Chemistry Note

Matter

For chemists, the most elemental building block of matter is the atom. While it is certainly true that

the atom can be divided into even more elemental building blocks, it is at the level of the atom that

the first distinctive "chemical" properties start appearing. There are many different types of atoms,

as you can see in the periodic table of the elements, each with their own distinctive chemical

properties. From these atoms molecules can be assembled. Molecules are groups of atoms held

together by forces called chemical bonds.

States of Matter

Chemists classify matter into three different states:

SOLID - has a definite volume, a definite shape, and is rigid.

LIQUID - has a definite volume but no definite shape.

GAS - has no distinct volume, no distinct shape and can be easily compressed to occupy a

smaller volume.

Mixtures

A substance containing only one type of atom or one type of molecule is a pure substance. Most of

the matter around us, however, consists of mixtures of pure substances. Air, wood, rocks and

dirt are examples of such mixtures. Mixtures can be classified

asHomogeneous and Heterogeneous.

Homogeneous Mixtures

Homogeneous Mixtures are uniformly mixed on an atomic or molecular level. These types of

mixtures are also called solutions. Below are a few examples of homogeneous mixtures.

Air is a homogeneous mixture (gaseous solution) of N2, O2, H2O, and CO2 gases. In contrast, a

container of each gas by itself would be a pure substance. Only when they are mixed on an

molecular level are they a homogeneous mixture (or gaseous solution).

Brass is a homogeneous mixture (solid solution) of copper and zinc. Again each metal by itself is

a pure substance. Only when they are mixed on an atomic level are they a homogeneous mixture

(or solid solution).

Beer is a homogeneous mixture (liquid solution) of H2O, C2H5OH, and a few other substances.

(Sorry, there is no beer molecule!)

Heterogeneous Mixtures

Heterogeneous mixtures are not uniformly mixed on an atomic or molecular level. For example,

Salt and pepper, chocolate chip cookies, or a Twix™ candy bar, ...

are all examples of heterogeneous mixtures, where substances are not mixed on a molecular level.

All mixtures, heterogeneous and homogeneous, can be separated into pure substances using physical methods.

Physical Change

Any change of matter that does not change the type of atoms and molecules within the matter is

called a physical change. Water boiling is an example of a physical change. When water boils it

is changing from a liquid state to a gaseous state. Chemists would represent this process as follows:

H2O(l) → H2O(g)

Here (l) stands for liquid and the (g) stands for gas. As this is a physical change, the H2O molecule

does not change.

Chemical Change

Molecules, on the other hand, can split or combine together to make other types of molecules. The

process where a molecule is transformed into a different molecule is called a chemical change.

To help us understand the concept of chemical change let's examine why light bulbs are made the

way they are. A lightbulb works by passing electrical current through a tungsten wire inside the

bulb. The tungsten wire is sealed inside a glass bulb since if you did this in air the light bulb would

burn out very quickly. This is because the tungsten wire undergoes a chemical reaction with

the O2 gas in the air to form tungsten oxide.

2W + 3O2 → 2WO3

To prevent this reaction from happening, all oxygen is removed from the air sealed inside the bulb.

If oxygen leaks into the bulb, then the tungsten wire reacts to become tungsten oxide and the oxidized wire will no longer pass electrical current readily. If you try to pass electrical current through the oxidized wire it will quickly heat up and break. That is, the lightbulb burns out.

Measurement Uncertainty

Precision and Accuracy

When we make a measurement in the laboratory we need to know how good it is. To this end, we

introduce two concepts: Precisionand Accuracy.

Precision indicates degree of reproducibility of a measured number, and

Accuracy indicates how close your measurements are to the true value.

Let's look at throwing darts and trying to hit the bullseye as an illustration of these two concepts.

When you make measurements in science you want them to be both precise and accurate.

Two students, Raffaella and Barbara, measured the temperature of boiling water, which by definition

should be 100°C under 1 atmosphere of pressure. Each student made 10 temperature

measurements, shown below as red (Raffaella) and blue (Barbara) dots.

The average of Raffaella's temperature measurements is 100.1°C and the average of Barbara's is

also 100.1°C. So, the accuracy of their measurements is identical. On the other hand, you can see

from the figure that the precision of Raffaella's measurements was better than Barbara's. The way

this is expressed in science is to include an uncertainty with measured values. In this case Raffaella

would report a boiling point of 100.1 ± 0.3°C, and Barbara would report 100.1 ± 1.4°C. This

uncertainty is also called a random error, and is different from a systematic error, which is the

difference between the average value and the true value. Here, both Raffaella and Barbara had

systematic errors of 0.1°C, since the true boiling point of water is 100°C.

If, for whatever reason, the measurement uncertainty cannot be specified, then at the very least, the precision in a measured number can be approximately specified through the number of significant figures. In the example above, Rafaella would report a boiling point of 1.001 x 10 2 °C, whereas Barbara would report 1.00 x 102 °C. That is, Rafaella's result has four significant figures while Barbara's has only three.

Significant Figures

Significant Figures are the number of digits that express the result to the true measured

precision. For example, let's consider the number:

92.154

Determining the number of significant figures in this number is very simple. Start from the left and

count the digits. In this case we find that there are 5 significant figures in this number.

Zeroes, however, are special digits. Sometimes they count as a significant digit, and sometimes they

do not count and simply act as place holders. For example, the number

0.092067

has 5 significant figures. The first two zeroes are not significant, they are just place holders.

Therefore, when you're counting significant digits always start from the left and don't start counting

until you come to the first non-zero digit, then everything after that counts, including zeroes!

A good way to avoid the problem of significant and non-significant zeroes it is to use scientific

notation. For example,

0.092067 is written 9.2067 x 10-2and has 5 sig. figs.



When using scientific notation all digits, including zeroes, are significant.

Rounding to the Correct Number of Significant Figures

In lab you were asked to weigh a crucible and cover three times. Let's say you measured

12.4337 grams, 12.4334 grams, and 12.4335 grams.

These numbers have six significant figures, but clearly the precision is poorer for the last digit. If I

calculate the average I get

(12.4337 + 12.4334 + 12.4335) / 3 = 12.4335333333 ← calculator display

When I report this number it makes no sense to write down all the numbers on my calculator display

because the average can have no more significant figures than what I get from a single

measurement. We need to round off the number to the correct number of significant figures. In this

case

12.4335333333 rounded to six sig. figs. is 12.4335

Generally, when insignificant digits are dropped from a number, the last digit retained should be

rounded for the best accuracy. For example,

To determine how the last digit should be rounded, the digits to be dropped are turned into a

decimal fraction. In the example above, our decimal fraction would be 0.333333. We can now follow

the rules for rounding numbers:

Rules for Rounding numbers to the correct number of significant figures.

1. If the decimal fraction is greater than 1/2, then add one to the last digit retained.

2. If the decimal fraction is less than 1/2, then leave the last digit retained alone.

3. If the decimal fraction is exactly 1/2, then add one to the last digit retained only if it is odd.

Let's look at some examples.

Round the numbers 9.473, 9.437, 9.450, and 9.750 to two significant figures.

For 9.473 the last digit retained is 4, and the decimal fraction is 0.73. So we use rule #1

above and 9.473is rounded to 9.5 For 9.437 the last digit retained is 4, and the decimal fraction is 0.37. So we use rule #2



above and 9.750is rounded to 9.8

Significant Figures

Generally, the more rigorous approach for determining the precision of a calculated result is to

propagate the uncertainty in all your measured quantities through the calculation. The topic of error

propagation through calculations, however, is outside the scope of this course. Thus, we will present

the simpler rules below for determining the approximate number of significant figures in a

calculated result.

Addition and SubtractionRule:

With Addition and Subtraction keep only the number of decimals in the result that occur in

the least precise number.

For example,

Note that only the final answer is rounded. There is no rounding numbers in the intermediate steps

of the calculation.

Multiplication and DivisionRule:

With Multiplication and Division, the final result should only have as many significant figures

as the term with the least number of significant figures.

For example,

Mixing Operations

It can be tricky keeping track of the number of significant figures in a calculation that combines

addition, subtraction, multiplication, and division. For example,

My calculator gives 0.013099698, but how many significant figures should be in the final answer? All

numbers in the calculation have 3 significant figures, but if you break up the calculation into steps

Exact numbers

Exact numbers are known with infinite precision. For example if there are 10 students in a

classroom, that number is an exact number. We treat exact numbers as numbers known with infinite

precision

What is the average of 10.2, 11.4, and 10.9?

Powers of 10 and Logarithms

When taking 10 to the power of a number the final answer will have the same number of significant

figures as the fractional part of the number.

Calculate 10-9.2067 with the correct number of significant figures.

We split the calculation into the integer and fractional part:

10-9.2067 = (10-9)(10-0.2067) = (10-9)(0.62129806352) → 0.6213 x 10-9 = 6.213 x 10-10

When taking the logarithm of a number the final answer will have the same number of significant

figures as the logarithm argument.

Express the logarithm of 6.213 to the correct number of significant figures.

log(6.213) = 0.7933013536 → 0.7933

When taking the logarithm of numbers expressed in scientific notation, remember that the exponent

of 10 is an exact number and therefore has infinite precision.

Express the logarithm of 6.213 x 10-10 to the correct number of significant figures.

Recalling that the log(AB) = log(A) + log(B) we can write:

log(6.213 x 10-10) = log(6.213)+log(10-10) = 0.7933013536 + (-10) = -9.2066986464 → -9.2067

You should only round off numbers when reporting your final result. Do not round off numbers in the

middle of a calculation.

Quantities and Units

In science a Quantity is expressed as the product of a number and a unit. A number without units

often has no meaning to a scientist. A set of standard (agreed upon) units are essential not only in

science, but also in commerce. The most widely accepted system of measurement in the sciences is

the International System, otherwise known as

S.I. Units.

S.I. stands for Le Systéme International. This is also called the metric system, with which you may

already be familiar.

The metric system is easy to learn and use because subdivisions and multiples of base units employ only factors of 10. Prefices indicate the size of the unit relative to a base unit.

Prefix Symbol Multiple of Unit

mega- M 1,000,000 or 106

kilo- k 1,000 or 103

deci- d 0.1 or 10-1

centi- c 0.01 or 10-2

milli- m 0.001 or 10-3

micro- μ 0.000001 or 10-6

nano- n 10-9

pico- p 10-12

The SI system defines seven base quantities, from which all other scientific quantities can be

derived. These base quantities are length, mass, time, electric current, thermodynamic temperature,

amount of substance, and luminous intensity. All other quantities can be derived from these base

quantities using the equations of physics and chemistry. Let's look at the base units for some base

and derived quantities.

SI Base Quantities

http://www.bipm.org/en/si/si_brochure

Length

The SI base unit for length is the meter. The abbreviated symbol for meters is m. A meter is slighly

longer than a yard (i.e., 1 m = 39.37 inches).

Mass

The SI base unit for mass is the kilogram. The abbreviated symbol for the kilogram is kg. 1 kg

weighs approximately 2.2 pounds. The choice of the kilogram instead of the gram as the base unit

for mass is only for historical reasons.

Time

The SI base unit for time is the second. The abbreviated symbol for the second is s.

Electric Current

The SI base unit for electric current is the ampere. The abbreviated symbol for the ampere is A.

Amount of Substance

The SI base unit for amount of substance is the mole. The abbreviated symbol for the mole is mol.

We will learn more about themole later in this course.

Luminous Intensity

The SI base unit for luminous intensity is the candela. The abbreviated symbol for the candela is cd.

Thermodynamic Temperature

The SI base unit for temperature is the Kelvin. The abbreviated symbol for the Kelvin is K. The

coldest temperature theoretically possible is 0 K. You simply cannot go any lower. In fact, it is

experimentally impossible to even reach 0 K.

For historical reasons it is also common to define temperature in terms of its difference from a

reference temperature T0=273.15 K, the freezing point of water. This is called the Celsius

temperature, and the abbreviated symbol for Celsius is °C. It is related to the Thermodynamic

Temperature by the equation

T(in Celsius) = T(in Kelvin) - 273.15 K

On the Celsius scale the freezing point of water is set at 0 °C, and the boiling point is 100 °C. Thus,

the coldest temperature theoretically possible is -273.15 °C.

To give you some other reference points:

System Temperature

Sun's Interior (thermonuclear fusion of 108 K

http://www.grandinetti.org/Teaching/Chem121/Lectures/TheMole

System Temperature

hydrogen to helium)Sun's Surface 6000 KEarth's Core 4600 K

Liquid N2 (boiling pt.) 77 KLiquid He (boiling pt.) 4.2 K

Some SI Derived Quantities

Volume

Volume is a derived quantity and has the dimensions of length3. Thus, the SI unit for volume is the

cubic meter which has the abbreviated symbol m3. In practice, the cubic meter is not a convenient

unit for everyday use, so the cubic decimeter (dm3) is used instead. Because it is so often used 1

dm3 is given the name liter. The abbreviated symbol for a liter is L or l.

1 liter = 1 dm3

When working with even smaller volumes the cubic centimeter is more commonly used. Since

1 L = 1 dm3 = (10 cm) (10 cm) (10 cm) = 1000 cm3,

we often use 1 mL (mL is the symbol for milliliter) in place of 1 cm3. You will sometimes see a mL

called a cc, which stands for cubic centimeter.

Pressure

Pressure is a derived quantity and has the dimensions of mass/(length•time2). The SI unit for

pressure is kg/m-s2. This product of base units is given the name pascal and the symbol Pa.

1 Pa = 1 kg/m-s2

Energy

Energy is a derived quantity and has the dimensions of mass/(length•time)2. The SI unit for energy is

kg/m2-s2. This product of base units is given the name joule and the symbol J.

1 J = 1 kg/m2-s2

Electric Charge

Electric Charge is a derived quantity and has the dimensions of current•time. The SI unit for electric

charge is A-s. This product of base units is given the name coulomb and the symbol C.

1 C = 1 A-s

Density

Density is a derived quantity and has the dimensions of mass/length3, and is defined as the mass

per unit volume.

density = mass/volume

The S.I. units for density are kg/m3, but these are not very convenient units for everyday use. More

common units are

g/cm3 for solids

g/ml for liquids

g/L for gases

If we know the mass and volume of a sample, then we can calculate its density.

A cube of lead 3.00 cm on a side has a mass of 305.0 g. What is the density of lead?

To answer this question we need to put the mass and volume of the sample into the equation above.

The mass of lead is given at 305.0 g. Although we are not given the volume directly, we do know

that the sample is a cube (i.e., all sides of equal length), and that length of a side is 3.00 cm . The

volume of this cube would be

Volume = ( 3.00 cm)3 = 27.0 cm3.

Thus, the density of the cube of lead is density = mass/volume = 305.0 g / 27.0 cm3 = 11.3 g/cm3.

Notice that the density is the same no matter the size or shape of the sample.

Density

Density is a derived quantity and has the dimensions of mass/length3, and is defined as the mass

per unit volume.

density = mass/volume

The S.I. units for density are kg/m3, but these are not very convenient units for everyday use. More

common units are

g/cm3 for solids

g/ml for liquids

g/L for gases

If we know the mass and volume of a sample, then we can calculate its density.

A cube of lead 3.00 cm on a side has a mass of 305.0 g. What is the density of lead?

To answer this question we need to put the mass and volume of the sample into the equation above.

The mass of lead is given at 305.0 g. Although we are not given the volume directly, we do know

that the sample is a cube (i.e., all sides of equal length), and that length of a side is 3.00 cm . The

volume of this cube would be

Volume = ( 3.00 cm)3 = 27.0 cm3.

Thus, the density of the cube of lead is density = mass/volume = 305.0 g / 27.0 cm3 = 11.3 g/cm3.

Notice that the density is the same no matter the size or shape of the sample.

Dimensional Analysis

When doing calculations we always write each number with its associated units. As you do the

calculation the units should cancel so that the final number you calculate also has the correct units.

Let's look at some examples.

Donuts cost $2.79 a dozen. How much do 3 dozen donuts cost?

Convert 0.34 cm to mm (micrometers).

In these two examples the conversion factors are exact numbers. That is, they have infinite

precision. Conversions factors, however, are not always exact numbers. Let's look at an example

using density as a conversion factor to convert between volume andmass.

What volume will 50.0g of ether occupy if the density of ether is 0.71g/mL?

A pitcher throws a baseball at 90 miles/hour. What is the speed in feet/second?

If the distance between the pitcher's mound and homeplate is 60.5 feet, how long does it

take the ball to travel this distance?

The Atom

Atomic Structure

An atom is composed of three types of subatomic particles: the proton, neutron, and electron.

Particle Mass (g) Charge

Proton 1.6727 x 10-24 +1Neutron 1.6750 x 10-24 0Electron 9.110 x 10-28 -1

Here, charge is given in multiple of 1.602 x10 -19 coulombs.

Protrons and neutrons have similar masses and electrons are much lighter (over 1,000 times

lighter).

Protons and electrons have equal and opposite charges while neutrons have no charge.

We have the following simple picture of the atom.

The atom is comprised of a positively charged nucleus composed of protons and neutrons. This

small nucleus is surrounded by orbiting electrons. Because the protons and neutrons are so much

more massive than the electrons, virtually all the mass of the atom is located in the nucleus. The

light negatively charged electrons move around in an orbit in the space around the nucleus.

We use the following symbol to describe the atom:

A= Z + N, where N is the number of neutrons.

If you add or subtract a proton from the nucleus, you create a new element.

If you add or subtract a neutron from the nucleus, you create a new isotope of the same element

you started with.

In a neutral atom, the number of positively charged protons in the nucleus is equal to the number

of orbiting electrons.

