
central atom: Br (period 4) · bonded atoms: 2 Cl · valence electrons: 22

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

22 − 2 bonds × 2 − 2 Cl × 6 = 6 left for Br
the 1− charge adds one electron · three lone pairs on each Cl use 12 · 3 lone pairs on Br
Dr. Karmach

Worked example 4: solution

BrCl₂⁻
central atom: Br (period 4) · bonded atoms: 2 Cl · valence electrons: 22
Step 1 · Draw the Lewis structure and find the central atom
22 − 2 bonds × 2 − 2 Cl × 6 = 6 left for Br
the 1− charge adds one electron · three lone pairs on each Cl use 12 · 3 lone pairs on Br
Step 2 · Count the electron domains Step 3 · Match the count
2 bonded groups + 3 lone pairs = 5 domains → sp³d
10 e⁻ on Br, allowed in period 4 · one 4s + three 4p + one 4d → five hybrids, trigonal bipyramidal
Dr. Karmach

Worked example 4: solution

BrCl₂⁻
central atom: Br (period 4) · bonded atoms: 2 Cl · valence electrons: 22
Step 1 · Draw the Lewis structure and find the central atom
22 − 2 bonds × 2 − 2 Cl × 6 = 6 left for Br
the 1− charge adds one electron · three lone pairs on each Cl use 12 · 3 lone pairs on Br
Step 2 · Count the electron domains Step 3 · Match the count
2 bonded groups + 3 lone pairs = 5 domains → sp³d
10 e⁻ on Br, allowed in period 4 · one 4s + three 4p + one 4d → five hybrids, trigonal bipyramidal
The three lone pairs take the equatorial hybrids and the two Cl sit on the axis: a linear ion. Read as 2 domains, it would wrongly be sp.

Dr. Karmach

Worked example 5: hexachlorostannate ion, SnCl₆²⁻

SnCl₆²⁻
central atom: Sn (period 5) · bonded atoms: 6 Cl · valence electrons: 4 + 6(7) + 2 = 48

Tin bonds to six chlorine atoms, and the ion carries a 2− charge. Assign the hybridization of the tin.

Dr. Karmach

Worked example 5: solution

SnCl₆²⁻
central atom: Sn (period 5) · bonded atoms: 6 Cl · valence electrons: 48

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

48 − 6 bonds × 2 − 6 Cl × 6 = 0 left for Sn
the 2− charge adds two electrons to the 46 from the atoms · lone pairs on Sn: 0
Dr. Karmach

Worked example 5: solution

SnCl₆²⁻
central atom: Sn (period 5) · bonded atoms: 6 Cl · valence electrons: 48
Step 1 · Draw the Lewis structure and find the central atom
48 − 6 bonds × 2 − 6 Cl × 6 = 0 left for Sn
the 2− charge adds two electrons to the 46 from the atoms · lone pairs on Sn: 0
Step 2 · Count the electron domains Step 3 · Match the count
6 bonded groups + 0 lone pairs = 6 domains → sp³d²
12 electrons on Sn, allowed in period 5 · one 5s + three 5p + two 5d → six hybrids, octahedral
Dr. Karmach

Worked example 5: solution

SnCl₆²⁻
central atom: Sn (period 5) · bonded atoms: 6 Cl · valence electrons: 48
Step 1 · Draw the Lewis structure and find the central atom
48 − 6 bonds × 2 − 6 Cl × 6 = 0 left for Sn
the 2− charge adds two electrons to the 46 from the atoms · lone pairs on Sn: 0
Step 2 · Count the electron domains Step 3 · Match the count
6 bonded groups + 0 lone pairs = 6 domains → sp³d²
12 electrons on Sn, allowed in period 5 · one 5s + three 5p + two 5d → six hybrids, octahedral
Six hybrids point at an octahedron's corners, every neighbor 90° away. The charge changed the electron count, not the rule: count the domains, then match.

Dr. Karmach

Practice 3

XeF₃⁺ · FNO
central atoms: Xe in XeF₃⁺ · N in FNO (skeleton F–N–O)

What are the hybridizations of the central atoms in XeF₃⁺ and FNO, respectively?

  1. sp³d² and sp²
  2. sp² and sp²
  3. sp³d and sp³
  4. sp³d² and sp³
  5. sp³d and sp²
Dr. Karmach

Practice 3 answer: E

XeF₃⁺: 3 bonded F + 2 lone pairs = 5 domains → sp³d · FNO: 2 bonded atoms + 1 lone pair = 3 domains → sp² → answer E
XeF₃⁺: 8 + 3(7) − 1 = 28 valence e⁻ · 3 bonds use 6, 9 F lone pairs use 18 · 4 left = 2 lone pairs on XeFNO: 5 + 7 + 6 = 18 · F–N and N=O use 6 · 5 lone pairs on F and O use 10 · 2 left = 1 lone pair on N

A added the 1+ charge: 8 + 21 + 1 = 30 leaves 6 electrons, 3 lone pairs, 3 + 3 = 6 domains, sp³d². A positive charge removes an electron. B counted only the three bonded F on Xe: 3 domains, sp². C counted N's double bond as two domains, 1 + 2 + 1 = 4, sp³. D made the slips of A and C together.

Dr. Karmach

Practice 3 answer: E

XeF₃⁺: 3 bonded F + 2 lone pairs = 5 domains → sp³d · FNO: 2 bonded atoms + 1 lone pair = 3 domains → sp² → answer E
XeF₃⁺: 8 + 3(7) − 1 = 28 valence e⁻ · 3 bonds use 6, 9 F lone pairs use 18 · 4 left = 2 lone pairs on XeFNO: 5 + 7 + 6 = 18 · F–N and N=O use 6 · 5 lone pairs on F and O use 10 · 2 left = 1 lone pair on N
Count each center on its own. The charge sets the electron total; the double bond adds electrons to N but no domain, while N's lone pair adds one.

