Chapter 2 Chemical Bonding

The interaction between atoms that leads to a rearrangement of the electrons to a more stable state is

what we define as chemical bonding.

All atoms, except those of the noble gases*, readily engage in chemical bonding either with atoms of

their own kind (in elements) or with atoms of a different kind (in compounds).

*) This statement is not absolutely true. Some of the heavier noble gases such as xenon and krypton can

be persuaded to form compounds with very strong oxidizers. Compounds of helium and neon are

presently not known.

Three types (extremes) of chemical bonds exist:

– Ionic bond - electrostatic forces hold together oppositely charged ions.

– Covalent bond - two atoms sharing electron pair(s). Electrons are mutually attracted by

the nuclei of adjacent atoms.

– Metallic bond -is a complicated type of bond. At the simples level metal atoms are

envisaged of being metal cations in a “sea” of electrons. The electrons are shared

between all the ions simultaneously. A better description of metals is obtained from the

band theory.

Ionic bond

Ionic bonds form between oppositely charged ions, generally, but not always, between metals and non-

metals.

Metals versus Nonmetals

Metals are conductors of heat and electricity, have a lustre (shiny) appearance, are ductile (can be

drawn in to wirers) and are malleable (can be hammered out into thin foils). With the exception of

hydrogen, which is a non-metal, metals are located to the left in the periodic table.

Non-metals are generally non conductors, do not have a metallic lustre and are brittle. They are located

to the right in the periodic table.

The elements near the “step ladder” have both, metallic and non-metallic properties and are called

metalloids or semi metals.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 1

2

H

He

2 3 4

5 6 7 8 9 10

Li Be

B C N O F Ne

3 11 12

13 14 15 16 17 18

Na Mg

Al Si P S Cl Ar

4 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 55 56 57* 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

7 87 88 89** 104 105 106 107 108 109

Fr Ra Ac Rf Db Sg Bh Hs Mt

* 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

** 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

To understand chemical bonding trends in three atomic properties must be considered: electron

affinity, ionization energy and atomic radius

Atomic radius

Atomic radius is difficult to define – there is a small probability of finding an electron at an infinite

distance from the nucleus. Often atomic radius is defined in terms of internuclear distance in a

compound.

Methods of representing Size

covalent radii: half of the internuclear distance in a homonuclear X-X single bond.

ionic radii: The ionic radius, rion, is the size of an ion (due to changes in effective nuclear charge; for

cations, rion < rcov; whilst for anions, rion > rcov). These values are estimated from electron

density maps determined from crystallographic data.

van der Waals Radii: half of the distance of closest approach of two non-bonded atoms. The forces

holding the atoms together are weak London Dispersion forces (more to

come…).

Source: petrucci

Periodic Trends in Atomic Radii

Trends in atomic radii can be rationalized if we consider the shielding or screening effect of the core

electrons.

Effective nuclear charge:

Zeff = Z – S

Z = atomic number, S = number of inner electrons that screen an outer electron

The more electronic shells in an atom, the larger the atom. Atomic radius increases from top

to bottom in a group.

The atomic radius decreases from left to right through a period of elements.

Class exercise: Arrange the following atoms in order of increasing atomic radius: Mg, N, Rb, Be

Source Petrucci

Ions

Ions are charged atoms or groups of atoms

• Ionic Radius: Cations are smaller than the atoms from which they are formed. Isoelectronic

cations have the same number of electrons in identical configurations. Na+ and Mg2+ both have

closed shells (both have 10 electrons but the magnesium nucleus has more protons to attract

the 10 electrons then the sodium nucleus). For isoelectronic cations, the more positive the

ionic charge, the smaller the ionic radius.

• Anions are larger than the atoms from which they are formed. For isoelectronic anions, the

more negative the charge, the larger the ionic radius.

Source Petrucci

Class exercise: Arrange the following ions in order of increasing ionic radius: Al3+, Cl-, Na+, F-

Electron affinity and ionization energy:

As we have seen an atom can gain one electron or more. This fundamental atomic property is known as

electron affinity and is usually exothermic.

H + e- → H- + energy

An atom can also lose one electron (or more). This fundamental atomic property is called ionization or

ionization energy. Atoms are stable species. It requires energy to release an electron from an atom so

ionization is an endothermic process and ionization energies are positive

H + energy → H+ + e-

It is important to understand that electron gain and ionization are truly atomic properties. They are

properties of isolated gas phase atom (not bulk phase atoms as in a chunk of copper).