The Hydrogen Atom

Let's look at the simplest example of an atom, the hydrogen atom.

The atom consists of a proton and an electron held together by the electromagnetic

force between the positively charged proton and the negatively charged electron.

The electron orbits around the proton because it is the lighter particle, sort of like the earth orbits

around the sun, There are, however, big differences in the picture of the earth going around the sun

and the electron going around the nucleus. This is because protons, neutrons and electrons exist on

a length scale so small that quantum mechanics is required to understand the electron's orbit

around the nucleus. We will learn more about the quantum theory of the atom later.

When we add neutrons to the nucleus of 11H we can make the isotopes of hydrogen. Here are

three common isotopes of hydrogen.

If we add a proton to the hydrogen nucleus we would get helium (a different element). Here are two

common isotopes of helium.

Another example is carbon.

Because the element symbol and atomic number are redundant, you will often see isotopes written

without the atomic number. For example, you would see 12C only.

How many electrons, protons and neutrons are contained in the isotope 3517Cl?

The number of protons is given by the atomic number, the bottom number, so the number of

protons is 17. This is a neutral atom, so there will be an equal amount of negatively charged

electrons to balance out the positively charged protons, thus making the number of electrons 17

also.

We know that the atomic mass is

A = Z + N

where N is the number of neutrons. Rearranging the equation we get

N = A - Z

Plugging in the numbers we already know, we get

N = 35 - 17 = 18 = the number of neutrons

Now you might think that an atomic nucleus with lots of protons (like 12C ) would fly apart from the

electrical repulsions between positively charged protons. It turns out that these forces of electrical

repulsions are overcome by an attractive force between protons and neutrons called the strong

nuclear force. At small distances inside a nucleus, this force is stronger than the electromagnetic

forces of repulsion, but at larger distances it becomes much weaker.

Atomic Mass

Grams is not a very convenient unit for atomic masses, so a new unit called the atomic mass unit

(amu) is defined.

1 amu = 1.660551 x 10 -24 g

Reexpressing the subatomic particle masses in terms of atomic mass units we have

Particle Mass(g) Mass(amu)

proton 1.6727 x 10-24 1.007316neutron 1.6750 x 10-24 1.008701electron 9.110 x 10-28 0.000549

Using an instrument called a mass spectrometer we can very accurately measure the mass of

atoms and molecules. Here are some measured isotope masses using a mass spectrometer.

Isotope Mass(amu)2H 2.0140

4He 4.002608Be 8.00530512C 12.00000016O 15.994915

24Mg 23.985042

That 12C has a mass of exactly 12.000000 amu is not a coincidence. A mass spectrometer can

only measure mass differencesaccurately. To solve this problem the 12C isotope is defined to have a

mass of exactly 12.000000 amu's. Then everything else is measured relative to 12C.

As you might expect different isotopes of the same element will have different masses. If you look at

the periodic table, however, you'll notice that there is only one number listed for the mass of each

element. How can you only have one mass if there is more than one isotope of each element?

The answer is that the mass under each element is the weighted average of all of the isotope

masses for that element. In thisweighted average, the weights are the percent abundance that

each isotope occurs in nature.

For example, if you analyzed a lump of pure carbon from the planet Earth, you would find that

98.89 % of all carbon atoms on earth are 12C atoms, and

1.11 % of all carbon atoms on earth are 13C atoms.

So the weighted average mass of carbon is

(0.9889)(12.000000 amu 12C ) + (0.0111)(13.0039 amu 13C ) = 12.011 amu

It is possible that on a planet in a far away galaxy, the natural abundances of carbon isotopes may

be different, and therefore they would have slightly different numbers under carbon in their periodic

table. The masses of the isotopes, however, are the same everywhere.

The natural abundance of 63Cu is 69.09 % and for 65Cu is 30.19 %. If the atomic weight

of 63Cu is 62.93 amu and 65Cu is 64.93 amu, what is the average atomic weight for

natural copper?

(0.6909)(62.93 amu 63Cu ) + (0.3091)(64.93 amu 65Cu ) = 63.55 amu

Periodic Table

Even before we had this nice understanding of the atomic structure, scientists had identified certain

substances as elements. So there were many attempts to arrange the known elements so that

there were some correlations between their known properties. The first reasonably successful

attempt was made by Dimitri Mendeleev in 1869. He had the idea of arranging elements in order

ofincreasing atomic mass , and, most importantly, found that elements with similar chemical and

physical properties occuredperiodically. He placed these similar elements under each other in

columns.

In 1914, Henry Moseley determined that a better arrangement was in order of increasing atomic

number, giving us the periodic table we have today. We can define the periodic table as an

arrangement of elements in order of increasing atomic number placing those with similar chemical

and physical properties in columns.

http://wild-turkey.mit.edu/Chemicool/

http://wild-turkey.mit.edu/Chemicool/

Features of the Periodic TableGroups

Vertical columns are called groups. Elements within a group have similar chemical and physical

properties. Groups are designated at the top by the numbers 1-8 and by the letters A and B. (Note:

group labeling is somewhat arbitrary, so watch out for other designations, particularly with A and B.)

A group elements- Representative or main group elements

B group elements- Transition elements

In addition to the number-letter designation, some groups have their own name.

1A → alkali metals

2A → alkaline earths

7A → halogens

8A → noble gases or rare gases

Periods

Horizontal rows are called periods. Periods are designated by the numbers on the left in the

periodic table. The two long rows placed just below the main body of the table are the inner

transition elements

.

Elements 58-71 are the Lanthanide Series

Elements 90-103 are the Actinide Series

The Three Categories of Elements

There are three broad categories of elements called

1. Metals

2. Non-metals

3. Metalloids

To separate the metals and non-metals we draw a stairstep line to the left of and

below B, Si, As, Te, and At.

This classification or group is useful because certain properties are associated with each category.

Metals

solids at room temperature (except Hg)

metallic luster

malleable and ductile

good conductors of heat and electricity

Non-metals

gases or solids at room temperature (except Br2)

variety of color and appearance

brittle solids

insulators (poor conductors)

Metalloids

intermediate in properties between metals and non-metals

solids at room temperature

many have more that one structure (one metallic, the other non-metallic)

some are semi-conductors

Ions

Earlier we learned that the atom was comprised of a very small positively charged core of protons

and neutrons surrounded by a much large "cloud" of orbiting electrons.

While chemistry doesn't usually involve changes in the number of protons or neutrons in an atom,

changes in the number of electrons in an atom is central to the science of chemistry.

If electrons are removed or added to a neutral atom, a charged particle called an ion is formed.

There are two types of ions:

Cation - a positively charged ion

Anion - a negatively charged ion

For example, a neutral sodium atom has a nuclear charge of +11 and contains 11 electrons. If we

strip off one electron we form a sodium cation:

This process can also be represented in short-hand notation.

A neutral chlorine atom has a nuclear charge of +17 and contains 17 electrons. If we add one

electron we form a chlorine anion:

Another example. Zn likes to lose 2 electrons to make a divalent cation:

You can use the periodic table to predict how many electrons an element will lose or gain when it becomes an ion. For example, here are the most stable ionic charges on monoatomic ions:

Group Charge

1A +12A +27A -16A -2

Generally, Aluminum likes to form the cation Al3+ and Zinc likes to form the cation Zn2+.

Here is a "rough rule" you can use to figure out how many electrons an element will gain or lose:

Elements tend to gain or lose electrons to achieve the same number of electrons as the

nearest noble gas.

For groups in the middle of the periodic table, it is not as simple. After you learn about quantum

mechanics, however, you will have a better idea of how to predict the stable ion charges for these

groups.

A final note: you will often hear chemists use the term proton interchangeably with hydrogen

cation, H+. Can you explain why?

Chemical Bonds

Chemical bonds are the forces that hold atoms together in compounds. They are formed because

atoms are not happy with the number of electrons that they have. Only the noble gases (column 8A)

are content with the number of electrons. They have the optimum number of electrons and don't

like to form chemical bonds. The desire of atoms to gain or lose electrons to get a noble gas number

of electrons is what leads to chemical bonding.

(It is quite common for chemists to personify the atoms and molecule with which they work! Saying

an atom wants an another electron is akin to saying a ball wants to roll down a hill. Later will we

examine the energetic and thermodynamic bases of this personification.)

There are three types of chemical bonds:

Covalent Bonds - atoms are held together by sharing electrons

Electrostatic (Ionic) Bonds - cations and anions are held together by electrostatic

attractions

Metallic Bonds - occurs in metals (similar to covalent bonds)

Later, you will be able to determine just how ionic or covalent a bond will be, but for now here are

some guidelines to follow:

Bonds amongst non-metal atoms are covalent. (For example, a P-S bond is a covalent

bond.)

Bonds between a non-metal and a metal are ionic (For example, a Na+Cl- bond is ionic.)

Bonds amongst metal atoms are metallic

Metalloid--Non-metal bonds are usually covalent

Metalloid--Metal bonds are usually ionic

Covalent Bonding

In covalent bonding atoms share electrons.

Take for example the H2 molecule. Each hydrogen atom says, "I only need one more electron to be

like a noble gas (helium) ." Since each hydrogen has only one electron, when two hydrogens get

together they can share their electrons.

So each hydrogen atom now sees 2 electrons when it is covalently bonded to another hydrogen

atom. Pure hydrogen exists as H2molecules. The same is true for all of the halogens in column 7A:

Pure chlorine exists as Cl2 Pure bromine exists as Br2

Pure iodine exists as I2

Chemists often use the symbol "-" to represent a bond. For example, H-H is a "hydrogen molecule"

and Cl-Cl is a "chlorine molecule." The line in between the two atoms means that they are sharing

two electrons between them. Let's take oxygen as another example. Oxygen atoms like to combine

to form O2. In this case, each oxygen atom wants 2 more electrons, so when the two oxygen atoms

get together they share a total of 4 electrons. We write O2 as:

Chemists call this a double bond. By forming a double bond between them, each oxygen atom can

then see as many electrons as a Ne atom has.

Now let's look at nitrogen. It also likes to combine to form a diatomic molecule, in this case N2. Each

nitrogen atom, however, wants 3 electrons, so two nitrogen atoms share a total of 6 electrons.

We call this a triple bond.

Of course, you can form molecules from more than one type of atom. Let's look at

water. H2O consists of two hydrogen atoms sharing their electrons with one oxygen atom.

Another example is hydrogen peroxide, H2O2.

Think about hydrogen peroxide and decide on your own if all of the atoms are happy with the

number of electons around them.

Here is one final example. Carbon atoms want to share 4 electrons, so it is very happy if it can get

together with 4 hydrogens to form methane, CH4.

In this example, carbon is sharing 4 electrons with 4 hydrogens and each hydrogen is sharing one

electron with carbon.

Structural and Empirical FormulasStructural Formula

To avoid confusion, chemists often write the structural formula when identifying a molecule.

The structural formula tells you how many of each type of atom are in a molecule and also how

they are connected. For example, here is the structural formula of ethanol.

Chemical Formula

You will also see the term chemical formula. The chemical formula tells you how many of each

type of atom are in a molecule. For example, the chemical formula for ethanol is

C2H6O.

Notice that this is less information than the structural formula (but more compact). You must be

careful not to confuse substances that have the same chemical formula. For example, ethanol and

dimethyl ether have the same chemicial formula (i.e. C2H6O).

Their chemical formulas are identical, but their structural formulas and their physiological effects

are markedly different.

Empirical Formulas

An empirical formula (simplest formula) tells us the simplest whole number ratio of atoms in a

molecule. When identifying an unknown pure substance, chemists will often start by performing

experiments to determine the empirical formula of the substance.

For example, hydrogen peroxide's chemical formula is H2O2, but its empirical formula is HO.

The chemical formula for glucose is C6H12O6, but its empirical formula is CH2O, and

its structural formula is

Now, let's try some sample quiz questions on Empirical Formulas:

Molecular Cations and Anions

Molecules can also lose or gain electrons to become cations or anions. For example,

the NO3 molecule will gain an electron to form the nitrate anion.

If you count up all of the electrons you'll find that all of the atoms feel like neon.

Here is the ammonium ion, an example of a molecular cation.

The ammonium ion has given up an electron to become a cation.

Ionic Bonds

Ionic bonds are generally formed when you bring atoms which really want to lose electrons

together with atoms which really want to gain electrons.

The Na+ cation and the Cl- anion are held together by electrostatic or ionic bonds. There is no

sharing in ionic bonding. The anion takes the electron for itself and the cation is happy to get rid of

its electron. The ions in ionic compounds are arranged in three-dimensional structures. There are no

discrete molecules of NaCl. We can only write an empirical formula of NaCl.

Here are some other examples.

NH4Cl : here NH4+ = cation

here Cl- = anion

BaCl2 : here Ba2+ = cation

here Cl- = anion

All substances are electrically neutral. We can use this fact to obtain the chemical formula of an ionic compound.

Ba2+ and SO42- form BaSO4

Na+ and S2- form Na2S

Notice that in Na2S, two sodium cations were needed to balance the -2 charge of S2-, making things

electrically neutral.

Chemical Nomenclature

Chemical Nomenclature is the systematic naming of chemical compounds. Here we examine just

a few rules for chemical substances that can be named relatively easily.

Naming Ionic Compounds1. Name the cation (the more electropositive one) first and the anion (the more

electronegative one) second.

2. Monoatomic cations take their name from the element name.

3. Monoatomic anions take their names from the first part of the element name and then

add "-ide".

Cation Name

H+ HydrogenNa+ SodiumAl3+ Aluminum

Anion Name

H- HydrideCl- ChlorideO2- Oxide

What are the names for NaCl, MgO, and HBr?

NaCl = sodium chloride

MgO = magnesium oxide

HBr = hydrogen bromide

4.

When elements form more than one type of ionic compound Roman numerals are used to indicate the charge on the cation.

Fe2+ and O2- give FeO = Iron(II) OxideFe3+ and O2- give Fe2O3 = Iron(III) OxideSn2+ and Cl- give SnCl2 = Tin(II) ChlorideSn4+ and Cl- give SnCl4 = Tin(IV) Chloride

6. There is an older system of naming these compounds that uses the Latin name of the metal

with the suffices of "-ic" and "-ous" to designate the higher and lower charge of the metal,

respectively.

6. So the examples above would be, under the older system,

FeO = Ferrous OxideFe2O3 = Ferric OxideSnCl2 = Stannous ChlorideSnCl4 = Stannic Chloride

6. Here are some other examples

Ion Latin Name Systematic Name

Au+ Aurous Gold(I)Au3+ Auric Gold(III)Cu+ Cuprous Copper(I)Cu2+ Cupric Copper(II)

For polyatomic ions, you should memorize the names below.

Ion Name

NH4+ ammonium

OH- hydroxideCN- cyanide

C2O42- oxalate

Cr2O72- dichromate

NO3- nitrate

SO42- sulfate

PO43- phosphate

ClO- hypochloriteClO3

- chlorateMnO4

- permanganate

Ion Name

HSO4- hydrogen sulfate or bisulfate

Hg22+ Mercury(I)

C2H3O2- acetate

SCN- thiocyanateCrO4

2- chromateNO2

- nitriteSO3

2- sulfiteCO3

2- carbonateAsO4

3- arsenateClO2

- chloriteClO4

- perchlorateHSO3

- hydrogen sulfite or bisulfite

HCO3- hydrogen carbonate or

bicarbonate

Naming Binary Covalent Compounds

When a pair of elements form more than one type of covalent compound, Greek prefixes are used to indicate how many of each element are in a compound. For example:

Compound Name

N2O dinitrogen monoxideNO nitrogen monoxide

N2O3 dinitrogen trioxideN2O5 dinitrogen pentoxide

Some of the Greek prefixes are given in the table below:

PrefixNumber of Particular

Element

mono 1di 2tri 3

tetra 4penta 5hexa 6hepta 7

PrefixNumber of Particular

Element

octa 8

Rules for Binary Covalent Compounds

1. The prefix mono is never used for naming the first element of a compound.

2. The final o or a of a prefix is often dropped when the element begins with a vowel.

For example, for CO the name will be carbon monoxide, and the final o of mono is dropped.

Remember, it's only the final o ora. So, the name of ClO2 will be chlorine dioxide, and no vowels are

dropped.

How do you know which element goes first? The element that comes first in the following list "goes"

first.

B, Si, C, Sb, As, P, N, H, Te, Se, S, I, Br, Cl, O, F

Finally, H2O, which according to the rules should be called dihydrogen monoxide is always called

water, and NH3, or nitrogen trihydride, is always called ammonia.

Naming Acids, Oxyacids and Their Salts1. If the anion does not contain oxygen, then the acid is named with the prefix hydro- and the

suffix -ic.o For example, when gaseous HCl is dissolved in H2O, it forms hydrochloric acid.

o HCN in H2O is hydrocyanic acid.

Before we learn the rule for naming oxyacids, let's learn the rules for naming oxyanions.

What are oxyanions? They are anions formed from oxygen and a nonmetal. Here are some

examples: ClO4-, ClO3

-, ClO2-, ClO-, SO4

2-, SO32-.

There are two rules for naming these:

2. If there are only two members in the same series, then the anion with the least number of

oxygens ends in -ite, and the anion with the most ends in -ate.o For example, SO3

2- is sulfite and SO42- is sulfate.

3. When there are more than two oxyanions in a series, hypo- (less than) and per- (more than)

are used as prefixes. Here are some examples:o ClO- is hypochlorite

o ClO2- is chlorite

o ClO3- is chlorate

o ClO4- is perchlorate

Finally, here are the rules for naming acids of oxyanions.