Dr. Karmach

Practice 4: sulfur and oxygen with fluorine

SF₆: sulfur hexafluoride · OF₂: oxygen difluoride
SF₆: central S (period 3), 6 F, 6 + 6(7) = 48 valence e⁻ · OF₂: central O (period 2), 2 F, 6 + 2(7) = 20 valence e⁻

Sulfur and oxygen share a group, and here each bonds only to fluorine. What hybridization do the sulfur in SF₆ and the oxygen in OF₂ use, respectively?

  1. S sp³d², O sp³: six bonds on S; two bonds + two lone pairs on O
  2. S sp³d², O sp: six bonds on S; only the two F count on O
  3. S sp³, O sp³: S capped at four, like O; two bonds + two lone pairs on O
  4. S sp³, O sp: S capped at four, like O; only the two F count on O
Dr. Karmach

Practice 4 answer: A

SF₆: 6 bonded atoms + 0 lone pairs = 6 domains → sp³d² · OF₂: 2 bonded atoms + 2 lone pairs = 4 domains → sp³ → answer A
SF₆: six S–F bonds use 12 e⁻ and 18 F lone pairs use 36, none left on S · OF₂: bonds and F lone pairs use 16 of 20, 4 left = 2 lone pairs on O

B counted only the two bonded F on O: 2 domains, sp. Its two lone pairs count too, 2 + 2 = 4, sp³. C held S to four hybrids as if it sat in period 2; S has 3d orbitals, so its six domains take sp³d². D made both slips.

Dr. Karmach

Practice 4 answer: A

SF₆: 6 bonded atoms + 0 lone pairs = 6 domains → sp³d² · OF₂: 2 bonded atoms + 2 lone pairs = 4 domains → sp³ → answer A
SF₆: six S–F bonds use 12 e⁻ and 18 F lone pairs use 36, none left on S · OF₂: bonds and F lone pairs use 16 of 20, 4 left = 2 lone pairs on O
Same group, different limits. S in period 3 reaches six domains; O in period 2 has no d orbitals and stops at four, which is why OF₆ does not exist.

Dr. Karmach

Practice 5: the tetrafluorobromate ion

BrF₄⁻: tetrafluorobromate ion
central atom: Br (period 4) · bonded atoms: 4 F · charge: 1−

The bromine and all four fluorines lie in one plane. What is the hybridization of the bromine atom?

  1. sp²: a flat ion, like a trigonal planar atom
  2. sp³: four fluorines bonded to the bromine
  3. sp³d: four bonds and one lone pair on the bromine
  4. sp³d²: four bonds and two lone pairs on the bromine
Dr. Karmach

Practice 5 answer: D

BrF₄⁻: 4 bonded F + 2 lone pairs = 6 domains → sp³d² → answer D
7 + 4(7) + 1 = 36 valence e⁻ · 4 bonds use 8 and 12 F lone pairs use 24 · 4 left = 2 lone pairs on Br · 12 e⁻ on Br, allowed in period 4

A read the hybridization from the flat shape. Hybrids follow domains, not the plane the atoms sit in. B counted only the four bonded F: 4 domains, sp³. The two lone pairs count too. C subtracted the charge: 7 + 28 − 1 = 34 leaves 2 electrons, one lone pair, 4 + 1 = 5 domains, sp³d. A 1− charge adds an electron.

Dr. Karmach

Practice 5 answer: D

BrF₄⁻: 4 bonded F + 2 lone pairs = 6 domains → sp³d² → answer D
7 + 4(7) + 1 = 36 valence e⁻ · 4 bonds use 8 and 12 F lone pairs use 24 · 4 left = 2 lone pairs on Br · 12 e⁻ on Br, allowed in period 4
The two lone pairs sit opposite each other in an octahedral frame, so the four F form a square plane. Six domains, six hybrids: sp³d².

Dr. Karmach

Practice 6: xenon trioxide

XeO₃: xenon trioxide
central atom: Xe (period 5) · bonded atoms: 3 O · each Xe–O drawn as a single bond

Which row is fully correct for the xenon atom (electron geometry | molecular shape | hybridization)?

  1. trigonal planar | trigonal planar | sp²
  2. tetrahedral | tetrahedral | sp³
  3. tetrahedral | trigonal pyramidal | sp³
  4. tetrahedral | trigonal pyramidal | sp²
Dr. Karmach

Practice 6 answer: C

XeO₃: 3 bonded O + 1 lone pair = 4 domains → tetrahedral | trigonal pyramidal | sp³ → answer C
8 + 3(6) = 26 valence e⁻ · 3 bonds use 6 and 9 O lone pairs use 18 · 2 left = 1 lone pair on Xe

A missed the lone pair on Xe: 3 domains carried through every column, as in SO₃. B named the electron geometry as the shape. The lone pair is a domain but not an atom, so the three O form a pyramid. D took the hybridization from the three bonded atoms while the geometry used all four domains.

Dr. Karmach

Practice 6 answer: C

XeO₃: 3 bonded O + 1 lone pair = 4 domains → tetrahedral | trigonal pyramidal | sp³ → answer C
8 + 3(6) = 26 valence e⁻ · 3 bonds use 6 and 9 O lone pairs use 18 · 2 left = 1 lone pair on Xe
One domain count sets both the electron geometry and the hybridization; only the shape leaves out the lone pair. Xe sits in period 5, yet four domains need only sp³.

Dr. Karmach

Practice 7: 1,2-butadiene

CH₂=C=CH–CH₃: 1,2-butadiene
carbons C1 to C4 from the left · no lone pairs on any carbon

With carbons numbered from the left, what are the hybridizations of C1, C2, C3 and C4, in that order?

