Lewis Structures with Double and Triple Bonds

Hello again. In the previous lesson, you learned to draw Lewis structures with single bonds and lone pairs by treating the total number of valence electrons as an electron budget. That same method still works here. The new idea is simple but important: if a single-bond skeleton leaves an atom without a full outer shell, you may need to turn a lone pair into an additional shared pair.

This lesson focuses on neutral molecules with double and triple covalent bonds. By the end, you should be able to draw complete Lewis structures for molecules such as $\mathrm{C}{\mathrm{O}}_2$, ${\mathrm{C}}_2{\mathrm{H}}_4$, $\mathrm{HCN}$, and ${\mathrm{C}}_2{\mathrm{H}}_2$, including all required lone pairs.

---

When one shared pair is not enough

A single bond is one shared electron pair. It contains two electrons.

A double bond is two shared electron pairs between the same two atoms. It contains four electrons.

A triple bond is three shared electron pairs between the same two atoms. It contains six electrons.

Bond type	Lewis-line form	Shared electrons
Single bond	$\mathrm{A}-\mathrm{B}$	2
Double bond	$\mathrm{A}=\mathrm{B}$	4
Triple bond	$\mathrm{A}\equiv \mathrm{B}$	6

The key point is that a multiple bond does not add electrons to the molecule. It rearranges electrons that are already part of the total valence-electron budget.

For example, an oxygen atom in a single bond often has three lone pairs:

• one single bond contains 2 electrons;
• three lone pairs contain 6 electrons;
• total around oxygen: 8 electrons.

If that oxygen forms a double bond, it has only two lone pairs:

• one double bond contains 4 electrons;
• two lone pairs contain 4 electrons;
• total around oxygen: still 8 electrons.

So the octet is maintained; the electrons have simply been redistributed.

---

The extended Lewis-structure method

Use the single-bond method from the last lesson first. Do not guess a double or triple bond immediately.

1. Count all valence electrons. This is your fixed electron budget.

2. Draw a skeleton with single bonds. Hydrogen is never central. Carbon is usually central when present.

3. Complete the octets of terminal atoms first, except hydrogen, which needs only one bond.

4. Place any remaining electrons on the central atom.

5. Check every atom. Hydrogen should have two electrons; carbon, nitrogen, oxygen, and halogens should normally have eight.

6. If a central atom lacks an octet, move a lone pair from an adjacent atom into the bond between them.

   • Moving one lone pair creates a double bond.
   • Moving two lone pairs creates a triple bond.

7. Audit the final drawing. Count the total electrons again, include lone pairs, and check the usual bonding patterns.

The crucial diagnostic is this:

If you have used every valence electron but an atom still lacks an octet, a multiple bond may be required.

Never solve this problem by inventing extra electrons.

---

Worked example: carbon dioxide, $\mathrm{C}{\mathrm{O}}_2$

Carbon dioxide is an excellent example because both oxygen atoms form double bonds with carbon.

1. Count valence electrons

Carbon has four valence electrons and each oxygen has six:

$$4+(2\times 6)=16$$

There are 16 valence electrons available.

2. Draw the single-bond skeleton

Carbon is central:

O-C-O

Two single bonds use four electrons, so 12 electrons remain.

3. Complete the outer oxygen atoms

Give each oxygen three lone pairs. This uses the remaining 12 electrons.

At this stage, each oxygen has an octet, but carbon has only four electrons around it from its two single bonds. Carbon is incomplete.

4. Form multiple bonds

Move one lone pair from the left oxygen into the carbon–oxygen bond. Then do the same on the right.

..     ..
O = C = O
..     ..

Each oxygen now has:

• one double bond;
• two lone pairs;
• eight electrons around it.

Carbon has two double bonds, so it also has eight electrons around it.

5. Electron-budget check

• Two double bonds: $2\times 4=8$ electrons
• Four lone pairs total: $4\times 2=8$ electrons

$$8+8=16$$

The final Lewis structure uses exactly 16 electrons.

A correct final structure is therefore:

$$\mathrm{O}=\mathrm{C}=\mathrm{O}$$

with two lone pairs on each oxygen.

---

Worked example: ethene, ${\mathrm{C}}_2{\mathrm{H}}_4$

Ethene contains a carbon–carbon double bond.

1. Count valence electrons

$$(2\times 4)+(4\times 1)=12$$

There are 12 valence electrons.

2. Draw the skeleton with single bonds

Each carbon is connected to two hydrogens, and the carbon atoms are connected to each other:

H   H
|   |
C-C
|   |
H   H

There are five single bonds, using:

$$5\times 2=10\text{ electrons}$$

That leaves two electrons.

3. Check the carbon atoms

With only single bonds, each carbon has:

• two $\mathrm{C}-\mathrm{H}$ bonds;
• one $\mathrm{C}-\mathrm{C}$ bond.

That is only three bonding pairs, or six electrons around each carbon. Carbon needs an octet.

The remaining pair is used to create another bond between the carbons.

H   H
 \ /
  C=C
 / \
H   H

More conventionally, this is written:

$${\mathrm{H}}_2\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$$

There are no lone pairs in ethene.

4. Check

Each carbon has:

• two single bonds to H;
• one double bond to C.

That gives each carbon four bonds in total and eight electrons around it. Each hydrogen has its usual one bond.

---

Worked example: hydrogen cyanide, $\mathrm{HCN}$

Hydrogen cyanide contains a carbon–nitrogen triple bond.