Metals tend to lose electrons and form cations

Mg(g) → Mg+(g) + e- I1 = 738 kJ/mol

Mg+(g) → Mg2+

(g)+ e- I2 = 1451 kJ/mol

I1 is called the first ionization energy, I2 the 2nd ionization energy.

Trends of ionization energy within a group in the periodic table: As you go down a group (chemical

family) the ionization energy decreases. It is not difficult to understand the rationale behind that. With

every new row (period) in the periodic table a new electronic shell is being filled with electrons (atoms

get bigger!) The further away the electron is from the nucleus, the smaller is its attractive interaction

with the nucleus and the easier it is to remove it.

Chemically speaking, the easier it is to remove the electron the greater the reactivity of the element

toward an electron accepting element (an oxidizer).

The table below lists the first ionization energies of the alkaline metals (group 1 in the periodic table)

Element Radius in pm I1 in kJ/mol

Lithium 152 520.2

Sodium 186 495.8

Potassium 227 418.8

Rubidium 248 403.0

Cesium 265 375.7

Trends of ionization energy within a period: As you go from the left to the right ionization energy

increases. Electrons enter the same shell within a perioid. This leads to an increase in repulsion between

the electrons in the outermost electronic shell. At the same time as the number of electrons increases

so does the number of protons in the nucleus and the greater the positive charge of the nucleus the

greater the attraction towards the electrons. It is this increased attraction that dominates over electron

repulsion. Thus ionization energies generally increase from the left to the right within a period. We will

take a closer look at the exceptions later. This nucleus having a “tighter grip” on the outermost electrons

also means that atomic radius decreases from the left to the right in a period.

Strictly speaking this is only true for gas phase atoms. In the bulk phase elements show different

bonding arrangements that might stabilize the atoms and make them less prone to react e.g. the

reactivity of the 3rd row elements toward O2 decreases from Na to Si but hugely increases at P! Why?

The table below lists ionization energies of the 3rd row elements in KJ/mol

Na Mg Al Si P S Cl Ar

I1 496 738 578 787 1012 1000 1251 1521

I2 4562 1451 1817 1577 1903 2251 2297 2666

I3 7733 2745 3232 2912 3361 3822 3931

I4 11580 4356 4957 4564 5158 5771

I5 16090 6274 7013 6542 7238

I6 21270 8496 9362 8781

I7 27110 11020 12000

Consider the reactivity of the following elements Na and K as well as Na and Mg!

There are two more points worth noting in this table. First there is a huge increase in ionization energy

after the last valence electron has vacated the valence shell and a core electron from the inner

electronic shell is removed e.g. it takes 496 kJ /mol to remove one electron from sodium but it takes

4562 kJ/mol to remove a second electron which is why no compounds are known that contain Na2+ ions

but essentially all compounds of sodium contain Na+ ions. Second, by the time you get to the middle of

the 3rd period (e.g. Si) ionization energies become quite large and it is less feasible for atoms to lose

electrons.

Nonmetals which are located to the right in the periodic table do not lose but tend to gain electrons and

form anions. We called this atomic property electron affinity and the following trend is observed within

the periodic table: Electron affinity generally increases from the left to the right within a period and

decreases from the top the bottom in a group.

Chemically speaking this means the element most eager to gain an electron is the most reactive non

metal element which is fluorine. Fluorine reacts with most elements in the periodic table.

Class exercise: Arrange the following atoms in order of increasing ionization energy: Si, F, Na, O

Arrange the following atoms in order of increasing electron affinity: Cl, As, P, Ba

Octet Rule

In taking a closer look at the number of electrons that metals lose and nonmetals gain it becomes

apparent that many try to obtain the same electron count as their nearest noble gas, especially the

elements close to the right or the left. So in compounds group 1 elements such as sodium or potassium

always lose just one electron, never two or more. In doing so they obtain the same number of electrons

as the nearest noble gas. Similarly in compounds with metals group 17 elements such as fluorine or

bromine always gain just one electron never two or more to achieve a noble gas electronic configuration

(more to come).