  1. sp², sp², sp², sp³
  2. sp², sp, sp², sp³
  3. sp², sp³, sp², sp³
  4. sp³, sp³, sp³, sp³
Dr. Karmach

Practice 7 answer: B

C1: 2 + 1 = 3 → sp² · C2: 1 + 1 = 2 → sp · C3: 1 + 1 + 1 = 3 → sp² · C4: 3 + 1 = 4 → sp³ → answer B
C1: 2 H + C2 · C2: C1 + C3, no H · C3: H + C2 + C4 · C4: 3 H + C3 · each double bond is one domain

A labeled every double-bonded carbon sp². C2 carries two double bonds but has only two neighbors: 2 domains, sp. C counted C2's two double bonds as four domains, 2 + 2 = 4, sp³. D counted every bond line as its own domain: C1 2 H + 2 lines = 4, C3 1 + 1 + 2 = 4, sp³ on every carbon.

Dr. Karmach

Practice 7 answer: B

C1: 2 + 1 = 3 → sp² · C2: 1 + 1 = 2 → sp · C3: 1 + 1 + 1 = 3 → sp² · C4: 3 + 1 = 4 → sp³ → answer B
C1: 2 H + C2 · C2: C1 + C3, no H · C3: H + C2 + C4 · C4: 3 H + C3 · each double bond is one domain
Count each atom on its own. The middle carbon matches the carbon of CO₂: two double bonds, two domains, sp, linear.

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. Why can a period-3 atom reach six electron domains while a period-2 atom stops at four? Name the hybridization at each limit.

Each hybrid orbital either overlaps end-to-end to form one sigma bond or holds a lone pair. The extra lines of a double or triple bond are pi bonds, built from the p orbitals an sp or sp² atom leaves 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

What a Lewis structure already shows

H–O–H: 2 lines · 2 bonded pairs  ·  N≡N: 3 lines · 1 bonded pair
each line is one shared electron pair · N≡N puts 3 − 1 = 2 extra lines between the same two atoms

Each line in a Lewis structure is one shared electron pair. Lines and bonded pairs match only when every bond is single. A double or triple adds extra lines between two atoms.

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. There are 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 Step 3 · Count the extra lines as π

No double or triple bonds means no extra lines.

σ: 4 + 0 + 0 = 4  ·  π: 0 + 2×0 = 0
four C–H bonds, one σ each · 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 Step 3 · Count the extra lines as π
σ: 4 + 0 + 0 = 4  ·  π: 0 + 2×0 = 0
four C–H bonds, one σ each · 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 ✗
Two σ cannot share one axis. The C=C holds one σ, and its second line is a 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

Step 2 · Give each bond one σ Step 3 · Count the extra lines as π

Five bonded pairs (four C–H and one C=C), one σ each. The double's second line is the 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
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.

memory hook: one σ per bond, the rest are π
count the bonded pairs for σ · every extra line of a double or triple 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. σ: 4 · π: 0
  2. σ: 3 · π: 1
  3. σ: 2 · π: 2
  4. σ: 4 · π: 1
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.

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
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 Step 3 · Count the extra lines as π

The triple bond has two extra lines beyond its σ.

σ: 2 + 1 = 3  ·  π: 2×1 = 2
the triple counts as one σ · it adds two π, and its 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 Step 3 · Count the extra lines as π
σ: 2 + 1 = 3  ·  π: 2×1 = 2
the triple counts as one σ · it adds two π, and its 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

HC≡C–CH=O: propynal
wanted: total σ and π

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

  1. σ: 8 · π: 0
  2. σ: 3 · π: 5
  3. σ: 5 · π: 3
  4. σ: 8 · π: 3
Dr. Karmach

Practice 2 answer: C

2 C–H + C–C = 3 single · C≡C and C=O add one σ each: 3 + 1 + 1 = 5 σ · π: 2 + 1 = 3 → answer C
one σ per bonded pair · the triple adds 2 π, the double 1 π

A counted every line as a σ: 3 + 3 + 2 = 8 σ, and no π. B dropped the σ inside both multiple bonds: 3 σ and 3 + 2 = 5 π. D counted every line as a σ, then added the π: 8 σ and 3 π.

Dr. Karmach

Practice 2 answer: C

2 C–H + C–C = 3 single · C≡C and C=O add one σ each: 3 + 1 + 1 = 5 σ · π: 2 + 1 = 3 → answer C
one σ per bonded pair · the triple adds 2 π, the double 1 π
Five σ and three π. Three single bonds plus one σ in each multiple bond make five; the triple adds two π and the double one.

Dr. Karmach

Guided example: matching a σ and π count

H–C(=O)–O–H  ·  CH₃–C≡N  ·  H₂N–C≡N
given: three Lewis structures · wanted: the one molecule with exactly 4 σ and 2 π

Formic acid, acetonitrile and cyanamide. Run the three method steps on each molecule. Then keep the one that matches both counts.

Dr. Karmach

Guided example: solution

H–C(=O)–O–H  ·  CH₃–C≡N  ·  H₂N–C≡N
given: three Lewis structures · wanted: the one molecule with exactly 4 σ and 2 π

Step 1 · Read the structure

Formic acid: 3 single bonds and 1 double. Acetonitrile: 4 single and 1 triple. Cyanamide: 3 single (2 N–H and N–C) and 1 triple.

Dr. Karmach

Guided example: solution

H–C(=O)–O–H  ·  CH₃–C≡N  ·  H₂N–C≡N
given: three Lewis structures · wanted: the one molecule with exactly 4 σ and 2 π
Step 1 · Read the structure Step 2 · Give each bond one σ Step 3 · Count the extra lines as π
σ: 3 + 1 = 4 · 4 + 1 = 5 · 3 + 1 = 4  ·  π: 1 · 2×1 = 2 · 2×1 = 2
formic acid 4 σ, 1 π · acetonitrile 5 σ, 2 π · cyanamide 4 σ, 2 π

Each molecule gets its single bonds plus one σ for its multiple bond. Its extra lines are the π.