4. If the anion name ends in -ate, then the acid name ends in -ic or -ric.

5. If the anion name ends in -ite, then the acid name ends in -ous.

Here are examples of the last three rules:

Acid Anion Acid Name

HClOhypochlorit

ehypochlorou

s acidHClO

2chlorite

chlorous acid

HClO3

chlorate chloric acid

HClO4

perchlorateperchloric

acid

The Mole

Once we know the mass of a sample we can use the mass of the atoms or molecules in the sample

to determine how many atoms or molecules are in the sample.

How many 12C atoms are in a 1.00 kg block of pure 12C isotope?

Obviously atoms and molecules are not a convenient unit of measure when we're working with

macroscopic (i.e., human size) objects. For this reason chemist define a new unit of measure

called the mole.

1 mole is defined as the number of carbon atoms in exactly 12.000000 grams of pure 12C.

From our definition of the atomic mass unit we get

1 mole = 6.0220943 x 1023 chemical units

The number 6.0220943 x 1023 is called Avagadro's number.

How many moles are in a 1.00 kg block of pure 12C isotope?

Of course we know a 1.00 kg block of naturally occuring carbon will contain a mixture of 12C, 13C,

and even some 14C isotopes. So the number of moles of carbon atoms in a 1.00 kg block of naturally

occuring carbon is

We can also calculate the mass of a mole of molecules.

What is the mass of a mole of methane (CH4) molecules?

mass of 1 mole of C is 12.011 g

mass of 4 moles of H is 4 X 1.00g

mass of 1 mole of CH4 is 16.043 g

In other words the molecular weight of CH4 is 16.043g.

Molecular weight = the mass in grams of 1 mole of a molecule.

For ionic compounds, which do not exist as individual molecules we use the term Formula weight

Formula weight = The mass in grams of 1 mole of the chemical formula.

What is the formula weight of CaCO3?

mass of 1 mole of Ca is 40.08 g

mass of 1 mole of C is 12.011 g

mass of 3 moles of O is 3 X 15.999 g

Formula weight of CaCO3 is 100.09 g

Let's look at another example.

How many hydrogen atoms are there in 2.50 g of NH3?

Now, let's try some sample quiz questions on:

Mass Percent

The Mass Percent of a component is defined

What is the mass percents of carbon, hydrogen, and oxygen in pure ethanol C2H6O?

First we calculate the mass of one mole of C2H6O...

mass of 2 moles of C is 2 X 12.011 g

mass of 6 moles of H is 6 X 1.008 g

mass of 1 mole of O is 15.999 g

Molecular weight of C2H6O is 46.069 g

Next we calculate the mass percents

Note that the mass percentages should add up to 100%.

Combustion Analysis

When chemists make new compounds one of the first things they often do is determine the mass %

for the different elements in the compound. To analyze the mass percent of carbon and hydrogen

chemist use a combustion device.

The sample is burned in the presence of excess oxygen which converts all the carbon to carbon

dioxide and all the hydrogen to water. The CO2 and H2O produced are absorbed in two different

stages and their masses determined by measuring the increase in weight of the absorbers.

Ascorbic acid (vitamin C) contains only C, H, and O. Combustion of 1.000 g of Ascorbic

acid produced 40.9% C and 4.5% H. What is the empirical formula for Ascorbic Acid?

First we need to calculate the mass percent of Oxygen. Since the sample contains C, H, and O, then

the remaining

100% - 40.9% - 4.5% = 54.6% is Oxygen

Now we need to express the composition in grams and determine the number of moles of each

element:

Next we divide by the smallest number of moles to obtain the mole ratio which is also the atom ratio. In this case carbon has the smallest number of moles, so...

C: 0.0340 moles/0.0340 moles = 1

H: 0.045 moles/0.0340 moles = 1.32 ~ 1 1/3

O: 0.0341 moles/0.0340 moles ~ 1

Finally we calculate the smallest whole integer ratios by multiplying each number above by 3 to get

C: 3 H:4 O:3

thus we obtain the empirical formula C3H4O3

Remember the empirical formula has the smallest whole integer ratios, the molecular formula can

be different, e.g., C6H8O6, orC9H12O9, or C12H16O12, ... are all possible molecular formulas.

Now let's look at a related question.

What is the molecular formula if the molecular weight of Ascorbic Acid was formed to be

176 g/mole?

In this case we need to find the multiplication factor between the molecular formula and the

empirical formula:

factor = (molecular weight)/(empirical formula weight)

The empirical formula weight of Ascorbic Acid is

mass of 3 moles of C is 3 X 12.011 g

mass of 4 moles of H is 4 X 1.008 g

mass of 3 moles of O is 3 X 15.999 g

mass of 1 mole of CH4 is 88.062g

Therefore the multiplicative factor is (176 g/mole)/(88.062 g/mole) ~ 2, and the molecular formula

for Ascorbic Acid is C6H8O6

Chemical Reactions

As we learned earlier the most fundamental building blocks of matter for chemists are atoms (that

is, the elements you see in the periodic table). We also learned that atoms can combine with other

atoms by chemical bonding to form molecules. Recall that the process where a molecule is

transformed into a different molecule is called a chemical change. This process of chemical change

is represented by a chemical reaction. For example,

It is important to keep in mind that while in a chemical reaction molecules are destroyed and

created by breaking and forming chemical bonds, atoms are neither created nor destroyed in a

chemical reaction. In other words - there must be the same number of each type of atom on the

product and reactant sides of the arrow. Making sure that this rule is obeyed is called "balancing the

chemical equation". In the above example the equation would be unbalanced without the 2 in front

of the O2 and H2O. We can make a table to confirm that the number of atoms on each side of the arrow are the same:

Reactants Products Balanced?

1 C 1 C yes4 H 4 H yes4 O 4 O yes

Notice that we never change the chemical formula of any product or reactant when trying to balance

a chemical equation.

A balanced equation is essential to known the stoichiometry for the chemical reaction.

Stoichiometry - Relationship between quantities of matter that participate in chemical reactions.

-or-

How much of this, plus how much of that gives how much of something else?

Let's consider the following example of water being converted into hydrogen and oxygen gas using a

Hoffman Apparatus:

2 H2O(l) → 2 H2(g) + O2(g)

The balanced chemical equation tells us that two molecules of water in the liquid state react to form

two molecules of H2 in the gas state and one molecule of O2 in the gas state. The Hoffman

apparatus allows us to trap the gaseous H2(g) and gaseous O2(g) in separate volumes. By examining

these volumes and using Avagadro's law we can see the stoichiometry of the chemical equation.

Avagadro's Law: The volume of a gas is directly proportional to the number of molecules in the

gas.

In this example there is twice the amount of H2 gas and O2 gas, as predicted by the stoichiometry of

the chemical equation. These integer relationships between gas volumes were some of the earliest

proof we had that matter existed in discrete packages called atoms and molecules.

Just as any good chef wants to add together the correct amount of each ingredient for the best

recipe so does the chemist also want to add together the correct number of molecules called for by

a chemical reaction to produce the desired product. For example, to separate water molecules

into H2 and O2 gas our Hoffmann apparatus had to put electrical energy into the reaction. We could

have written the chemical reaction for this as

2 H2O(l) + 571.6 kJ → 2 H2(g) + O2(g)

That is, we use 2 moles of H2O and 571.6 kJ of energy to make 2 moles of H2 gas and 1 mole

of O2 gas.

Now let's say we want to do the reverse, that is, convert H2 gas and O2 gas into water and energy.

2 H2(g) + O2(g) → 2 H2O(l) + 571.6 kJ

The stoichiometry tells us that combining two moles of H2(g) with one mole of O2(g) will give 2 moles

of H2O(l) plus 571.6 kJ of energy.

So, we take a balloon that is filled with only H2 gas. There will be some O2 molecules in the air

outside the balloon but we will take no care in making sure that there will be stoichiometric amounts

of H2 and O2 gases together when we start the reaction.

Next let's take a balloon that is filled with the correct amount of H2 and O2 gases according to the

stoichiometric ratio of 2:1 from the chemical equation.

When I start the reaction in the H2 only balloon I won't get the maximum abount of H2O and energy

(released in part as sound) because the stoichiometry isn't in the best ratio.

When I start the reaction in the 2:1 H2 to O2 balloon I expect to get the maximum amount of H2O

and energy (released in part as sound).

Thanks to Amadeus Avagadro we know the volume of a gas is directly proportional to the number of

molecules in the gas (or the number of moles). Thus to get the correct stoichiometry when working

with gases we can simply adjust the volume of each gas to match the stoichiometry of the reaction

(as I did with the second balloon). Generally, however, equal volumes of different compounds in the

solid and liquid state do not necessarily contain the same number of molecules, atoms or moles.

When working with solids or liquids it would be nice to have a scale or device that easily measures

the number of moles. In practice, however, the starting point for chemists is the mass of the

reactants, not the moles. Therefore we need to convert between mass (easily measured) and moles

(not so easily measured). Let's look at a few examples.

What mass of oxygen will react with 96.1 grams of propane?

First, write the balanced reaction:

C3H3(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

Second, convert the mass to moles. Note that the molecular weight of propane is 44.1 g/mole.

Baking soda (NaHCO3) is used as an antacid. It neutralize excess HCl secreted by the stomach.

NaHCO3 + HCl(aq) → NaCl(aq) + H2O(l) + CO2(aq)

How many moles of HCl are neutralized per gram of NaHCO3?

We set this problem up similar to the previous one:

Now try this one at home. Milk of Magnesia (Mg(OH)2) is also an antacid, and the chemical reaction

is

Mg(OH)2(s) + 2HCl(aq) → 2 H2O(l) + MgCl2(aq)

Try this one on your own: Which neutralizes more HCl per gram, NaHCO3 or Mg(OH)2?

Limiting Reagents

A Limiting Reagent is the reagent that limits the amount of products that can be formed.

For example, nitrogen gas is prepared by passing ammonia gas over solid copper(II) oxide at high

temperatures. The other products are solid copper and water vapor.

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(g)

If 18.1 g of NH3 are reacted with 90.4 g of CuO, which is the limiting reagent? How many

grams ofN2 will be formed?

First we compute the number of moles of NH3 (M.W. = 17.031 g/mole) and the number of moles

of CuO (M.W. = 79.5 g/mole).

To determine which reagent is limiting we use the mole ratio from the chemical equation to convert

moles NH3 to moles CuO.

So, only 1.14 moles of CuO is available, therefore CuO is the limiting reagent. That is, CuO will run

out before theNH3 does.

The mass of N2 produced will be

Solution Chemistry

In the first week of class we learned that a homogeneous mixture of two or more substances is

called a solution. If one of the substances is present in much greater quantities than all the other

substances then it is called the solvent. The other substances in solution are known as solutes. For

example, when a small amount of NH4Cl is dissolved in a large quantity of water we refer to water

as the solvent and NH4Cl as the solute. Another example is Napthalene (used in mothballs) can be

dissolved in benzene. In this example benzene is the solvent and napthalene is the solute.

Solutes dissolved in water (solvent) are called aqueous solutions. Not all substances are soluble in

water. Why do some substances dissolve in water and others don't? It has to do with the structure of

the water molecule.

Oxygen has a greater attraction for electrons, so the shared electrons (bonding electrons) spend

more time close to oxygen then to either of the hydrogens. This gives oxygen a slightly excess

negative charge and hydrogen a slightly more positive charge.

This unequal charge distribution makes water a polar molecule, and gives water its ability to

dissolve compounds. When an ionic solid dissolves in water, the positive ends of the water molecule

are attracted to the negatively charged anions and the negative ends of the water molecule are

attracted to the positively charged cations. For example, when NaCl is dissolved in water we find

So when an ionic substance (salt) dissolves in water, it is broken up into individual cations and

anions which are surrounded by water molecules. For example, when NH4 NO3 is dissolved in water

it breaks up into separate ions.

NH4+ and NO3

- ions are floating around in H2O essentially independent of each other.

Water also dissolves non-ionic substances. For example, C2H5OH (ethanol) is very soluble in H2O.

This is because C2H5OH has a polar OH bond that the water molecules like to hang around.

Many substances do not dissolve in water and that is because they are non-polar and do not interact

well with water molecules. A common example is oil and water. Oil contains molecules that are non-

polar, thus they do not dissolve in water.

How do we know that ionic solids dissolve in water and form cations and anions that float around

separately? One clue comes from conductivity experiments. Anions and Cations should act as

charge carriers in solution. Therefore a solution with dissolved ions should conduct electricity. Let's

look at a few examples. Pure (distilled) water contains no dissolved ions. Therefore pure water will

not conduct electricity. In a simple conductivity experiment as shown below we would not expect the

light to be on.

An aqueous NaCl solution, however, will have dissolved ions present and therefore will conduct

electricity. Therefore the light in our conductivity experiment will be on if dipped in an

aqueous NaCl solution.

NaCl ionizes completely when dissolved in water. It's helpful to think of this process as two steps:

Substances that exist in solution almost completely as ions are called strong electrolytes.

Substances that do not form ions when they dissolve in water are called non-electrolytes. And

example of a non-electrolyte is sugar. Sugar will readily dissolve in water but doesn't form cations

and anions in solution. That is, there are no charge carriers formed.

Substances that only partially ionize into ions when dissolved in water are called weak

electrolytes. For example, Acetic Acid (HC2H3O2) dissolves in water, but only partially dissociates

into ions.

Be careful not to confuse how soluble a substance is in water with whether it is a weak, strong, or

non-electrolyte. For example, sugar dissolves completely in water but it is a non-electrolyte. Another

example are salts that can be very insoluble in water but the small amount of salt that does dissolve

in water is a strong electrolyte.

Acids and Bases

With our understanding of strong, weak, and non-electrolytes we can now examine the Arrhenius

Definition of Acids and Bases:

Acids:

Substance that produces H+ ions when it is dissolved in H2O.

Bases:

Substance that produces OH- ions when it is dissolved in H2O.

For example, HCl is an acid,

and NaOH is a base,

Acids and Bases that are strong electrolytes are called Strong Acids and Strong Bases,

respectively. Acids and Bases that are weak electrolytes are called Weak Acids and Weak Bases,

respectively.

Strong Acids

HCl is an example of a strong acid:

HCl(aq) → H+(aq) + Cl-(aq)

Other examples include HBr, HI, HClO4, HClO3, H2SO4, and HNO3.

Weak Acids

HF is an example of a weak acid:

Other examples include HC2H3O2, H2CO3, H2SO3, H3PO3, and H3PO4. The last four are examples

of polyprotic acids. These are acids that can produce more than one H+ ions when dissolved in

water. H2CO3 and H2SO3 are called diprotic acids, andH3PO3 and H3PO4 are called triprotic

acids. HF, HCl, HBr, and HC2H3O2 are examples of monoprotic acids. The dissociation of

polyprotic acids usually occurs in steps. For example, only after H3PO4 loses its first H+ ion will it

lose its second H+ ion, and then it will lose its third.

In this example, all three species H3PO4, H2PO4-, and HPO4

2- are weak electrolytes so H3PO4 is

considered to be a weak acid.H2SO4 is another example of a diprotic acid. In the case of H2SO4 the

first H+ ion is produced readily so the species H2SO4 is considered to be a strong electrolyte

(i.e. strong acid). The species HSO4-, however, is a weak electrolyte.

Strong Bases

NaOH is an example of a strong Base:

NaOH(aq) → Na+(aq) + OH-

(aq)

Other examples include LiOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2.

Weak Bases

Ammonia, NH3, is a good example of a weak base.

Other examples include C6H5N (pyridine) and C6H5NH2 (aniline).

Solution Reactions

Chemical reactions in solution can be written in three different forms:

Molecular Equation

Gives the overall reaction stoichiometry but not necessarily the actual form of the reactants and

products in solution. For example, when you mix HCl with aqueous NaOH, a reaction occurs that

forms water and the salt NaCl.

HCl(aq) + NaOH(aq) → H2O(aq) + NaCl(aq)

This is a molecular equation. It doesn't give a clear picture of what actually occurs in solution. Each

reactant and product, HCl(aq), NaOH(aq), H2O(aq), and NaCl(aq), actually exist in solution split into their

respective cations and anions.

Complete Ionic Equation

Represents all solution phase reactants and products that are strong electrolytes as ions.

H+(aq) + Cl-(aq) + Na+

(aq) + OH-(aq) → H2O(aq) + Na+

(aq) + Cl-(aq)

This is a more accurate representation of what actually exists in the solution than the molecular

equation shown above. In a complete ionic equation all substances that are strong electrolytes are

represented as ions. From the complete ionic equation we see that Na+ and Cl- ions don't really

participate in the reaction. They are examples of what we call spectator ions: Ions that appear on

both sides of the equation.

Net Ionic Equation

Includes only those species undergoing a change in the reaction. That is, spectator ions are not

included. Thus, for the reaction above the net ion equation would be

H+(aq) + OH-

(aq) → H2O(aq)

As you might imagine, it is possible to mix substances and have no net ionic equation. Consider the

following molecular equation

CaCl2(aq) + 2 HNO3(aq) → Ca(NO3)2(aq) + 2 HCl(aq)

The complete ionic equation is

Ca2+(aq) + 2 Cl-(aq) + 2 H+

(aq) + 2 NO3-(aq) → Ca2+

(aq) + 2 NO3-(aq) + 2 H+

(aq) + 2 Cl-(aq)

Removing all the spectator ions from this equation leaves nothing! All the ions are spectator ions, so

there is no net ionic reaction.

Processes that lead to a Net Ionic Equation

We will now consider the following four processes that will lead to an actual net ionic equation:

Formation of an insoluble solid (Precipitation Reaction)

Formation of a gas that escapes from solution

Formation of either a weak electrolyte or non-electrolyte

An oxidation/reduction reaction

The first three processes are generally reactions where a cation meets up with an anion and forms a

bond that has enough covalent character that they can no longer be separated into individual

cations and anions any more.

