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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

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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%

• xygen, and 12% carbon by mass.

Law of Definite Composition

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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)

Law of Definite Composition Slide 8 / 121

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

Sample Size Sample location % mass composition 1 500.0 g San Fransisco Lab 88.9% O 11.1 % H 2 330.0 g Toronto Lab 88.9% 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.

Law of Definite Composition 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.

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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

• xygen, 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

• f hydrogen would be present in a pure 230. gram

sample of hydrogen peroxide?

Answer

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1 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams

• f hydrogen would be present in a pure 230. gram

sample of hydrogen peroxide?

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Answer 13.6 g

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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?

Answer

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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?

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Answer 30.7 g

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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

• f 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

Answer

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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

• f 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

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Answer NO

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This law allows us to classify the different types of matter

MATTER Does the material obey the Law of Definite composition? Yes No Pure 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

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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

• f 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 Law of Multiple Proportions and Atomic Theory

The 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)

• xygen (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

• f oxygen required

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

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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
• xide 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
• xide 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

• f the oxygen atoms in oxide A compared to oxide B.

Law of Multiple Proportions and Atomic Theory

g Cu reacted g oxygen reacted g of copper oxide made

• xide A

4.3 g ? 4.84 g

• xide B

7.5 g ? 9.38 g

Slide 20 / 121 Law of Multiple Proportions and Atomic Theory

The 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 = Cu

2O Oxide B = CuO

Formula Set Two: (half the oxygen atoms) Oxide A = CuO Oxide B = CuO

2

It took many more experiments to determine which formulas were

• correct. We will leave those until later.

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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 Answer

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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

Answer

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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

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Answer C

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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

Answer

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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

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Answer C

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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 Answer

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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

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Answer A

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worked solution

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

• xygen gas

grams of

• xide

produced

• xide A

2.3 4.6

• xide B

1.4 3.5

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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

• f the mass (88.9%) is due to oxygen. One oxygen atom MUST be a

lot heavier than a hydrogen atom! water

• xygen = 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

• ther 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.

Determination of Atomic Masses 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

• xygen atom?

Determination of Atomic Masses 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)

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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 Intensity 100 A B C

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"

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0 10 20 30 40 Intensity 100 A B C

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 mass

Using 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

• f 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 B

What 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

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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: Intensity A B

Peak A: 28.0 u , Intensity = 100 Peak B: 44.0 u , Intensity = 23

0 10 20 30 40 50 m/z 100 Peak A = N

2 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

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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

Answer

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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

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Answer E

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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

Answer

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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

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Answer D

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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

Answer

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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

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Answer D

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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 Answer

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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

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Answer B

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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

Intensity m/z 10 20 B

100 A

Peak A: 12.0 u , Intensty = 100 Peak B: 13.0 u , Intensity = 1.08

Answer

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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

Intensity m/z 10 20 B

100 A

Peak A: 12.0 u , Intensty = 100 Peak B: 13.0 u , Intensity = 1.08

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Answer C

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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

Intensity 0 10 20 30 40 50 60 70 80 100 A B ?

m/z

Peak A: 78.92 u , Intensity = 100 Peak B: 80.92 , Intensity = ?

Answer

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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

Intensity 0 10 20 30 40 50 60 70 80 100 A B ?

m/z

Peak A: 78.92 u , Intensity = 100 Peak B: 80.92 , Intensity = ?

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Answer C

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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

• 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

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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 nm 486 nm 434 nm 410 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.

1 2 3 4 5 N 1 2 3 4 5 Energy

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

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n = 1 n = 2 n = 3

+

Hydrogen atom

n = 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 atom

n = 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
• rbits, 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 atom

n = 4 n = 1 n = 2 n = 3

+

Hydrogen atom

n = 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

• rbits. Therefore an electron will always emit the same

frequencies it absorbs.

Bohr Model of Atom

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15 Which of the following is equal to the mass number

• f 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

Answer

Slide 55 (Answer) / 121

15 Which of the following is equal to the mass number

• f 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

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Answer A

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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

Answer

Slide 56 (Answer) / 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

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Answer B

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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

• rbits becomes greater as the value of N increases

Answer

Slide 57 (Answer) / 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

• rbits becomes greater as the value of N increases

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Answer E

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18 In order to move hydrogen's lone electron from the ground state (n=1) to an excited state (n=3), a photon

• f 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 Answer

Slide 58 (Answer) / 121

18 In order to move hydrogen's lone electron from the ground state (n=1) to an excited state (n=3), a photon

• f 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

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Answer frequency = 2.78 x 1014 1/s wavelength = 108 nm UV

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19 What must be the difference in energy between Bohr

• rbits n=3 and n=2 if an electron emits light with a

wavelength of 656 nm when undergoing this transition?

Answer

Slide 59 (Answer) / 121

19 What must be the difference in energy between Bohr

• rbits n=3 and n=2 if an electron emits light with a

wavelength of 656 nm when undergoing this transition?

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Answer

3.03 x 10-19 J

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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

Answer

Slide 60 (Answer) / 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

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Answer C

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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 Intensity 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?

Photoelectron Spectroscopy

Three questions arise:

1.31 2.37

Slide 63 / 121

Let's answer each of these questions.

Energy Intensity 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.

Photoelectron Spectroscopy

1.31 2.37

Slide 64 / 121

Let's answer each of these questions.

