ap chemistry chapter 5 models of the atom
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Democritus was the early (around 400BC) Greek philosopher who is credited with the concept of the atom (atomos) –which means invisible
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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
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.
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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.
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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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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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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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Nucleus of the atom—discovered by Lord Ernest Rutherford
Gold foil experiment—actually done by Hans Geiger and Ernest Marsden
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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 15
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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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
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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
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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 2
9 6 3
10 6 4
11 6 5
12 6 6
13 6 7
14 6 8
15 6 9
16 6 10
17 6 11
18 6 12
19 6 13
20 6 14
Isotopes of Carbon
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Isotopes of CarbonMass Number of Carbon
IsotopesName of Isotopes
8 carbon-8
9 carbon-9
10 carbon-10
11 carbon-11
12 carbon-12
13 carbon-13
14 carbon-14
15 carbon-15
16 carbon-16
17 carbon-17
18 carbon-18
19 carbon-19
20 carbon-20 22
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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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) +…..
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Example: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
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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. 31
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 2
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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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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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Section 1
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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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)
1
s
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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 ν56
Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?
ν =c
λ
ν =3.00 × 108 m/s
413 nm
WAIT, This won’t work!
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Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?
ν =c
λ
ν =3.00 × 108 m/s
413 nm
413 nm 1 m=
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Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?
ν =c
λ
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
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Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?
ν =c
λ
ν =3.00 × 108 m/s
4.13 × 10-7 m
ν = 7.26 × 1014 Hz
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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 ν65
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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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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Section 2
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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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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Section 3 Electron Configurations
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 97
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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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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How does a letter get to you?
5501 Haltom Rd
Haltom City, TX 76137
Very general – includes many cities
Still general – includes a handful of cities
Specific, but includes many places
Very specific – specifies only 1 place
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Quantum numbers are mathematical “addresses” of electrons for an atom – no two electrons can have the same exact address
(n, l, ml, ms)
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n = principle quantum number
• energy level
• relates to size
• possible values are all positive integers (currently 1 - 7)
n = 1, 2, 3, 4, 5, 6, 7
(seven periods on the periodic table)
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l = azimuthal quantum number
• sublevel
• relates to shape
• possible values are 0 to n-1 (currently 0-3)
s = 0p = 1d = 2f = 3
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ml = magnetic quantum number
• orbitals
• possible values are integers from –l to l
if l = 0 , then s = 0
if l = 1, then p = -1, 0, 1
if l = 2, then d = -2, -1, 0, 1, 2
if l = 3, then f = -3, -2, -1, 0, 1, 2, 3
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example – Ti (22 electrons)
orbital notation
___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
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example – Ti (22 electrons)
orbital notation
___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
1st arrow (1, 0, 0, ½)
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example – Ti (22 electrons)
orbital notation
___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
1st arrow (1, 0, 0, ½)
2nd arrow (1, 0, 0, -½)
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example – Ti (22 electrons)
orbital notation
___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
1st arrow (1, 0, 0, ½)
2nd arrow (1, 0, 0, -½)
Can be combined into (1, 0, 0, ±½)
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___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
3rd and 4th arrows = (2, 0, 0, ±½)
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___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
3rd and 4th arrows = (2, 0, 0, ±½)
for 2p: (2, 1, -1, ±½) and (2, 1, 0, ±½) and (2, 1, 1, ±½)
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___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
3rd and 4th arrows = (2, 0, 0, ±½)
for 2p: (2, 1, -1, ±½) and (2, 1, 0, ±½) and (2, 1, 1, ±½)
for 3s: (3, 0, 0, ±½)
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___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
3rd and 4th arrows = (2, 0, 0, ±½)
for 2p: (2, 1, -1, ±½) and (2, 1, 0, ±½) and (2, 1, 1, ±½)
for 3s: (3, 0, 0, ±½)
for 3p: (3, 1, -1, ±½) and (3, 1, 0, ±½) and (3, 1, 1, ±½)
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___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
3rd and 4th arrows = (2, 0, 0, ±½)
for 2p: (2, 1, -1, ±½) and (2, 1, 0, ±½) and (2, 1, 1, ±½)
for 3s: (3, 0, 0, ±½)
for 3p: (3, 1, -1, ±½) and (3, 1, 0, ±½) and (3, 1, 1, ±½)
for 4s: (4, 0, 0, ±½)
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___ ___ ___ ___ ___ ___ ___ ___ __ ___ ___ ___ ___ ___ ___
1s 2s 2p 3s 3p 4s 3d
3rd and 4th arrows = (2, 0, 0, ±½)
for 2p: (2, 1, -1, ±½) and (2, 1, 0, ±½) and (2, 1, 1, ±½)
for 3s: (3, 0, 0, ±½)
for 3p: (3, 1, -1, ±½) and (3, 1, 0, ±½) and (3, 1, 1, ±½)
for 4s: (4, 0, 0, ±½)
for 3d: (3, 2, -2, ½) and (3, 2, -1, ½)
notice -- no more arrows
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