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www.njctl.org

The Atom

AP Chemistry




Slide 3 / 121

Deducing the structure of the atom took a lot of brainpower. . .

and even more funky hair styles

Slide 4 / 121

Birth of Atomic Theory

A number of key discoveries led to the current understanding that the atom is the basic building block of matter.

1. Law of Definite Composition

2. Law of Multiple Proportions

Slide 5 / 121

If a compound is pure, it will always consist of the same composition no matter where the sample was taken or the size of the sample.

Law of Definite Composition

Example: calcium carbonate

Slide 6 / 121

Sample Location Size Analysis Composition

1 Eastern Pennsylvania 50.0 g

20.0 g Ca

24.0 g O

6.0 g C

40% calcium

48% oxygen

12% carbon

2 Wyoming 250.0 g

100.0 g Ca

120.0 g O

30.0 g C

40 % calcium

48% oxygen

12% carbon

Example: calcium carbonate

If it's calcium carbonate, it's guaranteed to be 40% calcium, 48% oxygen, and 12% carbon by mass.


Slide 7 / 121

Some substances are not pure and do not obey the law of definite composition. These are called mixtures.

Example: Pure water (pure substance) vs. Salt water (mixture)


Slide 8 / 121

Example: Pure water (pure substance) vs. Salt water (mixture)

Sample Size Sample location

% mass composition

1 500.0 gSan

Fransisco Lab

88.9% O

11.1 % H

2 330.0 g Toronto Lab88.9% O

11.1% H

Sample Size Sample location

% mass composition

1 500.0 g Atlantic Ocean

85.3% O

10.7 % H

1.6% Na

2.4 % Cl

2 330.0 g Indian Ocean

79.5 % O

10.0 % H

4.2 % Na

6.3 % Cl

Pure water Salt water

Since the % composition of the water doesn't change, we know it is a pure substance. Since salt water varies in its % composition,

it violates the law of definite composition and is a mixture.


Slide 9 / 121

400 grams of water x 0.889 g O = 355 g O

1 g water

Example 1: Water is known to be 88.9% oxygen and 11.1 % hydrogen by mass. How many grams of oxygen would be present in a 400 gram sample of pure water?

Using the Law of Definite Composition

The law of definite composition can be used mathematically to see how much of a given substance is present in a sample.

Slide 10 / 121

The law of definite composition can be used mathematically to see how much of a given substance is present in a sample.

Example 2: If a sample of water was found to contain 34.1 grams of oxygen, how many grams of hydrogen and water would be present?

34.1 g O x 0.111 g H = 4.26 g H

0.889 g O

34.1 g O x 1 g water = 38.4 g water

0.889 g O

Using the Law of Definite Composition

Slide 11 / 121

1 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams of hydrogen would be present in a pure 230. gram sample of hydrogen peroxide?

Ans

wer

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2 Calcium oxide is 71.4% calcium by mass with the rest being oxygen. How many grams of calcium would be present if a sample of calcium oxide sample was found to contain 12.3 grams of oxygen?

Ans

wer

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3 Calcium carbonate is known to be 40% Ca, 48% O, and 12% carbon by mass. When a 200 gram sample of what is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the sample. Is this substance pure?

Yes

No Ans

wer

Slide 14 / 121

This law allows us to classify the different types of matterMATTER

Does the material obey the Law of Definite composition?

Yes NoPure Substance Mixture

Compounds Elements

Can the material be broken down into different elements with

distinct properties?

Yes No

Law of Definite Composition and the Classification of Matter

Slide 15 / 121

The easiest way to explain the law of definite proportions is if we picture matter as being made of atoms.

If a given substance consists of a certain number of each type of atom, this would explain why the % composition

would always be the same for that substance.

For example, since water always has two oxygens for every hydrogen, it's % composition by mass must always

be the same.

Law of Definite Composition and Atomic Theory

Slide 16 / 121

Things got particularly interesting when they examined pure substances that were made out of the same elements.

Water Hydrogen peroxide

88.9 % by mass O 94.1

11.1 % by mass H 5.9

Since the % composition of each substance was different, atomic theory would suggest this was due to each substance consisting

of different numbers of atoms of each element.

Law of Definite Composition and Atomic Theory

Slide 17 / 121

When we examine the RATIO of these % amounts of each element, we find something very interesting.

Water Hydrogen peroxide

88.9 % by mass O 94.1

11.1 % by mass H 5.9

88.9/11.1 = 8/1 ratio of O/H 94.1/5.9 = 16/1

WHOA!!!! Hydrogen peroxide has exactly twice the ratio of oxygen by mass compared to water.

The easiest way to explain this is if we assume that matter is composed of atoms, and hydrogen peroxide has either exactly twice the number of oxygen atoms or half the number of hydrogen atoms

as water does!

Law of Multiple Proportions and

Atomic Theory

Slide 18 / 121


Atomic TheoryThe law of multiple proportions states that when a fixed amount of one

element is reacted with another to form different compounds of the same two elements, the ratio of the masses of the second element that reacts to form each compound, can always be expressed as a whole

number.

