Bonding and Molecular Structure

Chapter 08

Lewis Dot Symbols

Lewis dot symbols are diagrams that illustrate the number of valence electrons about an atom. Dots represent electrons. Below are the Lewis dot symbols for the first 10 elements of the periodic table

Noble gases, with the exception of helium, has a total of 8 valence electrons. This is referred to as an octet of electrons and is considered to be a “full valence shell” for main group elements. This leads to the octet rule which is a rule of thumb which states that main group elements tend to bond in such a way that each atom has with electrons in its valence shell. The duplet rule (bonding that results in 2 electrons in its valence shell) applies for hydrogen and helium. For transition metals, the 18-electron rule (18 valence electrons) applies.

We will see examples of these rules later in these notes.

Chemical bonds

Chemical reactions occur due to the rearrangement of electrons between atoms, molecules, and ions (i.e. things happen because of what electrons do). Some chemical reactions lead to the formation of new chemical bonds, Lewis structures are diagrams that illustrate bonding between atoms in molecules and lone electron pairs.

Covalent bonding

Covalent bonds are chemical bonds that involve the sharing of valence electrons between atoms.

For example, two hydrogen atoms, each with one valence electron, form H₂ due to the sharing of one electron from each individual hydrogen atom. This shared electron pair between the atoms is represented by a single dash (called a single bond). This fulfills the duplet rule for each hydrogen in H₂ meaning, each hydrogen is stable when it has 2 electrons in its valence shell.

The formation of F₂ from two fluorine atoms results in a single bond and each fluorine atom has an octet of electrons in the molecule fulfilling the octet rule.

Two electrons in a bond

A “dash” in a Lewis structure is a chemical bond that is formed from two electrons.

Sometimes more than one electron pair is shared between atoms in order to fulfill the octet rule. Consider the case of carbon dioxide. Here, four electron pairs are shared between carbon and two oxygen atoms. Four electrons in a chemical bond leads to a double bond represented by a double-dash between atoms.

Once again, each atom in CO₂ contains eight electrons in the valence shell fulfilling the octet rule.

Below is an example of a triple bond involving 6 shared electrons (or three shared electron pairs) between two nitrogen atoms.

The length of a bond (bond length) can qualitatively be compared by the type of bond it is. Single bonds are typically longer than double bonds and double bonds are typically longer than triple bonds.

Ionic bonding

Ionic bonds form between ions due to the transfer of electrons from a reducing agent to an oxidizing agent.

Take the formation of sodium chloride for example. Sodium has a low ionization energy and readily loses an electron whereas chlorine has a high electron affinity and readily gains an electron. This results in two ions of opposite charge that are attracted to each other to form the ionic compound, NaCl. Each atom has a full octet (lone pairs for sodium not shown for clarity).

Bond polarity and electronegativity

So far we have seen examples of covalent and ionic bonds. Covalent bonds are bonds where electrons are shared whereas ionic bonds are formed when electrons are transferred (not shared). We can predict the type of bond that forms between atoms based on the extent at which electrons are shared by examining the difference in electronegativity of the atoms involved in the bonding.

Electronegativity (χ; unitless) is the tendency for an atom of a given element to attract shared electrons (or electron density) to itself when forming a chemical bond. An atom’s atomic number and the distance of the valence electrons from the nucleus both play a role in determining the electronegativity of an atom. This property allows one to easily predict a bond energy as well as the polarity of a bond.

Electronegativity tends to increase as you move

• up (a group)
• right (across a period)

Click to view in 3D

When thinking of bond polarity, envision the electrons around the atoms as an electron cloud of negative charge (i.e. electron density). An electron cloud can be shifted/morphed depending on its chemical environment.

Bonds between atoms of identical electronegativity are classified as being non-polar, meaning there is no net positive or negative charge in either end of the molecule. The electrons (or electron cloud) is shared equally between the two atoms. The resulting covalent bond is completely covalent.

A polar covalent bond occurs between two different atoms because electron density is not equally shared since the atoms involved have different electronegativities. The electron cloud morphs and is “pulled” or “transferred” toward the atom with a larger electronegativity giving the bond some ionic character. This results in an increase of negative charge (partial negative charge; δ^–) around the more electronegative atom and makes the more electropositive atom electron deficient giving rise to a partial positive charge (δ⁺).

