new chm 151_unit_4_power_points
TRANSCRIPT
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Chapter 7
Quantum Theory and Atomic Structure
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Goals & Objectives
• See the following Learning Objectives on pages 283 and 322.
• Understand these Concepts:• 7.1-7, 9-12; 8.1-8.
• Master these Skills:• 7.1-2, 5; 8.1-2.
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The Wave Nature of Light
Visible light is a type of electromagnetic radiation.
The wave properties of electromagnetic radiation are described by three variables:- frequency (n), cycles per second- wavelength (l), the distance a wave travels in one cycle- amplitude, the height of a wave crest or depth of a trough.
The speed of light is a constant:
c = n x l= 3.00 x 108 m/s in a vacuum
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Figure 7.1 The reciprocal relationship of frequency and wavelength.
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Figure 7.2 Differing amplitude (brightness, or intensity) of a wave.
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Figure 7.3 Regions of the electromagnetic spectrum.
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Sample Problem 7.1 Interconverting Wavelength and Frequency
PROBLEM: A dental hygienist uses x-rays (l= 1.00Å) to take a series of dental radiographs while the patient listens to a radio station (l = 325 cm) and looks out the window at the blue sky (l= 473 nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x108 m/s.)
PLAN: Use the equation c = nl to convert wavelength to frequency. Wavelengths need to be in meters because c has units of m/s.
wavelength in units given
wavelength in m
frequency (s-1 or Hz)
use conversion factors1 Å = 10-10 m
n =lc
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Sample Problem 7.1
SOLUTION:
For the x-rays: l = 1.00 Å x = 1.00 x 10-10 m1 Å
10-10 m
= = 3.00 x 1018 s-1n =
lc 3.00 x 108 m/s
1.00 x 10-10 m
For the radio signal: = 9.23 x 107 s-1=n =
lc 3.00 x 108 m/s
325 cm x10-2 m 1 cm
For the blue sky: = 6.34 x 1014 s-1=n =
lc 3.00 x 108 m/s
473 nm x10-9 m 1 cm
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Figure 7.4 Different behaviors of waves and particles.
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Figure 7.5 Formation of a diffraction pattern.
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Energy and frequency
A solid object emits visible light when it is heated to about 1000 K. This is called blackbody radiation.
E = nhn
The color (and the intensity ) of the light changes as the temperature changes. Color is related to wavelength and frequency, while temperature is related to energy.
Energy is therefore related to frequency and wavelength:
E = energyn is a positive integerh is Planck’s constant
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Figure 7.6 Familiar examples of light emission related to blackbody radiation.
Lightbulb filament
Smoldering coal Electric heating element
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The Quantum Theory of Energy
Any object (including atoms) can emit or absorb only certain quantities of energy.
Energy is quantized; it occurs in fixed quantities, rather than being continuous. Each fixed quantity of energy is called a quantum.
An atom changes its energy state by emitting or absorbing one or more quanta of energy.
DE = nhn where n can only be a whole number.
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Figure 7.7 The photoelectric effect.
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Sample Problem 7.2
SOLUTION:
Calculating the Energy of Radiation from Its Wavelength
PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation?
PLAN: We know l in cm, so we convert to m and find the frequency using the speed of light. We then find the energy of one photon using E = hn.
=E = hn =
hcl
(6.626 x 10-34) J∙s)(3.00 x 108 m/s)
(1.20 cm)10-2 m
1 cm( )
= 1.66 x 10-23 J
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Figure 7.8A The line spectrum of hydrogen.
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Figure 7.8B The line spectra of Hg and Sr.
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= RRydberg equation -1
l
1
n22
1
n12
R is the Rydberg constant = 1.096776x107 m-1
Figure 7.9 Three series of spectral lines of atomic hydrogen.
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
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The Bohr Model of the Hydrogen Atom 1913
Bohr’s atomic model postulated the following:
• The H atom has only certain energy levels, which Bohr called stationary states.
– Each state is associated with a fixed circular orbit of the electron around the nucleus.