Example: Carbon dioxide and Carbon monoxide

Substance carbon (g) oxygen (g)

carbon dioxide 1 2.66

carbon monoxide 1 1.33

Exactly twice the amount of oxygen was needed to make carbon dioxide compared to carbon monoxide!

Notice that the ratio of oxygen required

to form the compounds was 2.66/1.33 = 2/1.

Slide 19 / 121

Example: Below is some data from a laboratory experiment in which two different oxides of copper were produced. Demonstrate they obey the

law of multiple proportions.

1. Use the law of conservation of mass to find the g of O

oxide A = 4.84 - 4.3 = 0.54 g O oxide B = 9.38 - 7.5 = 1.88 g O

2. Find the ratio of Cu to O for both

oxide A = 4.3/0.54 = 8:1 oxide B = 7.5/1.88 = 4:1

The ratio of copper found in both oxides is clearly a whole number multiple 8:4 or 2:1. There are exactly twice the copper atoms or half

of the oxygen atoms in oxide A compared to oxide B.


Atomic Theory

g Cu reacted g oxygen reacted g of copper oxide made

oxide A 4.3 g ? 4.84 g

oxide B 7.5 g ? 9.38 g

Slide 20 / 121


Atomic TheoryThe law of multiple proportions allowed scientists to hypothesize as

to how many atoms of each kind may be in a compound.

Let's look at the oxides of copper we just examined:

Oxide A had either twice the copper atoms or half the oxygen atoms. Based on this, one can propose a series of possible formulas!

Formula Set One:(twice the copper atoms)

Oxide A = Cu2O Oxide B = CuO

Formula Set Two: (half the oxygen atoms)

Oxide A = CuO Oxide B = CuO2

It took many more experiments to determine which formulas were correct. We will leave those until later.

Slide 21 / 121

4 Two samples of a material are taken and the composition of each sample is given below. Is this material a pure substance?

Yes

No

Sample A Sample B

45 % Cu, 12% Si, 43 % O 34% Cu, 19% Si, 47 % O

Ans

wer

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5 A certain material is found to vary in composition by mass. What kind of matter is this?

A Element

B Compound

C Mixture

D Pure Substance

E A, B, and D

Ans

wer

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6 Which of the following is TRUE regarding a pure compound?

A It will not obey the law of definite composition

B It can be broken down into different elements

C The amounts of each element by mass in the compound will not vary

D It must contain the same equal mass % of each element in the compound

E None of these are true

Ans

wer

Slide 24 / 121

7 Which set of compounds demonstrate the law of multiple proportions?

A N2O, NO

B CO2, CS2

C CH4, CF4

D IBr, HBr

E NH3, PH3

Ans

wer

Slide 25 / 121

wor

ked

solu

tion

8 A student analyzes two oxides of sulfur. Does the data they collected demonstrate the law of multiple proportions?

Yes

No

Sample

grams of S reacted with

excess oxygen gas

grams of oxide

produced

oxide A 2.3 4.6oxide B 1.4 3.5

Slide 26 / 121

The masses of atoms were first determined on a relative basis. For example in water, we know that water has the formula H2O yet most of the mass (88.9%) is due to oxygen. One oxygen atom MUST be a

lot heavier than a hydrogen atom!

water oxygen = 88.9% 2 x hydrogen = 11.1 %

Since each hydrogen would contribute just (11.1/2) or 5.55 % of the mass...each oxygen must be 88.9/5.55 = 16 x heavier than a

hydrogen atom.

All we have here though is a relative mass, not an absolute mass! This does NOT tell us how heavy each atom actually is.

Determination of Atomic Masses

Slide 27 / 121

Historically, hydrogen was assigned a mass value of 1 and all of the other elements were compared to it. The usage of hydrogen as a

standard made sense as it was known to be the lightest of the elements.

Later, it was proposed that the standard be defined as exactly 1/16 the mass of an oxygen atom. This standard was abandoned after the

discovery of isotopic oxygen in 1929.


Slide 28 / 121

Class Question:

Why would the discovery of isotopes be problematic for those trying to create a standard mass scale based on the mass of a hydrogen or

oxygen atom?


Slide 29 / 121

As we know, the kind of element an atom is - is defined by it's number of protons or atomic number. However, atoms of the same element need not have the same number of neutrons - these are isotopes and since they differ in the number of neutrons, they will

differ in mass.

O - 17: 8 protons, 9 neutrons

O - 18: 8 protons, 10 neutrons

As a result, each sample of oxygen is a mixture of it's isotopes so we can only really say that a mixture of oxygen atoms is about 16 x

more massive than an equal mixture of H atoms!

How can we figure out the relative mass of a particular atom, not a mixture of them?

Isotopes and Atomic Masses

Slide 30 / 121

To determine a standard to be used to determine the mass of a particular atom, it was necessary to agree on a particular isotope to

use.

It is now agreed that the standard for all atomic masses is equal to exactly 1/12 the mass of a C-12 atom - that is the isotope of carbon

with 6 protons and 6 neutrons.

All of the masses on the periodic table have been normalized to this number.

Unified Atomic Mass Unit (u)

Slide 31 / 121

Determining the relative atomic masses of individual atoms or molecules is possible now with the advent of mass spectrometry.