Below is an example of a polar covalent bond between hydrogen and fluorine in hydrogen fluoride. Flourine is more electronegative than hydrogen and attracts electron density towards itself and pulls some electron density away from hydrogen. This leads to a partial positive charge on hydrogen and a partial negative charge on fluorine.

Hydrogen fluoride is a polar molecule due to this separation of partial charges. The bigger the difference in electronegativity of the atoms in the bond, the greater the extend of the polarity of the bond and the more polar a diatomic molecule will be.

An ionic bond results in the complete transfer of electron density that is not shared. These bonds are polar.

Summary

The type of bond can be determined by analyzing the electronegativity difference (Δchi;) in the atoms involved in the bond.

• Pure covalent – Δχ < 0.4
• Polar covalent – Δχ between 0.4 to 1.8
• Ionic – Δχ > 1.8

Comparision of bonds and their ionic character.

Polarizability

The polarizability of an atom is the ease at which an electron cloud about an atom or ion can be distorted by an external electric field. Polarizability is generally inversely proportional to an atom’s electronegavitity.

Fluroine is not very polarizable as it is a very electronegative atom with a “tight” electron cloud. Iodine is very polarizable as it is much less electronegative and has a very large (or diffuse) electron cloud.

The more polarizable an atom or ion, the more its electron cloud can distort, leading to the creation of partial charges within the atom or ion.

A ion or molecule can act as an electric field which can deform the electron clouds of a neighboring particle. An induced dipole is formed when the electron cloud deforms.

Lewis structures

Drawing Lewis structures to represent molecules can be performed by following a series of steps.

Step 1: Determine total number of valence electrons in the molecule or ion.
Step 2: Determine the arrangement of atoms within a molecule by usually
choosing the least electronegative atom as the central atom and surrounding it with the remaining atoms (called terminal atoms). Draw single bonds between the central atom and each terminal atom.
Step 3: Place remaining electrons as pairs around terminal atoms to fulfill the octet or duplet rule. Remaining electrons are placed on the central atom.
Step 4: Move electron pairs on terminal atoms into bonds (double or triple) to fulfill the octet rule (if possible) of the central atom if it has not been fulfilled in Step 4.
Step 5: If the particle is an ion, draw square brackets around the Lewis structure and include the charge in the top right.

Determining the central atom

1. The central atom is usually the atom with the lowest electronegativity.
2. For simple compounds, the first atom in the molecular formula is usually the central atom (shown in bold for the following examples: SO₂, NH₃, CF₄, PO₄^3–). Some compounds do not follow this rule such as acids (e.g. HNO₃).
3. C, N, P, and S are typically the central atom for simple compounds.
4. Atoms that form single bonds (e.g. hydrogen and halogens) are typically terminal atoms. Central atoms must be able to form multiple bonds.
5. Oxygen is a central atom in water but usually a terminal atom elsewhere.
6. Hydrogen is rarely a central atom. Hydrogen can be a central atom in transition metal hydrides or complex anions (such as B₂H₆).

Note: The examples below illustrate the steps laid out above. However, the concept of formal charge when writing Lewis structures will be relgated to a later section (see: Exceptions to the Octet Rule).

Lewis structure for simple compounds

Example: Trifluoromethane, CF₃H

Step 1: Determine total number of valence electrons in the molecule or ion.

• C = 4 valence e^–
• 3F = 21 valence e^–
• H = 1 valence e^–
• Total = 26 valence e^–

Step 2: Determine the arrangement of atoms within a molecule by usually choosing the least electronegative atom as the central atom and surrounding it with the remaining atoms (called terminal atoms). Draw single bonds between the central atom and each terminal atom.

• C: χ = 2.5
• H: χ = 2.2
• F: χ = 4.0

Even though hydrogen is the least electronegative atom, hydrogen is a terminal atom since it can only form one bond. Therefore, carbon is the central atom.

So far, this structure has accounted for 8 of the 26 total valence electrons.

Step 3: Place remaining electrons as pairs around terminal atoms to fulfill the octet or duplet rule. Remaining electrons are placed on the central atom.

