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AP Chemistry Chapter 3 The Structure of the Atom

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AP Chemistry Chapter 3 The Structure of the Atom. Beginning of the Atomic Model. Democritus was the early (around 400BC) Greek philosopher who is credited with the concept of the atom ( atomos ) –which means invisible. - PowerPoint PPT Presentation
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Page 1: AP Chemistry Chapter 3 The Structure  of the Atom

AP Chemistry

Chapter 3

The Structure of the Atom

1

Page 2: AP Chemistry Chapter 3 The Structure  of the Atom

Beginning of the Atomic Model

2

Page 3: AP Chemistry Chapter 3 The Structure  of the Atom

Democritus was the early (around 400BC) Greek philosopher who is credited with the concept of the atom (atomos) –which means invisible

3

Page 4: AP Chemistry Chapter 3 The Structure  of the Atom

Dalton (around 1800AD) is an English school teacher who proposed the law of conservation of mass, the law of definite proportions, and the law of multiple proportions.His many experimentswith gases proved these laws are true, ifatoms exist. Dalton is also known as the Father of the (Modern) Atomic Theory 4

Page 5: AP Chemistry Chapter 3 The Structure  of the Atom

Dalton’s atomic theory:1.All matter is composed of very small

particles called atoms2.Atoms of a given element are

identical in size, mass, and other properties; atoms of different elements differ in these properties.

5

Page 6: AP Chemistry Chapter 3 The Structure  of the Atom

3.Atoms cannot be subdivided, created, or destroyed

4.Atoms of different elements combine in simple whole-number ratios to form chemical compounds.

5.In chemical reactions, atoms are combined, separated, or rearranged.

6

Page 7: AP Chemistry Chapter 3 The Structure  of the Atom

Two aspects of Dalton’s atomic theory proven to be incorrect:

a.We now know atoms are divisible.

b. Atoms of the same element can have different masses.

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Page 8: AP Chemistry Chapter 3 The Structure  of the Atom

J. J. Thomson is the man credited with the discovery of the electrons in the late 1800’s, using cathode ray tubes

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Millikan calculated the mass of the electron (very, very small)

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Knowledge of electrons led to two inferences about atomic structure:

1.Because atoms are electrically neutral, they must contain positive charge to balance the negative electrons.

2. Because electrons have so little mass, atoms must contain other particles to account for most of their mass

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Page 11: AP Chemistry Chapter 3 The Structure  of the Atom

Nucleus of the atom—discovered by Lord Ernest Rutherford

Gold foil experiment—actually done by Hans Geiger and Ernest Marsden

11

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Observations:a.Majority of the alpha (α) particles

penetrated foil undeflected.

b.About 1 in 20,000 were slightly deflected

c.About 1 in 20,000 were deflected back to emitter

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Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus

2.Most of the atom is empty

3.Magnitude of charge on the nucleus is different for different atoms 14

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4. Number of electrons outside the nucleus = number of units of nuclear charge (to account for the fact that the atom is electrically neutral)

Atoms are electrically neutral because they contain equal numbers of protons and electrons

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Page 16: AP Chemistry Chapter 3 The Structure  of the Atom

A couple years later Rutherford presented evidence for a neutral particle which was also in the nucleus and contained a similar mass to that of a proton – called a neutron

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Mass of one proton = mass of neutron = mass of 1837 electrons

Thus the total mass of an atom is basically the sum of the protons and neutrons, called the atomic mass or mass number, abbreviated A

17

Page 18: AP Chemistry Chapter 3 The Structure  of the Atom

Atomic number—the number of protons in the nucleus of the atom.