Precipitation Reactions

Precipitate - an insoluble solid formed by a reaction in solution.

Consider the following molecular equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The net ionic equation of the above molecular equation is:

Ag+(aq) + Cl-(aq) → AgCl(s)

In order to know which ions will combine to form precipitates, we have the solubility rules

Solubility Rules

1. All salts containing NH4+ and group IA cations (Li+, Na+, K+, Rb+, Cs+, ) are soluble.

2. All salts containing NO3-, C2H3O2

-, HClO3-, and ClO4

- are soluble.

3. All salts containing Cl-, Br-, and I- are soluble, except those with Ag+, Hg22+, and Pb2+

4. All salts containing SO42- are soluble, except PbSO4, BaSO4, HgSO4, CaSO4,

and AgSO4.

5. Most salts containing O2-, OH-, PO43-, CO3

2-, and S2- are insoluble, except those

containing NH4+ and group IA cations.

Determine the net ionic equation for the following reaction:

Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3

Rule number 4 tells us that BaSO4 is insoluble. So we can write the complete ionic equation as

Ba2+(aq)+2 NO3

-(aq) +2 Na+

(aq)+SO42-

(aq) → BaSO4(s)+2 Na+(aq)+NO3

-(aq)

Removing the spectator ions leaves us with the net ionic equation:

Ba2+(aq) + SO4

2-(aq) → BaSO4(s)

Gas Forming Reactions

Sometimes a gas will be involved as one of the reactants or products in a solution reaction. For

example,

2 HCl(aq) + Na2S(aq) → H2S(gas) + 2 NaCl(aq)

In this example the complete ionic equation would be:

2 H+(aq) + 2Cl-(aq) + 2Na+

(aq) + S2-(aq) → H2S(g) + 2Na+

(aq) + 2Cl-(aq)

Removing the spectator ions we obtain the net ionic equation:

2 H+(aq) + S2-

(aq) → H2S(g)

H2S is just one example of a gaseous substance that can form in a solution reaction. Another way

gases can form in solution is through the decomposition of weak electrolytes. For

example, H2CO3 readily decomposes into H2O and CO2 gas,

H2CO3(aq) → H2O(l) + CO2(g)

So, any solution reaction that leads to the production of H2CO3, such as

HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2CO3(aq)

will ultimately lead to the production of H2O and CO2 gas as the H2CO3 decomposes, giving us a

new equation:

HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2O(l) + CO2(g)

If we eliminate all the spectator ions we would write the net ionic equation as

H+(aq) + HCO3

-(aq) → H2O(l) + CO2(g)

Two other substances that will decompose and form gases are H2SO3 and NH4OH:

H2SO3(aq) → H2O(l) + SO2(g)

NH4OH(aq) → H2O(l) + NH3(g)

Weak Electrolyte Formation

Because weak electrolytes do not completely dissociate into ions, whenever two ions can get

together to form a weak electrolyte there will be a net ionic equation. For example,

HCl(aq) + NaC2H3O2(aq) → HC2H3O2(aq) + NaCl(aq)

Writing out the complete ionic equation we obtain:

H+(aq) + Cl-(aq) + Na+

(aq) + C2H3O2-(aq) → HC2H3O2(aq) + Na+

(aq) + Cl-(aq)


H+(aq) + C2H3O2

-(aq) → HC2H3O2(aq)

Water is also considered to be a weak electrolyte. That is, only a small fraction of the H2O molecules

in water dissociate to form H+and OH- ions. Therefore, any reaction that leads to the formation of

water will have a net ionic equation. Let's look at a classic example of an acid reacting with a base.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)


H+(aq) + OH-

(aq) → H2O(l)

Reduction/Oxidation Reactions

Any reaction where electrons are transferred between reactants is called an reduction-oxidation

reaction or redox reaction. These are reactions where one substance wants an electron so badly

that it takes it away from another substance. Whether or not it succeeds depends on who it meets.

For example, consider the reaction

FeCl2(aq) + CeCl4(aq) → FeCl3(aq) + CeCl3(aq)

What happened here? If we remove the spectator ions and write the net ionic equation we find:

Fe2+(aq) + Ce4+

(aq) → Fe3+(aq) + Ce3+

(aq)

Ce4+ took an electron from Fe2+! This is an oxidation/reduction reaction. In this

example, Fe2+ is oxidized and Ce4+ isreduced. The charge of Fe went from +2 to +3, that is,

it lost an electron. This process is called oxidation.

Oxidation: The loss of an electron by a substance.

Likewise, the charge of Ce went from +4 to +3, that is, it gained an electron. This process is

called reduction.

Reduction: The gain of an electron by a substance.

Half-Reactions

Oxidation/reduction reactions are important because we can exploit them as a way of generating

electrical current. For example, we know that Ce4+ will pull an electron away from Fe2+ when we

mix the two in solution. The trick to making a battery is to find a way to make Ce4+ pull an electron

from Fe2+ when they are not mixed together in a single solution.

In order to do this we set up two 1/2 reactions in separate beakers and connect them with a salt

bridge. The salt bridgeelectrically connects the two beakers, but prevents Fe2+ and Ce4+ from

mixing.

You need two 1/2 reactions to make a reaction, so, we add them together.

Fe2+ → Fe3+ + e-

Ce4+ + e- → Ce3+

__________________________________

Fe2+ + Ce4+ →Fe3+ + Ce3+

Notice that electrons on both sides of the half-reactions must cancel each other out when added

together.

What about other atoms and molecules. How do you know if one chemical substance is strong

enough to take an electron from another? We simply refer to a list known as the activity series.

Oxidation States

Oxidation numbers provides a means of keeping track of electrons in redox reactions. For some

elements, the oxidation number is just another way of stating what its most stable cation or anion

will be. In other cases, it is not so obvious, so we have rules.

Rules for assigning oxidation numbers1. The oxidation number for elements is always zero. For example, Na(s), O2(g), C(s) all

have zero oxidation numbers.

2. The oxidation number of monoatomic ions is the same as their charge. You already

know this one. This means that for Na+, the oxidation number is +1 and for Cl-, the

oxidation number is -1.

3. Oxygen is assigned a -2 oxidation number in covalent compounds. This refers to

compounds such as CO, CO2, SO2, and SO3. There is an exception to this rule, and it is in

peroxides, such as H2O2. Here, each O in the O22- group has a -1 oxidation number.

4. Hydrogen is assigned a +1 oxidation number in covalent compounds. This refers to

compounds such as HCl, NH3, and H2O.

5. In binary compounds, the element with the greatest attraction to electrons gets

the negative oxidation number. In other words, the most electronegative of the pair

gets the negative number. For example, in HF, F is more electronegative and thus has a -1

oxidation number. In NH3, the N atom is more electronegative and has a -3 oxidation

number.

6. The sum of the oxidation numbers is zero for a neutral compound and equal to

the ion's charge for an ionic species. For example, in H2O, a neutral species, H is +1

and O is -2, and the sum of the two is 0. For CO32-, each O is -2 and C is +4, and the sum is

-2.

Assign oxidation numbers to the atoms in SF6.

Since this is a binary compound let's first start with rule 5. We know that F has a greater attraction

to electrons than S does, therefore we give it the negative oxidation number, which in this case will

be -1 for F. To assign sulfur its oxidation number, we go to rule 6. This is a neutral compound, so the

sum of the total oxidation numbers must be zero. Since there are 6 F atoms, each with a -1

oxidation state then the sulfur must have a +6 oxidation number to balance out the fluorine atoms.

Solution Concentration

For chemical reactions that take place in solution the chemist needs to express the amount of

material (solute) in a given amount of dissolving material (solvent). Generally, concentration is the

ratio of the amount of solute to amount of solvent. There are several ways to express concentration.

We will examine the most common ones used in chemistry.

Molarity

Molarity is define as the number of moles of solute per liter of solvent.

What is the Molarity of a solution prepared by dissolving 8.0 grams of NaOH in H2O so

that the final volume is 250 ml?

First convert grams to moles:

Then calculate the molarity

Mass Percent

Another way to express concentration is using Mass Percent.

Calculate the the mass percent of ethanol in a solution prepared by mixing 1.00 g of

ethanol C2H6Owith 100.0 g of H2O.

Dilution

Often times solutions of a given concentration are prepared by diluting solutions of higher

concentration. Here's an example of how to calculate how much a concentrated solution needs to be

diluted to prepare a given concentration.

How many milliliters of 16 M HNO3 must be used to prepare 1.0 liters of a 0.10

M HNO3 solution?

So 6.2 ml of 16 M HNO3 diluted to 1 liter will give 0.10 M HNO3.

Solution Stoichiometry

Now let's use molarity in some stoichiometric calculations.

Calculate the minimum amount (in liters) of a 0.050 M BaCl2 solution that is required to

precipitate all the SO42-

(aq) in 0.10 liter of a 0.10 M Na2SO4 solution.

Ba2+(aq) + SO4

2-(aq) → BaSO4

In order to use the chemical equation we need to work in terms of moles. The concentration of SO42-

(aq) is converted into moles by multiplying by the volume.

Titrations

A useful way to determine a solute's concentration in a solution is to react the solution with a solute

in another solution of known concentration. This is known as a Titration.

Titration:

Experiment which determines the concentration of a solute (reactant) using its reaction of

known stoichiometry with another solution (reactant) of known concentration.

For example, if I have a solution of sulfuric acid, H2SO4 (aq), but don't know its concentration, then I

can react it with a NaOHsolution of known concentration.

2 NaOH(aq) + H2SO4 (aq) → Na2SO4 (aq) + 2 H2O(l)

In a titration the titrant is added dropwise until the reaction is complete.

Equivalence Point:

Point at which stoichiometrically equivalent quantities are brought together.

40.0 mL of 0.20 M NaOH is needed to neutralize (reach equivalence point) for 20.0 mL

of H2SO4solution. What is the concentration of H2SO4 solution?

2 NaOH + H2SO4 → Na2SO4 + 2 H2O

How do we know when the reaction is complete? We add a tiny amount of indicator to the analyte

that will change color when the solution has excess titrant (e.g. excess OH-). For example,

phenolphthalein molecules are colorless in neutral and acidic solutions, but are reddish purple in

basic (i.e., excess OH-) solutions.

Indicator:

A material which (by changing colors or other means) signals that the equivalence point has

been attained.

End Point:

When the indicator indicates that the equivalence point have been reached (e.g., changes

color).

Indicators need to be chosen carefully, so they don't change color too soon or too late.

Thermodynamics

So far, we have been looking at chemistry from a microscopic scale upwards, starting with the

electron, proton, and neutron, and working our way up to the molecules. In this fashion we learned

to understand and predict what happens on a macroscopic scale. Inthermodynamics we will look

at matter from the another point of view. We will consider only the macroscopic properties of

matter. Thermodynamics is a unique theory because it looks only at the macroscopic properties of

matter and, on that basis alone, tries to predict what other macroscopic behavior exists. (e.g.,

whether a particular reaction will occur or not occur under certain conditions). It is based on a few

basic tenets and is a general theory. In fact, if the entire atomic theory of matter were overthrown

(i.e., electrons, neutrons, protons, atoms, molecules), the foundations of thermodynamics would still

be sound. There are many things, however, that thermodynamics doesn't tell us. For example, while

thermodynamics tells us that diamonds at atmospheric pressure will transform into graphite, it

doesn't tell us how long for that transformation to occur.

To best understand the application of thermodynamics to chemistry we will first review some

important general concepts. Scientists like to divide or cut the whole universe into smaller parts, and

then study (and hopefully understand) the smaller parts. In the science of thermodynamics things

are no different and we begin by distinguishing between our system of interest and its surroundings.

System: The part of the universe under study.

Surroundings: Everything else that can interact with the system.

In chemistry, the system is often the reactants and products of the chemical reaction, and

surroundings will be some kind of container and everything outside the container. The surroundings

may even include a solvent in which the reactants and products are dissolved.

Associated with a system are intensive and extensive properties.

Extensive Properties are linearly dependent on amount of substance. For example, Mass,

Volume, Energy

Intensive Properties don't depend on amount of substance. For example, Temperature, Pressure,

and Density

Remember, when two identical systems are brought together extensive properties will double in

value, and intensive properties will stay the same.

The First Law of Thermodynamics

The first law of thermodynamics is also known as the "Law of Conservation of Energy".

First Law: Energy can be converted from one form to another but can be neither created nor

destroyed.

Energy is classified into one of two forms:

Potential Energy: Depends on object's position or composition

Kinetic Energy: Depends on object's motion, that is, Ekinetic = ½mv2, where m and v are the

object's mass and velocity, respectively.

Consider the example of a marble rolling in a bowl.

At any instant in time, t, the marble has a potential energy given by

Epotential = mgh(t)

Here g is the acceleration constant due to gravity. The kinetic energy, at any instant in time t is

given by

Ekinetic = ½ mv2(t)

When the marble is at the maximum height, hmax, its potential energy will be at a maximum, and its

kinetic energy at a minimum (i.e., v = 0). As the marble rolls down the side of the bowl, its potential

energy gets converted into kinetic energy (i.e., its velocity increases from zero). As the marble

passes through h=0, that is, the bottom of bowl, the marble's potential energy will be at a

minimum, and its kinetic energy at a maximum (i.e., maximum velocity). In the absence of friction,

the marble would continue rolling up and down forever with its energy converting back and forth

between potential and kinetic, and the total energy would remain constant.

Etotal = mgh(t) + ½mv2(t)

In the real world, where there is friction between the marble and the bowl, the marble eventually

stops rolling. Since energy must be conserved, where did the energy go? The answer is that it

gets dissipated into the marble and the bowl, that is, it is transferred to the internal

energy associated with random atomic motion inside the marble and the bowl. Therefore, if we

want to correctly describe this situation and obey the first law of thermodynamics, then we need to

include the internal energy, U, of the system (marble and bowl) in our expression for the total

energy of the system:

Etotal = U(t) + mgh(t) + ½mv2(t)

Note: While most scientists, including myself, prefer to use the symbol U to represent internal

energy, be aware that some texts (including the online quizzes used here) also use E for the internal

energy.

Work

By lifting the marble up to start it rolling we put energy into the marble. This type of energy transfer

into our system is called work. Work is not a form of energy, but rather it is a process in which

energy is transferred between the system and its surroundings.

Work is an energy transfer process.

In physics we learn that work can be calculated given the Forces applied on an object over a given

distance.

Work = (Force) X (distance applied)

As we just noted, the friction between the marble and the bowl causes the energy that we initially

transferred into the system (marble and bowl) as work (i.e., lifting the marble and starting it rolling)

to be eventually transferred (i.e., dissipated) to the internal energy of the marble and the bowl. So,

when the system comes to equilibrium (i.e., the marble stops rolling) we will find that the internal

energy of the system (marble and bowl) has increased.

Because energy must be conserved, the difference in the internal energy of the system (marble and

bowl) before we lift the marble and start it rolling, and after it comes to equilibrium (i.e., stops

rolling), must be equal to the work we performed on the system.

ΔU = Ufinal - Uinitial = w ← work performed on the system.

In this example, the initial and final states of the system look the same to the naked eye, that is, a

marble sitting on the bottom of the bowl and not rolling. However, on closer inspection, one would

noticed that the marble and bowl of the final state will have a slightly higher temperature due to the

increased internal energy. Temperature is a measure of the degree of random motion of the atoms

and molecules in a particular substance.

Heat

Another way we could obtain the same change in internal energy of the system (marble and bowl) is

to heat the system. That is, by placing it in contact with an object that has a higher temperature,

such as a hot plate, until we get the same change in temperature of the system that we obtained by

performing work on the system.

Heat is energy transfer by means of a temperature difference between system and surroundings.

ΔU = Ufinal - Uinitial = q ← heat is energy transferred

Just like work, Heat is not a form of energy, but rather, is an energy transfer process.

Heat is not a substance but you will often hear or read (erroneously) about it as though it is. "Putting

heat into a substance" really means putting energy into a substance by the energy transfer process

of heat.

In summary, there are only two forms of Energy: (1) Kinetic and (2) Potential, and there are only

two ways to transfer energy: (1) Work and (2) Heat. Any change in energy of a system arises from

the heat, q, and work, w, done on the system.

ΔU = q + w

State Functions

We've just seen two different ways to obtain identical changes in the internal energy of a system.

That is, we can increase the internal energy using work, or using heat. Either way, the internal

energy of final state is the same. When a property of the system does not depend on the history of

the system (i.e., whether heat, work or both were employed), but rather, only its initial and final

states, then it is called a state function.

State Function: Property of a system that does not depend on the previous history of the system,

only its present condition.

The internal energy is an example of a state function. For example, the difference in internal

energy between a liter of water in equilibrium at 10°C and 1 atmosphere, and a liter of water in

equilibrium at 75°C and 2 atmospheres is the same no matter how many times the liter of water was

heated and cooled down and the pressure changed while moving between the two equilibrium

states. In contrast, heat and work are not state functions. Without knowing a system's history we cannot know how much energy was lost or gained by a system in the form of heat or work. Other state functions are

ΔV = Vf - Vi← Change in Volume

ΔT = Tf - Ti ← Change in TemperatureSign Conventions

We define a positive ΔU > 0 to mean that the system has gained energy from its surroundings, and

a negative ΔU < 0 to mean that the system lost energy to its surroundings.

More specifically we define:

q > 0 to mean energy was added to the system as heat.

w > 0 to mean energy was added to the system as work.

q < 0 to mean energy was lost from the system as heat.

w < 0 to mean energy was lost from the system as work.