Dr. Karmach

Guided example: solution

H–C(=O)–O–H  ·  CH₃–C≡N  ·  H₂N–C≡N
given: three Lewis structures · wanted: the one molecule with exactly 4 σ and 2 π
Step 1 · Read the structure Step 2 · Give each bond one σ Step 3 · Count the extra lines as π
σ: 3 + 1 = 4 · 4 + 1 = 5 · 3 + 1 = 4  ·  π: 1 · 2×1 = 2 · 2×1 = 2
formic acid 4 σ, 1 π · acetonitrile 5 σ, 2 π · cyanamide 4 σ, 2 π
Only cyanamide, H₂N–C≡N, matches both counts. Formic acid matches only the σ; acetonitrile only the π.

Dr. Karmach

Practice 3

Which molecule has exactly 9 σ bonds and 2 π bonds?

  1. acetone, CH₃–C(=O)–CH₃
  2. propargyl alcohol, HC≡C–CH₂–OH
  3. methyl acetate, CH₃–C(=O)–O–CH₃
  4. methyl propiolate, HC≡C–C(=O)–O–CH₃
  5. pyruvic acid, CH₃–C(=O)–C(=O)–O–H
Dr. Karmach

Practice 3 answer: E

pyruvic acid: 7 single + 2 C=O → σ: 7 + 2 = 9 · π: 1 + 1 = 2 → answer E
single: 3 C–H, 2 C–C, C–O, O–H · C₃H₄O₃ has 10 atoms and no ring: 10 − 1 = 9 bonded pairs ✓

A matched only the σ: acetone is 8 + 1 = 9 σ but 1 π. B counted every line as a σ and then added the π: propargyl alcohol's 6 + 3 = 9 lines read as 9 σ, plus 2 π. One σ per bonded pair gives 6 + 1 = 7 σ. C dropped the σ in the C=O: methyl acetate reads as 9 σ and 2 π, but it has 9 + 1 = 10 σ and 1 π. D missed the C=O's π: methyl propiolate is 9 σ and 2 + 1 = 3 π.

Dr. Karmach

Practice 3 answer: E

pyruvic acid: 7 single + 2 C=O → σ: 7 + 2 = 9 · π: 1 + 1 = 2 → answer E
single: 3 C–H, 2 C–C, C–O, O–H · C₃H₄O₃ has 10 atoms and no ring: 10 − 1 = 9 bonded pairs ✓
Both counts must match. Three choices have 9 σ and two have 2 π; only pyruvic acid has both.

Dr. Karmach

Practice 4

H₂N–CH₂–C(=O)–O–H: glycine
the simplest amino acid in proteins

How many σ bonds and how many π bonds are in one glycine molecule?

  1. σ: 10 · π: 0
  2. σ: 8 · π: 2
  3. σ: 9 · π: 0
  4. σ: 9 · π: 1
  5. σ: 10 · π: 1
Dr. Karmach

Practice 4 answer: D

8 single + 1 σ in the C=O = 9 σ · π: 1 → answer D
single: 2 N–H, N–C, 2 C–H, C–C, C–O, O–H · the C=O's second line is the π

A counted every line as a σ: 8 + 2 = 10 σ, and no π. B dropped the σ inside the C=O, reading both of its lines as π: 8 σ and 2 π. C read the C=O as a single bond and missed its π: 9 σ and 0 π. E counted the double as 2 σ and then added its π: 10 σ and 1 π.

Dr. Karmach

Practice 4 answer: D

8 single + 1 σ in the C=O = 9 σ · π: 1 → answer D
single: 2 N–H, N–C, 2 C–H, C–C, C–O, O–H · the C=O's second line is the π
Nine σ and one π. Nine bonded pairs give nine σ; the only extra line is the C=O's π.

Dr. Karmach

Practice 5

CH₂=C(C≡N)–C(=O)–O–CH₂–CH₃: ethyl cyanoacrylate
the bonding agent in instant glue

What are the total σ and π bond counts for ethyl cyanoacrylate?

  1. σ: 19 · π: 0
  2. σ: 12 · π: 7
  3. σ: 13 · π: 2
  4. σ: 19 · π: 4
  5. σ: 15 · π: 4
Dr. Karmach

Practice 5 answer: E

12 single + 3 multiple (one σ each) = 15 σ · π: 1 + 2 + 1 = 4 → answer E
single: 2 C–H, C–CN, C–C(=O), C–O, O–C, 2 C–H, C–C, 3 C–H · multiple: C=C, C≡N, C=O

A counted every line as a σ: 12 + 2 + 3 + 2 = 19 σ, and no π. B dropped the σ inside all three multiple bonds: 12 σ and 2 + 3 + 2 = 7 π. C skipped the (C≡N) branch, losing a C–C and the triple: 11 + 2 = 13 σ and 2 π. D counted every line as a σ, then added the π: 19 σ and 4 π.

Dr. Karmach

Practice 5 answer: E

12 single + 3 multiple (one σ each) = 15 σ · π: 1 + 2 + 1 = 4 → answer E
single: 2 C–H, C–CN, C–C(=O), C–O, O–C, 2 C–H, C–C, 3 C–H · multiple: C=C, C≡N, C=O
Fifteen σ and four π. C₆H₇NO₂ has 16 atoms and no ring, so 16 − 1 = 15 bonded pairs, one σ each.

Dr. Karmach

Practice 6

CH₃–C(=O)–O–H: acetic acid
the acid in vinegar · the bond asked about joins the C=O carbon to the OH oxygen

Which orbital overlaps make up the C–O single bond in acetic acid?