A mass spectrometer is an instrument designed to determine the relative mass and abundance of charged ions of a particular

substance.

For example, a mass spectrometer can effectively determine the mass and abundance of the isotopes of a given element!

Relative Atomic Masses

Slide 32 / 121

1. The sample is ionized (made a charged ion)

2. The now ionized sample is accelerated into a magnetic field.

3. Based on where the ion hits the detector, the relative mass to charge ratio and abundance of each ion can be determined.

heavier ions are deflected less

lighter ions are deflected more

How does a Mass Spectrometer Work?

Slide 33 / 121

For a given sample, a mass spectrum will be produced showing the intensity of signal (abundance) of the ion on "y" axis and the relative

mass (u) of the ion on the "x" axis.

Example: The following is the mass spectrum of a sample of neon.

0 10 20 30 40

Inte

nsity

100

0

A

BC

Peak A: Mass (u) = 19.99 , Intensity = 100

Peak B: Mass (u) = 21.99 , Intensity = 10.22

Peak C: Mass (u) = 20.99 , Intensity = 0.29

m/z (mass/charge) = u

The three peaks represent the three isotopes of neon. They are not all the same intensity because of the different

natural abundances of each isotopes.

How to read a "mass spectrum"

Slide 34 / 121

0 10 20 30 40

Inte

nsity

100

0

A

BC

Peak A: Mass (u) = 19.99 , Intensity = 100

Peak B: Mass (u) = 21.99 , Intensity = 10.22

Peak C: Mass (u) = 20.99 , Intensity = 0.29

The intensities can be converted into % abundances of each isotope.

Ne -19.99 = 100/110.51 = 90.5 %

Ne -21.99 = 10.22/110.51 = 9.25%

Ne -20.99 = 0.29/110.51 = 0.25%

How to read a "mass spectrum"

Slide 35 / 121

Determining an average atomic massUsing the % abundances and masses from the mass spectrometer,

we can determine the average atomic mass of a given element. Remember, the average atomic mass is a the sum of the products of

the mass of each isotope and it's abundance.

Let's use our data for neon. Ne -19.99 = 100/110.51 = 90.5 %

Ne -21.99 = 10.22/110.51 = 9.25%

Ne -20.99 = 0.29/110.51 = 0.25%

19.99(.905) + 21.99(.0925) + 20.99(0.0025) = 20.18 u

This is the mass you see on the periodic table!

*Note - the mass written on the periodic table is the average of the stable isotopes of an element!!

Slide 36 / 121

Class Practice: Determining an average atomic mass

The following is a mass spectrum of an element.

A BWhat element is this?

Calculate the % abundance of both isotopes:

Find the average atomic mass.

Peak A: 10.01 u , Intensity = 24.8

Peak B: 11.01 u, Intensity = 100

Slide 37 / 121

Mass spectrometry has uses that go beyond isotopic study. Finding the precise mass of a molecule can aid in determining

it's identity and abundance .

Using Mass Spectrometry to Identify Unknown

Slide 38 / 121

Let's look at the mass spectrum below for a mixture of carbon dioxide gas and nitrogen gas:

Inte

nsity A

B0

Peak A: 28.0 u , Intensity = 100

Peak B: 44.0 u , Intensity = 23

0 10 20 30 40 50m/z

100

Peak A = N2 gas and = 100/123 = 0.81 x 100 = 81% of the mixture.

Peak B = CO2 gas and = 23/123 = 0.19 x 100 = 19% of the mixture.

*Note: These mass spectra are simplified. In many cases, compounds are split into pieces by the impact of electrons when

ionized. This creates many smaller peaks on the spectrum but this is beyond the scope of the course.

Using Mass Spectrometry to Identify Unknown

Slide 39 / 121

9 The unified mass unit is defined as:

A The mass of a hydrogen atom

B The mass of a proton

C Exactly 1/16 of an oxygen atom

D Exactly 1/12 of a perfect mixture of carbon atoms

E Exactly 1/12 of a C-12 atom

Ans

wer

Slide 40 / 121

10 Which of the following would be TRUE of isotopes of an element?

A They are found in equal abundances

B They differ only in the number of protons

C They will have the same mass

D They will have the same atomic number

E None of these are true

Ans

wer

Slide 41 / 121

11 In order to be an isotope of boron, an atom must:

A Have a mass of 10.81

B Have a mass of 5

C Have 6 neutrons

D Have 5 protons

E Both B and C

Ans

wer

Slide 42 / 121

12 The mass spectrum of an element reveals two peaks, A and B. What is the average atomic mass of this element?

A 35.18 u

B 35.45 u

C

D 36.00 u

E 35.78 u

Peak A: 34.97 u , Intensity = 100

Peak B: 36.97 u , Intensity = 32

36.18 u

Ans

wer

Slide 43 / 121

13 What is the relative abundances of both stable isotopes of C given the following mass spectrum?

A 100 % C-12, 1.08 % C-13

B 1.08 % C-13, 100 % C-12

C 98.9% C-12, 1.1% C-13

D 97.6 % C-12, 2.4% C-13

E Cannot be determined from the data In

tens

ity

m/z 0 10 20

0B

100

A

Peak A: 12.0 u , Intensty = 100

Peak B: 13.0 u , Intensity = 1.08 Ans

wer

Slide 44 / 121

14 Bromine consists of two isotopes, Br-79 (78.92 u) and Br-81 (80.92 u). If the relative intensity of Br-79 was 100 on the mass spectrum, what would be the relative intensity of the Br-81 peak? Use the average atomic mass of bromine from the periodic table.