There are 18 remaining valence electrons to place. Six electrons are placed on each fluorine atom.

All valence electrons have now been accounted for in the Lewis structure.

Step 4: Move electron pairs on terminal atoms into bonds (double or triple) to fulfill the octet rule (if possible) of the central atom if it has not been fulfilled in Step 4.

Skip this step. The octet rule for all non-hydrogen atoms (and the duplet rule for hydrogen) have been satisfied.

Step 5: If the particle is an ion, draw square brackets around the Lewis structure and include the charge in the top right.

Skip.

The final Lewis structure for CF₃H is given in Step 3.

Example: Formaldehyde, CH₂O

Step 1: Determine total number of valence electrons in the molecule or ion.

• C = 4 valence e^–
• 2H = 2 valence e^–
• O = 6 valence e^–
• Total = 12 valence e^–

Step 2: Determine the arrangement of atoms within a molecule by usually choosing the least electronegative atom as the central atom and surrounding it with the remaining atoms (called terminal atoms). Draw single bonds between the central atom and each terminal atom.

• C: χ = 2.5
• H: χ = 2.2
• O: χ = 3.5

Even though hydrogen is the least electronegative atom, hydrogen is a terminal atom since it can only form one bond. Therefore, carbon is the central atom.

So far, this structure has accounted for 6 of the 12 total valence electrons.

Step 3: Place remaining electrons as pairs around terminal atoms to fulfill the octet or duplet rule. Remaining electrons are placed on the central atom.

There are 6 remaining valence electrons to place. Six electrons are placed on the oxygen atom.

All valence electrons have now been accounted for in the Lewis structure.

Step 4: Move electron pairs on terminal atoms into bonds (double or triple) to fulfill the octet rule (if possible) of the central atom if it has not been fulfilled in Step 4.

The octet rule for the central carbon atom has not been fulfilled (it currently has 6 valence electrons). Move a lone electron pair from the terminal oxygen atom into a bond with carbon to form a double bond.

All non-hydrogen atoms now fulfill the octet rule (hydrogen atoms fulfill duplet rule).

Step 5: If the particle is an ion, draw square brackets around the Lewis structure and include the charge in the top right.

Skip.

The final Lewis structure for CH₂O is given in Step 4.

Example: Ammonium ion, NH₄⁺

Step 1: Determine total number of valence electrons in the molecule or ion.

• N = 5 valence e^–
• 4H = 4 valence e^–
• +1 charge = –1 valence e^–
• Total = 8 valence e^–

Step 2: Determine the arrangement of atoms within a molecule by usually choosing the least electronegative atom as the central atom and surrounding it with the remaining atoms (called terminal atoms). Draw single bonds between the central atom and each terminal atom.

• N: χ = 3.0
• H: χ = 2.2

Even though hydrogen is the least electronegative atom, hydrogen is a terminal atom since it can only form one bond. Therefore, nitrogen is the central atom.

This structure has accounted for 8 of the 8 total valence electrons. The octet rule is fulfilled for nitrogen and the duplet rule is fulfilled for each hydrogen atom.

Step 3: Place remaining electrons as pairs around terminal atoms to fulfill the octet or duplet rule. Remaining electrons are placed on the central atom.

Skip.

Step 4: Move electron pairs on terminal atoms into bonds (double or triple) to fulfill the octet rule (if possible) of the central atom if it has not been fulfilled in Step 4.

Skip.

Step 5: If the particle is an ion, draw square brackets around the Lewis structure and include the charge in the top right.

Example: Hydroxide, OH^–

Step 1: Determine total number of valence electrons in the molecule or ion.

• O = 6 valence e^–
• H = 1 valence e^–
• –1 charge = +1 valence e^–
• Total = 8 valence e^–

Step 2: Determine the arrangement of atoms within a molecule by usually choosing the least electronegative atom as the central atom and surrounding it with the remaining atoms (called terminal atoms). Draw single bonds between the central atom and each terminal atom.

There is no central atom in a diatomic. Simply draw the to atoms and connect them with a single bond.

This structure has accounted for 2 of the 8 total valence electrons.

Step 3: Place remaining electrons as pairs around terminal atoms to fulfill the octet or duplet rule. Remaining electrons are placed on the central atom.