--number of protons identifies the element and is equal to the number of

electrons (of a neutral atom)

--symbol is Z

18

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Isotopes are atoms of the same element that have different masses because they have different numbers of neutrons but they still have similar chemical properties

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Mass Number of Isotope

Number of Protons Number of Neutrons

8 6 29 6 3

10 6 411 6 512 6 613 6 714 6 815 6 916 6 1017 6 1118 6 1219 6 1320 6 14

Isotopes of Carbon

20

Page 21: AP Chemistry Chapter 3 The Structure  of the Atom

Isotopes of CarbonMass Number of Carbon

IsotopesName of Isotopes

8 carbon-89 carbon-9

10 carbon-1011 carbon-1112 carbon-1213 carbon-1314 carbon-1415 carbon-1516 carbon-1617 carbon-1718 carbon-1819 carbon-1920 carbon-20 21

Page 22: AP Chemistry Chapter 3 The Structure  of the Atom

Nuclide is the general term for any isotope of any element

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Nuclide is the general term for any isotope of any element

Atomic Mass Unit (amu) is exactly 1/12 the mass of a carbon-12 atom

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Page 24: AP Chemistry Chapter 3 The Structure  of the Atom

Average atomic mass is the weight average of the atomic masses of the naturally occurring isotopes of an element.

Ave. Atomic mass = %abundace(mass of isotope 1) + %abundance(mass of isotope 2) +…..

24

Page 25: AP Chemistry Chapter 3 The Structure  of the Atom

Example 1:Element Sciencium has two isotopes.

Sciencium-301 has an abundance of 59.5%, and Sciencium-304 is the other. What is the average atomic mass?

301 amu x .595 = 179 amu 304 amu x .405 = 123 amu

302 amu

Page 26: AP Chemistry Chapter 3 The Structure  of the Atom

Example 2:Element Pepsium has an average

atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?

327 amu x 0.305 = 99.7 amu ? amu x 0.695 = ? amu

335 amu

1 – 0.305 = 0.695

Page 27: AP Chemistry Chapter 3 The Structure  of the Atom

Example 2:Element Pepsium has an average

atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?

327 amu x 0.305 = 99.7 amu ? amu x 0.695 = ? amu

335 amu

Step 1: Find the missing weighted mass

Page 28: AP Chemistry Chapter 3 The Structure  of the Atom

Example 2:Element Pepsium has an average

atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?

327 amu x 0.305 = 99.7 amu ? amu x 0.695 = 235 amu

335 amu

335 – 99.7 = 235.3

Page 29: AP Chemistry Chapter 3 The Structure  of the Atom

Example 2:Element Pepsium has an average

atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?

327 amu x 0.305 = 99.7 amu ? amu x 0.695 = 235 amu

335 amu

Step 2: Find the missing mass

Page 30: AP Chemistry Chapter 3 The Structure  of the Atom

Example 2:Element Pepsium has an average

atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?

327 amu x 0.305 = 99.7 amu 338 amu x 0.695 = 235 amu

335 amu

235 ÷ 0.695 = 338

Page 31: AP Chemistry Chapter 3 The Structure  of the Atom

Mass Spectrometry

How we know isotopes.

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A mass spectrometer has three parts:1. ionizer2. magnetic field3. detector

And might look like this

Page 33: AP Chemistry Chapter 3 The Structure  of the Atom

If a sample of a pure element is placed in the spectrometer, then all the ions formed will have the same charge. For example Cl → Cl-

Cl Cl-

Ionizer

Page 34: AP Chemistry Chapter 3 The Structure  of the Atom

The ions then pass through a magnetic field that will change their paths. Which will change direction more, something heavy or something light?

Cl Cl-

Magnetic Field

-

+

Cl-

Detector

Page 35: AP Chemistry Chapter 3 The Structure  of the Atom

The computer attached to the detector gives a readout like this:

Page 36: AP Chemistry Chapter 3 The Structure  of the Atom

The locations tell the masses – one group of Cl had a mass of 35 amu’s and the other had a mass of 37 amu’s.

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The size of the bars indicate the relative amounts of each isotope.

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As the bar for 35Cl is 3 times bigger than 37Cl (measuring with a ruler) thus 35Cl is about 75% abundant and 37Cl is about 25% abundant.

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Unstable Nuclei and Radioactive Decay

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1) When referring to nuclear reactions people commonly think of nuclear fission (the splitting of large atoms into smaller pieces)

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1) and nuclear fusion (the combining of small atoms into one large one), but on earth these reactions do not occur naturally.

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2) Naturally occurring nuclear reactions result from the unusual number of neutrons of an isotope which makes it unstable (unusually high in energy). This often results in the isotope changing from one element into another element in an attempt to become more stable (lower in energy).

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A)These reactions are called nuclear reactions, as they involve changes in the nucleus.