– The higher the energy level, the farther the orbit is from the nucleus.
– When the H electron is in the first orbit, the atom is in its lowest energy state, called the ground state.
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• The atom does not radiate energy while in one of its stationary states.
• The atom changes to another stationary state only by absorbing or emitting a photon.– The energy of the photon (hn) equals the difference
between the energies of the two energy states.– When the E electron is in any orbit higher than n = 1,
the atom is in an excited state.
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Figure 7.10 A quantum “staircase” as an analogy for atomic energy levels.
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Figure 7.11 The Bohr explanation of three series of spectral lines emitted by the H atom.
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Figure B7.1 Flame tests and fireworks.
copper 29Custrontium 38Sr copper 29Cu
The flame color is due to the emission of light of a particular wavelength by each element.
Fireworks display emissions similar to those seen in flame tests.
Tools of the Laboratory
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Figure B7.2 Emission and absorption spectra of sodium atoms.
Tools of the Laboratory
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Light Waves
Atomic Emission Spectra
Light in discrete lines;
individual wavelengths, not continuum
http://jersey.uoregon.edu/elements/Elements.html
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Tools of the Laboratory
Figure B7.3 Components of a typical spectrometer.
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Figure B7.4 Measuring chlorophyll a concentration in leaf extract.
Tools of the Laboratory
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The Wave-Particle Duality of Matter and Energy
Matter and Energy are alternate forms of the same entity.
E = mc2
All matter exhibits properties of both particles and waves. Electrons have wave-like motion and therefore have only certain allowable frequencies and energies.
Matter behaves as though it moves in a wave, and the de Broglie wavelength for any particle is given by:
h
mul = m = mass
u = speed in m/s
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Figure 7.12 Wave motion in restricted systems.
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Table 7.1 The de Broglie Wavelengths of Several Objects
Substance Mass (g) Speed (m/s) l (m)
slow electron
fast electron
alpha particle
one-gram mass
baseball
Earth
9x10-28
9x10-28
6.6x10-24
1.0
142
6.0x1027
1.0
5.9x106
1.5x107
0.01
25.0
3.0x104
7x10-4
1x10-10
7x10-15
7x10-29
2x10-34
4x10-63
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CLASSICAL THEORY
Matter particulate,
massive
Energy continuous,
wavelike
Since matter is discontinuous and particulate, perhaps energy is discontinuous and particulate.
Observation Theory
Planck: Energy is quantized; only certain values allowed
Blackbody radiation
Einstein: Light has particulate behavior (photons)Photoelectric effect
Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit.
Atomic line spectra
Figure 7.15
Major observations and theories leading from classical theory to quantum theory
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Since energy is wavelike,perhaps matter is wavelike.
Observation Theory
deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons
Davisson/Germer: Electron beam is diffracted by metal crystal
Since matter has mass, perhaps energy has mass
Observation Theory
Einstein/deBroglie: Mass and energy are equivalent; particles have
wavelength and photons have momentum.
Compton: Photon’s wavelength increases (momentum decreases)after colliding withelectron
Figure 7.15 continued
QUANTUM THEORY
Energy and Matterparticulate, massive, wavelike
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Heisenberg’s Uncertainty Principle
Heisenberg’s Uncertainty Principle states that it is not possible to know both the position and momentum of a moving particle at the same time.
The more accurately we know the speed, the less accurately we know the position, and vice versa.
D x∙m D u ≥ h
4px = position
u = speed
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Sample Problem 7.5
SOLUTION:
Applying the Uncertainty Principle
PROBLEM: An electron moving near an atomic nucleus has a speed 6x106 m/s ± 1%. What is the uncertainty in its position (Dx)?
PLAN: The uncertainty in the speed (Du) is given as ±1% (0.01) of
6x106 m/s. We multiply u by 0.01 and substitute this value
into Equation 7.6 to solve for Δx.