1. Count valence electrons

$$1+4+5=10$$

There are 10 valence electrons.

2. Draw the skeleton

Hydrogen must be at an end:

H-C-N

The two single bonds use four electrons, leaving six.

3. Complete nitrogen’s outer shell

Put the six remaining electrons as three lone pairs on nitrogen. Nitrogen has an octet, but carbon has only four electrons around it.

Carbon is incomplete, and there are no extra electrons available.

4. Convert nitrogen lone pairs into bonding pairs

Move one nitrogen lone pair into the carbon–nitrogen bond. Carbon still lacks an octet, so move a second nitrogen lone pair into that bond too.

The result is:

$$\mathrm{H}-\mathrm{C}\equiv \mathrm{N}:$$

The colon represents one lone pair on nitrogen.

5. Check

• Hydrogen has one single bond and therefore a duet.
• Carbon has one single bond plus one triple bond: four bonds total.
• Nitrogen has one triple bond and one lone pair: eight electrons around it.
• The electron count is correct:

$$2+6+2=10$$

The 2 electrons come from the $\mathrm{H}-\mathrm{C}$ bond, 6 from the triple bond, and 2 from nitrogen’s lone pair.

---

Triple bonds in ${\mathrm{N}}_2$ and ${\mathrm{C}}_2{\mathrm{H}}_2$

Some common two-atom or carbon-chain molecules also require triple bonds.

Nitrogen, ${\mathrm{N}}_2$

Each nitrogen has five valence electrons:

$$2\times 5=10$$

The completed Lewis structure is:

$$:\mathrm{N}\equiv \mathrm{N}:$$

Each nitrogen has:

• one triple bond;
• one lone pair;
• eight electrons around it.

Ethyne, ${\mathrm{C}}_2{\mathrm{H}}_2$

Ethyne has the skeleton $\mathrm{H}-\mathrm{C}-\mathrm{C}-\mathrm{H}$. Its final structure is:

$$\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$$

Each carbon has one single bond to hydrogen and one triple bond to carbon. That is four bonds per carbon, matching carbon’s usual bonding pattern.

Unlike $\mathrm{HCN}$, ethyne has no lone pairs.

---

Patterns worth recognising quickly

For the simple neutral molecules in this course, these completed patterns are useful checks:

Molecule	Multiple bond(s)	Lone pairs in final structure
$\mathrm{C}{\mathrm{O}}_2$	Two $\mathrm{C}=\mathrm{O}$ double bonds	2 on each O
$\mathrm{C}{\mathrm{H}}_2\mathrm{O}$	One $\mathrm{C}=\mathrm{O}$ double bond	2 on O
${\mathrm{C}}_2{\mathrm{H}}_4$	One $\mathrm{C}=\mathrm{C}$ double bond	None
${\mathrm{C}}_2{\mathrm{H}}_2$	One $\mathrm{C}\equiv \mathrm{C}$ triple bond	None
$\mathrm{HCN}$	One $\mathrm{C}\equiv \mathrm{N}$ triple bond	1 on N
${\mathrm{N}}_2$	One $\mathrm{N}\equiv \mathrm{N}$ triple bond	1 on each N

A quick bonding-pattern check can catch many mistakes:

Atom	Common neutral pattern
$\mathrm{H}$	1 bond
$\mathrm{C}$	4 bonds total
$\mathrm{N}$	3 bonds and 1 lone pair
$\mathrm{O}$	2 bonds and 2 lone pairs
Halogen	1 bond and 3 lone pairs

When counting carbon’s “four bonds,” a double bond counts as two bonds and a triple bond counts as three.

For instance:

• carbon in $\mathrm{C}{\mathrm{O}}_2$ has two double bonds, so $2+2=4$;
• carbon in ethene has two single bonds and one double bond, so $1+1+2=4$;
• carbon in ethyne has one single bond and one triple bond, so $1+3=4$.

---

Common mistakes to avoid

Adding electrons rather than moving them

A double bond does not give you two new electrons. It uses a lone pair already included in the valence-electron total.

Forgetting lone pairs after drawing a multiple bond

In $\mathrm{C}{\mathrm{O}}_2$, each oxygen has a double bond and two lone pairs. Writing only $\mathrm{O}=\mathrm{C}=\mathrm{O}$ is a correct displayed formula but incomplete as a Lewis structure unless the lone pairs are shown.

Giving carbon only three bonds

A carbon with three single bonds has only six electrons around it. In simple neutral molecules, this is a warning sign that a double bond may be needed.

Drawing multiple bonds to hydrogen

Hydrogen is complete with one single bond. It cannot form a double or triple bond in the simple Lewis structures you are drawing.

Forgetting the final electron count

Use the same audit every time:

$$\text{electrons in bonds}+\text{electrons in lone pairs}=\text{total valence electrons}$$

If this equality fails, the structure cannot be correct.

---

Key takeaways

A double bond is two shared electron pairs, while a triple bond is three shared electron pairs. You begin every Lewis structure by counting valence electrons and drawing single bonds. Only after completing outer atoms and checking octets do you convert lone pairs into additional bonds where necessary.

For simple neutral molecules:

• use a double bond when one extra shared pair is needed;
• use a triple bond when two extra shared pairs are needed;
• show every remaining lone pair;
• confirm both the octet rule and the total electron count.

Next, you will use this same process for polyatomic ions, where brackets and an overall charge change the electron budget.