A 98.2

B 79.9

C 97.3

D 1.8

E 2.7

Inte

nsity

0 10 20 30 40 50 60 70 80

100

0

A

B

?

m/zPeak A: 78.92 u , Intensity = 100

Peak B: 80.92 , Intensity = ?

Ans

wer

Slide 45 / 121

As we know, atoms are composed of subatomic particles.

proton neutron electron

mass (kg) 1.67 x 10-27 1.67 x 10-27 9.11 x 10-31

mass (u) 1.007 1.008 0.000549

charge (C) 1.60 x 10-19 0 -1.60 x 10-19

The element to which an atom belongs is determined by it's atomic number (Z) or number of protons.

The mass number (A) is equal to the sum of neutrons and protons.

Subatomic Particles

Slide 46 / 121

Ernest Rutherford - via his scattering experiments - deduced that the protons

and neutrons are concentrated in a very

small area of high density within the atom called the

nucleus.

Rutherford (Nuclear Model)

Slide 47 / 121

alpha particle emitter

gold foil most alpha particles went

straight through indicating the atom is mostly empty space

Some alpha particles were deflected significantly leading scientists to believe that the positive charge within an atom

was congregated within a small area - the nucleus.

An alpha particle consists of 2 protons

and neutrons

electrons (-)

protons(+) and neutrons

Rutherford Model

Rutherford (Nuclear Model)

Slide 48 / 121

The key to understanding the behavior of the electrons in an atom came from examining the interaction of light with atoms.

Recall that light is a wave and that the wavelength, frequency, and energy of light are related by the following equations.

c = v E = hv

wavelength( ) and frequency(v) are inversely related

Energy (E) and frequency(v) are directly related

c = 3.00 x 108 m/s h = 6.626 x 10-34 J*s

Absorption and Emission Spectra

Slide 49 / 121

Atoms absorb and emit light only at specific frequencies or wavelengths. These frequencies can be seen in absorption and

emission spectra.

Visible Light Emission Spectrum of Hydrogen

656 nm486 nm434 nm410 nm

Visible Absorption Spectrum for Hydrogen

Absorption and Emission Spectra

Slide 50 / 121

Neils Bohr and other scientists realized that the wavelengths of the emission spectral lines were proportional to some variable "n".

No one knew what "n" was. Bohr proposed that "n" referred to a particular orbit around the nucleus where an electron could be.

Balmer Series (spectral lines in the visible and UV range)

Lyman Series (spectral lines in the UV range)

Paschen Series (spectral lines in the infrared range)

Interpreting Emission Spectra

Slide 51 / 121

Bohr Model of Atom

Bohr proposed that the electrons are found in orbits of discrete energy found at specific distances away from the nucleus.

12

3 4 5N

1

2345

Ene

rgy

Notice that the difference in energy between the Bohr orbits becomes less as one moves farther from the nucleus.

Slide 52 / 121

n = 1

n = 2

n = 3

+

Hydrogen atomn = 4

Bohr reasoned that each spectral line was being produced by an electron "decaying" from a high energy Bohr orbit to

a lower energy Bohr orbit.

Only specific wavelengths of light are seen in an emission spectrum because electrons can only transition between

certain specific Bohr orbits.

Bohr Model of Atom

Slide 53 / 121

n = 1

n = 2

n = 3

+

Hydrogen atomn = 4 Due to coulombic attractions,

hydrogen's lone electron will exist in the orbit closest to the nucleus. We call this lowest energy state the

ground state.

Absorption spectra could be easily explained by this model also. Since electrons are only permitted in discrete energy orbits, they require specific amounts of energy to transition

from one to another.

Let's examine hydrogen's only electron.

Bohr Model of Atom

Slide 54 / 121

n = 1

n = 2

n = 3

+

Hydrogen atomn = 4

n = 1

n = 2

n = 3

+

Hydrogen atomn = 4

GROUND STATE EXCITED STATE

photon

In order to move to a higher energy state (excited state), a photon of light with the proper energy must be absorbed.

An electron will not remain in an excited state very long. Due to coulombic attractions, it will decay but only between allowed

orbits. Therefore an electron will always emit the same frequencies it absorbs.