Du = (0.01)(6x106 m/s) = 6x104 m/s
Dx∙mDu ≥ h
4p
≥ 1x10-9 mDx ≥ h
4pmDu 4p (9.11x10-31 kg)(6x104 m/s)
6.626x10-34 kg∙m2/s≥
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The Quantum-Mechanical Model of the Atom
The matter-wave of the electron occupies the space near the nucleus and is continuously influenced by it.
The Schrödinger wave equation allows us to solve for the energy states associated with a particular atomic orbital.
The square of the wave function gives the probability density, a measure of the probability of finding an electron of a particular energy in a particular region of the atom.
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The Shrodinger Atom (1926)
• Quantum Mechanical Model
– Wave equation is solved to obtain– Wave function (orbital) – Square of wave function 2 gives probability
of finding an electron in a certain space
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The Schrödinger Equation
HY = EY
d2Y
dy2
d2Y
dx2
d2Y
dz2+ +
8p2mQ
h2(E-V(x,y,z)Y(x,y,z) = 0+
how y changes in space
mass of electron
total quantized energy of the
atomic system
potential energy at x,y,z
wave function
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Figure 7.16
Electron probability density in the ground-state H atom.
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The Shrödinger Atom (1926)
• Wave function for each electron in an atom contains four quantum numbers
– n Principle– l Angular momentum– ml Magnetic
– ms Spin
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Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
The principal quantum number (n) is a positive integer.The value of n indicates the relative size of the orbital and therefore its relative distance from the nucleus.
The magnetic quantum number (ml) is an integer with values from –l to +l The value of ml indicates the spatial orientation of the orbital.
The angular momentum quantum number (l) is an integer from 0 to (n -1).The value of l indicates the shape of the orbital.
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Quantum Numbers and The Exclusion Principle
An atomic orbital can hold a maximum of two electrons and they must have opposing spins.
Each electron in any atom is described completely by a set of four quantum numbers.The first three quantum numbers describe the orbital, while the fourth quantum number describes electron spin.
Pauli’s exclusion principle states that no two electrons in the same atom can have the same four quantum numbers.
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Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol(Property) Allowed Values Quantum Numbers
Principal, n(size, energy)
Angular momentum, l
(shape)
Magnetic, ml
(orientation)
Positive integer(1, 2, 3, ...)
0 to n – 1
-l,…,0,…,+l
1
0
0
2
0 1
0
3
0 1 2
0
0-1 +1 -1 0 +1
0 +1 +2-1-2
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Sample Problem 7.6
SOLUTION:
Determining Quantum Numbers for an Energy Level
PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3?
PLAN: Values of l are determined from the value for n, since l can take values from 0 to (n-1). The values of ml then follow from the values of l.
For n = 3, allowed values of l are = 0, 1, and 2
For l = 1 ml = -1, 0, or +1
For l = 2 ml = -2, -1, 0, +1, or +2
There are 9 ml values and therefore 9 orbitals with n = 3.
For l = 0 ml = 0
PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3?
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Sample Problem 7.7
SOLUTION:
Determining Sublevel Names and Orbital Quantum Numbers
PLAN: Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals.
n l sublevel name possible ml values # of orbitals
(a) 3 2 3d -2, -1, 0, 1, 2 5
0(b) 2 2s 0 1
(c) 5 1 5p -1, 0, 1 3
(d) 4 3 4f -3, -2, -1, 0, 1, 2, 3 7
PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers:
(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3
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Sample Problem 7.8
SOLUTION:
Identifying Incorrect Quantum Numbers
PROBLEM: What is wrong with each of the following quantum numbers designations and/or sublevel names?
(a) A sublevel with n = 1 can only have l = 0, not l = 1. The only possible sublevel name is 1s.
n l ml Name
(a)
(b)
(c)
1
4
3
1
3
1
0
+1
-2
1p
4d
3p
(b) A sublevel with l = 3 is an f sublevel, to a d sublevel. The name should be 4f.
(c) A sublevel with l = 1 can only have ml values of -1, 0, or +1, not -2.
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Figure 7.17 The 1s, 2s, and 3s orbitals.
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Figure 7.18 The 2p orbitals.