In chemistry we define the reactants and products as our thermodynamic system. Thus we define a

chemical reaction according to its sign of q.

Exothermic Reaction: Reaction that gives off energy as heat to its surroundings, that is, q < 0.

Endothermic Reaction: Reaction that absorbs energy as heat from its surroundings, that is, q > 0.

Work

A common type of work associated by a chemical process is through gas expansion or compression.

An example that you have experienced is the energy produced from the combustion of gasoline.

Gasoline combustion is used to create expanding gases in the cylinders of your car's engine that

push out the pistons. This motion is then translated into the motion of the car. Let's look at the work

associated with moving a piston.

Now remember that work is defined as a Force applied over a distance is

|Work| = |F · Δh|,

where Δh = hfinal - hinitial. We write the absolute value of work since we still need to make sure the

sign of work agrees with our earlier definitions. Using the definition of p = F/A, and recognizing A · Δh as the change in volume of the cylinder, we can write the work associated with moving a piston

a distance Δh as

|Work| =p · A · Δh = p · ΔV

where ΔV = Vfinal - Vinitial. If ΔV is positive then the gas is expanding and doing work on the

surroundings. So work should be negative

Work = - p · ΔV

This equation is true in general, not just for pistons.

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant

external pressure of 15 atm.

First we calculate

ΔV = 64 L - 46 L = 18 L

Using P=15 atm, we can then calculate

Work = - p · ΔV = (15 atm) (18 L) = -270 atm-L

Using the conversion between atm-L and Joules of

1 L-atm = 101.325 J

Thus, we obtain

So, when the gas expands it does 27.4 kJ of work on its surroundings (i.e., 27.4 kJ of energy flows out

of the system so the work is negative).

Heat

An important concept for understanding how heat is measured is the concept of heat capacity.

Heat Capacity: amount of heat needed to raise an object's temperature by 1 degree Kelvin.

More specifically we define

Molar Heat Capacity: amount of heat required to raise the temperature of one mole of a

substance by one degree Kelvin.

Specific Heat Capacity: amount of heat required to raise the temperature of one gram of a

substance by one degree Kelvin.

The heat capacity can vary quite a bit, particularly for gasses, depending on whether the pressure or

volume is held constant as heat is added. Therefore, we further distinguish the heat capacities with

the following symbols:

CV: Molar Heat Capacity at constant volume.

CP: Molar Heat Capacity at constant pressure.

CS: Specific Heat Capacity at constant pressure.

Since most processes discussed in this course are carried out at a constant 1 atmosphere of

pressure, we will emphasize CS andCP.

If we have n moles of a substance, the heat transferred and the corresponding change in

temperature are given by

Alternatively, if we have m grams of a substance, the heat transferred and the corresponding

change in temperature are given by

Below are the specific heats of some common substances

CompoundTemperature

(°C)Specific Heat (J/g-

°C)

H2O(l) 15 4.1814H2O(s), ice -11 2.03CaCO3(s) 0 0.85MgO(s) 0 0.87SiO2(s) 25 0.739O2(g) 25 0.917

The specific heat of water is one of the highest known specific heats. It is several times higher than

that of compounds like limestone (CaCO3) and quartz (SiO2), which are the major constituents of

rocks. Water can absorb or give up large amounts of heat with a smaller change in temperature than

rocks (CaCO3, SiO2, MgO). That's why temperature fluctuations are smaller near large bodies of

water like the ocean or Lake Erie than someplace like the desert in Arizona.

A piece of iron with a mass of 72.4 grams is heated to 100°C and plunged into 100 grams

of water that is initially at 10.0°C. Calculate the final temperature that is reached

assuming no heat loss to the surroundings. The heat capacities of iron and water are

Cs(H2O) = 4.18 J/°C and Cs(Fe) = 0.449 J/g-°C.

First of all, we note that the heat gained by the cooler body + the heat lost by the warmer body = 0

In other words

qH2O = - qFe

thus we write

mH2O Cs(H2O) ΔTH2O = - mFe Cs(Fe) ΔTFe

at equilibrium we have

Tf(H2O) = Tf(Fe)

so we solve for Tf

mH2O Cs(H2O) [ Tf - Ti (H2O)] = - mFe Cs(Fe) [Tf - Ti(Fe)]

and obtain

The heat flow associated with a chemical reaction is experimentally measured using a device called

a calorimeter.

Calorimetry:

the science of measuring heat flow (based on observing the temperature change when a

body absorbs or discharges heat).

Consider a reaction that gives off energy to its surroundings

A + B → C + D + Energy

We know the energy given off by this reaction will be transferred to the surroundings as heat and

work.

ΔU= q + w

Now, from

Work = - p · ΔV

we know that if we seal the reaction in a vessel strong enough then that the volume can't change,

and thus there can be no work done by the reaction since ΔV = 0. In this case we will have

ΔU=q at constant volume.

That is, all the energy will be transferred as heat into the surroundings. Therefore, if we surround our

sealed chemical reaction container (i.e., our system) with substances of known heat capacity, then

all we need to do is measure the temperature change of these surroundings to determine q (i.e., the

heat evolved) and thus the total change in energy.

This is done in what is called a Bomb Calorimeter.

Bomb Calorimeter:

Calorimeter designed to measure heat flow with constant volume (no work is possible).

1.00 grams of N2H4, hydrazine, is burned in a bomb calorimeter

and the temperature of the calorimeter increases by 3.51°C. The bomb calorimeter has a

heat capacity of 5.510 kJ/°C. What is the quantity of heat evolved?

What is the heat evolved per mole of N2H4?

Thus, we can finally write

We have measured an energy change of 618 kJ for this reaction (i.e., the heat evolved at constant

volume). Remember that ΔU is a state function (unlike q), so this number is good regardless of the

history of the reactants and products. That is, 1 mole of hydrazine will always react with 1 mole

oxygen and give 618 kJ of energy.

Enthalpy

Many chemical reactions occur at constant pressure rather than constant volume: any reaction in an

open beaker will be under the constant 1 atmosphere of pressure. Gases expanding out of open

reaction vessel are doing p ΔV work on the surroundings, but this work is lost from a practical point

of view, since it is dissipated into the surroundings. When we do a calorimetry experiment at

constant (atmospheric) pressure and measure the heat given off by the system, we don't know

the p ΔV work done by the system on the surroundings.

ΔU = q + w = q - p ΔV

We want to measure ΔU for the reaction because it is a state function, with is history independent

(unlike heat and work). To get around this problem we do a mathematical trick. We define a new

state function called Enthalpy:

H = U + p V

Since internal energy, pressure, and volume are all state functions, then enthalpy, H, must also be a

state function.

In a constant pressure experiment the change in enthaply is given by

ΔH = Δ(U + p V) = ΔU + p ΔV

Remember p is constant. Thus, we obtain

ΔH = qp

where qp is the heat transfer measured at constant pressure.

Enthalpy is the perfect state function to use for measuring the energy flowing out of a reaction as

heat at constant pressure. Also, enthalpy, like internal energy is an extensive parameter.

50.0 mL of 1.0 M HCl at 25°C is mixed with 50.0 mL of 1.0 M NaOH also at 25°C in a

constant atmospheric pressure calorimeter. After the reactants are mixed the

temperature increases to 31.9°C. What is ΔH for this reaction? Assume the density

of H2O is 1.0 g/mL.

H+(aq) + OH-

(aq) → H2O(l)

First let's calculate the heat evolved?

q = - m Cs Δ T

We have a negative sign in this equation because heat transfer is out of the system. Next we

calculate ΔH per mole of H2O(l) produced.

Thus we have

H+(aq) + OH-

(aq) → H2O(l) + 58 kJ

There are 5.8 kJ of heat evolved at constant pressure. In other words, ΔH = - 58 kJ.

Enthalpy Characteristics

Because enthalpy is a state function we can always write for a chemical reaction that

ΔH = Hproducts - Hreactants

Some other characteristics of enthalpy are...

1. If the reaction is reversed then the sign of ΔH is also reversed. For example, for the reaction

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

the ΔH is -802 kJ, that is, it is exothermic.

In contrast, the reverse reaction

CO2(g) + 2 H2O(g) → CH4(g) + 2 H2O(g)

has a ΔH of +802 kJ, that is, it is now endothermic.

2. ΔH is extensive. If the coefficients in a balanced reaction are multiplied by an integer, then

the value of ΔH is multiplied by the same integer. For example,

Xe(g) + 2 F2(g) → XeF4(g) ΔH = -251 kJ.

but...

2 Xe(g) + 4 F2(g) → 2 XeF4(s) ΔH = -502 kJ.

Hess' Law

Since enthalpy is a state function, the change in enthalpy in going from some initial state to some

final state is independent of the pathway. This leads us to Hess' Law...

Hess' Law

In going from a particular set of reactants to a particular set of products, the change in

enthalpy is the same whether the reaction takes place in one step or in a series of steps.

For example, consider the following reaction

N2(g) + 2O2(g) → 2 NO2(g) ΔH = 68 kJ.

This reaction can also be carried out in two distinct steps,

N2(g) + O2(g) → 2 NO(g) ΔH1 = 180 kJ.

2 NO(g) + O2(g) → 2 NO2(g) ΔH2 = - 112 kJ.__________________________________ __________________________________

N2(g) + 2O2(g) → 2 NO2(g) ΔHtotal = ΔH1 + ΔH2 = 68 kJ.

Determine ΔH for

Cgraphite(s) → Cdiamond(s)

given that

Cgraphite(s) + O2(g) → CO2(g) ΔH = - 394 kJ

Cdiamond(s) + O2(g) → CO2(g) ΔH = - 396 kJ

To solve this problem we simply switch the direction of the 2nd reaction (the diamond reaction) and add the two reactions together. That is,

Cgraphite(s) + O2(g) → CO2(g) ΔH1 = - 394 kJ

CO2(s) → Cdiamond(s) + O2(g) ΔH2= + 396 kJ

_____________________ _____________________

Cgraphite(s) → Cdiamond(s) ΔHtotal = ΔH1 + ΔH2 = 2 kJ.

Hess' Law is great because it means we can get the ΔH for a reaction even if we can't easily

measure it with a calorimeter. All we need are reactions of known ΔH that will add up to our

unknown reaction ΔH. Before we start making tables of all known reaction enthalpies we need to

adopt a reference state for the enthalpies of substances.

Standard State:

Stable form of an element or compound at 25°C under a pressure of 1 atmosphere.

Thus, we define

Standard Enthalpy of formation (ΔHf°):

change in Enthalpy that accompanies the formation of 1 mole of a compound from

its elements with all substances in their standard states at 25°C.

For example, the formation of CO2(g) from it's elements, carbon and oxygen, in their standard states is written

Cgraphite(s) + O2(s) → CO2(s) ΔH = - 393.5 kJ

therefore, we define the ΔH for this reaction as the Standard Enthalpy of formation for CO2(g)

ΔHf° (CO2(g)) = -393.5 kJ/mole.

Another example: the formation of H2O(l) from it's elements, hydrogen and oxygen, in their standard

states is written

H2(g) + 1/2 O2(g) → H2O(l) ΔH = - 286 kJ

therefore, the Standard Enthalpy of formation for H2O(l) is

ΔHf° (H2O(l)) = - 286 kJ/mole.

We use the coefficient 1/2 in front of O2 because we interested in heat of formation for 1 mole

of H2O(l).

Obviously, any element that is already in its standard state will have a ΔHf° of zero.

Here are some examples of ΔHf° values.

Substance Name ΔHf°(kJ/mole)

C3H8(g) propane - 103.85CO2(g) carbon dioxide - 393.5H2O(g) water - 241.8

Using all the ΔHf° for the reaction's reactants and products you can calculate ΔH° for a reaction

using

ΔH°rxtn = Σ n ΔHf° (products) - Σ n ΔHf° (reactants)

where n and m are the number of moles of each product and reactant, respectivelty, involved in

the reaction.

What is ΔH°rxtn for

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

Using the expression above we obtain:

ΔH°rxtn = [3ΔHf° (CO2(g)) + 4ΔHf° (H2O(g))] - [ΔHf° (C3H8(g)) + 5ΔHf° (O2(g))]

Since O2(g) is the stable form of oxygen we know that ΔHf° (O2(g)) = 0, and so we get

ΔH°rxtn = [3 (-393.5 kJ/mole) + 4 (-241.8 kJ/mole) ] - [ (-103.85 kJ/mole)]= -2220 kJ

The heat of formation of CO2(g) and H2O(l) are -394 kJ/mole and -285.8 kJ/mole,

respectively. Using the data for the following combustion reaction, calculate the heat of

formation of C3H4(g).

C3H4(g) + 4 O2(g) → 3 CO2(g) + 2 H2O(l) ΔH = -1939.1 kJ

To solve this problem we can start by reversing the reaction above to make C3H4(g) a product and

then add together the reactions for the heat of formation of CO2(g) and H2O(l)

3 CO2(g) + 2 H2O(l) → C3H4(g) + 4 O2(g) ΔH1 = + 1939.1 kJ

3 C(s) + 3 O2(g) → 3 CO2(g) 3 ΔHf = 3 (-394 kJ/mole)

2 H2(g) + 2 (1/2) O2(g) → 2 H2O(l) 2 ΔHf° = 2 (-285.8 kJ/mole)

_____________________________ _________________________

3 C(s) + 2 H2(g) → C3H4(g) ΔHf° = 185.5 kJ/mole

Wave-Particle Duality of Light

Quantum theory tells us that both light and matter consists of tiny particles which have wavelike

properties associated with them. Light is composed of particles called photons, and matter is

composed of particles called electrons, protons, neutrons. It's only when the mass of a particle gets

small enough that its wavelike properties show up.

To help understand all this let's look at how light behaves as a wave and as a particle.

Wave-like Behavior of Light

One behavior of waves is Diffraction

Another behavior of waves is Interference

Although I said that light is composed of particles called photons, one can easily show that light can

be thought of as an electromagnetic wave that travels at the speed of light. The frequency of light is

related to its wavelength according to

Let's look at an example calculation.

The light blue glow given off by mercury street lamps has a wavelength of λ = 436nm.

What is its frequency?

The unit s-1 is so common when talking about waves that it was given the name Hertz. That is, 1 s-

1 = 1 Hz. Thus, we would say that light with a wavelength of 436 nm corresponds to a frequency of

6.88 × 1014 Hertz.

The region from λ ≈ 400-750 nm is visible to the human eye and is therefore called the visible

region of the electromagnetic radiation. As we saw in the example above, blue light is near the high

frequency limit of our eyes. Red light, with wavelengths near 750 nm are at the low frequency limit

of our eyes. Light that contains all frequencies in the visible region will appear as white light.

More generally, the different regions of the electromagnetic spectrum are given different names.

Below are the names given to the different regions (frequency ranges) of light according to their

frequency range.

Particle-Like Behavior of Light

At this point you may think that it's pretty obvious that light behaves like a wave. But where's the

proof that light is really composed of particles called photons? The proof comes from an experiment

that is called the photoelectric effect.

An important feature of this experiment is that the electron is emitted from the metal with a specific

kinetic energy (i.e. a specific speed).

Now anyone who is familiar with the behavior of waves knows that the energy associated with a

wave is related to its amplitude or intensity. For example, at the ocean the bigger the wave, the

higher the energy associated with the wave. It's not the small waves that knock you over it's the big

waves! So everyone who thought light is just a wave was really confused when the intensity of the

light was increased (brighter light) and the kinetic energy of the emitted electron did not change.

What happens is that as you make the light brighter more electrons are emitted but all have the

same kinetic energy.

Well, they thought the kinetic energy of the emitted electron must depend on something. So they

varied the frequency of the light and this changed the kinetic energy of the emitted electron.

However, there is a critical frequency for each metal, ν0, below which no electrons are emitted. This

tells us that the kinetic energy is equal to the frequency of the light times a constant (i.e., the slope

of the line). That constant is called Plank's Constant and is given the symbol h.

h = 6.63 × 10-34 J · s ← Planck's Constant

Now we can write an equation for the kinetic energy of the emitted electron.

This result is not consistent with the picture of light as a wave. An explanation that is consistent with

this picture is that light comes in discrete packages, called photons, and each photon must have

enough energy to eject a single electron. Otherwise, nothing happens. So, the energy of a single

photon is:

Ephoton = h ν

When this was first understood, it was a very startling result. It was Albert Einstein who first

explained thephotoelectric effect and he received the Nobel Prize in Physics for this work.

So, in summary-light is a particle, but has some wave-like behavior.

Wave-Particle Duality of Matter

Wave-like Behavior of Matter

In 1925, Louis DeBroglie hypothesized that if light, which everyone thought for so long was a wave,

is a particle, then perhaps particles like the electron, proton, and neutron might have wave-like

behaviors. He went further and reasoned that since waves are described by their wavelength λ

and particles are described by their momentum, p

then we can relate these two variables by recalling that the Quantum Theory says

E = h ν = hc/λ,

and the theory of Relativity says

E = m c2 = p c.

Then let's equate these two equations to get the DeBroglie relationship between momentum (a

particle property) and wavelength(a wave property)

p = h/λ

The first real experimental proof of this relationship came from Davisson and Germer in 1925, who

found that electrons will diffract and interfere like waves, just like X-ray photons (light). For example,

an electron with a velocity of 5.97 X 106 m/s (mass of an electron =9.11 X 10-28g) has a wavelength

of:

So, matter and light are composed of particles that have wave-like properties. The wave-like

behavior is only observed on the subatomic length scales where the masses are small enough for

the wavelengths to be detectable.