Bohr Model of Atom

Slide 55 / 121

15 Which of the following is equal to the mass number of an atom?

A Protons and neutrons

B Protons, neutrons, and electrons

C Protons and electrons

D Neutrons and electrons

E The mass number is independent of the number of subatomic particles

Ans

wer

Slide 56 / 121

16 Which of the following observations, IF TRUE, would have rendered the nuclear model implausible?

A The observation that atoms produce emission spectra

B The observation that none of the alpha particles shot at the gold foil were deflected

C The observation that most of the alpha particles went through the gold foil

D The observation that some of the alpha particles were deflected

E The observation that atoms emit and absorb photons of light of the same frequency

Ans

wer

Slide 57 / 121

17 Which of the following is NOT a characteristic of the Bohr Model?

A Electrons orbit the nucleus in discrete energy levels

B Photons of light are emitted when electrons transition from a higher energy orbit to a lower one

C Only certain frequencies of light can be absorbed and emitted by an atom

D The nucleus consists of positive protons and neutral neutrons

E The difference in energy between adjacent Bohr orbits becomes greater as the value of N increases

Ans

wer

Slide 58 / 121

18 In order to move hydrogen's lone electron from the ground state (n=1) to an excited state (n=3), a photon of light with energy = 1.84 x 10-18 J must be absorbed. What would be the frequency and wavelength of this light?

Bonus: What kind of electromagnetic radiation would this be? Recall your EM spectrum

Ans

wer

Slide 59 / 121

19 What must be the difference in energy between Bohr orbits n=3 and n=2 if an electron emits light with a wavelength of 656 nm when undergoing this transition?

Ans

wer

Slide 60 / 121

20 Which of the following transitions would produce light of the longest wavelength?

A 3 --> 1

B 4 --> 3

C 5 --> 4

D 2 --> 1

E They would all produce the same wavelength of light

Ans

wer

Slide 61 / 121

The Bohr model of the atom was corroborated by evidence gained through photoelectron spectroscopy.

As we know, electrons are bound to an atom by coulombic forces. This can be referred to as the binding energy. Since the

electrons in Bohr orbit (n=1) are closer to the nucleus, the binding energy is high.

PES spectroscopy involves determining how much energy is required to remove an electron from an atom.

Photoelectron Spectroscopy

Slide 62 / 121

Let's start by looking at a PES spectrum of hydrogen

and helium.

Energy

Inte

nsity

He

H

1. Why is more energy required to remove an electron from helium than hydrogen?

2. Why is the signal stronger for helium than for hydrogen?

3. Why is there only one signal for helium even though it has two electrons?


Three questions arise:

1.31

2.37

Slide 63 / 121

Let's answer each of these questions.

Energy

Inte

nsity

He

H

1. Why is more energy required to remove an electron from helium than

hydrogen?

Helium's electrons experience stronger coulombic attractions due to the extra proton found in helium's nucleus.


1.312.37

Slide 64 / 121


Energy

Inte

nsity

He

H

2. Why is the signal stronger for helium than for hydrogen?


Helium has two electrons to hydrogen's one.

1.312.37

Slide 65 / 121


Energy

Inte

nsity

He

H

This was the interesting one. This indicated to everyone that both of helium's electrons were in the same shell or orbit!


3. Why is there only one signal for helium even though it has two

electrons?

Slide 66 / 121

Now, let's examine the PES spectrum for lithium - the third element on the periodic table.

Energy

Inte

nsity

He

H

Li

Li

1. Why does lithium have two peaks?

2. What can explain the difference in energy between the two peaks?

3. How do we compare the energy of the peaks for lithium to those of hydrogen and helium?


Again, questions arise:

Slide 67 / 121

1. Why does lithium have two peaks?

This tells us that lithium has electrons in two different Bohr orbits. The less energetic peak represents the electrons in the

N=2 orbit. This also tells us that the maximum number of

electrons in the 1st orbit is two.

Energy

Inte

nsity

He

HLi

Li

n= 1n= 2

Photoelectron SpectroscopyLet's answer each of these questions.

H He Li

Slide 68 / 121

2. What can explain the difference in energy between the two peaks?

Energy

Inte

nsity

He

HLi

Li

n= 1n= 2

Less energy is required to remove electrons from the n=2 orbit as it is farther from the nucleus thereby reducing the

coulombic attraction.


Slide 69 / 121

The n=1 peak is higher for lithium because of the additional proton in lithium's nucleus thereby creating a stronger coulombic attraction.

The n=2 peak is lower than that of hydrogen because this electron is farther from the nucleus than hydrogen's electron.

Energy

Inte

nsity

He

HLi

Li

n= 1n= 2


3. How do we compare the energy of the peaks for

lithium to those of hydrogen and helium?

Slide 70 / 121

PES and Limitations of Bohr Model

Examining the PES spectrum of boron revealed some limitations of the Bohr model. Bohr believed that the n=2 peak's energy

should increase with atomic number, that is the n=2 signal should get bigger and more energetic as this orbit added electrons.

Well, it did and it didn't .....

energy

inte

nsity n=1n=2n=?

Slide 71 / 121


energy

inte

nsity n=1n=2n=?

The peak to the left of the n=2 peak was unexpected. Based on the known periodic table and other calculations, Bohr anticipated that 8 electrons would exist in the 2nd Bohr orbit. The third peak implies that there are suborbits within the Bohr orbits - that the

atom is more complicated than it appears.