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

1

+/-1

H

2 1+ 2+

3- 2- 1-

Li Be

N O F

3 1+ 2+

3+ 3- 2- 1-

Na Mg

Al P S Cl

4 1+ 2+ 3+ 4+ 2,3,4,

5+ 3, 6+

2,4, 7+

2,3, 6+ 2+ 2+ 1,2+ 2+ 2- 1-

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Se Br

5 1+ 2+ 1+ 2+ 2,4+ 2- 1-

Rb Sr Ag Cd Sn Te I

6 1+ 2+ -1 to +7

3,4,6, 8+ 3,4+ 2,3,4+ 1,3+ 1,2+ 1,3+ 2,4+ 3,5+

Cs Ba Re Os Ir Pt Au Hg Tl Pb Bi

Common oxidation states/oxidation numbers of selected elements

Since metals tend to lose electrons and non-metals tend to gain electrons combining a metal and a no-

metal leads to a chemical reaction in which electrons are completely transferred from the metal to the

non-metal resulting in metal cations and non-metal anions that feel a strong electrostatic attraction.

This might be represented by the equation below, which uses Lewis dot formulas or Lewis structures

including the number of valence electrons (electrons in the outermost electronic shell) as dots around

the element symbol.

Li F Li ++ F

Compounds between metals and non-metals are called salts (e.g. sodium chloride, magnesium bromide)

or ionic compounds (e.g. magnesium oxide, copper sulfide). Many important materials are ionic

including ordinary table salt, ceramics, many ores and gypsum.

There is a fundamental difference between an atom and an ion. Sodium ions cannot be isolated and

placed in a jar on a shelf like sodium atoms which come in form of bars or chunks of metal under

mineral oil in a tin can. They are associated with a counter ion (chloride, fluoride, sulfate) to preserve

overall electrical neutrality. A sodium ion is a stable species abundant in large quantity in the human

body. Sodium metal is a reactive element that vigorously reacts with water to produce hydrogen and

sodium hydroxide, a strong base.

Class exercise

Write balanced chemical equations for the reaction of:

sodium with bromine

magnesium with chlorine

aluminum with sulfur

strontium with fluorine

hydrogen with calcium

nitrogen with lithium

Remember that electrical neutrality must be preserved in the final product and that the charges of the

ions cancel properly.

Energetic of ionic bonding

As we have seen it requires 496 kJ/mol to remove electrons from sodium (first ionization energy of

sodium). We get 349 kJ/mol back by giving the electrons to chlorine (electron gain is exothermic).

But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium

chloride is so exothermic!

What is as of yet unaccounted for is the electrostatic attraction between the newly formed ions. Negative anions are attracted to positive cations. The result is an ionic bond and a three-dimensional

crystal lattice of anions and cations.

The binding energy might be calculated from the following experimental data (Born Haber Cycle:

Na(s) + 0.5Cl2(g)

Na(g) + 0.5Cl2(g)

Na(g) + Cl(g)

Na+(g) + e

- + Cl(g)

Na+(g) + Cl

-(g)

NaCl(s)

-411 kJ

+107 kJ

+122 kJ

+496 kJ

-349 kJ

-787 kJStart

End

Step 1 (Start) Sublime one mole of sodium atoms

Na(s) → Na(g) +107 kJ/mol

Step 2 Dissociate half a mole of chlorine molecules

0.5 Cl2(g) → Cl(g) +122 kJ/mol

Step 3 Ionize one mole of sodium atoms

Na(g) → Na+(g) +496 kJ/mol

Step 4 Convert one mole of chlorine atoms to one mole of chlorine ions

Cl(g) + e- → Cl-(g) -349 kJ/mol

Step 5 Combine one mole of chloride ions with one mol of sodium ions to form one mole of sodium

chloride

Na+(g) + Cl-(g) → NaCl(s) (lattice energy unknown)

Step 6 (End) Measure the heat of formation for the overall reaction

Na(s) + 0.5 Cl2(g) → NaCl(s) -411 kJ/mol

The energy change between reactants and products is independent of the path that the reaction takes

(more to come in thermochemistry). The heat of the reaction that we obtain by directly combining the

elements should thus be equal to the energy change if we take the reactants clockwise through the

Born-Haber cycle because we end up with the same product in the same state.