  1. σ (C sp² – O sp³)
  2. σ (C sp³ – O sp³)
  3. σ (C sp² – O sp)
  4. σ (C sp² – O sp²) + π (C 2p – O 2p)
Dr. Karmach

Practice 6 answer: A

C: 3 + 0 = 3 domains → sp² · O: 2 + 2 = 4 domains → sp³ → answer A
C: CH₃, =O and OH, no lone pair · OH oxygen: 2 bonds, 2 lone pairs · one σ (C sp² – O sp³)

B counted every bond line as a domain: the carbon's 1 + 2 + 1 = 4 lines make it sp³. A double bond is one domain, so the carbon has 3. C counted only the oxygen's 2 bonded atoms and dropped its 2 lone pairs: 2 domains, sp. D moved the C=O's overlaps onto the C–O: an sp² oxygen plus a π. A single bond is one σ and nothing more.

Dr. Karmach

Practice 6 answer: A

C: 3 + 0 = 3 domains → sp² · O: 2 + 2 = 4 domains → sp³ → answer A
C: CH₃, =O and OH, no lone pair · OH oxygen: 2 bonds, 2 lone pairs · one σ (C sp² – O sp³)
Two hybrids meet end-on to make the σ. The carbon's leftover 2p forms the π of the C=O, not of the C–O.

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

Bond order in a Lewis structure

H–H  ·  O=O  ·  N≡N
1 shared pair · 2 shared pairs · 3 shared pairs → bond order 1, 2, 3

A Lewis structure shows bond order: the number of electron pairs two atoms share. Molecular orbital theory counts the same pairs from bonding and antibonding electrons. It also finds half bonds.

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

Bonding must beat antibonding; bond order counts the 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 · memory hook: a star means subtract, then halve for 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

Worked examples 1 and 2: the route

N₂: (8 − 2)/2 = 3  ·  O₂: (8 − 4)/2 = 2
found: N₂ 10 valence electrons, bond order 3 · O₂ 12 valence electrons, bond order 2

Dr. Karmach

Worked examples 1 and 2: the route

N₂: (8 − 2)/2 = 3  ·  O₂: (8 − 4)/2 = 2
found: N₂ 10 valence electrons, bond order 3 · O₂ 12 valence electrons, bond order 2


Same four steps, same verdict: both bonds hold. O₂'s two extra electrons go into antibonding π*2p and cost one bond. ✓
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. Six antibonding electrons cancel six of eight bonding: one net pair. ✓
Dr. Karmach

Guided example: neon, Ne₂

bond order = (bonding − antibonding) ÷ 2
given: Ne₂, two neon atoms with 8 valence electrons each · wanted: the bond order

Run all four method steps in order and name each move as you make it. Decide whether Ne₂ holds together.

Dr. Karmach

Guided example: solution

Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding

Sixteen electrons fill every valence orbital, σ*2p last. Then sort them: unstarred orbitals are bonding, starred are antibonding.

Ne₂: σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴ σ*2p²
2 × 8 = 16 valence electrons · bonding 2 + 2 + 4 = 8 · antibonding 2 + 4 + 2 = 8
Dr. Karmach

Guided example: solution

Step 1 · Fill the orbitals lowest-first Step 2 · Count bonding and antibonding

Ne₂: σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴ σ*2p²
2 × 8 = 16 valence electrons · bonding 2 + 2 + 4 = 8 · antibonding 2 + 4 + 2 = 8
Step 3 · Bond order = (bonding − antibonding) ÷ 2 Step 4 · Read the bond
bond order = 8 − 82 = 0
Bond order 0: no net bond. Neon stays as separate atoms, the same verdict as He₂. ✓

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
Li₂: bonding 2, antibonding 0 · Be₂: bonding 2, antibonding 2 · B₂: bonding 4, antibonding 2

Three second-period diatomics are listed with their molecular-orbital electron counts. Which one is predicted not to exist as a stable molecule?

  1. Be₂
  2. Li₂
  3. B₂
  4. None of them: each has bonding electrons
Dr. Karmach

Practice 1 answer: A

Li₂: (2 − 0)/2 = 1 · Be₂: (2 − 2)/2 = 0 · B₂: (4 − 2)/2 = 1
bond order 0 means no net bond: Be₂ does not hold together
Be₂: bond order = 2 − 22 = 0 → answer A

B judged by size: Li₂ has few electrons, but none are antibonding, so (2 − 0)/2 = 1, a single bond like H₂. C saw antibonding electrons and stopped; B₂ still has a surplus, (4 − 2)/2 = 1. D counted only the bonding electrons; Be₂'s two antibonding electrons cancel its two bonding ones.

Dr. Karmach

Practice 1 answer: A

Li₂: (2 − 0)/2 = 1 · Be₂: (2 − 2)/2 = 0 · B₂: (4 − 2)/2 = 1
bond order 0 means no net bond: Be₂ does not hold together
Be₂: bond order = 2 − 22 = 0 → answer A
Be₂ repeats the He₂ pattern: equal bonding and antibonding counts, bond order 0, no molecule. ✓

Dr. Karmach

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.

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 ✗

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

N₂: bonding 8, antibonding 2 · N₂⁺: one electron fewer
wanted: the bond order of N₂⁺

Nitrogen gas loses one electron to form N₂⁺. What is the bond order of N₂⁺?

  1. 3.5
  2. 2.5
  3. 5
  4. 3
Dr. Karmach

Practice 2 answer: B

N₂⁺: σ2s² σ*2s² π2p⁴ σ2p¹
10 − 1 = 9 valence electrons · bonding 8 − 1 = 7, antibonding 2
bond order = 7 − 22 = 2.5 → answer B

A took the electron from antibonding σ*2s, as O₂⁺ loses one: (8 − 1)/2 = 3.5. But N₂ fills σ2p last, so the electron leaves that bonding orbital. C forgot to halve: 7 − 2 = 5. D kept the counts of N₂, as if no electron left: (8 − 2)/2 = 3.