Slide 72 / 121


Carefully examine the actual PES spectra for the first 18 elements in the link below and answer the

questions.

http://www.chem.arizona.edu/chemt/Flash/photoelectron.html

1. What evidence do you see that chlorine has electrons in three distinct Bohr "orbits"?

2. What evidence do you see that 2s electrons are farther from the nucleus than 1s electrons?

3. What do you see in oxygen's PES spectra that indicates a more complex atom than Bohr envisioned?

Slide 73 / 121

A more complicated atom was also supported by the existence of additional "hyperfine" spectral lines not previously seen indicating that there are many more possible "orbits" for electrons than Bohr

had first proposed.

The Bohr Model explained a great many things but it would take seeing the electron a different way to develop

a model consistent with all of the data.


Slide 74 / 121

21 Which of the following influence the intensity of a PES peak?

A The atomic mass

B The atomic number

C The number of electrons in a specific orbit

D The energy used to ionize the atom

E Both B and C

Ans

wer

Slide 75 / 121

22 Which of the following influences the energy of a PES peak?

A Atomic mass

B Atomic number

C Which Bohr orbit electron is in

D The number of electrons in a particular Bohr orbit

E Both B and C

Ans

wer

Slide 76 / 121

23 Which of the following correctly explains why lithium demonstrates two peaks in it's PES spectrum?

A Lithium has more protons than either hydrogen or helium

B Lithium has a larger atomic mass than either hydrogen or helium

C Lithium has electrons in two different Bohr orbits

D Both A and B

E Both B and C

Ans

wer

Slide 77 / 121

24 Below is the PES spectrum for oxygen. Which of the following would be TRUE?

A The "C" peak should be of a higher energy than the "C" peak in nitrogen.

B More electrons are present in the orbit producing the "A" peak than in the others.

C Oxygen must have three orbits where electrons are present

D Orbits "B" and "C" must have identical numbers of electrons in each.

E All of these are true

energy

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B C

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As you will recall from earlier, a major shortcoming of the Bohr model was explaining why the electrons could

exist in a stable orbit despite the coulombic attraction of the nucleus.

de-Broglie proposed that if we view the electron as a wave, the stability of

the Bohr orbits could be explained.

The Beginnings of the Quantum Model

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In 1927 Davisson and Germer provided experimental evidence

for de-Broglie's work as they demonstrated that electrons

produce an interference pattern like X-rays (known to behave as a

wave) when shot at a crystal.

The Beginnings of the Quantum Model

This demonstrated that electrons had wave-like properties and these could be used to refine the model of the atom.

Slide 80 / 121

As we may recall from our first year course, if we view an electron as a standing wave, a set of 4 quantum numbers can be used to describe

the quantum state of that electron.

Name Symbol Orbital Meaning

Range of Values Value Examples

Principal Quantum Number n shell 1 # n n = 1, 2, 3, …

Azimuthal Quantum Number

(Angular Momentum)#

subshell (s orbital is listed as 0, p orbital as 1 etc.)

0 # # # (n # 1) for n = 3:# = 0, 1, 2 (s, p, d)

Magnetic Quantum Number

(Project ion of Angular Momentum)

m#

energy shif t (orientat ion of the subshell's shape)

## # m# # #for # = 2:

m# = # 2, # 1, 0, 1, 2

Spin Project ion Quantum Number ms

spin of the electron (# ½ = "spin down",

½ = "spin up")#s # ms # s

for an electron s = ½so ms = # ½ or ½

The Quantum Model

Slide 81 / 121

As we know, these quantum numbers describe possible quantum states for an electron. There is no guarantee that an electron will

be in any one of these at a given time.

Regions described by these numbers are therefore called orbitals to distinguish them from Bohr orbits.

The Quantum Model

Slide 82 / 121

The Quantum Model and the Aufbau Principle

Electrons will occupy the lowest energy quantum state available. Since the quantum states correlate to orbitals around an atom, we can

rephrase as electrons will occupy the lowest energy orbital possible.

Lowest energy quantum states

n =1 < 2 < 3..etc.

l= 0 (s orbital) < 1 (p orbital) < 2 (d orbital) < 3 (f orbital)

Therefore we know a 2s orbital will fill before a 3s orbital

Likewise a 3s orbital will fill before a 3p orbital

Slide 83 / 121

Quantum Numbers and the Pauli Exclusion Principle

The Pauli-Exclusion principle dictates that no two electrons can occupy the same quantum state.

Example: Lithium (3 electrons)

The lowest energy quantum states for the lithium electrons are: 1, 0, 0, 1/2 and 1,0,0,-1/2

(1s orbital) (1s orbital)

As adding another electron to the 1s orbital would violate the Pauli Exclusion principle, the next lowest available quantum

state for the third electron is:

2,0,0,1/2 or -1/2

The Pauli Exclusion principle limits the number of electrons in each orbital to two!

Slide 84 / 121

Quantum Numbers and Hund's Rule

2p (3 orientations or suborbitals)

2s (1 orientation or suborbital)

1s ( 1 orientation or suborbital)

The energy is lowered by having each electron in

the same spin state. This would not be possible if

they paired up.

Hund's rule dictates that the an atom is most stable when it has the most electrons in a particular spin state (+1/2 or -1/2).

The practical implication of this is that electrons will occupy as many equal energy suborbitals as possible before pairing up as this allows

them to share the same spin state.