B)During these nuclear reactions, rays and particles are given off, which is called radiation. 43

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C) Sometimes an unstable nucleus will change into many different elements as it tries to become more stable. This is called radioactive decay.

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3) When radioactive decay occurs, there are three different types of radiation that can be given off. Each type has a different mass, and sometimes a charge.

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A) The first type of radiation to be discovered was called alpha radiation and came from alpha particles.

226 Ra → 222 Rn + ?? ??88 86 ??

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i) Because the mass numbers must be equal, 226 = 222 + x. So the mass of the alpha particle must be 4.

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ii) Because the atomic numbers must be equal, 88 = 86 + x. So the atomic number of the alpha particle must be 2.

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iii) The element Helium has a mass of 4 and an atomic number of 2, so the alpha particle is just like a helium atom without any electrons;

4He or 4

2 249

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B) The second type of radiation to be discovered was called beta radiation and came from beta particles.

14 C → 14 N + ?? ??6 7 ??

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i) Because the mass numbers must be equal, 14 = 14 + x. So the mass of the beta is zero.

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ii) Because the atomic numbers must be equal, 6 = 7 + x. So the atomic number must be -1.

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iii) The electron has a mass of zero and a charge of -1, so the beta particle is just like an electron;

0-1 β

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C) The last type of radiation to be discovered was called gamma radiation and came from gamma particles.

238 U → 234 Th + 4 He + ?? ??92 90 2 ??

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i) Because the mass numbers must be equal, 238 = 234 + 4 + x. So the mass of the gamma particle must be zero.

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ii) Because the atomic numbers must be equal, 92 = 90 + 2 + x. So the atomic number must also be zero.

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iii) The gamma particle was the last to be found because it has no mass and no charge;

00 γ

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The Development of a New Atomic Model

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Previously, Rutherford reshaped our thoughts of the atom by showing the protons were located in the nucleus of the atom, but he could not model for us where the electrons were, other than outside the nucleus somewhere. Fortunately, studies into the properties of light and the effects of light on matter soon gave clues to where electrons actually are.

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Page 60: AP Chemistry Chapter 3 The Structure  of the Atom

Light is a small part of all the radiation (something that spreads from a source) called electromagnetic radiation. Electromagnetic radiation is energy in the form of waves (of electric and magnetic fields). Electromagnetic radiation includes radio waves, microwaves, infrared, visible light, X-rays, and Gamma rays. All these together are considered the Electromagnetic Spectrum.

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As all the forms of electromagnetic radiation are waves, they all have similar properties.• All electromagnetic radiation travels at the

speed of light (c), 299,792,458 m/s (3 x 108) in a vacuum

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•The crest is the top of the waves, the trough is the bottom of the waves, and the amplitude is a measurement from the rest or zero line to a crest or trough

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•The wavelength (λ – lambda) is the distance between successive crests/troughs and is measured in meters (often nm = 1 x 10-9 m)

•The frequency (ν – nu) is the number of waves that pass a point in one second and is measured

in (per second – can be written as s-1) or

Hz (Hertz)

1s

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How many hertz is the first wave?1 wave per second = 1 Hz

How many hertz is the second wave?2 waves per second = 2 Hz

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The speed of a wave is directly proportional to the wavelength and the frequency; c = λν is the formula

c

λ ν67

Page 68: AP Chemistry Chapter 3 The Structure  of the Atom

Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?

68

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Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?

ν =cλ

ν =3.00 × 108 m/s

413 nm

WAIT, This won’t work!

69

Page 70: AP Chemistry Chapter 3 The Structure  of the Atom

Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?

ν =cλ

ν =3.00 × 108 m/s

413 nm

413 nm 1 m=

70

Page 71: AP Chemistry Chapter 3 The Structure  of the Atom

Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?

ν =cλ

ν =3.00 × 108 m/s

4.13 x 10-7 m

413 nm 1 m= 4.13 x 10-7 m

1 x 109 nm

71

Page 72: AP Chemistry Chapter 3 The Structure  of the Atom

Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?