Dr. Karmach

Practice 2 answer: B

N₂⁺: σ2s² σ*2s² π2p⁴ σ2p¹
10 − 1 = 9 valence electrons · bonding 8 − 1 = 7, antibonding 2
bond order = 7 − 22 = 2.5 → answer B
Losing a bonding electron weakens the bond: N₂⁺ at 2.5 sits below N₂ at 3. O₂⁺ moved the other way because its electron left an antibonding orbital. ✓

Dr. Karmach

Practice 3

Ne₂²⁺
wanted: the bond order

What bond order does molecular orbital theory predict for Ne₂²⁺?

  1. 0.5
  2. 2
  3. 0
  4. 1
Dr. Karmach

Practice 3 answer: D

Ne₂²⁺: 2 × 8 − 2 = 14 valence electrons · σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴
bonding 2 + 2 + 4 = 8, antibonding 2 + 4 = 6 · σ*2p is empty
bond order = 8 − 62 = 1 → answer D

A removed only one electron for the 2+ charge: Ne₂⁺ has 2 × 8 − 1 = 15, (8 − 7)/2 = 0.5. B forgot to halve: 8 − 6 = 2. C ignored the charge: neutral Ne₂ has 16, (8 − 8)/2 = 0.

Dr. Karmach

Practice 3 answer: D

Ne₂²⁺: 2 × 8 − 2 = 14 valence electrons · σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴
bonding 2 + 2 + 4 = 8, antibonding 2 + 4 = 6 · σ*2p is empty
bond order = 8 − 62 = 1 → answer D
Taking two electrons off Ne₂ empties σ*2p, so the bond order rises from 0 to 1: a single bond, like F₂, which has the same 14 electrons. ✓

Dr. Karmach

Practice 4

B₂⁺: σ2s² σ*2s² π2p¹
valence molecular orbitals, filled lowest-first · wanted: the bond order

The valence molecular orbitals of B₂⁺ are filled as shown. What is its bond order?

  1. 1
  2. 0.5
  3. 2.5
  4. 1.5
Dr. Karmach

Practice 4 answer: B

B₂⁺: σ2s² σ*2s² π2p¹
bonding: σ2s + π2p = 2 + 1 = 3 · antibonding: σ*2s = 2
bond order = 3 − 22 = 0.5 → answer B

A forgot to halve: 3 − 2 = 1. C added the antibonding electrons: (3 + 2)/2 = 2.5. D left the antibonding electrons out: 3/2 = 1.5.

Bond order 0.5: half a bond. B₂ has bond order 1, and losing one bonding π2p electron halves it. ✓
Dr. Karmach

Practice 5

cyanide ion, CN⁻
wanted: the bond order · the π bonds in its Lewis structure

What bond order does molecular orbital theory give CN⁻, and how many π bonds does its Lewis structure show?

  1. Bond order 3; 0 π bonds
  2. Bond order 2.5; 2 π bonds
  3. Bond order 6; 2 π bonds
  4. Bond order 3; 3 π bonds
  5. Bond order 3; 2 π bonds
Dr. Karmach

Practice 5 answer: E

CN⁻: 4 + 5 + 1 = 10 valence electrons, the same count as N₂
σ2s² σ*2s² π2p⁴ σ2p² · bonding 8, antibonding 2 · Lewis structure [:C≡N:]⁻
bond order = 8 − 22 = 3; C≡N = 1 σ + 2 π → answer E

A counted all three lines of C≡N as σ: 3 σ, 0 π. B ignored the charge: CN has 4 + 5 = 9, (7 − 2)/2 = 2.5. C forgot to halve: 8 − 2 = 6. D read the triple bond as pure π and dropped its σ.

MO bond order 3 matches the Lewis triple bond: one σ on the axis plus two π. ✓
Dr. Karmach

Practice 6

O₂²⁺ · N₂⁻ · F₂²⁺ · Be₂⁺
wanted: the order of decreasing bond order

Using molecular orbital theory, arrange these four species in order of decreasing bond order.

  1. N₂⁻ > O₂²⁺ > F₂²⁺ > Be₂⁺
  2. F₂²⁺ > N₂⁻ > O₂²⁺ > Be₂⁺
  3. O₂²⁺ > N₂⁻ > F₂²⁺ > Be₂⁺
  4. Be₂⁺ > F₂²⁺ > N₂⁻ > O₂²⁺
  5. N₂⁻ > O₂²⁺ > Be₂⁺ > F₂²⁺
Dr. Karmach

Practice 6 answer: C

valence electrons: O₂²⁺ 12 − 2 = 10 · N₂⁻ 10 + 1 = 11 · F₂²⁺ 14 − 2 = 12 · Be₂⁺ 4 − 1 = 3
bond order: (8 − 2)/2 = 3 · (8 − 3)/2 = 2.5 · (8 − 4)/2 = 2 · (2 − 1)/2 = 0.5
3 > 2.5 > 2 > 0.5: O₂²⁺ > N₂⁻ > F₂²⁺ > Be₂⁺ → answer C

A ignored the charges: neutral N₂ 3 > O₂ 2 > F₂ 1 > Be₂ 0. B ranked by electron count, 12 > 11 > 10 > 3. D listed increasing bond order, the reverse. E moved each charge the wrong way: N₂⁻ as 9 electrons gives (7 − 2)/2 = 2.5, O₂²⁺ as 14 gives (8 − 6)/2 = 1, Be₂⁺ as 5 gives (3 − 2)/2 = 0.5, F₂²⁺ as 16 gives (8 − 8)/2 = 0.

More electrons do not mean more bond. F₂²⁺ has the most, but four of them sit in antibonding orbitals. ✓
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.

Dr. Karmach

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.

Dr. Karmach

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.