The remaining 6 electrons are placed on oxygen, completing its octet. Hydrogen’s duplet is already fulfilled.

Step 4: Move electron pairs on terminal atoms into bonds (double or triple) to fulfill the octet rule (if possible) of the central atom if it has not been fulfilled in Step 4.

Skip.

Step 5: If the particle is an ion, draw square brackets around the Lewis structure and include the charge in the top right.

Example: Hydronium ion, H₃O+

Step 1: Determine total number of valence electrons in the molecule or ion.

• O = 6 valence e^–
• 3H = 3 valence e^–
• +1 charge = –1 valence e^–
• Total = 8 valence e^–

Step 2: Determine the arrangement of atoms within a molecule by usually choosing the least electronegative atom as the central atom and surrounding it with the remaining atoms (called terminal atoms). Draw single bonds between the central atom and each terminal atom.

Oxygen is the only atom that can form more than one bond. Oxygen is the central atom.

This structure has accounted for 6 of the 8 total valence electrons.

Step 3: Place remaining electrons as pairs around terminal atoms to fulfill the octet or duplet rule. Remaining electrons are placed on the central atom.

The duplet rule for the terminal hydrogen atoms is fulfilled. Place the remaining 2 electrons around the central oxygen atom to fulfill its octet.

Step 4: Move electron pairs on terminal atoms into bonds (double or triple) to fulfill the octet rule (if possible) of the central atom if it has not been fulfilled in Step 4.

Skip.

Step 5: If the particle is an ion, draw square brackets around the Lewis structure and include the charge in the top right.

Formal Charges

A formal charge is a hypothetical charge on an atom in a molecule or ion that results from the equal sharing of all bonding electrons. It allows us to estimate the distribution of electric charge within a molecule. The following equation is used to determine the formal charge of an atom in a molecule/ion:

\[\mathrm{formal~charge} = N\mathrm{V}e^- - [\mathrm{lone}~e^- + \frac{1}{2}(\mathrm{bonding}~e^-)]\]

where

• N Ve^– = number of valence electrons of atom of interest
• lone e^– = number of lone electrons around atom of interest
• bonding e^– = number of electrons in bonds around atom of interest

A simplified version of this equation is given below and works well when analyzing a Lewis structure.

\[\mathrm{formal~charge} = N\mathrm{V}e^- - \mathrm{dots} - \mathrm{dashes}\]

When constructing a Lewis structure, use formal charges to determine the best arrangement of atoms and bond types by minimizing the formal charges on the atoms. The formal charges for all atoms in a particle must sum to the total charge of the particle.

Example: Carbon dioxide, CO₂

Writing out a Lewis structure for carbon dioxide can result in three different structures when fulfilling the octet rule.

We know from experiment that the carbon-oxygen bonds are equivalent. The middle Lewis structure minimizes the formal charges across all the atoms and gives the appropriate structure.

Example: Nitric oxide, NO

Nitric oxide gives an odd number of valence electrons (11 Ve^–). Nitrogen is assigned the unpaired electron (rather than oxygen) due to the minimization of formal charges.

Example: Nitrogen dioxide, NO₂

Nitrogen dioxide gives an odd number of valence electrons (17 Ve^–). Nitrogen is assigned the unpaired electron (rather than oxygen) due to the minimization of formal charges.

Limitations of formal charge

Formal charges are useful but has its limitations. Formal charges treat electrons as being shared equally between atoms.

Example: CO

• 10 Ve^–

Carbon monoxide has a triple bond.

While the structure with the double bond has formal charges of zero, the octet for carbon has not been fulfilled. A triple bond fulfills the octet rule. Also, the formal charges sum to the overall charge of zero.

Example: BF₄^–.

The fluorine atoms have a formal charge of 0 whereas the boron atom has a formal charge of –1. One would imagine that a more electronegative fluorine atom should have the –1 formal charge instead. Furthermore, the B–F bonds are polar with the partial negative charge on fluorine. This is an example of a case where formal charge and bond polarity are not in agreement.