Energy Intensity He H

• 2. Why is the signal stronger for

helium than for hydrogen?

Photoelectron Spectroscopy

Helium has two electrons to hydrogen's one.

1.31 2.37

Slide 65 / 121

Let's answer each of these questions.

Energy Intensity He H

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

Photoelectron Spectroscopy

• 3. Why is there only one signal for

helium even though it has two electrons?

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

Energy Intensity 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?

Photoelectron Spectroscopy

Again, questions arise:

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• 1. Why does lithium have two peaks?

This tells us that lithium has electrons in two different Bohr

• rbits. 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 Intensity He H Li Li

n= 1 n= 2

Photoelectron Spectroscopy

Let's answer each of these questions. H He Li

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• 2. What can explain the difference in

energy between the two peaks?

Energy Intensity He H Li Li

n= 1 n= 2 Less energy is required to remove electrons from the n=2

• rbit as it is farther from the nucleus thereby reducing the

coulombic attraction.

Photoelectron Spectroscopy

Let's answer each of these questions.

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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 Intensity He H Li Li

n= 1 n= 2

Photoelectron Spectroscopy

Let's answer each of these questions.

• 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

• f 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 intensity n=1 n=2 n=?

Slide 71 / 121 PES and Limitations of Bohr Model

energy intensity n=1 n=2 n=? 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 PES and Limitations of Bohr Model

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?

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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.

PES and Limitations of Bohr Model 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

Answer

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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

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Answer C

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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

Answer

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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

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Answer E

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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

Answer

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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

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Answer C

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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

• f electrons in each.

E All of these are true

energy intensity A B C

Answer

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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

• f electrons in each.

E All of these are true

energy intensity A B C [This object is a pull tab]

Answer E

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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 Slide 79 / 121

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.

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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

• rbital as 1 etc.)

0 # # # ( n # 1)

for n = 3: # = 0, 1, 2 (s, p, d) Magnetic Quantum Number (Projection of An gular Momentum ) m#

energy shift (orientation of the subshell's shape) ## # m# # #

for # = 2: m# = # 2, # 1, 0, 1, 2 Spin Projection Quantum Number ms

spin of the electron (# ½ = "spin down", ½ = "spin up") #s # ms # s

for an electron s = ½ so ms = # ½ or ½

The Quantum Model

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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

• ccupy 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)

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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

• f the Bohr orbits

D The failure of the Bohr model to explain the stability

• f the nucleus

E None of these

Answer

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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

• f the Bohr orbits

D The failure of the Bohr model to explain the stability

• f the nucleus

E None of these

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Answer C

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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

Answer

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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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Answer C

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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

Answer

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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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Answer C

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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

Answer

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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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Answer D

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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

• rbitals 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 move for answer *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

Answer

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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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Answer C

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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

Answer

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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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Answer C

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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

Answer

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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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Answer E

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32 How many TOTAL electrons would exist in "p"

• rbitals in a ground state chlorine atom?

A 5 B 10 C 11 D 12 E 17

Answer

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32 How many TOTAL electrons would exist in "p"

• rbitals in a ground state chlorine atom?

A 5 B 10 C 11 D 12 E 17

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Answer C

Slide 98 / 121 Core vs. Valence Electrons

Core 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

• utermost 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 Diagrams

Unlike 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.

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33 Which of the following noble gases represents the core electrons of vanadium (V)? A Ne B O C Ar D Kr E Ca

Answer

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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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Answer C

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34 How many valence electrons are present in a zinc atom? A 12 B 10 C 8 D 6 E 2

Answer

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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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Answer E

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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

Answer

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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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Answer E

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36 How many unpaired electrons are present in the nitride ion (N3-)? A 0 B 1 C 2 D 3 E 5

Answer

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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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Answer A

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37 Which of the following atoms would have 2 valence electrons? A Zn B Mg C Li D Si E Both A and B

Answer

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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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Answer E

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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 Model

As we earlier saw, PES data indicated a more complex view

• f 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 intensity 1s2 2s2 2p1

Both have same intensity b/c of same #

• f 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 intensity 1s2 2s2 2p1

0.80 1.36 19.3

Slide 109 / 121 PES Support of the Quantum Model

Examine 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 move for answer

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

energy intensity A B C

Answer

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

energy intensity A B C

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Answer B

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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

Answer

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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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Answer B

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40 The PES spectrum of sulfur (S) should have how many distinct peaks? A 1 B 3 C 4 D 5 E 6

Answer

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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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Answer D

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41 Which of the following elements could have produced the PES spectrum below? A He B Li C Na D Mg E Ne

energy intensity

Answer

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41 Which of the following elements could have produced the PES spectrum below? A He B Li C Na D Mg E Ne

energy intensity

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Answer D

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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

Answer

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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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Answer A

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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

Answer

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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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Answer E

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

Answer

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

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Answer FALSE

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45 Which of the following would NOT be paramagnetic? A Sc3+ B Rh C W D O E They are all paramagnetic

Answer

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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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Answer A

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Now that we have a strong understanding of the structure

• f the atom and how it was determined, we will turn our

attention this next unit to the periodic table, it's

• rganization, and how we can use it to predict various

properties and behaviors of atoms.

Preview of Coming Events! Slide 121 / 121