Dr. Karmach

What a Lewis structure already shows

O=O with two lone pairs on each O
2 bonding pairs + 4 lone pairs = 6 pairs = 12 valence electrons · not one drawn alone

A Lewis structure counts each bond as a shared pair and draws every electron paired. Molecular orbitals keep that count as the bond order, and they can find an electron left alone.

Dr. Karmach

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.

Dr. Karmach

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.

Dr. Karmach

Reading magnetism from the diagram

Every electron spins and acts as a tiny magnet. Paired electrons spin opposite ways and cancel. An unpaired electron leaves a net magnet for the field to grab: paramagnetic. All paired: diamagnetic, not pulled.

memory hook: unpaired means paramagnetic, pulled into the field
hear the di in diamagnetic as two: every electron sits in a pair, nothing for the field to grab
Dr. Karmach

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
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
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; read the diagram 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
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
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

Guided example: diboron, B₂

Step 1 · Count the valence electrons

B₂: each B contributes 3 valence electrons
2 × 3 = 6 valence electrons to place

Boron sits early in the second period. Pick its orbital order, then run Steps 2 to 4 and name each move as you make it. Find the magnetism and the bond order.

Dr. Karmach

Guided example: solution

B₂: 6 valence electrons
2 × 3 = 6 to place · boron sits early: π2p lies below σ2p

A common first attempt

✗ oxygen's order: σ2s² σ*2s² σ2p² → 0 unpaired
σ2p placed below π2p · electrons 5 and 6 pair there · wrongly predicts diamagnetic

That order belongs to O₂ and F₂; boron sits early in the period.

Dr. Karmach

Guided example: solution

B₂: 6 valence electrons
2 × 3 = 6 to place · boron sits early: π2p lies below σ2p
A common first attempt
✗ oxygen's order: σ2s² σ*2s² σ2p² → 0 unpaired
σ2p placed below π2p · electrons 5 and 6 pair there · wrongly predicts diamagnetic
Step 2 · Fill the MOs from the bottom up Step 3 · Read the magnetism Step 4 · Find the bond order

σ2s and σ*2s take four electrons. The last two reach the degenerate π2p pair and take one orbital each. Bonding σ2s and π2p hold 2 + 2 = 4; antibonding σ*2s holds 2.

σ2s² σ*2s² π2p² → paramagnetic · bond order = (4 − 2)/2 = 1
2 + 2 + 2 = 6 placed · π2p: ↑   ↑, 2 unpaired · σ2p stays empty
Dr. Karmach

Guided example: solution

B₂: 6 valence electrons
2 × 3 = 6 to place · boron sits early: π2p lies below σ2p
A common first attempt
✗ oxygen's order: σ2s² σ*2s² σ2p² → 0 unpaired
σ2p placed below π2p · electrons 5 and 6 pair there · wrongly predicts diamagnetic
Step 2 · Fill the MOs from the bottom up Step 3 · Read the magnetism Step 4 · Find the bond order
σ2s² σ*2s² π2p² → paramagnetic · bond order = (4 − 2)/2 = 1
2 + 2 + 2 = 6 placed · π2p: ↑   ↑, 2 unpaired · σ2p stays empty
B₂ is in fact paramagnetic: the magnet test shows π2p lies lowest for boron. ✓
Dr. Karmach

Guided example: the route

B₂: 2 × 3 = 6 valence electrons
given: B₂ · found: 2 unpaired, paramagnetic · bond order 1

Dr. Karmach

Guided example: the route

B₂: 2 × 3 = 6 valence electrons
given: B₂ · found: 2 unpaired, paramagnetic · bond order 1

The order fork decides the magnetism. Boron's π2p-first order keeps the last two electrons apart; σ2p first would pair them and miss the magnet test.
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

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 1 answer: A

O₂⁻ → bond order 1.5, paramagnetic → answer A
π*2p holds 3 electrons: one orbital paired, one single · (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.

Dr. Karmach

Practice 1 answer: A

O₂⁻ → bond order 1.5, paramagnetic → answer A
π*2p holds 3 electrons: one orbital paired, one single · (8 − 5)/2 = 1.5
An odd electron count guarantees at least one unpaired electron. The extra antibonding electron drops the bond order below O₂'s 2.

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

Worked example 3: the route

O₂⁺: 12 − 1 = 11 valence electrons
given: O₂ minus one electron · found: 1 unpaired, paramagnetic · bond order 2.5

Dr. Karmach

Worked example 3: the route

O₂⁺: 12 − 1 = 11 valence electrons
given: O₂ minus one electron · found: 1 unpaired, paramagnetic · bond order 2.5

O₂⁺ runs the same σ2p-first order as O₂, one electron short. Only the π*2p count changes, and with it both results: 1 unpaired, bond order up from 2 to 2.5.
Dr. Karmach

Practice 2

Be₂⁻
wanted: the molecular orbital that holds an unpaired electron

Which molecular orbital of Be₂⁻ holds an unpaired electron?

  1. σ*2s
  2. π2p
  3. σ2p
  4. None: every electron is paired
Dr. Karmach

Practice 2 answer: B

Be₂⁻: 2 × 2 + 1 = 5 · σ2s² σ*2s² π2p¹ → answer B
beryllium sits early: π2p below σ2p · the 5th electron enters the π2p pair alone · 1 unpaired

A applied the charge backwards: 2 × 2 − 1 = 3, σ2s² σ*2s¹, the lone electron in σ*2s. C used oxygen's order: σ2p below π2p puts the 5th electron in σ2p. Beryllium sits early, so π2p fills first. D dropped the charge: neutral Be₂ has 2 × 2 = 4, σ2s² σ*2s², all paired.