Quantized States

You may be wondering how does all this about wave-particle duality relates to the electronic

structure of the atom. To begin answering that question let's look at an analogy between the

standing waves I can set up with a vibrating string, and an electron trapped between two walls.

An important property of standing waves is you can't have any frequency you want because ends

are fixed. When the ends are fixed only certain discrete wavelengths (frequencies) are allowed

(which depend on the length of the string).

nodes are where the wave function changes from positive to negative (ends don't count!).

The lowest frequency is called thefundamental or first harmonic and has no nodes. The next

frequency is the second harmonic and has one node; it is twice the fundamental frequency. You can

keep going, increasing the number of nodes, and increasing the frequency in multiples of the

fundamental frequency.

Now we use the analogy of the standing waves of a string because the wavefunction of an electron

trapped between two walls would in the same way be constrained to be a standing wave with only

discretely allowed wavelengths. That is, an electron trapped between two walls has its wavelength

determined by the distance between the walls and the number of nodes.

As in the case of the photon, the energy of the trapped electron is proportional to its frequency. The

lowest energy is called theground state and it has the fundamental frequency. The higher

energy states are called the excited states and occur at harmonics (multiples) of the fundamental

frequency.

As you can see, the energy states of the trapped electron, just like its frequency,

are discrete or quantized. Quantized means that there are only certain "allowed" energy levels

or frequencies, and nothing in between. Trapped electrons have quantized energy levels for the

same reason that the standing wavelengths we set up with our string are quantized...hence the

name Quantum Theory.

What is the physical reality behind this wavefunction?

So we have this picture where electrons, protons, neutrons, and photons are all particles with a

wave-like behavior that causes them to constructively and destructively interfere with themselves,

and to set up standing waves when confined within boundaries. But what is the physical meaning of

this wavefunction that is associated with these subatomic particles?

One interpretation is that the square of the wave function tells us the probability of finding an

electron at that point x in space.

Probability that electron is at x is [ψ(x)]2

So if we square the wavefunctions above we can obtain the probability of finding a trapped electron

between two walls as a function of position and the state of the wavefunction (i.e., the number of

nodes in its standing wave).

Notice how the probability equals zero at the nodes, since the wave function is always zero at those

points. n, the number of nodes plus one, is called the principal quantum number since it fully

describes (labels) each state in this one dimensional example.

Quantum Numbers - Specifying the electron state

Quantum Numbers for Electron Orbital

Now it turns out that if you consider an electron trapped inside a sphere (instead of in a one-

dimensional box) it will have standing waves that are very similar to an electron bound to a

positively charged nucleus by electrostatic attraction. Normally we operate in 4 dimensions (3 space

and 1 time dimension), so in general we will need 4 quantum numbers to fully specify an electron

state (i.e., standing wave).

Let's look at the quantum numbers needed to label the possible standing waves or states of an

electron trapped by its electrostatic attraction to a positively charged nucleus. For an electron

trapped by its electrostatic attraction to positively charged nucleus we use the three following

quantum numbers to describe the electron state (orbital).

n : the principal quantum number

The principal quantum number has integral values of n = 1, 2, 3... As n increases, the electron

orbital becomes larger and the electron spends more time farther from the nucleus.

ℓ : azimuthal quantum number

The azimuthal quantum number has integral values of ℓ = 0 to ℓ = n - 1 for each value of n. This

quantum number defines the shape of the orbital. There are special letters assigned to each ℓ value

(see table below).

ℓ symbol

1 p2 d3 f

For an electron with n = 1 and ℓ = 0 we say that the electron is in the 1s state or 1s orbital

For an electron with n = 2 and ℓ = 0 we say ... 2 s orbital

For an electron with n = 2 and ℓ = 1 we say ... 2 p orbital

For an electron with n = 3 and ℓ = 2 we say ... 3 d orbital

mℓ : magnetic quantum number

The magnetic quantum number has integral values of mℓ = - ℓ to + ℓ including 0.

For example, if n = 3 and ℓ = 2 then the possible values of mℓ are -2, -1, 0, +1, +2

ms, : spin quantum number

The spin quantum number has only two possible values of +1/2 or -1/2. If a beam of hydrogen

atoms in their ground state (n = 1,ℓ = 0, mℓ = 0) or 1s is sent through a region with a spatially

varying magnetic field, then the beam splits into two beams.

Clearly the three quantum numbers, n, ℓ, mℓ, are not enough to completely describe the state of the

H-atom, Another quantum number is required to describe whether it goes up or down in a spatially

varying magnetic field. This property is called spin because if electrons were balls of charge

spinning about their own axes, they would behave in this way in a magnetic field. Actually, this is

not a correct picture. The spin quantum number shows up when the wave function of quantum

mechanics is modified to include the effects of relativity.

The electron spin is very important in understanding the electronic structure of atoms containing

many electrons (not just one like Hydrogen) because of the Pauli Exclusion Principle.

Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum

numbers n, ℓ, mℓ, ms.

Since there are only two values of ms, then any orbital can only hold two electrons with opposite

vales of ms. Any more would violate Pauli's Principle.

In summary, you can think of electron orbitals as standing waves of the electron wave function when

it is bound to a postively charged nucleus. Remember, however, that the electron itself is

a particle, not a wave. That wave function squared gives the probability of finding the electron at

any particular position.

Orbital Energies

We saw earlier that the energy of the electron in a hydrogen atom depends only on the principal

quantum number, n. The nucleus of a hydrogen atom has a charge of +1, however, if the electron is

bound to a nucleus of arbitrary charge +Z, then the energy of the electron is

This expression is for a single electron orbiting a single nucleus of charge +Z. If I had a mole of

atoms like this, then I could multiply this expression by Avogadro's Number to get the total energy

for all the atoms:

This equation is so popular that the number 1312 is named the Rydberg Constant and given the

symbol RH = 1312 kJ/mole.

Let's look carefully at this equation:

As n increases (holding Z constant), then the energy increases (becomes less negative). In

the limit that n goes to infinity then the energy goes to zero.

As Z increases (holding n constant), then the energy decreases (becomes more negative).

This makes sense, since a higher Z means a more positively charged nucleus, which holds

the electron tighter.

Hydrogen Atom Energy Levels

Let's look at the energy levels of the hydrogen atom.

For the hydrogen atom Z=1 so En= - RH/n2

Notice that the energy level spacing decreases as n increases, that the number of orbitals

(i.e. l values) increase with n, and all orbitals with the same n have the same energy (degenerate).

(H-atom only).

Spectrum of Hydrogen

We can look at either absorption or emission spectra.

Using our equation for the energy of the hydrogen levels we can write an equation for the change in

energy of an electron that charges orbitals and emits or absorbs a photon.

With this equation we can calculate the frequency of light emitted or absorbed when an electron

moves between orbitals of different principal quantum numbers.

Using the equation above we can calculate the wavelengths for various transitions in the H-atom.

ni → nf Wavelength

3 → 2 λ=657 (red)

4 → 2 λ=487 (green)

5 → 2 λ=435 (blue)

∞ → 2 λ=365 (purple)

The energy required to promote an electron to n = ∞ is called the ionization energy.

(Because this is the energy required to make an ion).

Electron Orbital Shapes

Now, let's look closer at these electron orbitals and their shapes. Remember, we used a two-

dimensional plot of the wave function versus x to visualize the standing wave of an electron trapped

in one dimension. To visualize the standing waves (or orbitals) of electrons bound to a positively

charged nucleus in three dimensions, we will need a four-dimensional plot of the wave function

vs. x,y, and z. This can be a bit tricky since our visual perception is limited to three spatial

dimensions. So we will need a few tricks to help us visualize the four-dimensional standing waves of

the electron in 3 dimension.

1s-orbital:

The lowest energy orbital of the hydrogen atom.

2s-orbital:

The second harmonic state.

Remember that at the node, the probability of finding the electron is zero. In general, an orbital with

high n (principal quantum number) (e.g. n = 2, 3, 4...) means that the electron will extend out

from the nucleus further, and so will be held less tightly than a1s electron.

2p-orbital:

When n = 2, we have 2 possible values for ℓ. The first is ℓ = 0, or 2s orbital, which we just

discussed above. The second possibility is ℓ = 1 or the 2 p orbital. For a given value of ℓ there are 2 ℓ + 1 possible ml values. So for ℓ = 1, we have mℓ= -1, 0, +1. These three values

of mℓ correspond to three different p-orbitals.

P-orbitals look like dumbbells along each axis. Instead of a radial node, we have an angular node,

which lies along the plane perpendicular to the axis in which the orbital lies. Since the energy, E, of

each orbital is a function of only n, then all the n = 2orbitals (2s, 2px, 2py, 2pz) have the same

energy.

n = 3 orbitals:

For the n=3 orbitals the possible quantum numbers are:

n=3 ℓ=0 mℓ=0 3s orbitaln=3 ℓ=1 mℓ=-1, 0, +1 3p orbitalsn=3 ℓ=2 mℓ=-2, -1, 0, +1, +2 3d orbitals

For much nicer three-dimensional renderings of all the atomic orbitals visit Mark Winter's Orbitron

site .

http://winter.group.shef.ac.uk/orbitron

http://winter.group.shef.ac.uk/orbitron

Multielectron Atoms

While Quantum Theory gives exact equations describing the H-atom, which has only one electron, it

runs into problems trying to give exact equations of atoms with many electrons. This is because in

addition to the electrostatic attraction between the electron and the positively charged nucleus,

there are electrostatic repulsions between electrons. The problem starts to get complicated quickly.

In spite of this problem, approximate solutions can be obtained, which can, in fact, be quite

accurate. For a multi-electron atom the energy of a particular electron in the atom is given by

Which looks the same as for a single electron atom except that now we use an effective charge for

the positively charged nucleus. The effective charge is reduced from the full charge due to the

shielding of the nuclear charge by other electron in the atom.

The effective nuclear charge equates the number of protons in the nucleus, Z, minus the average

number of electrons, S, between the nucleus and the electron of interest.

In a multi-electron atom it turns out that the effective charge, Zeff, decreases with increasing value

of ℓ, the azmuthial quantum number. This is because electrons in the s-orbital have a greater

probability of being near the nucleus than a p-orbital, so the s-orbital is less shielded than a p-

orbital. Likewise, a p-orbital is less shielded than a d-orbital.

In a multi-electron atom, the energy of an orbital increases with increasing value of ℓ for a given

value of n.

Electron shell: All the orbitals with the same value of n

Electron subshell: All the orbitals with the same value of n and ℓ. Electrons in the same

subshell are degenerate (i.e., have same energies).

Aufbau Principle

The energy structure of a many-electron atom is obtained by filling the orbitals one-electron at a

time, in order of increasing energy starting with the lowest energy. This is called the Aufbau

principle.

The ordering of orbital energy levels is

These are two ways to indicate the electronic structure of an atom: the Electronic Configuration and

the Orbital Diagram.

1. Electronic Configuration

e.g., for the H-atom, the electronic configuration is 1s1. This notation is compact, and describes how

the electrons are distributed with principal, n, and azimuthal, ℓ, quantum numbers, but does indicate

the magnetic, mℓ and spin, ms quantum numbers for the electrons.

2. Orbital Diagram

Orbital Diagrams give a more complete indication of the electron quantum numbers. Each orbital

represented by a box and each electron by a half-arrow.

For example Hydrogen has 1 electron. That electron goes into the lowest energy orbital, that is the

1s orbital. Thus we write...

For Boron we start with 5 electrons. Again we start by filling the lowest energy orbital 1s, then the

2s orbital, and finally putting one electron in the 2p orbital...

Now let's look at Carbon, which has 6 electrons.

Hund's Rules: Electrons occupy different orbital or a given subshell with spins in the same

direction before spin pairing occurs.

Therefore for carbon we have

Another example, Neon has 10 electrons.

For Neon, both the n=1 and n=2 shells are completely full.

Let's look at Sodium, which has 11 electrons.

Na has one electron its outermost shell. For convenience we often represent the electron

configuration of the closed shells with the corresponding noble gas symbol. So for sodium we would

write:

[Ne] 3 s1 where [Ne] represents 1s2 2s2 2p6

In fact we distinguish between electrons depending on whether they part of the closed shell or in the

outermost shell.

Valence Electrons: Electrons in the outermost shell (principal quantum level) of an atom.

Core Electrons: Electrons in the inner closed shells.

If you look at the periodic table you will notice that elements in the same group have the same

number of valence electrons. That is the main reason why elements in the same group have such

similar chemical and physical properties.

Let's look at Argon, which has 18 electrons. It has the configuration

1s2 2s2 2p6 3s2 3p6

Now you might be tempted for Potassium (the next element) to put the extra electron in the 3d

orbital, however, it turns out that the 4s orbital is slightly lower in energy than the 3d, so the

electron configuration of Potassium is:

K is [Ar] 4s1

After the 4s orbital is filled, then you can start to fill the 3d orbital.

For example, Titanium has 22 electrons. It's configuration is

The triangular diagram below can be used to simplify memorizing the order in which the orbitals are

filled.

Atomic Radii Trends in the Periodic Table

In order to talk about the radius of an atom, we have to make an arbitrary decision about where

the edge of the atom is. It is arbitrary because the electron orbitals do not end sharply.

Nevertheless, we can do like we did with the 3D contour plots of the orbitals and just arbitarily

choose the radius that the electron spends 90% of its time inside.

The electrons spend 90% of the time inside the black line.

Using this definition consistently, we can look at the trends of the atomic radii as a function of

position in the periodic table.

That trend is...

In general the size of the atom depends on how far the outermost valence electron is from the

nucleus. With this in mind we understand two general trends...

Size increases down a group:

The increasing principle quantum number of the valence orbitals means larger orbitals and

an increase in atomic size.

Size generally decreases across a period from left to right:

To understand this trend it is first important to realize that the more strongly attracted the

outermost valence electron is to the nucleus then the smaller the atom will be. While the

number of positively charged protons in the nucleus increases as we move from left to right

the number of negatively charged electrons between the nucleus and the outer most

electron also increases by the same amount. Thus you might expect there to be no change

in the radius of the outermost electron orbital since the increasing charge of the nucleus

would be canceled by the electrons between the nucleus and the outermost electron. This,

however, is not the case. The ability of an particular inner electron to cancel the charge of

the nucleus for the outermost electron depends on the orbital of that inner electron.

Remember that electrons in the s-orbital have a greater probability of being near the

nucleus than a p-orbital, so the s-orbital does a better job of canceling the nuclear charge

for the outermost electron than an electron in a p-orbital. Likewise, an electron in a p-orbital

is does a better job than a d-orbital. Thus, as we move across a given period the ability of

the inner electrons to cancel the increasing charge of the nucleus diminishes and the

outermost electron is more strongly attracted to the nucleus. Hence the radius decreases

from left to right.

Ionization Energy Trends in the Periodic Table

The ionization energy of an atom is the amount of energy required to remove an electron from the

gaseous form of that atom or ion.

1st ionization energy - The energy required to remove the highest energy electron from a neutral

gaseous atom.

For Example:

Na(g) → Na+(g) + e- I1 = 496 kJ/mole

Notice that the ionization energy is positive. This is because it requires energy to remove an

electron.

2nd ionization energy - The energy required to remove a second electron from a singly charged

gaseous cation.

For Example:

Na+(g) → Na2+

(g) + e- I2 = 4560 kJ/mole

The second ionization energy is almost ten times that of the first because the number of electrons

causing repulsions is reduced.

3rd ionization energy - The energy required to remove a third electron from a doubly charged

gaseous cation.

For Example:

Na2+(g) → Na3+

(g) + e- I3 = 6913 kJ/mole

The third ionization energy is even higher than the second!

Succesive ionization energies increase in magnitude because the number of electons, which cause

repulsion, steadily decrease. This is not a smooth curve There is a big jump in ionization energy after

the atom has lost its valence electrons. An atom that has the same electronic configuration as a

noble gas is really going to hold on to its electrons. So, the amount of energy needed to remove

electrons beyond the valence electrons is significantly greater than the energy of chemical reactions

and bonding. Thus, only the valence electrons (i.e., electrons outside of the noble gas core) are

involved in chemical reations.

The ionization energies of a particular atom depend on the average electron distance from the

nucleus and the effective nuclear charge

These factors can be illustrated by the following trends:

1st ionization energy decreases down a group.

This is because the highest energy electrons are, on average, farther from the nucleus. As the

principal quantum number increases, the size of the orbital increases and the electron is easier to

remove.

Examples:

I1(Na) > I1(Cs)

I1(Cl) > I1(I)

1st ionization energy increases across a period.

This is because electrons in the same principal quantum shell do not completely shield the

increasing nuclear charge of the protons. Thus, electrons are held more tightly and require more

energy to be ionized.

Examples:

I1(Cl) > I1(Na)

I1(S) > I1(Mg)

The graph of ionization energy versus atomic number is not a perfect line because there are

exceptions to the rules that are easily explained.

Filled and half-filled subshells show a small increase in stability in the same way that filled shells

show increased stability. So, when trying to remove an electron from one of these filled or half-filled

subshells, a slightly higher ionization energy is found.

Example 1:

I1(Be) > I1(B)

It's harder to ionize an electron from beryllium than boron because beryllium has a filled "s"

subshell.

Example 2:

I1(N) > I1(O)

Nitrogen has a half-filled "2p" subshell so it is harder to ionize an electron from nitrogen than

oxygen.

Which element has a higher ionization energy, Zinc or Gallium?

Electron Affinity Trends in the Periodic Table

Electron Affinity is the energy associated with the addition of an electon to a gaseous atom.

Example:

Cl(g) + e- → Cl-(g) E.A. = -349 kJ/mole

Notice the sign on the energy is negative. This is because energy is usually released in this process,

as apposed to ionization energy, which requires energy. A more negative electron affinity

corresponds to a greater attraction for an electron. (An unbound electron has an energy of zero.)