ν =cλ

ν =3.00 × 108 m/s4.13 × 10-7 m

ν = 7.26 × 1014 Hz

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Page 73: AP Chemistry Chapter 3 The Structure  of the Atom

Unfortunately, thinking of light as waves lead to a problem. It was noticed that if light strikes a metal, then sometimes it could cause electrons to be emitted (leave the atoms entirely – like in a solar panel); called the photoelectric effect. If light was a wave, then all amounts of light energy should cause this to happen, but this was not the case. It always took some minimum amount of energy to get the electrons to be emitted.

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This lead Max Planck to theorize that light must carry energy in basic minimum amounts that he called quanta. Like a delivery person cannot correctly deliver half a box, the electrons in atoms cannot gain a fraction of a quantum of energy (it has to be in whole numbers).

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He proposed that this energy was directly proportional to the frequency of the electromagnetic radiation and a constant, now called Planck’s constant. E = h ν

E = energy in Joules (J)

h = Planck’s constant = 6.626 × 10-34 Js

ν = frequency in Hz or 1/s

E

h ν76

Page 77: AP Chemistry Chapter 3 The Structure  of the Atom

Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?

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Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?

Note: wavelength is not in the energy equation, but frequency is. So first, you must solve for the frequency. As seen in the earlier example, a wavelength of 413 nm gives a ν = 7.26 × 1014 Hz.

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Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?

ν = 7.26 × 1014 Hz

E = h × ν

E = 6.626 × 10-34 Js × 7.26 × 1014 1/s

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Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?

ν = 7.26 × 1014 Hz

E = h × ν

E = 6.626 × 10-34 Js × 7.26 × 1014 1/s

E = 4.81 × 10-19 J

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In 1905 Einstein used Plancks work to propose that electromagnetic radiation had a dual wave-particle nature. As a particle, electromagnetic radiation carries a quantum of energy of energy, has no mass, and is called a photon.

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So to get an electron to emit from a metal, it must be struck with a photon having quantum energy big enough, or nothing will happen. Each metal requires a different quantum energy, thus each metal can be identified by the frequency of light needed to emit electron.

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This idea was expanded upon to develop an idea of where the electrons were in an atom. It was found that low pressure gases could be trapped in a tube and electrified, and would then glow a color particular to the gas inside.

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Furthermore this light could be passed into a prism, and instead of getting the entire spectrum (rainbow) of colors, only certain wavelengths of light would be seen as small bars of color, called a line-emission spectrum.

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This would indicated that the electrons in an atom were only absorbing specific amounts of energy from the electricity, causing the electrons to move from their ground state (normal position close to the nucleus) to an excited state (higher energy position further away from the nucleus). The electrons do not stay in the excited state for long and fall back to their ground state, losing the energy equal to what they gained.

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Niels Bohr used this to develop a model of the atom where the electrons could only be in certain, specific energy level (n) orbits around the nucleus. Just as you cannot go up half a rung on a ladder, the electron could not go up a partial energy level. The electrons gained or lost enough energy to move a whole number amount of energy levels (n) away from or closer to the nucleus, or it did not move.

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He calculated the amount of energy needed for an electron of hydrogen to move between each energy level (n) (which was not constant) and his calculations agreed with experimental results.

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The Balmer series of hydrogen spectral lines refer to the four lines seen in the visible light region (the four colored bars). If the electron was excited to energy level (n) 6, 5, 4, or 3 and fell to energy level (n) 2, the resulting energy given off would have a frequency in the visible region of electromagnetic radiation. (One line for dropping from 6 to 2, one for 5 to 2, one for 4 to 2, and one for 3 to 2).

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However, there are other possibilities. If the electrons drop from n=6, 5, or 4 to n=3, then the energy given off is not big enough to be seen as it is in the infrared region. These three lines in the infrared region are referred to as the Paschen series. If the electrons drop to n=1, then the five lines given off are too high in energy to be seen, as they are in the ultraviolet region. These lines are referred to as the Lyman series.

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Model of Atom Review:

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1. Thomson’s Plum Pudding Model – the atom is a ball of evenly spread positive stuff with random negative particles (electrons).

2. Rutherford’s Nuclear Model – the atom has a central nucleus containing the positive particles (protons) with the electrons outside.