-411 kJ/mol = 107 kJ/mol + 122 kJ/mol + 496 kJ/mol +(-349 kJ/mol) +x kJ/mol

x = -787 kJ/mol

As you can see it is the hugely negative lattice energy that is largely responsible for the exothermic

reaction between the elements sodium and chlorine that releases 411 kJ/mol.

Relationship between Bonding and Properties

Now why is an understanding of chemical bonding important? There is a relationship between structure

and properties of compounds (materials!) If you know and understand the structure of a material, the

bonding at the atomic level, you can understand and often predict its physical properties. This is why

chemists have been so successful in engineering materials with properties tailored to specific

applications.

Ionic compounds have very largely negative lattice energies. This tells us that they are very stable with

respect to the elements from which they have formed and explains why ionic compounds are usually

hard materials (ceramics, “rock salt”). Another property of ionic compounds is their very high melting

and boiling points which is related directly to the high lattice energy. Although anions and cations can

and do form ion pairs in solution and in the gas phase in the solid state they form crystalline solids which

can be described by a three dimensional crystal lattice. The lattice energy is defined as the energy given

off when oppositely charged isolated gaseous ions come together to form one mole of ionic solid. The

greater the lattice energy the harder it is to vaporize an ionic compound.

Alumina and magnesia have such high lattice energies that they make excellent materials for furnace

bricks or crucibles. Zirconium oxide on top of having high melting and boiling points has exceptional

fracture toughness along with very good chemical resistance. Another ceramic, zirconium nitride is a

cement like refractory material. A coating of zirconium nitride is commercially applied on parts that are

subject to high wear and tear, corrosive environments or to improve refraction. Ta4HfC5 is presently

holding the record with the highest melting point of any compound at 4215 °C.

Class exercise

Given below are calculated lattice energies as well as measured melting points, boiling points and

aqueous solubility of selected ionic compounds from Handbook of Chemistry and Physical 90th Edition

2009.

Rationalize the trends in lattice energies, melting and boiling points.

Compound Lattice energy (kJ/mol)

Melting point (°C)

Boiling Point (°C)

Solubility (g/100mL H2O)

LiF 1,049 848 1673 0.13

NaCl 787 801 1465 36

KBr 691 734 1435 67

RbI 632 656 1300 165

Compound Lattice energy (kJ/mol)

Melting point (°C)

Boiling Point (°C)

Solubility (g/100mL H2O)

Na2O 2481 1134 decomposition decomposition

MgO 3795 2825 3600 insoluble

Al2O3 15916 2054 2977 insoluble

ZrN 7633 2952 _____ insoluble

Picture source: Wikipedia

The sodium chloride structure in which each sodium ion is surrounded by six chloride ions and each

chloride ion is surrounded by six sodium ions in octahedral geometry is adopted by many other

compounds which have a stoichiometry of 1:1 and where anions and cations don’t differ to greatly in

ionic radius.

Covalent bond

For some elements it is not feasible to gain the number of electrons required to reach an octet.

e.g. carbon would have to gain 4 electrons to have the same number of electrons as neon.

Carbon (and many nonmetals) can share valence electrons so that they are surrounded by 8 electrons

(hydrogen by 2 electrons). This sharing of electrons is called covalent bonding.

It is not immediately obvious why the sharing of electrons leads to a more stable state. Electrons after

all are negatively charged particles and as such repel another. Let’s consider the simplest case, a

dihydrogen molecule. Hydrogen possesses only one electron. To obtain the same number of electrons as

its nearest noble gas (helium) it would have to gain one electron. In reactions with active metals this is

exactly what hydrogen does! It acquires the electron of a sodium or potassium atom and forms ionic

metal hydrides which contain the hydride ion H-. Dihydrogen though is not a combination of two hydride

ions which would give the H22- ion but a neutral molecule. In H2 each of the two electrons is shared

equally between the two atoms. There are several electrostatic interactions in this bond: Attractions between electrons and nuclei

Repulsions between electrons

Repulsions between nuclei

Atractive Interactions

Repulsive Interactions

If the electrons are shared, which means they physically spend a lot of time in between the two nuclei,

the attractive interactions outnumber the repulsive interaction. Furthermore, the electrons effectively

shield the nuclei from their positive charge. The net effect is that the electrons pull the two nuclei closer

together resulting in a covalent bond which in the case of dihydrogen is 436 kJ/mol more stable than the

isolated hydrogen atoms.