Trends:

As with ionization energy, there are two rules that govern the periodic trends of electron affinities:

Electron affinity becomes less negative down a group.

As the principal quantum number increases, the size of the orbital increases and the affinity for the

electron is less. The change is small and there are many exceptions.

Electron affinity decreases or increases across a period depending on electronic

configuration.

This occurs because of the same subshell rule that governs ionization energies.

Example:

Since a half-filled "p" subshell is more stable, carbon has a greater affinity for an electron than

nitrogen.

Obviously, the halogens, which are one electron away from a noble gas electron configuration, have

high affinities for electrons:

(More negative energy = greater affinity)

Element Electron AffinityI -295.2 kJ/mole

Br -324.5 kJ/moleCl -348.7 kJ/moleF* -327.8 kJ/mole

*Fluorine's electron affinity is smaller than chlorine's because of the higher electron - electron

repulsions in the smaller 2p orbital compared to the larger 3p orbital of chlorine.

Predicting Stable Anions and Cations

Main group elements (Group A) will gain or lose minimum number of electrons to have a filled

shell (noble gas configuration).

Example:

Na 1s2 2s2 2p6 3s2 → Na+ 1s2 2s2 2p6 3s2 + e-

Cl 1s2 2s2 2p6 3s2 3p5 + e- → Cl- 1s2 2s2 2p6 3s2 3p6

Transition group elements (Group B) generally lose outer shell s - electrons. Many also lose

electrons to have either a filled or half-filled subshell.

Example:

Cu [Ar] 4s2 3d9 → Cu2+ [Ar] 3d9 + 2 e-

The Cu2+ lost its outer shell s electrons.

Fe [Ar] 4s2 3d6 → Fe3+ [Ar] 3d5 + 3 e-

The Fe3+ has a half-filled 3d subshell.

Ionic sizes

A cation is smaller than its parent atom. An anion is larger than its parent atom.

Generally, ion size increases going down a group. Change in ion size horizontally is complicated

because we change from cations on the left to anions on the right. However, it is helpful to look at

the relative size of isoelectronic ions.

Isoelectronic Ions-- ions containing the same number of electrons.

For example, O2-, F-, Na+, Mg2+, and Al3+ all have the electron configuration of Neon. But the charge

of the nucleus increases from +8 on Oxygen to +13 for Aluminum. So the same number of electrons

is Aluminum (Al3+) will be bound much closer to the nucleus than the electrons of O2-.

For a series of isoelectronic ions, the size decreases as the nuclear charge, z, increases.

Arrange the following ions in order of increasing size: Se2-, Br-, Rb+, Sr2+

smallest Sr2+ < Rb+ < Br- < Se2- largest38 protons 37 protons 35 protons 34 protons

36 electrons 36 electrons 36 electrons 36 electrons

Big Picture: Chemistry is all about transferring electrons between different atoms and/or molecules. The different chemical properties of each atom in the periodic table is a result of the different degrees of which each atom wants to gain or lose electrons.

Electronegativity and Bond Polarity

If the bonding electrons are not shared equally in a covalent bond, then the bond will be polar. To

quantify how polar a bond will be we introduce the concept of Electronegativity.

Electronegativity - the ability of an atom in a molecule to attract electrons to itself.

For example, in the HF molecule, the bonding electrons spend more time on the fluorine than the

hydrogen because the fluorine has a higher electronegativity than hydrogen. We indicate this slight

excess of negative charge on the fluorine with the symbol δ-:

As a result of this unequal sharing of bonding electrons the HF molecule will have an electric

dipole moment. The electric dipole moment is a vector quantity, and is represented by the

symbol μ. In the case of HF, μ will lie along the direction of the H-F bond:

The strength of the electric dipole moment across a bond will be proportional to the difference in electronegativity of the two atoms forming the bond. There are a number of ways to quantify atom electronegativities. Below are a few numbers based on an approach by Linus Pauling.

H: 2.1

Li: 1.0 Be: 1.5 B: 2.0 C: 2.5 N: 3.0 O:3.5 F: 4.0

Na: 0.9 Mg: 1.2 Al: 1.5 S: 1.8 P: 2.1 S: 2.5 Cl: 3.0

The general trend of Electronegativity in the periodic table is

Bond Polarity

Using the Electronegativities we can predict whether a given bond will be non-polar, polar covalent,

or ionic. The greater the difference in electronegativity the more polar the bond.

Order the following bonds according to polarity: H-H, O-H, Cl-H, S-H, and F-H.

In practice no bond is totally ionic. There will always be a small amount of electron sharing.

So what is and ionic compound? There's no clear cut line. Therefore we use the following practical

definition for an ionic compound

Ionic Compound is any solid that conducts an electric current when melted or dissolved in water.

Or you can simply say that a "salt" is an ionic compound.

Once again, the picture we're forming is that, in virtually every case, the atom in a stable compound

has a noble gas arrangment of electrons. When two non-metals react to form a covalent bond, they

share electrons in a way that completes the valence electron configurations of both atoms (i.e., both

non-metals attain a noble gas configuration).

Lewis Dot Structures

During chemical bonding it is the valence electrons which move amongst different atoms. In order to

keep track of the valence electrons for each atom and how they may be shared in bonding we use

the Lewis Dot Structure for atoms and molecules. In this approach we represent the valence

electrons as dots around the element symbol. For example, oxygen has 6 valence electrons, so we

write the symbol O for oxygen and surround it with 6 dots:

The unpaired electrons are represented as single dots, and the paired electrons as double dots. The

placement of the single or double dots around the symbol is not critical. Alternatively, we can

represent the paired electrons as a line. That is, we replace the double dots as shown below:

Let's consider other examples. A sodium atom has 11 electrons, but only one is a valence electron.

The other 10 are inside a closed shell with a Neon electron configuration. Thus, we draw the Lewis

structure for a sodium atom as the symbol Na with a single dot:

A chlorine atom has 17 electrons, but only 7 of these are valence electrons. Thus, we draw the Lewis

structure as:

In Ionic Bonds valence electrons are completely transferred (not shared). Thus, we write the Lewis

structure for NaCl as:

As you can see Chlorine is now surrounded by 8 electrons in the n=3 shell and Sodium has lost its

one valence electron in the n=3shell. Of course, Sodium, is still surrounded by the 8 electrons of

the n=2 shell, but we do not show electrons in the inner closed shells.

For period 2 elements, where all the valence electrons of an atom are in s and p orbitals, we find

that the Lewis dot structure of molecules will often follow the Octet Rule:

Octet Rule - Atoms tend to gain, lose, or share electrons until they are surrounded by eight

electrons (4 electron pairs).

Using Lewis dot structures and the octet rule, we can predict and represent the electronic structure

of covalently bonded molecules. For example, when two chlorine atoms, each with 7 valence

electrons, come together to form a diatomic chlorine molecule, the Lewis structure shows that there

will be a sharing of two electrons between the two chlorine atoms which allows both chlorine to be

surrounded by 8 electrons.

Of course, hydrogen is a period 1 element, with only has a 1s orbital, so it has a maximum of two

electrons allowed in its valence shell. When two hydrogen atoms come together into a

diatomic H2 molecule the Lewis structure shows that there will be a sharing of two electrons

between the two hydrogen, allowing both hydrogen to be surrounded by a closed n=1 shell of 2

electrons:

We can represent the electronic structure and reaction of hydrogen and chlorine atoms to form HCl

with Lewis structures:

For diatomic oxygen, the Lewis dot structure predicts a double bond.

While the Lewis diagram correctly predict that there is a double bond between O atoms, it

incorrectly predicts that all the valence electrons are paired (i.e., it predicts that each valence

electron is in an orbital with another electron of opposite spin). Later we will examine a more

advanced theoretical approach called Molecular Orbital Theory, which correctly predicts both the

double bond of O2and its unpaired valence electrons. Generally, Lewis-dot structures have the

advantage that they are simple to work with, and often present a good picture of the electronic

structure. Let's consider another example. For diatomic nitrogen, the Lewis-dot structure correctly

predicts that there will be a triple bond between nitrogen atoms:

This triple bond is very strong. The strength of the triple bond makes the N2 molecule very stable

against chemical change, and, in fact, N2 is considered to be a chemically inert gas.There is a relationship between the number of shared electron pairs and the bond length.

Bond Bond Length

N-N 1.47 ÅN=N 1.24 ÅN≡N 1.10 Å

The distance between bonded atoms decrease as the number of shared electron pairs increase.

Rules for drawing Lewis dot structures1. Count the number of valence e- each atom brings into the molecule. For ions, the

charge must be taken into account.

How many valence electrons in BeCl2?

How many valence electrons in NO2- and NO2

+?

1.

Put electron pairs about each atom such that there are 8 electrons around each

atom (octet rule), with the exception of H, which is only surrounded by 2

electrons. Sometimes it's necessary to form double and triple bonds. Only C, N, O,

P and S (rarely Cl) will form multiple bonds.

Draw the Lewis dot structure for CF4.

The number of valence electrons is 4 + 4 ( 7 ) = 32 electrons.

So, we obtain:

Draw the Lewis dot structure for CO.

The number of valence electrons is 4 + 6 = 10 electrons or 5 pairs. Since both C and O

allow multiple bonds we can still follow the octet and write:

2. If there is not enough electrons to follow the octet rule, then the least

electronegative atom is left short of electrons.

Draw the Lewis dot structure for BeF2.

In BeF2 number of valence e- = 2+ 2(7) = 16 e- or 8 pairs. Since neither Be or F form

multiple bonds readily and Be is least electronegative we obtain:

3. If there are too many electrons to follow the octet rule, then the extra electrons

are placed on the central atom.

Draw the Lewis dot structure for SF4.

In SF4 the number of valence electrons is 6 + 4 ( 7 ) = 34 electrons or 17 pairs. Placing the

extra electrons on S we obtain:

How can the octet rule be violated in this last example? The octet rule arises because the s and p

orbitals can take on up to 8 electrons. However, once we reach the third row of elements in the

periodic table we also have d-orbitals, and these orbitals help take the extra electrons. Note that you

still need to know how the atoms are connected in a polyatomic molecule before using the Lewis-Dot

structure rules.

Resonance Structures

The Lewis structure for certain molecules or ions can be drawn in more than one way. For example,

for NO2- the number of valence eletrons is 5 + 2 (6) + 1 = 18 e- (or 9 pairs), and we find that there

are two equally valid Lewis structures that can be drawn:

Which one is correct? Well, you would expect that the doubly bonded oxygen-nitrogen distance to

be slightly less than the singly bonded distance. Actually, what is found experimentally is that both

N-O distances are equivalent. The true structure of the molecule is a combination of the two.

Anytime you have more than one valid structure for a molecule or ion, you have what are known

asresonance structures. So in this case, both resonances structures contribute equally to the final

structure of the molecule. Sometimes you will have multiple resonance structures which do not

contribute equally to the final structure of the molecule. In these cases it can be helpful to know

which structure has the greatest contribution to the final structure. If you have many possible

resonance forms, you choose the most likely resonance form by calculating the formal charge on

each atom in each resonance form. In these situations it is helpful to calculate the formal

charge on each atom in each possible resonance structure, and use the formal charges to

determine the most representative structure.

Formal charge = Group number - number of nonbonding e-- (number of bonding e-) / 2.

In the example below, we calculate the formal charge on each atom in a Lewis structure.

What are the formal charges on each atom in NO2-?

The sum of the formal charges must equal that of the compund or ion. So, 0 + 0 - 1 = -1 as

expected for NO2-.

To use the formal charge to determine most representative resonance forms we follow:

Rules for determining most representative resonance form

1. The resonance form(s) with the least number of atoms with formal charge is (are) the most

preferred.

2. Resonance forms with low formal charges are favored over high formal charge. (e.g., ±1 is

favored over ±2).

3. Resonance forms with negative formal charge or most electronegative atoms are favored.

4. Resonance forms with the same charge on adjacent atoms are not favored.

For example, N2O has number of 2 ( 5 ) + 6 = 16 valence electrons or 8 pairs. We can draw the

three valid Lewis structures below, labeled A, B, and C:

For each structure we can calculate the formal charges below on each atom:

A B C

N1 5 - 2 - ( 6 ) / 2 = 0 5 - 4 - ( 4 ) / 2 = - 1 5 - 6 - ( 2 ) / 2 = -2N2 5 - 0 - ( 8 ) / 2 = + 1 5 - 0 - ( 8 ) / 2 = + 1 5 - 0 - ( 8 ) / 2 = + 1

A B C

O 6 - 6 - ( 2 ) / 2 = - 1 6 - 4 - ( 4 ) / 2 = 0 6 - 2 - ( 6 ) / 2 = + 1

Examining the formal charges above we see that Formula C is less representative because it has a -

2 charge, and formula B is less representative because it has a -1 charge on N and 0 charge on O.

Oxygen is more electronegative and should get -1 charge, therefore Formula A is most

representative.

Valence Shell Electron Pair Repulsion Model

In the last section we learned about a simple approach called the Lewis-Dot structure that gives a

good approximation of how the valence electrons are distributed in a molecule. What Lewis-Dot

strucutres do not tell us is the shape of the molecule. For example, why do XeF4 and CF4 have

different shapes even though the central atom is coordinated by four Fluorine atoms in both cases?

One way to answer these questions about molecular structure is to use a simple approach that

builds on the Lewis-Dot structure approach called the Valence Shell Electron Pair Repulsion

Model, or VSEPR model.

The idea behind this approach is that the structure around a given atom is determined principally by

minimizing electron repulsions. That is, the bonding and non-bonding electrons around a given atom

will be positioned as far apart as possible. For example, BeCl2has the Lewis Structure.

There are only two pairs of electrons around Be. The arrangement that puts the bonding electron

pairs as far apart as possible is a linear arrangement:

...as far apart as possible - a very simple model.

Using this guiding principle let's look at the possible arrangements that arise when there are

different numbers of electron domains (i.e., bonding and non-bonding electrons) surrounding an

atom.

Given the arrangements above we use the following rules for predicting the geometry around an

atom.

VESPR Rules for Determining Structure1. Draw the Lewis Structure.

2. Add together the number of atoms bound to the central atom and the number of lone pair

electrons and choose the appropriate arrangement. (i.e., linear, triangular planar,

tetrahedral, trigonal bipyramidal, or octahedral).

3. Draw the structure, placing the appropriate number of bonds and lone electron pairs (i.e.,

electron domains) about the central atom according to the chosen arrangement.

What's the structure of BF3?

The Lewis Dot Structure for BF3 is:

There are only three atoms around B and no lone pairs so we use a Trigonal Planar

arrangement.

The name of structure for BF3 is also Trigonal Planar.

1.

What's the structure of SO2?

The Lewis Dot Structure of SO2 is:

There are only 2 bonds and one lone pair around S; a total of three "domains". Therefore we

use a Trigonal Planar arrangement.

However, we write "bent" for the structure because lone pairs don't count for the name of

the shape.

2. If more that one structure is possible, then minimize repulsions, keeping in mind that:

1. 90° repulsions > 120° repulsions > 180° repulsions

2. Lone Pair-Lone Pair repulsions > Bond-Lone Pair repulsions > Bond-Bond repulsions

3. Lone Pair-Lone Pair repulsions at 90° > Bond-Lone Pair repulsions at 90° > Bond-

Bond repulsions at 90°

What's the structure of XeF4?

The Lewis Dot Structure is

There are 4 bonds + 2 lone pairs → 6 domains → Octahedral Arrangement. Using this arrangement

we find there are two possibilities:

A is better, since B has a 90° lone pair-lone pair repulsions, and A has a 180° L.P.-L.P. repulsions.

Therefore, the structure of XeF4, ignoring the lone pair electrons is "square planar".

Finally, what's the structure of ICl2?

The Lewis Dot Structure is:

Around Iodine we have 2 bonds + 3 lone pairs → 5 domains so we use a trigonal bipyramidal

arrangement. We find there are three possibilities for the structure:

A and B have 90° L.P.-L.P. repulsions. C has none, so C is the most likely structure. Therefore the

shape of ICl2 is linear.

Using the rules above we name the shape using only the positions of atoms, ignoring the lone pair

electrons. Below are the names for different geometries.

Net Electric Dipole Moments for Molecules

The net electric dipole moment for a molecule is the vector sum of the electric dipole moments of all

its bonds. For example, from the bent structure of a water molecule we can see how the electric

dipole moment vectors for the two O-H bonds will combine into an overall electric dipole moment

vector for the water molecule:

Whereas, the electric dipole moment vectors for the two C-O bonds in CO2 will combine into an

overall zero net electric dipole moment vector for the CO2 molecule:

Using our new understanding of molecular geometries we can now predict whether molecules with

polar bonds will have a net electric dipole moment.

Molecular Orbital Theory

The Lewis Structure approach provides an extremely simple method for determining the electronic

structure of many molecules. It is a bit simplistic, however, and does have trouble predicting

structures for a few molecules. Nevertheless, it gives a reasonable structure for many molecules and

its simplicity to use makes it a very useful tool for chemists.

A more general, but slightly more complicated approach is the Molecular Orbital Theory. This

theory builds on the electron wave functions of Quantum Mechanics to describe chemical bonding.

To understand MO Theory let's first review constructive and destructive interference of standing

waves starting with the full constructive and destructive interference that occurs when standing

waves overlap completely.

The H2+ molecule has only one valence electron.

The H2 molecule has two valence electrons.

The He2+ molecule has three valence electrons.

The He2 molecule has four valence electrons.