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Figure 7.19 The 3d orbitals.
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Figure 7.20 The 4fxyz orbital, one of the seven 4f orbitals.
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The Shrödinger Atom (1926)
• Magnetic quantum number ml
– Value = –l to + l– Properties of orbital:
orientation in space
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The Shrödinger Atom (1926)
• Spin quantum number ms
– Value = +1/2 or – 1/2– Two electrons can occupy an orbital
one has spin +1/2, the other has spin – 1/2
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The Shrödinger Atom (1926)
• Subshell capacities in each shell (n)
– s (l=0) 1 orbital, 2 electrons– p (l=1) 3 orbitals, 6 electrons (n≥2) – d (l=2) 5 orbitals, 10 electrons (n≥3)– f (l=3) 7 orbitals, 14 electrons (n≥4)
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Table 8.1 Summary of Quantum Numbers of Electrons in Atoms
Name Symbol Permitted Values Property
principal n positive integers (1, 2, 3, …) orbital energy (size)
angular momentum
l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.)
magnetic mlintegers from -l to 0 to +l orbital orientation
spin ms+½ or -½ direction of e- spin
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Orbitals: Order of Filling
• Lowest energy orbitals are filled first• In subshells with multiple orbitals, start
filling at lowest ml.• In subshells with multiple orbitals, each
orbital receives one electron (ms = +1/2) until all are filled,
• then second electron electron goes in
(ms = –1/2).
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Electron Configurations and the Periodic Table
•3Li 1s22s1
•11Na 1s22s22p63s1
•19K 1s22s22p63s23p64s1
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Electron Configurations and the Periodic Table
•9F 1s22s22p5
•17Cl 1s22s22p63s23p5
•35Br 1s22s22p63s23p64s23d104p5
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Electron Configurations and the Periodic Table
•10Ne 1s22s22p6
•18Ar 1s22s22p63s23p6
•36Kr 1s22s22p63s23p64s23d104p6
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Electron Configurations and the Periodic Table
•17Cl– 1s22s22p63s23p6
•18Ar 1s22s22p63s23p6
•19K+ 1s22s22p63s23p6
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Some Notes on the d Subshell and the Transition Metals
• There is special stability associated with filled or half-filled subshells
• (n) s subshell fills first, then (n–1) d subshell
• (n) s subshell empties first, then (n–1) d subshell
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Some Notes on the d Subshell and the Transition Metals
• Transition metals often form several stable ions of differing charge:
• Fe 1s22s22p63s23p64s23d6
• Fe2+ 1s22s22p63s23p64s13d5
• Fe3+ 1s22s22p63s23p64s03d5
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Some Notes on the d Subshell and the Transition Metals
• Some transition metals have anomalous ground state electron configurations:
•24Cr 1s22s22p63s23p64s13d5
•29Cu 1s22s22p63s23p64s13d10
•46Pd 1s22s22p63s23p64s23d104p65s04d10
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Exercises
• Give the electronic configurations:
(III-2)–
5B ________________________
–23V ________________________
–16S ________________________
–16S2– _________________________
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n l#
as s, p, d, f
Electron Configurations and Orbital Diagrams
Electron configuration is indicated by a shorthand notation:
# of electrons in the sublevel
Orbital diagrams make use of a box, circle, or line for each orbital in the energy level. An arrow is used to represent an electron and its spin.
↑↓↑↓ ↑↓
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Figure 8.6 A vertical orbital diagram for the Li ground state.
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Building Orbital Diagrams
The aufbau principle is applied – electrons are always placed in the lowest energy sublevel available.
The exclusion principle states that each orbital may contain a maximum of 2 electrons, which must have opposite spins.
H (Z = 1) 1s1
He (Z = 2) 1s2
1s
↑↓
1s
↑
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Building Orbital Diagrams
Hund’s rule specifies that when orbitals of equal energy are available, the lowest energy electron configuration has the maximum number of unpaired electrons with parallel spins.