Hypothetical “snapshot” of a hydrogen molecule. The same

photographic plate is exposed over and over again to show where it is

most likely to encounter the electrons in the molecule.

Electron Dot Formulas (Lewis structures)

We have written simple Lewis structures of elements and ions to describe electron transfer in ionic

compounds by placing the number of valence electrons (electrons in the outermost electronic shell) as

dots around the element symbol. It is essential that we become familiar in drawing Lewis structures of

molecules and polyatomic ions to describe covalent bonding between atoms. First let’s make sure we

can easily identify the number of valence electrons for an element. The number of valence electrons for

an element follows directly from its position in the periodic table of the elements.

The elements are organized into eighteen groups. Each group shares similar chemical behaviour. The 18

groups are organized into four blocks. H and He are unique and are not generally placed in a “block”

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1

2

3

4

s-

p-

5 block

block

d-

6

block

7

f-

block

Valence Electrons are referred to as the number of s and p electrons in the outermost electronic shell.

d and f electrons are not counted as valence electrons.

For s-block elements: The number of valence electrons is the group number.

For p-block elements: The number of valence electrons is the group number minus 10.

Class exercise: Determine the number of valence electrons for each of the following elements. Potassium

Selenium

Iodine

Barium

Class exercise: Write Lewis structures Electron dot formulas) for the following elements:

– Calcium

– Boron

– Bromine

– Sulfur

Lewis Structures of Molecules

Lewis structures are representations of molecules showing all valence electrons - bonding and

nonbonding. Each bonding electron pair is represented as a single line between the bound atoms. Each

nonbonding electrone pair (also known as lone pair) is represented as a pair of dots.

Drawing Lewis structures of simple molecules illustrated on the examples CH4, NH3, H2O, HF and O2

Draw each atom along with its valence electrons.

H FH O HNH

H

H O OCH

H

H

H

Make a bond between the central atom and each terminal atom by pairing up electrons.

H FH

O

HNH

H

H O OCH

H

H

H

Represent each shared electron pair with a single line between the bonded atoms. Make multiple bonds

if necessarily to complete octets for all atoms (maximum of triple bond and no more than four bonds

per atom). Hydrogen only requires two electrons.

H FH O HH N

H

H O OH C

H

H

H

Fluorine only needs to share one of its seven valence electrons to be surrounded by an octet, oxygen

needs to share 2 of its 6 valence electrons and nitrogen must share three of its 5 valence electrons to

obtain an octet. Carbon must share all four valence electrons to gain an octet (carbons four bond

requirement).

In the case of CH4, NH3 and H2O you might have wondered which atom to choose as the central atom.

Note that hydrogen would not make a very good central atom since it would have to form more than

one bond and thus be surrounded by more than 2 electrons. Hydrogens are never central atoms. How

about CO2, SO2 or N2O though?

The following guidelines and some practice will greatly aid you in drawing Lewis structures of hundreds

of species. Being able to draw them quickly will greatly aid you in succeeding in the chapters to come.

1. Usually, but not always, the least electronegative element is the central atom. Electronegativity

correlates very well with electron affinity. It decreases from the top to the bottom in a group

and it increases within a period. For example in all oxoanions such as [CO3]2-, [ClO4]

-, [PO4]3-or

[SO4]2- the non oxygen atom is the central atom.

2. Generally if there are four atoms or more, small compact structures are preferred over linear

arrangements. For example the molecule SO3 does not consist of a chain of four atoms but

carries a central sulfur atom (less electronegative then oxygen) with three oxygen atoms

surrounding it.

Class exercise:

Draw the Lewis structures for the following molecules: CH2O (formaldehyde), CO2, SO2 and HCN

(hydrogen cyanide).

Do the examples above illustrate every possible way how carbon can meet its four bond requirement?

Lewis structures of polyatomic ions

The concept of Lewis structures also applies to polyatomic ions. We just have to take into account the

extra electron(s) for anions, or the missing electron for cations. The following exercises will work you

through two example:

Practice example: Draw the Lewis structure for the ammonium ion.

Start by drawing the central nitrogen atom with all its valence electrons surrounded by four hydrogen

atoms with one valence electron each.