In the Lewis Structure Theory we had single, double, and triple bonds, in the Molecular Orbital

Theory we similarly define the bond order.

Bond order = 1/2 (# of electrons in bonding orbitals - # of electrons in anti-bonding orbitals).

The bond order in our four examples above are given in the table below.

Bond Bond Order

H2+ 1/2

H2 1He2

+ 1/2He2 0

The bond order must be positive non-zero for a bond to be stable. He2 has a bond order of zero and

that is why the He2 molecule is not observed.

For example, e.g. O2 has 6 + 6 = 12 valence electrons which can be placed in bonding and anti-

bonding orbitals.

Notice that Molecular Orbital Theory predicts that O2 has unpaired electrons, so it will be

paramagnetic.

Another example is F2. It has 14 electrons which are placed in the bonding and anti-bonding orbitals

starting with the lowest energy orbital first.

In the case of B2, C2, and N2 there is a slightly different ordering in orbital energies.

For example, B2 has 3 + 3 = 6 valence electrons.

Orbital Hybridization

We've learned how constructive and destructive interference of atomic orbitals explains the

formation of bonding and anti-bonding orbitals. We also leaned about two types of

bonding: σ and π bonding. So you might expect that for polyatomic molecules, all you need to do is

put the atoms of the molecule near each other in the right geometry and then see

what σ or π bonds form between all the atomic orbitals.

Well, it is almost that simple. The only problem is that for most molecular geometries the atomic

orbitals on an atom do not point in the right direction for a σ or π bond to form. Let's look at BF3 as

an example. From VSEPR we know the geometry around the Boron atom should be trigonal planar.

But for a Boron atom all the valence elelctrons are in the 2s, 2px, 2py, 2pz orbitals. Recall their

shapes:

The problem you'll find is that there's no way you can put three Fluorine atoms around the s and p

orbitals of Boron in a trigonal planar configuration and form 3 equivalent σ or 3 equivalent π bonds.

Yet, we know the B-F bonds are all equivalent because they all have the same bond dissociation

energy.

Actually, what happens is that as you bring the three Fluorine atoms near Boron, the atomic orbitals

on Boron change (or hybridize) so that they can form σ bonds in a trigonal planar shape.

In this example, one s, and two p orbitals, i.e., px, and py, hybridize to form 3 new orbitals that point

along the correct direction to form σ bonds with all 3 Fluorines. This is called sp2 hybridization;

Let's look at another example, BeF2. From the VSEPR model we know its structure is

In Be, the s and px orbitals hybridize to give two similar sp hybrid orbitals.

Remember that the atomic orbitals are standing waves associated with the electrons bound to a

nucleus. When you bring atoms together the boundary conditions for these standing waves change

and so the standing waves which were the atomic orbitals change. That is all hybridization is. It's

analogous to holding down and releasing a violin string while you're playing. There's one standing

wave (one frequency) while you're holding down the string, and another standing wave (another

frequency) when you release the string.

Let's conside another example, CH4.

To get these bonds you hybridize one s and three p orbitals. These are called sp3 hybrid orbitals.

Sometimes it is not necessary for all the valence electron orbitals to hybridize. For example,

ethylene has the following structure:

The bonds between C and H are all σ bonds between sp2 hybridized C atoms and the s-orbitals of

Hydrogen. The double bond between the two C atoms consists of a σ bond (where the electron pair

is located between the atoms) and a π bond (where the electron pair occupies the space above and

below the σ-bond).

You should remember that we learned about molecules where the central atom gets more than an

octet of electrons.

We learned earlier that the extra bonding electron pairs are possible if we include the d-orbitals of

phosphorous. This is done by forming hybrid orbitals from s, p, and now d orbitals. For trigonal

bipyramidal the central atom is bonded through dsp3 hybrid orbitals.

In the case of molecules with an octahedral arrangement of electron pairs, another d-orbital is used

and the hybridization of the central atom is d2sp3

In summaryTotal # of L.P. and B.P

about atomArrangement Hybridization

2 linear sp3 trigonal planar sp2

4 tetrahedral sp3

5trigonal

bipyramidald sp3

6 octahedral d2sp3

What is the hybridization of Xe in XeF4?

Starting with the # of valence electrons = 8 + 4 ( 7 ) = 36 e- (i.e., 18 pairs), and using the VSEPR

model we predict an octahedral arrangement of electron pairs about Xe:

Therefore we say that Xe has a d2sp3 hybridization.

Covalent Bond Strengths - Bond Enthalpies

In a chemical reaction we'll have bonds broken, bonds formed, and energy either absorbed or emitted by the reaction. Let's look closer at the enthalpy change associated with each individual bond broken or formed.

Process ΔH (kJ/mole)

CH4(g) → CH3(g) + H(g) 435CH3(g) → CH2(g) + H(g) 453CH2(g) → CH(g) + H(g) 425

CH(g) → C(g) + H(g) 339total = 1652

If we wanted to know what is the enthalpy change associated with breaking a C-H bond we find that

it is slightly dependent on what molecule it is in. Thus, we take an average change in enthalpy when

a C-H bond breaks as DC-H = 1652/4 kJ/mole = 413 kJ/mole.

Other average bond enthalpy changes in kJ/mole found this way are...

Bond ΔH (kJ/mole) Bond ΔH (kJ/mole)

H-H 432 C-H 413H-F 565 C-C 347H-Cl 427 C-N 305

C-O 358

Notice how multiple bonds are shorter and require more energy to break than single bonds.

Bond ΔH (kJ/mole) Bond Length

C=C 614 1.37ÅC≡C 839 1.20ÅC-N 305 1.43ÅC=N 615 1.38ÅC≡N 891 1.16Å

We can use these average bond enthalpy changes to calculate the approximate enthalpy change for

reactions. For example, to calculate the change in enthalpy for the following reaction:

H2(g) + F2(g) → 2 HF(g)

we identify and count all the bonds that are broken (shown in red) and formed (shown in blue).

H—H + F—F → 2 H—F

Substituting the average bond enthalpies: DH-H = 432 kJ/mole, DF-F = 154 kJ/mole, DH-F = 565 kJ/mole in the expression

ΔH = Σ (ΔH of bonds broken) - Σ (ΔH of bonds formed)

we obtain

ΔH = [(1 mole) (432 kJ/mole) + (1 mole) (154 kJ/mole)] - [(2 moles) (565 kJ/mole)] = -544 kJ

Ionic Solids

When cations and anions precipitate out of a saturated solution they crystallize into a lattice

arrangement that maximizes the attractive forces between cations and anions while minimizing the

repulsive forces between ions of the same charge. For example, shown below is the arrangement of

ions found in crystalline sodium chloride.

The chloride ions are represented as green, and the sodium ions as yellow in this figure. To better

understand the energetics of this arrangement let's examine Coulomb's Law for the potential energy

of interaction between two charges:

In NaCl we have three types of Coulombic interactions, an attractive interaction between sodium

cations and chloride anions, and repulsive interactions between sodium cations and between

chloride anions.

For a mole of solid NaCl, the sum of all these Coulombic energies for all cations and anions in a crystal is -861 kJ/mole.

Na+(g) + Cl-

(g)→NaCl(s)

ΔU = -861 kJ/mole

Lattice Energy- change in energy when completely gaseous atoms are brought together into 1

mole of a solid ionic substance.

As we see from Coulomb's law, the lattice energy is related to the charge and distance between ions

in the solid. Smaller the ions will have smaller distances between them in a solid, and thus have

higher lattice energies.

Born-Haber Cycle

Lattice Energies can also be measured experimentally using Hess's law in what is called the Born-

Haber Cycle. Recall that Hess's law tells us that the change in energy when going from reactant to

products is the same whether the reaction takes place in one step or in a series of steps. For

example, given the enthalpy changes for the following reactions we can obtain the enthalpy change

for the formation of the crystalline NaCl lattice.

Cl-(g) → Cl(g) + e- ΔH = -E.A.(Cl) = 349

kJ/mole

Na+(g) + e- → Na(g)

ΔH = -E.A.(Na) = -496 kJ/mole

Na(g) → Na(s) ΔH = -104.8 kJ/moleCl(g) → 1/2 Cl2(g) ΔH = -120.5 kJ/mole

Na(g) + 1/2 Cl2(g)

→ NaCl(s) ΔH = -411.2 kJ/mole

--------------------------------------------------------------------------------------------------------

Na+(g) + Cl-

(g) → NaCl(s) ΔH = -783.5 kJ/mole

That is pretty close to our calculated number of -861 kJ/mole (< 10% error). But, of course, our first

calculation was based on infinitely seperated ions at 0 K, and the Born-Haber cycle is based on

reaction enthalpy changes at room temperature and pressure, so we expect some differences.

Chemical Reactivity

Hydrogen

Unlike the rest of the Group 1A elements, which exist as metals, elemental hydrogen exists as

gaseous H2 molecules. Compounds formed between hydrogen and non-metals are molecular rather

than ionic. (i.e., hydrogen forms covalent bonds with non-metals). For example, hydrogen reacts

with halogens (Group VIIA) according to:

H2 (g) + X2 → 2 HX(g)

where X can be any halogen, such as F, Cl, Br, or I. Hydrogen in these compounds has an oxidation

state of +1 while the halogens are -1. Similarly, hydrogen reacts to other elemental non-metals in a

predictable fashion:

2 H2 (g) + O2(g) → 2 H2O(g)

8 H2 (g) + S8(s) → 8 H2S(g)

3 H2 (g) + N2(g) → 2 NH3(g)

Hydrogen can also form compunds with more active metals to form ionic hydrides. For example,

lithium hydride is formed according to:

2 Li(s) + H2 (g) → 2 LiH(s)

The metal (Li in this case) loses an electron to become a cation and H gains an electron to

become H- (hydride anion), which has an charge of -1. Here's another example:

Mg(s) + H2 (g) → MgH2(s)

By gaining an electron, the hydride ion obtains the stable electron configuration of a

closed n=1 shell, that is, the noble gas configuration of He.

Oxygen

Oxygen is a group 6A element. Elemental Oxygen is found in two forms: oxygen gas (O2) and and

ozone gas (O3). Different forms of an element in the same state are called Allotropes.

Reactions between Oxygen and Metals

When oxygen reacts with most metals a metal oxide is formed where oxygen has an oxidation state

of -2. For example, zinc oxide is formed when zinc metal reacts with oxygen gas:

2 Zn(s) + O2 (g) → 2 ZnO(s)

and aluminum oxide is formed when aluminum metal reacts with oxygen gas:

4 Al(s) + 3 O2 (g) → 2 Al2O3(s)

There are, however, some exceptions, which we consider next.

Group IA Metals - Alkali Metals

Because alkali metals are so active, the product of their reaction with oxygen gas is not what you

might expect. While lithium metal reacts with oxygen gas to form lithium oxide, as one might

expect:

4 Li(s) + O2 (g) → 2 Li2O(s) ,

when sodium metal reacts with oxygen gas under the same conditions it forms sodium peroxide:

2 Na(s) + O2 (g) → Na2O2(s)

and the very active alkali metals, potassium, rubidium, and cesium, reacts with oxygen gas to form

superoxides:

K(s) + O2 (g) → KO2(s)

Group IIA Metals - Alkaline Earth Metals

Oxygen reacts with most alkaline earth metals to form a metal oxide:

2 M(s) + O2 (g) → 2 MO(s)

For example,

Ca(s) + O2 (g) → CaO(s)

However, oxygen combines with barium metal, the most active of this group, to form a peroxide:

Ba(s) + O2(g) → BaO2 (s)

Reactions between Oxygen and Non-metals (except Group 7A and 8A)

When oxygen combines with non-metals in their elemental form, the product is a non-metal oxide.

For example, oxygen reacts with solid carbon to form carbon monoxide or carbon dioxide,

respectively, as shown below (reaction not balanced):

C(s) + O2 (g) → CO(g) or CO2(g)

Similarly, oxygen reacts with solid phosphorus to form tetraphorphorus heptoxide or

tetraphorphorus decoxide, respectively, as shown below (reaction not balanced):

P4(s) + O2 (g) → P4O6(g) or P4O10(g)

When reacted with solid sulfur, oxygen forms sulfur dioxide gas:

S8(s) + 8 O2 (g) → 8 SO2(g)

Oxide Reactions

Non-Metal Oxide reactions

The oxides of non-metals are acidic. If a non-metal oxide dissolves in water, it will form an acid.

Non-Metal Oxide + Water → Acid

For example,

SO3(g) + H2O(l) → H2SO4(aq)

N2O3(g) + H2O(l) → HNO2(aq)

The non-metal oxides can be neutralized with a base to form a salt and water.

Non-Metal Oxide + Base → Salt + Water

For example,

SO3(g) + Ba(OH)2(aq) → BaSO4(aq) + H2O(l)

P4O10(s) + 12 NaOH(aq) → 4 Na3PO4(aq) + 6 H2O(l)

Metal Oxide reactions

The oxides of metals are basic. If a metal oxide dissolves in water, it will form a metal hydroxide.

Metal Oxide + Water → Metal Hydroxide

For example,

BaO(s) + H2O(l) → Ba(OH)2(aq)

K2O(s) + H2O(l) → 2 KOH(aq)

Like any base, these bases can be neutralized by an acid to form a salt and water.

Metal Oxide + Acid → Salt + Water

Examples:

CuO(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2O(l)

Al2O3(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2O(l)

Generally, the more metallic character an element has, the more basic its oxide will be. Likewise,

the more non-metalic character an element has, the more acidic its oxide will be. The metalic

character of an element can be determined by its position on the periodic table:

Finally, we note that a salt can also be formed by the direct reaction of a metal and a non-metal.

Metal + Non-Metal → Salt

For example,

2 Al(s) + 3 Br2(l) → 2 AlBr3(s)

Question 1.

A brand of gasoline can be shown to be a mixture by:

a) reacting it with oxygen

b) burning it

c) separating it into its components

d) determining its density

e) smelling it

Question 2.

Which of the following is always true when a substance undergoes a physical change?

a) Its density remains the same.

b) It changes color.

c) Its composition remains the same.

d) It boils.

e) A new substance is formed.

Question 3.

Which of the following statements is correct?

a) Compounds cannot be decomposed into elements.

b) Neither elements nor compounds are pure substances.

c) Both elements and compounds are pure substances.

d) Compounds, but not elements are pure substances.

e) Elements, but not compounds are pure substances.

Question 1.

The term that is used to indicate the reproducibility of a measurement is:

a) precision

b) accuracy

c) quantitative

d) qualitative

Question 2.

A crucible is known to weigh 24.3162 g. Three students in the class determine the weight of the crucible by repeated weighings on a simple balance. Using the following information, which student has done the most accurate determination?

Trial 1 Trial 2 Trial 3 Trial 4 Trial 5A 24.8 24.0 24.2 24.1 24.3B 24.5 24.3 24.5 24.4 24.3C 24.8 24.9 24.8 24.9 24.8

a) Student A has done the most accurate work.

b) Student B has done the most accurate work.

c) Student C has done the most accurate work.

Question 3.

A crucible is known to weigh 24.3162 g. Three students in the class determine the weight of the crucible by repeated weighings on a simple balance. Using the following information, which student has done the most precise determination?

Trial 1 Trial 2 Trial 3 Trial 4 Trial 5A 24.8 24.9 24.8 24.9 24.8B 24.8 24.0 24.2 24.1 24.3C 24.5 24.1 24.5 24.1 24.3

a) Student A has done the most precise work.

b) Student B has done the most precise work.

c) Student C has done the most precise work.

Question 1.

How many significant figures are in the number 0.01020?

a) four

b) six

c) five

d) two

e) three

Question 2.

How many significant figures are in the number 0.010 x 105?

a) two

b) one

c) three

d) five

e) four

Question 3.

How many significant figures are in the number 1.0001 x 10-4?

a) five

b) one

c) two

d) four

e) three

Question 1.

Express the quantity 785.4 g in kg rounded to the nearest 0.01 kg.

a) 0.078

b) 0.79

c) 0.0785

d) 0.079

e) 0.78

Question 2.

Round off the quantity 0.042349 m to the nearest 0.01 cm.

a) 42.3

b) 4.23

c) 4.235

d) 42.35

e) 4.24

Question 3.

Round off the quantity 785.4 g to the nearest 0.1 kg.

a) 0.785

b) 0.8

c) 1.0

d) 7.8

e) 0.79

Question 1.

Calculate the following to the correct number of significant figures.

2123.1 + 1.2340 = _____

a) 2124.3340

b) 2124

c) 2124.33

d) 2124.334

e) 2124.3

Question 2.


212.1 - 89.342 = _____

a) 122.7

b) 122.758

c) 123

d) 122.76

e) 122.8

Question 3.


3.466 - 0.2342 = _____

a) 3.23

b) 3.2318

c) 3.7002

d) 3.700

e) 3.232

Question 1.


23.1 x 0.0012 = _____

a) 0.028

b) 0.0277

c) 0.2772

d) 0.027

e) 0.03

Question 2.


2.34 / 0.12 = _____

a) 2.0 x 101

b) 19

c) 19.5

d) 0.28

e) 0.281

Question 3.


2.301 / 0.0120 = _____

a) 192

b) 191.75

c) 191.7

d) 191.8

e) 191

Question 1.


23.1 x 0.12 = _____

a) 2.8

b) 28

c) 2.77

d) 2.8 x 102

e) 2.772

Question 2.


81.25 x 0.0014 = _____

a) 0.11

b) 0.1137

c) 0.114

d) 0.1138

e) 0.1

Question 3.


0.023 / 0.123 = _____

a) 0.19

b) 0.187

c) 0.186

d) 0.18

e) 0.1870

Chemistry Note

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