3. Bohr’s Orbital Model – The electrons circle the nucleus in specific energy orbits, like the planets orbit the sun. Unfortunately this only works for atoms with one electron…

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4. Quantum Mechanical Model – electrons are found in specific regions around the nucleus, but the exact location of the electrons inside the regions cannot be determined

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The quantum mechanical model starts with a Principal Quantum Number (n), which is the basic energy level of an electron, and often matches the period number. Possible values (currently) are 1-7.

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The quantum mechanical model starts with a Principal Quantum Number (n), which is the basic energy level of an electron, and often matches the period number. Possible values (currently) are 1-7.

Inside the principal quantum energy level are sublevels that correspond to different cloud shapes. The sublevels are designated as s (sharp), p (principal), d (diffuse), and f (fundamental).

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Inside the sublevels are orbitals, specific regions with a 90% probability of finding electrons.

• s –orbitals are spherically shaped clouds around the nucleus

• p -orbitals are bar-bell shaped clouds with the nucleus between the lobes

• d and f are much more complex in shape

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Each sublevel has room for a different amount of electrons, because an orbital can hold two electrons, then each sublevel has a different amount of orbitals

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• s –sublevel can hold 2 electrons, so it has 1 orbital (shape)

• p –sublevel can hold 6 electrons, so it has 3 orbitals (shapes)

• d –sublevel can hold 10 electrons, so it has 5 orbitals (shapes)

• f –sublevel can hold 14 electrons, so it has 7 orbitals (shapes)

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The s sublevel is simply a sphere centered on the nucleus.                                                                                           

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The p sublevel has three orbitals.  These are often referred to a dumbbell shape.

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The d sublevel has five orbitals:

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The f sublevel has seven orbitals

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To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2

if n=1, then

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To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2

if n=1, then 2 electrons will fit

if n=4,

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To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2

if n=1, then 2 electrons will fit

if n=4, then 32 electrons will fit

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In order to show on paper where electrons are likely to be located in an atom, orbital filling diagrams and electron configurations are drawn or written. When this is done, three rules must be followed:

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1. Aufbau principle – electrons fill lower energy levels first, thus 1 before 2 and s before p, etc.

a. orbitals within a sublevel are equal in energy (called degenerate)

b. the principal energy levels often overlap, making them seem a little out of order

c. boxes are used to represent orbitals

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Another way of writing the aufbau principle diagram:

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7s 7p 108

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2. Pauli Exclusion principle – an orbital (box) can hold a maximum of two electrons (arrows)

a. for two electrons to fit, they have to have opposite spins

b. for one electron in the orbital

c. for two electrons in the orbital (opposite spins)

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3. Hund’s Rule – when electrons occupy degenerate orbitals, one electron is placed into each orbital with parallel spins before doubling up

Ex. _____ _____ _____ NOT _____ _____ _____

3p 3p

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Please navigate to http://intro.chem.okstate.edu/WorkshopFolder/Electronconfnew.html

And click through all the elements. Make note of which ones are exceptions to the Aufbau principle, and where they are located in the periodic table.

When the d sublevel get close in energy to the s sublevel the electrons from the s sublevel repel each other and one ends up in the d, even though it is slightly higher in energy.

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We separate the orbitals from each other to be able to talk about the electrons’ locations easier, but in a real atom all the electrons and orbitals exist at once, which might look like this:

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Orbital Notation shows the arrows in the boxes to represent the electrons in an atom. To shorten this process, an electron configuration can be written. It leaves out the information about the number of orbitals in each sublevel, so it will be expect you remember that information.

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It has the general form nl°n = principal quantum number (1-7…)

l = sublevel letter (s, p, d, or f)

° = number of e- in that orbital (1-14)

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Ni = 28 e-

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Ni = 28 e-

1s2 2s2 2p6 3s2 3p6 4s2 3d8

2 + 2 + 6 + 2 + 6 + 2 + 8 = 28

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Sn = 50 e-

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Sn = 50 e-

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2

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If writing out the entire electron configuration is too much, we can use the previous (in the periodic table) noble gas to take the place of part of the electron configuration:

Polonium:1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p4

Xenon:1s22s22p63s23p64s23d104p65s24d105p6

Polonium: [Xe] 6s24f145d106p4

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When the electron configuration is written for an element using the noble gas configuration the electrons written after the noble gas are the ones that appear on the outside of the atom, called valence electrons..