Remove an electron from the central atom. Indicate the absence of that electron by placing a circled

positive charge sign on the nitrogen atom.

Make four bonds between the central nitrogen atom and the terminal hydrogen atoms.

Practice example: Draw the Lewis structure for the nitrite ion.

Start by drawing the central nitrogen atom with all its valence electrons surrounded by two oxygen

atoms with six valence electrons each.

Add one electron to a terminal atom. Indicate the presence of the extra electron with a circled negative

charge sign on that atom.

Make two bonds between the central and the terminal atom.

Form an additional bond to satisfy the octet for all atoms.

Resonance

The nitrite example raises an important point.

How did you decide which of the two oxygen atoms in the nitrite ion was to receive the extra electron.

The one on the right or the one on the left? And what would have happened had you chosen the other

oxygen instead. We could have come up with two different Lewis structures as shown below:

O N O ONO

But how are they different. Can we distinguish the right from the left oxygen atom? Of course we

cannot. They are equivalent which is why we call the two Lewis structures above equivalent Lewis

structures. They both represent extreme forms of what the ion could look like. When one actually

determines the bond distance between the nitrogen and the two oxygen in the nitrite ion the

observation is made that they are both identical within experimental error and that they are somewhere

in between the expected bond length for nitrogen to oxygen single and double bonds.

The observed structure is said to be a resonance hybrid with the two resonance structures contributing

equally (50 % each). The double headed arrow is used for resonance structures. Sometimes chemists like

to draw the resonance hybrids by using dashed lines for partial bonds that result from delocalized

electrons. Note that the charge of – 0.5 would reside on average on each oxygen in the resonance

hybrid depicted in the centre. In Chem 150 we will only draw full bonds (e.g. structure on the right and

left with a resonance arrow in between them!)

O N O ONO ONO

Formal charges:

In the example above one of the two oxygen atoms in the equivalent Lewis structures carries a charge of

-1. This is called a formal charge. Of course there must be a charge on some atom since we have an ion

but as we shall see formal charges can also arise in neutral molecules. Consider the molecule NOF3. The

least electronegative element is nitrogen which is surrounded by three fluorine atoms and one oxygen

atom.

F

O

N F

F

N

O

F F

F

N

O

F F

F

N

O

F F

F

a) b)

c)

d)

One might be persuaded to pair the two electrons between nitrogen and oxygen in b) and draw a

double bond as in c) but that would violate the octet rule for nitrogen now being surrounded by 10

electrons instead of 8. To get to the correct Lewis structure d) the electron from nitrogen is transferred

to oxygen. This will lead to formal charges since nitrogen now has ownership of 4 rather than 5 valence

electrons and oxygen now owns 7, one more than its normal 6 valence electrons. Nitrogen is said to

have a formal charge of +1 and oxygen has a formal charge of -1. Omitting formal charges invalidates a

Lewis structure.

Do not confuse formal charges with oxidation numbers (oxidation states) commonly used to balance

redox equations.

NOF3 N O F

Formal Charge +1 -1 O

Oxidation Number +5 -2 -1

The formal charge (FC) might be calculated using the equation:

FC = number of valence electrons – 0.5 x number of bonding electrons – number of lone pair electrons.

e.g. the formal charge of nitrogen in NOF3(structure d) is: FCN = 5 - (0.5 x 8) – (0.5 x 0) = +1

In neutral molecules the formal charges of all atoms must add up to zero. In polyatomic ions the

formal charges of all atoms must add up to the charge of the ion.

Class exercise: Draw the Lewis structure of the azide ion [N3]-

Let’s use a general algorithm applicable to any polyatomic ion or molecule.

1. Draw each atom along with the number of valence electrons

2. Make one bond between the central atom and each terminal atom by pairing electrons

3. If overall charge (n-) add n electrons to a terminal atom(s), if overall charge n+ remove n

electrons from the central atom.

4. Pair electrons to make multiple bonds (maximum of four bonds per atom) and complete octet

for all atoms.

5. If necessary complete octets by forming additional bonds or transferring electrons from one

atom to another atom. This will lead to formal charges on some atoms.

Non equivalent Lewis structures.