N (Z = 7) 1s22s22p3
↑↓
2s 2p
↑ ↑ ↑
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Sample Problem 8.1 Determining Quantum Numbers from Orbital Diagrams
PLAN: Identify the electron of interest and note its level (n), sublevel, (l), orbital (ml) and spin (ms). Count the electrons in the order in which they are placed in the diagram.
PROBLEM: Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom.
SOLUTION:
F (Z = 9) 1s22s22p3
For the 3rd electron: n = 2, l = 0, ml = 0, ms = +½
↑↓
2s 2p
↑↓ ↑ ↑↑↓
1s
For the 8th electron: n = 2, l = 1, ml = -1, ms = -½
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Partial Orbital Diagrams and Condensed Configurations
A partial orbital diagram shows only the highest energy sublevels being filled.
Al (Z = 13) 1s22s22p63s23p1
A condensed electron configuration has the element symbol of the previous noble gas in square brackets.
Al has the condensed configuration [Ne]3s23p1
↑↓
3s 3p
↑
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Table 8.2 Partial Orbital Diagrams and Electron Configurations* for the Elements in Period 3.
*Colored type indicates the sublevel to which the last electron is added.
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Electron Configuration and Group
Similar outer electron configurations correlate with similar chemical behavior.
Elements in the same group of the periodic table have the same outer electron configuration.
Elements in the same group of the periodic table exhibit similar chemical behavior.
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Figure 8.8 Condensed electron configurations in the first three periods.
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Figure 8.9 Similar reactivities in a group.
Potassium reacting with water. Chlorine reacting with potassium.
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Figure 8.11 Orbital filling and the periodic table.
The order in which the orbitals are filled can be obtained directly from the periodic table.
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Aid to memorizing sublevel filling order.
The n value is constant horizontally.The l value is constant vertically.n + l is constant diagonally.
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Categories of Electrons
Inner (core) electrons are those an atom has in common with the pervious noble gas and any completed transition series.
Outer electrons are those in the highest energy level (highest n value).
Valence electrons are those involved in forming compounds.For main group elements, the valence electrons are the outer electrons.
For transition elements, the valence electrons include the outer electrons and any (n-1)d electrons.
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Sample Problem 8.2 Determining Electron Configurations
PROBLEM: Using the periodic table on the inside cover of the text (not Figure 8.10 or Table 8.3), give the full and condensed electron configurations, partial orbital diagrams showing valence electrons only, and number of inner electrons for the following elements:
(a) potassium (K; Z = 19)
(b) technetium (Tc; Z = 43)
(c) lead (Pb; Z = 82)
PLAN: The atomic number gives the number of electrons, and the periodic table shows the order for filling orbitals. The partial orbital diagram includes all electrons added after the previous noble gas except those in filled inner sublevels.
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Sample Problem 8.2
SOLUTION:
(a) For K (Z = 19)
[Ar] 4s1 condensed configuration
1s22s22p63s23p64s1full configuration
There are 18 inner electrons.
partial orbital diagram ↑
4s 4p3d
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Sample Problem 8.2
SOLUTION:
(b) For Tc (Z = 43)
[Kr]5s24d5 condensed configuration
1s22s22p63s23p64s23d104p65s24d5full configuration
There are 36 inner electrons.
partial orbital diagram ↑↓
5s 5p4d
↑ ↑ ↑ ↑ ↑
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Sample Problem 8.2
SOLUTION:
(c) For Pb (Z = 82)
[Xe] 6s24f145d106p2condensed configuration
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2full configuration
There are 78 inner electrons.
partial orbital diagram ↑↓
6s 6p
↑ ↑
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Exercises
• For the following electronic configurations, give the four quantum numbers, and the ground state element:– 2p4 ________________________– 3d6 ________________________– 1s1 ________________________– 4p2 ________________________
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Exercises
• For the following sets of quantum numbers, give the electronic configuration and the ground state element:– n l ml ms
– 3 1 0 +1/2 _______,_______– 1 0 0 –1/2 _______, _______ – 4 2 2 +1/2 _______, _______
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