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When elements bond to form compounds, it is these electrons that are involved. The amount of valence electrons makes a big difference in how the element will bond, so to make it easy to predict, we draw electron dot diagrams.

A) In an electron dot diagram, we use the symbol of the element and dots to represent the number of valence electrons.

B) Only s and p electrons with the highest quantum number count for dot diagrams, even if there are d and f electrons after the noble gas.

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Lithium = 1s22s1

So Li

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Beryllium = 1s22s2

So Be

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Boron = 1s22s22p1

So B

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Carbon = 1s22s22p2

So C

125

Page 126: AP Chemistry Chapter 3 The Structure  of the Atom

Nitrogen = 1s22s22p3

So N

126

Page 127: AP Chemistry Chapter 3 The Structure  of the Atom

Oxygen = 1s22s22p4

So O

127

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Fluorine = 1s22s22p5

So F

128

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Neon = 1s22s22p6

So Ne

129

Page 130: AP Chemistry Chapter 3 The Structure  of the Atom

Periodic Law – properties of the elements are a periodic function of their atomic number

The moon’s phases and magazine subscriptions are also periodic…

Page 131: AP Chemistry Chapter 3 The Structure  of the Atom

The earliest and most successful arrangement of the elements was by Dmitri Mendeleev in the 1870s. He arranged the elements known at that time by their physical and chemical properties into groups.

Page 132: AP Chemistry Chapter 3 The Structure  of the Atom

His arrangements of the elements left some gaps, which he claimed were elements not yet discovered. With his table he predicted the characteristics of these missing elements, and was correct.

Page 133: AP Chemistry Chapter 3 The Structure  of the Atom

During the 1910s, Henry Moseley used x-ray spectra to determine the atomic number (number of protons) for elements and proved that each element had a different amount of protons.

Page 134: AP Chemistry Chapter 3 The Structure  of the Atom

Moseley rearranged the elements based on atomic number, as it is arranged today, and found gaps which he also claimed were undiscovered elements. Some of these elements were found quickly, while others were not found until after his death due to their highly radioactive (unstable) nature.

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There are three key items in understanding and explaining all the trends (and exceptions) on the periodic table.

Page 136: AP Chemistry Chapter 3 The Structure  of the Atom

There are three key items in understanding and explaining all the trends (and exceptions) on the periodic table.

1. Effective Nuclear Charge

Page 137: AP Chemistry Chapter 3 The Structure  of the Atom

There are three key items in understanding and explaining all the trends (and exceptions) on the periodic table.

1. Effective Nuclear Charge2. Energy Levels

Page 138: AP Chemistry Chapter 3 The Structure  of the Atom

There are three key items in understanding and explaining all the trends (and exceptions) on the periodic table.

1. Effective Nuclear Charge2. Energy Levels3. Coulombs Law

Page 139: AP Chemistry Chapter 3 The Structure  of the Atom

1. Effective Nuclear Charge – pull of the protons in the nucleus on the valence (outer) energy level electrons

Page 140: AP Chemistry Chapter 3 The Structure  of the Atom

1. Effective Nuclear Charge – pull of the protons in the nucleus on the valence (outer) energy level electrons

The greater the atomic number, the greater the number of protons, and the greater the effective nuclear charge.

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Effective Nuclear Charge has the greatest effect moving from left to right across a period (it increases).

Page 142: AP Chemistry Chapter 3 The Structure  of the Atom

Effective Nuclear Charge has the greatest effect moving from left to right across a period (it increases).

Which element has more effective nuclear charge, P or S?

Page 143: AP Chemistry Chapter 3 The Structure  of the Atom

Effective Nuclear Charge has the greatest effect moving from left to right across a period (it increases).

Which element has more effective nuclear charge, P or S?

Which element has more effective nuclear charge, S or Se? (Why won’t this work?)

Page 144: AP Chemistry Chapter 3 The Structure  of the Atom

2. Energy levels - the principal quantum level of the electrons, sometimes called shells

Page 145: AP Chemistry Chapter 3 The Structure  of the Atom

2. Energy levels - the principal quantum level of the electrons, sometimes called shells

As elements increase in atomic number they also increase in the number of electrons. These electrons occupy higher and higher energy levels.