Example: Nitrogen (7 electrons)

Slide 85 / 121

25 Which of the following problems of the Bohr model were explained by the de-Broglie wave hypothesis?

A The failure of the Bohr model to explain hyperfine spectral lines

B The failure of the Bohr model to explain complicated PES spectra

C The failure of the Bohr model to explain the stability of the Bohr orbits

D The failure of the Bohr model to explain the stability of the nucleus

E None of these

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Slide 86 / 121

26 Which of the following is/are NOT TRUE of the quantum model of the atom?

A The Pauli Exclusion principle limits the number of electrons in any suborbital to two

B The quantum state of an electron can be described by four quantum numbers

C An atom is most stable when there are few like spin state electrons

D The principal quantum number describes the main energy level of an electron

E All of these

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Slide 87 / 121

27 Which of the following would be the first orbital to fill according to the Aufbau principle?

A 3d

B 3p

C 3s

D 4s

E 4p

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Slide 88 / 121

28 Which of the following correctly describes the number of suborbitals/orientations in each orbital?

A s = 1

B p = 3

C d = 4

D Both A and B

E A, B, and C

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Slide 89 / 121

We can use our knowledge of quantum numbers to assign quantum states for all of the electrons in an atom.

Let's review a little first!

s orbitals (1 orientation) = fit 1x2 = 2 electrons

p orbitals (3 orientations) = fit 3x2 = 6 electrons

d orbitals (5 orientations) = 5x2 = 10 electrons

f orbitals (7 orientations) = 7x2 = 14 electrons

Each of the suborbitals are degenerate meaning they have the same energy.

Quantum Numbers and Electron Configurations

Slide 90 / 121

We can use the periodic table to allow us to determine which orbitals will fill first.

As we can see, orbitals will fill in the following order:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p....

Don't memorize this, use your periodic table!!

Quantum Numbers and Electron Configurations

Slide 91 / 121

What would be the electron configuration for the following:

Sodium (Na) 1s22s22p63s1

Selenium (Se) 1s22s22p63s23p64s23d104p4

Vanadium (V) 1s22s22p63s23p64s23d3

move for answer

move for answer

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*Note: If an atom is in an excited state, the aufbau rule will be violated and an electron will be occupying a higher than normal

quantum state. Example: 1s12s1

Electron Configurations of ground state atoms

Slide 92 / 121

Let's try a few!!

Sodium ion (Na+) 1s22s22p6

Selenide ion (Se2-) 1s22s22p63s23p64s23d104p6

Vanadium ion (V3+) 1s22s22p63s23p63d2 move for answer

move for answer

move for answer

Electron Configurations of ions

As we know from the first year course, atoms often gain or lose electrons to achieve a lower energy position. The electrons will always be lost or gained from the orbital farthest from the nucleus first - that is the orbital

with the highest principal quantum number (N)

Slide 93 / 121

Atoms or ions that have the same electron configuration are known as isoelectronic. Many ions are isoelectronic with the noble gases as

they are stable due to a full principal energy level.

Examples:

(Na+) , (O2-), (Mg2+), and (Ne) = 1s22s22p6

(Se2-), (Br-), (Sr2+), and (Kr) = 1s22s22p63s23p64s23d104p6

Isoelectronic Atoms and Ions

Slide 94 / 121

29 Which of the following atoms or ions would have an electron configuration ending as ….3p3?

A N

B N3-

C P

D P3-

E As

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Slide 95 / 121

30 Which of the following would represent the electron configuration for the Zr2+ ion?

A 1s22s22p63s23p64s23d104p65s24d2

B 1s22s22p63s23p64s23d104p65s2

C 1s22s22p63s23p64s23d104p64d2

D 1s22s22p63s23p64s23d104p65s24d4

E none of these

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31 Which of the following would NOT be isoelectronic with Li+?

A H-

B Be2+

C He

D B3+

E H

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Slide 97 / 121

32 How many TOTAL electrons would exist in "p" orbitals in a ground state chlorine atom?

A 5

B 10

C 11

D 12

E 17

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Core vs. Valence ElectronsCore electrons are the electrons found in all but the outermost principal energy level. These electrons are tightly bound to the

nucleus and are often not involved in chemical reactions.

The core electrons in an atom are often represented by the noble gas that is isoelectronic with the core electrons.

Example: Silicon 1s22s22p63s23p2 = [Ne]3s23p2

core electrons valence electrons

Valence electrons are electrons in the s and p orbitals of the outermost principal energy level. They are often involved in ion

formation and chemical reactions.

Slide 99 / 121

Shorthand Electron Configurations

Electron configurations in which only the outermost principal energy level is shown are known as shorthand electron configurations. The

appropriate noble gas is used to show the core electrons.

Some examples:

Na = [Ne]3s1

Fe = [Ar]4s23d6

Ba = [Xe]6s2

Slide 100 / 121

Orbital DiagramsUnlike an electron configuration which only shows the first two

quantum states of an electron, an orbital diagram shows all four quantum states.

For example, all suborbitals and spins are shown.

Example: Draw the correct orbital diagram for a ground state sulfur atom.

1s 2s 2p 3s 3p

*Note: Hund's rule is considered when filling the "3p" suborbitals.