Resonance Lewis structures

Resonance is a property of a molecule or ion that occurs when a single Lewis structure fails to accurately describe the electronic structure. Resonance in a Lewis structure can be identified by the ability to draw multiple Lewis structures where the octet rule is obeyed for all atoms without regard to the formal charges. We only consider the formal charges when determining the contribution of each resonance structure to the overall (or best) structure for the molecule/ion.

Example: Ozone, O₃

Consider the Lewis structure for ozone (O₃).

A double bond may be placed between the first two oxygen atoms thereby fulfilling the octet rule for all atoms. However, a double bond can also be placed between the last two oxygen atoms instead, still fulfilling the octet rule for all atoms. This is a case where one Lewis structure is not able to represent the bonding within an ozone molecule.

According to Lewis theory, a double bond is shorter than a single bond. Experimental observations demonstrate that ozone has two equivalent O–O bonds (the bond lengths are the same). In order to properly represent this, we must draw both equivalent Lewis structures and draw a double-headed arrow between them.

Each Lewis structure for ozone is called a resonance structure. These structures have identical energies and bonding patterns. In reality, the structure of ozone is a composite of the two resonance structures, sometimes called a hybrid or hybrid resonance structure. You can imagine that each O–O bond is a “single and a half” bond, not quite single bonds and not quite double bonds.

Resonance is due to the delocalization of electrons. In the case of ozone, a pair of electrons is delocalized over all three atoms.

Alternative schemes for drawing ozone include drawing the half-bond as a dashed line though this type of scheme should not be used for this class.

Each resonance structure is equivalent in terms of the distribution of formal charges, and therefore, each contribute equally to the overall Lewis structure for ozone.

Example: Carbon dioxide, CO₂

Three resonance structures for CO₂ exist. Notice how each resonance structure fulfills the octet rule.

Resonance structure A has the best set of formal charges (they are all zero) and therefore this structure contributes the most to the actual structure of CO₂.

Resonance structures B and C have a poor distribution of formal charges. Notice how one of the oxygens has a +1 formal charge while the other has a –1. Structures B and C have a smaller contribution to the actual structure of CO₂.

Example: Benzene, C₆H₆

Benzene is another case of resonance. Benzene is a cylic six-membered ring. The double bonds alternate between each pair of carbon atoms and can be written out in two different ways. This results in two resonance structures for benzene.

Experiment shows that all the carbon-carbon bonds are equivalent in energy and length.

It is common in organic chemistry to omit the explicit carbon atom labels. They are implied to be at the vertices (intersections) of the bonds.

The Lewis structure for benzene is sometimes abbreviated to be

Each resonance structure is equivalent in terms of the distribution of formal charges, and therefore, each contribute equally to the overall Lewis structure for ozone.

Example: OCN^–

Below are the three resonance structures for OCN^–. The formal charges are included. Notice how the octet rule is fulfilled for each resonance structure which allows us to identify that three resonance structures are present (we do not consider formal charges).

To determine the contribution of each structure to the overall structure, you only then consider the formal charges. The resonance structure that contributes the most is the one with the most appropriate distribution of formal charges. Here, resonance structure 1 contributes the most to the overall structure since the negative formal charge resides on the most electronegative atom, as expected. Resonance structure 3 contributes the least since the distribution of formal charge is the least expected (+1 on the more electronegative oxygen and –1 on the less electronegative nitrogen).

Example: Other resonance structures

Nitrite ion, NO₂^–

There are two resonance structures for the nitrite ion.

Each resonance structure is equivalent in terms of the distribution of formal charges, and therefore, each contribute equally to the overall Lewis structure for ozone.

Carbonate ion, CO₃^2–

There are three resonance structures for the carbonate ion.

Each resonance structure is equivalent in terms of the distribution of formal charges, and therefore, each contribute equally to the overall Lewis structure for ozone.

Exceptions to the Octet Rule

Hypovalent atoms

Hypovalent refers to central atoms that have fewer than 8 valence electrons.

Example: Borane, BH₃

Borane has a central atom with less than 8 valence electrons. It can only obtain six.

Interestingly, ammonia borane forms from a reaction with ammonia and borane. A single bond forms between the nitrogen and boron where the two electrons involved in this bond comes from nitrogen (and not one electron from nitrogen and one electron from boron). This type of covalent bond is known as a coordinate covalent bond or dative bond, a bond that is formed with a lone electron pair originates from only one of the atoms involved in the bond.