In the nitrite example, which we used to introduce the concept of resonance, we saw that two

equivalent Lewis structures contributed equally to the resonance hybrid. Consider now the molecule

N2O, dinitrogen monoxide, also known as nitrous oxide or laughing gas because of its use in general

anesthesia. Draw all valid Lewis structures for the molecule

Which one is the best Lewis structure? To answer this, consider the following rules:

The best Lewis structures has the lowest possible formal charges (e.g. zero formal charges!)

If formal charges are necessarily the best Lewis structure will be the one that has the negative formal

charge on the more electronegative atom and the positive charge on the less electronegative atom.

Formal charges of the same sign on adjacent atoms are very unlikely.

On the basis of these rules which of the three resonance structures above is the worst and which one is

the best?

When we have non-equivalent Lewis structures like in the nitrous oxide case above the different

resonance structures will not contribute equally to the resonance hybrid but the best Lewis structure

will contribute most strongly.

Would you expect the nitrogen-nitrogen bond in N2O resemble that of

a) a single bond

b) a double bond

c) a triple bond

d) in between a single and double bond

e) in between a double and a triple bond

Justify your answer!

Exceptions to the octet rule

Although the octet rule is very useful there are quite a few exceptions we need to be aware of.

1. Odd-electron species (e.g. NO) No matter how you try to share the 11 valence electrons in NO

you cannot obey the octet for both atoms simultaneously. Two even numbers cannot add up to

give an odd number.

2. Incomplete octets (e.g. BF3) Rare but important case. Even if it shares all of its three valence

electrons it can only form 3 bonds!

3. Expanded valence shells e.g. PCl5, SF6, SO2, [SO4]2- and [ClO4]

- (Most important!) The very

existence of compounds such as PCl5 with one central and five terminal atoms shows that

certain elements can exceed the octet rule. The bottom line is: Elements of the 2nd period can

never exceed the octet. Elements of the 3rd or higher periods can exceed the octet rule.

Draw as many valid Lewis structures for the sulfate ion as you can think of. Does any of them obey the

octet rule and if yes at what cost? Identify the best Lewis structure/structures. Identify equivalent Lewis

structures.

In Chem 150 we will always minimize formal charges if we can to get to the best Lewis structure.

Remember that this simplified two dimensional drawing of the sulfate ion is a poor representation of its

three dimensional shape which is tetrahedral. (Chapter 3)

Energetics of covalent bonding:

The energetics of covalent compounds are generally more complex than those of ionic solids.

Nevertheless we can make some qualitative predictions about material properties based on bond

energies and structure.

For example a double bond between two atoms is stronger (has a greater bond energy) than a single

bond and a triple bond is stronger yet. We can measure bond energies by measuring the energy it takes

to break a bond.

The following table gives a selection of bond energies obtained experimentally from a variety of sources

including calorimeter and spectroscopy and more recently theoretically using sophisticated

computational methods. The bond energies of diatomic molecules such as H-H, NΞN or Cl-Cl are actually

bond dissociation energies and are known much more precisely.

bond bond energy bond bond energy bond Bond energy (in kJ/mol)

H-H 436 C-C 347 N-N 163

H-C 414 C=C 611 N=N 418

H-N 389 CΞC 837 NΞN 946

H-O 464 C-N 305 N-O 222

H-S 368 C=N 615 N=O 590

H-F 565 CΞN 891 O-O 142

H-Cl 431 C-O 360 O=O 498

HBr 364 C=O 736 F-F 159

H-I 297 C-Cl 339 Cl-Cl 243

Br-Br 193 I-I 151__________________________________________

We can estimate the energy absorbed or given off by a reaction (mostly heat) using these bond energy

values. Consider the formation of water from the elements:

2H2 + O2 2H2O

For the reaction to proceed two H-H bonds as well as one O=O bond need to be broken and four O-H

bonds form.

Energy required for bond breaking: 2 mol of H-H bonds + 872 kJ

1 mol of O=O bonds + 498 kJ

Energy gained from bond formation: 4 mol of O-H bonds - 1856 kJ

Sum: - 486 kJ

486 kJ are released in this reaction which forms two mol of water indicating the exothermic character of

the combustion of hydrogen gas.

The energy (or heat) of formation for one mol of gaseous water would thus be - 243 kJ/mol indicating

that one mole of water is 243 kJ more stable than the elements it is composed of.

The experimentally determined value is 242 kJ/mol measured for gas phase formation (steam).

More to come in thermochemistry…