Dr. Karmach

Practice 2 answer: B

Be₂⁻: 2 × 2 + 1 = 5 · σ2s² σ*2s² π2p¹ → answer B
beryllium sits early: π2p below σ2p · the 5th electron enters the π2p pair alone · 1 unpaired
An odd count of 5 leaves one electron alone. It enters a bonding orbital: bond order (3 − 2)/2 = 0.5, up from Be₂'s 0. ✓

Dr. Karmach

Practice 3

F₂⁻ and C₂²⁻
wanted: the paramagnetic one · the higher bond order

Of F₂⁻ and the acetylide ion C₂²⁻, which one is paramagnetic, and which one has the higher bond order?

  1. C₂²⁻ is paramagnetic; C₂²⁻ has the higher bond order
  2. F₂⁻ is paramagnetic; F₂⁻ has the higher bond order
  3. C₂²⁻ is paramagnetic; F₂⁻ has the higher bond order
  4. F₂⁻ is paramagnetic; C₂²⁻ has the higher bond order
Dr. Karmach

Practice 3 answer: D

F₂⁻ paramagnetic, bond order 0.5 · C₂²⁻ diamagnetic, bond order 3 → answer D
F₂⁻: 2 × 7 + 1 = 15 · π*2p full, the 15th electron alone in σ*2p · (8 − 7)/2 = 0.5 · C₂²⁻: 2 × 4 + 2 = 10, all paired · (8 − 2)/2 = 3

A reversed the rule and called the all-paired ion paramagnetic. B added the antibonding electrons: F₂⁻ (8 + 7)/2 = 7.5 beats C₂²⁻ (8 + 2)/2 = 5. C made both slips.

Dr. Karmach

Practice 3 answer: D

F₂⁻ paramagnetic, bond order 0.5 · C₂²⁻ diamagnetic, bond order 3 → answer D
F₂⁻: 2 × 7 + 1 = 15 · π*2p full, the 15th electron alone in σ*2p · (8 − 7)/2 = 0.5 · C₂²⁻: 2 × 4 + 2 = 10, all paired · (8 − 2)/2 = 3
Count the charge before filling: the 2− adds two electrons to C₂, the − adds one to F₂, past the full π*2p into σ*2p. An odd count always leaves an electron unpaired. ✓

Dr. Karmach

Practice 4

B₂²⁺ · N₂²⁺ · O₂²⁻ · C₂²⁺
wanted: the paramagnetic species

Which of these species is paramagnetic?

  1. B₂²⁺
  2. N₂²⁺
  3. O₂²⁻
  4. C₂²⁺
Dr. Karmach

Practice 4 answer: D

C₂²⁺: 2 × 4 − 2 = 6 · σ2s² σ*2s² π2p² → answer D
the two π2p electrons take one orbital each · 2 unpaired · paramagnetic

A took both electrons from σ*2s, leaving σ2s² π2p². They leave the highest filled orbital, π2p: B₂²⁺ has 2 × 3 − 2 = 4, only the 2s orbitals filled, all paired. B applied the charge backwards, 2 × 5 + 2 = 12, O₂'s count with two unpaired. N₂²⁺ loses 2: 2 × 5 − 2 = 8, π2p full, all paired. C dropped the charge and filled neutral O₂, which is paramagnetic. O₂²⁻ has 2 × 6 + 2 = 14: both antibonding π orbitals full, 0 unpaired.

An even count can still leave electrons unpaired. The last two C₂²⁺ electrons enter the degenerate π2p pair, so Hund's rule keeps them apart. ✓
Dr. Karmach

Practice 5

B₂⁻
wanted: the valence MO configuration, lowest energy first

Which is the correct valence molecular-orbital configuration of B₂⁻, written from lowest to highest energy?

  1. σ2s² σ*2s² π2p³
  2. σ2s² σ*2s² σ2p² π2p¹
  3. σ2s² σ*2s² π2p² σ2p¹
  4. σ2s² σ*2s² π2p¹
Dr. Karmach

Practice 5 answer: A

B₂⁻: 2 × 3 + 1 = 7 · σ2s² σ*2s² π2p³ → answer A
boron sits early: π2p below σ2p · the π2p pair holds 3 of its 4 places

B used the oxygen ordering, σ2p below π2p. Boron sits early in the period, so the π2p pair fills before σ2p. C let the π2p pair take only 2, as if it were one orbital. It is two orbitals and holds 4. D applied the charge backwards: 2 × 3 − 1 = 5 electrons.

Seven is an odd count: one π2p electron stays unpaired, so B₂⁻ is paramagnetic, bond order (5 − 2)/2 = 1.5. ✓
Dr. Karmach

Practice 6

F₂²⁻ · C₂⁻ · N₂²⁻ · B₂²⁻
wanted: the species that matches O₂ in bond order and in magnetism

Which species has the same bond order and the same magnetic behavior as O₂?

  1. F₂²⁻
  2. C₂⁻
  3. N₂²⁻
  4. B₂²⁻
Dr. Karmach

Practice 6 answer: C

N₂²⁻: 2 × 5 + 2 = 12 · σ2s² σ*2s² π2p⁴ σ2p² π*2p² → answer C
bond order (8 − 4)/2 = 2 · 2 unpaired in π*2p · O₂: bond order 2, paramagnetic

A applied the charge backwards, 2 × 7 − 2 = 12, O₂'s count. F₂²⁻ gains 2: 2 × 7 + 2 = 16, every antibonding orbital full, (8 − 8)/2 = 0. B matched the magnetism only: C₂⁻ has 2 × 4 + 1 = 9, 1 unpaired in σ2p, but (7 − 2)/2 = 2.5. D matched the bond order only: B₂²⁻ has 2 × 3 + 2 = 8, (6 − 2)/2 = 2, all paired.

N₂²⁻ holds 12 electrons, the same count as O₂. The σ2p and π2p order differs, but both diagrams end with one electron in each π*2p orbital. ✓
Dr. Karmach

Check yourself

  1. N₂⁻ has 11 valence electrons, one more than N₂, and the extra one enters π*2p. How many electrons are unpaired, and what is the bond order?
  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

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