This bond is able to form due to an empty (unoccupied and unhybridized) p orbital on boron. The lone electron pair on nitrogen is donated into this atomic orbital to create a bonding molecular orbital.

Hypervalent atoms

Hypervalent refers to central atoms that have more than than 8 valence electrons. Only elements of the third or higher periods on the periodic table can have more than 8 valence electrons (sometimes called an expanded octet). This is because the availability of d orbitals that can expand their valency shells to contain five or six pairs of electrons. Additionally, the size of the outermost shell is large enough to accommodate more than 8 electrons.

Example: Sulfur dioxide, SO₂

Sulfur is a period 3 main group element and can be hypervalent. Sulfur, the central atom, has 10 neighboring electrons.

Sometimes you will see a pair of resonance structures written out as

Here, the octet rule has been fulfilled for each atom but the formal charges are not minimized.

If one accounts for the “hypervalent” character of main group elements beyond Period 2, you will get a structure that contains two double bonds. Notice how sulfur now has 10 valence electrons around it and the formal charges are all zero.

Other Examples

Notice how silicon and phosphorus are hypervalent and have 10 or 12 valence electrons around them.

Lewis structures for multi-center compounds

Many compounds have more than once central atom. Notice in the examples below how the duplet/octet rules are fulfilled. The skeletal structure of a multi-centered compound can usually be determined from the structural formula of the molecule.

Propane, C₃H₈

Propane (20 valence e^–), an alkane, has three central carbon atoms.

Structural formula: CH₃CH₂CH₃

Acetylene, C₂H₂

Acetylene (10 valence e^–), an alkyne, has two central carbon atoms.

Methanol, CH₃OH

Methanol (14 valence e^–), an alcohol, has two central atoms, C and H. Sometimes Lewis structures omit lone electron pairs. Also, methyl groups (–CH₃) and hydroxyl groups (–OH) are often condensed (single bonds omitted) as seen in the right side of the figure below.

Structural formula: CH₃OH

Ethanol, CH₃CH₂OH

Ethanol (20 valence e^–), an alcohol, has three central atoms, C and H. Sometimes Lewis structures omit lone electron pairs. CH₂– groups are often condensed (single bonds omitted) as seen in the right side of the figure below.

Structural formula: CH₃CH₂OH

Acetic acid, C₂H₄O₂

Acetic acid (ethanoic alcohol; 24 valence e^–), has three central atoms.

Glycine, C₂H₅NO₂

Glycine (30 valence e^–) has four central atoms.

Structural formula: NH₂CH₂COOH

Oxidation number and Lewis structures

Oxidation numbers can be determined by analyzing Lewis structures.

Recall that with formal charge, electrons in bonds are “shared” or assigned equally to each atom involved in the bond.

When it comes to determining oxidation numbers, electrons in bonds are assigned to the most electronegative atom involved in the bond. If both atoms involved in the bond are identical, split the electrons and assign equally. Lone pairs on atoms remain assigned to those atoms.

The oxidation state can then be determined as

\[\mathrm{Ox. Number} = \mathrm{V}e^- - \mathrm{assigned~}e^-\]

The sum of all the individual oxidation numbers should equal the charge on the molecule/ion.

Example: H₂

Rules for assinging oxidation numbers leads us to the following oxidation states for each atom in H₂.

We can reach the same conclusion by analyzing the Lewis structure for CO₂. Draw the Lewis structure and then assign the electrons in each bond to the neighboring atom that is the most electronegative. Here, since both atoms have the same electronegativity, the electrons are shared equally.

Next, determine the oxidation state using the equation above.

• H: 1 - 1 = 0
• H: 1 - 1 = 0

Example: CO₂

Rules for assinging oxidation numbers leads us to the following oxidation states for each atom in CO₂.

We can reach the same conclusion by analyzing the Lewis structure for CO₂. First, draw the Lewis structure.

Next, assign the electrons in each double bond to the neighboring atom that is the most electronegative. Here, the most electronegative atom is oxygen.

Next, determine the oxidation state using the equation above.

• O: 6 - 4 = –2
• C: 4 - 0 = +4