Page 146: AP Chemistry Chapter 3 The Structure  of the Atom

Higher numbered energy levels are farther away from the nucleus.

1s

1s2s1s2s3s

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Energy Levels have the greatest effect moving from top to bottom within a group (it increases).

Page 148: AP Chemistry Chapter 3 The Structure  of the Atom

Energy Levels have the greatest effect moving from top to bottom within a group (it increases).

Which element has more energy levels, S or Se?

Page 149: AP Chemistry Chapter 3 The Structure  of the Atom

Energy Levels have the greatest effect moving from top to bottom within a group (it increases).

Which element has more energy levels, S or Se?

Which element has more energy levels, P or S? (Why won’t this work?)

Page 150: AP Chemistry Chapter 3 The Structure  of the Atom

3. Coulombs Law states that the force of attraction between things is directly proportional to the size of the charge and inversely proportional to the square of the distance between them.

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For the following trends, you must know the general trends (memorize), but using the three key items you must be able to explain the trend as well (understand).

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1. Atomic Radiussize of the atom

a. increases from top to bottom within a group

Why does it increase from top to bottom?

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1. Atomic Radiussize of the atom

b. decreases from left to right within a period

Why does it decrease from left to right?

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Page 155: AP Chemistry Chapter 3 The Structure  of the Atom

2. Ionization Energy energy required to remove an

electron from a gaseous atoma. increases bottom to top

Why is it easier to remove electrons from atoms at the bottom of the P.T.?

Page 156: AP Chemistry Chapter 3 The Structure  of the Atom

2. Ionization Energy energy required to remove an

electron from a gaseous atomb. increases left to right

Why is it easier to remove electrons from atoms at the left of the P.T.?

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Page 158: AP Chemistry Chapter 3 The Structure  of the Atom

2. Ionization Energy energy required to remove an

electron from a gaseous atomc. This is sometimes called metallic

character, as metals tend to lose electrons easily.

Which element would exhibit the most metallic character?

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Page 160: AP Chemistry Chapter 3 The Structure  of the Atom

3. Electron Affinityenergy released when an electron is added to a gaseous atom

a. increases bottom to top

Why is more energy released when electrons are added to atoms at the top of the P.T.?

Page 161: AP Chemistry Chapter 3 The Structure  of the Atom

3. Electron Affinityenergy released when an electron is added to a gaseous atom

b. increases left to right

Why is more energy released when electrons are added to atoms at the right of the P.T.?

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Page 163: AP Chemistry Chapter 3 The Structure  of the Atom

3. Electron Affinityenergy released when an electron is added to a gaseous atom

c. This is sometimes called nonmetallic character, as nonmetals tend to gain electrons easily.

Which element would exhibit the most nonmetallic character?

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Page 165: AP Chemistry Chapter 3 The Structure  of the Atom

4. Electronegativity measure of an atom’s pull on another

atom’s electronsa. increases bottom to top

Why are atoms at the top of the P.T. able to pull stronger on a different atom’s electrons?

Page 166: AP Chemistry Chapter 3 The Structure  of the Atom

4. Electronegativity measure of an atom’s pull on another

atom’s electronsb. increases left to right

Why are atoms at the right of the P.T. able to pull stronger on a different atom’s electrons?

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Page 168: AP Chemistry Chapter 3 The Structure  of the Atom

5. ionscharged atoms

a. cations – positive ion formed from an atom losing an electron

The ionic radius is always smaller than the original atom.

Why?

Page 169: AP Chemistry Chapter 3 The Structure  of the Atom

5. ionscharged atoms

b. anions – negative ion formed from an atom gaining an electron

The ionic radius is always larger than the original atom.

Why?

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Page 171: AP Chemistry Chapter 3 The Structure  of the Atom

Let’s summarize the trends. Remember you must know the trend (memorize) but you also need to be able to explain why the trend exists.

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Atomic Radius

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Ionization Energy, Electron Affinity, Electronegativity

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Effective Nuclear ChargeEn

ergy

Lev

els

Page 175: AP Chemistry Chapter 3 The Structure  of the Atom

Metallic Character

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Nonmetallic Character