Slide 101 / 121

33 Which of the following noble gases represents the core electrons of vanadium (V)?

A Ne

B O

C Ar

D Kr

E Ca

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Slide 102 / 121

34 How many valence electrons are present in a zinc atom?

A 12

B 10

C 8

D 6

E 2

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Slide 103 / 121

35 After drawing the orbital diagram for manganese (Mn), how many unpaired electrons would be present?

A 0

B 1

C 2

D 3

E 5

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Slide 104 / 121

36 How many unpaired electrons are present in the nitride ion (N3-)?

A 0

B 1

C 2

D 3

E 5

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Slide 105 / 121

37 Which of the following atoms would have 2 valence electrons?

A Zn

B Mg

C Li

D Si

E Both A and B

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Slide 106 / 121

The quantum model is supported by a wealth of experimental evidence.

We will focus particularly on two of these pieces:

1. PES - Photoelectron spectroscopy

2. Para and diamagnetic behavior

Experimental Support for the Quantum Model

Slide 107 / 121

PES Support of the Quantum ModelAs we earlier saw, PES data indicated a more complex view of the atom with additional orbitals within each energy level.

This is exactly what quantum theory shows.

Let's look at the PES spectrum of boron again!

binding energy

inte

nsity 1s22s22p1

Both have same intensity b/c of same # of electrons

What used to be a mystery peak can now be identified as the 2p orbital. It has half the intensity of the other peaks because of only having 1 electron in it.

0.80 1.36 19.3

Slide 108 / 121

PES Support of the Quantum Model

Can you explain why the 2p1 peak is of a lower binding energy than the 2s2 peak?

The 2p orbital is farther from the nucleus so experiences less coulombic attractions.

move for answer

Note: Each PES spectra for an element will have the same number of peaks as it does unique orbitals of different energies.

binding energy

inte

nsity 1s22s22p1

0.80 1.36 19.3

Slide 109 / 121

PES Support of the Quantum ModelExamine once again the link below to actual PES spectra of the

first 21 elements and answer the following questions.

http://www.chem.arizona.edu/chemt/Flash/photoelectron.html

Examine the PES spectra for carbon. What evidence do you see that each of the "p" orientations are degenerate?

Examine the PES spectra for scandium. How would you expect the spectra for iron (Fe) to be different?

There is only one peak for the "p" electrons

All of the peaks would have a higher binding energy due to Fe's higher "Z". The 3d peak will have an intensity 6x than that of Sc.

move for answer

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Slide 110 / 121

38 Referring to the PES spectrum of oxygen below, which peak corresponds to the 2s orbital?

A

B

C energyin

tens

ity AB C

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Slide 111 / 121

39 Compared to the PES spectrum of flourine, the 2p peak in the PES spectrum of oxygen should be…

A Less intense and of a higher energy

B Less intense and of a lower energy

C more intense and of a higher energy

D more intense and of a lower energy

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Slide 112 / 121

40 The PES spectrum of sulfur (S) should have how many distinct peaks?

A 1

B 3

C 4

D 5

E 6

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Slide 113 / 121

41 Which of the following elements could have produced the PES spectrum below?

A He

B Li

C Na

D Mg

E Ne

energy

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Slide 114 / 121

Additional experimental support for the quantum model of the atom comes from the way atoms behave when an external

magnetic field is applied.

Atoms with unpaired electrons are drawn to a magnetic field and are called paramagnetic.

Example: Lithium

1s 2s

Atoms with all of their electrons paired are repelled by the magnetic field and are called diamagnetic.

Example: Beryllium

1s 2s

Paramagnetism and Diamagnetism

Slide 115 / 121

The quantum model allows us to predict accurately which elements should be para and which should be diamagnetic.

For example, all elements that finish with a full s, p, d, or f orbital should be diamagnetic.

If an element has any unfilled orbitals, it MUST be paramagnetic.

Circle which of the following you would predict to be diamagnetic:

Ba O2- Na Kr Zn Fe

Circle which of the following you would predict to be paramagnetic:

He Na+ F F- Mn Cd Pb

Paramagnetism and Diamagnetism

Slide 116 / 121

42 A particular atom is found to be attracted to an external magnetic field. Which of the following could be the element to which the atom belongs?

A Se

B Mg

C Ar

D He

E None of these

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Slide 117 / 121

43 Which of the following atoms or ions would be diamagnetic?

A S2-

B Zn2+

C Zn

D A and B

E A, B, and C

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Slide 118 / 121

44 An atom with only 1 unpaired electron can be diamagnetic.

True

False

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Slide 119 / 121

45 Which of the following would NOT be paramagnetic?

A Sc3+

B Rh

C W

D O

E They are all paramagnetic

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Slide 120 / 121

Now that we have a strong understanding of the structure of the atom and how it was determined, we will turn our

attention this next unit to the periodic table, it's organization, and how we can use it to predict various

properties and behaviors of atoms.

Preview of Coming Events!

Slide 121 / 121

Slide 1 / 121 This material is made freely available at …content.njctl.org/courses/science/ap-chemistry/the-atom/...This material is made freely available at Slide 1 / 121 and is

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