Transcript l - SCHOOLinSITES

Quantum Theory and the
Electronic Structure of Atoms
•To understand the electronic structure of atoms, one
must understand the nature of electromagnetic radiation.
Properties of Waves
Wavelength (l) is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
Properties of Waves
Frequency (n) is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s).
The speed (c or u) of the wave = ln
Waves
• The number of waves
passing a given point per
unit of time is the
frequency (n).
• For waves traveling at the
same velocity, the longer
the wavelength, the smaller
the frequency.
Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Electromagnetic
radiation is the emission
and transmission of energy
in the form of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
c = ln
Waves of electric and magnetic fields at right angles to each other.
Electromagnetic Radiation
• All electromagnetic
radiation travels at the
same velocity: the speed
of light (c), 3.00  108
m/s.
• Therefore,
c = ln
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region?
l
c = ln
n
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency
Two electromagnetic waves are represented in the image below. (a) Which wave has the higher frequency? (b) If
one wave represents visible light and the other represents infrared radiation, which wave is which?
PRACTICE EXERCISE
If one of the waves in the image above represents blue light and the other red light, which is which?
SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency
Two electromagnetic waves are represented in the image below. (a) Which wave has the higher frequency? (b) If
one wave represents visible light and the other represents infrared radiation, which wave is which?
Solution (a) The lower wave has a longer wavelength (greater distance between peaks). The longer the
wavelength, the lower the frequency (n = c/ l). Thus, the lower wave has the lower frequency, and the upper one
has the higher frequency.
(b) The electromagnetic spectrum indicates that infrared radiation has a longer wavelength than visible light.
Thus, the lower wave would be the infrared radiation.
PRACTICE EXERCISE
If one of the waves in the image above represents blue light and the other red light, which is which?
Answer: The expanded visible-light portion of the electromagnetic spectrum tells you that red light has a longer
wavelength than blue light. The lower wave has the longer wavelength (lower frequency) and would be the red
light.
SAMPLE EXERCISE 6.2 Calculating Frequency from Wavelength
The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is
the frequency of this radiation?
PRACTICE EXERCISE
(a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm.
Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a
frequency of 103.4 MHz (megahertz; MHz = 106 s–1). Calculate the wavelength of this radiation.
SAMPLE EXERCISE 6.2 Calculating Frequency from Wavelength
The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is
the frequency of this radiation?
Solution
Analyze: We are given the wavelength, l of the radiation and asked to calculate its frequency, n.
Plan: The relationship between the wavelength (which is given) and the frequency (which is the unknown) is
given by Equation 6.1. We can solve this equation for n and then use the values of l and c to obtain a numerical
answer. (The speed of light, c, is a fundamental constant whose value is given in the text or in the table of
fundamental constants on the back inside cover.)
Solve: Solving Equation 6.1 for frequency gives n = c/ l. When we insert the values for c and l, we note that
the units of length in these two quantities are different. We can convert the wavelength from nanometers to
meters, so the units cancel:
Check: The high frequency is reasonable because of the short wavelength. The units are proper because
frequency has units of “per second,” or s–1.
PRACTICE EXERCISE
(a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm.
Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a
frequency of 103.4 MHz (megahertz; MHz = 106 s–1). Calculate the wavelength of this radiation.
Answers: (a) 4.688  1014 s–1, (b) 2.901 m
Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Energy (light) is emitted or absorbed
in discrete units (quantum).
The Nature of Energy
• The wave nature of light
does not explain how an
object can glow when its
temperature increases.
• Max Planck explained it
by assuming that energy
comes in packets called
quanta.
E = hn
Planck’s constant (h)
h = 6.63 x 10-34 J•s
(and sub for freq.)
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
•
•
Einstein used this assumption to explain the
photoelectric effect.
He concluded that energy is proportional to
frequency:
E = hn
where h is Planck’s constant, 6.63  10−34 J-s.
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
Light has both:
1. wave nature
2. particle nature
http://media.pearsoncmg.com/ph/esm
/esm_brown_chemistry_10/ch06/Phot
oelectricEffect.html
hn
KE e-
When copper is bombarded with high-energy electrons,
X rays are emitted. Calculate the energy (in joules)
associated with the photons if the wavelength of the X
rays is 0.154 nm.
E=hxn
E=hxc/l
E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
E = 1.29 x 10 -15 J
SAMPLE EXERCISE 6.3 Energy of a Photon
Calculate the energy of one photon of yellow light whose wavelength is 589 nm.
SAMPLE EXERCISE 6.3 Energy of a Photon
Calculate the energy of one photon of yellow light whose wavelength is 589 nm.
Solution
Analyze: Our task is to calculate the energy, E, of a photon, given l = 589 nm.
Plan: We can use Equation 6.1 to convert the wavelength to frequency:
We can then use Equation 6.3 to calculate energy:
Solve: The frequency, n, is calculated from the given wavelength, as shown in Sample Exercise 6.2:
The value of Planck’s constant, h, is given both in the text and in the table of physical constants on the inside
front cover of the text, and so we can easily calculate E:
Comment: If one photon of radiant energy supplies 3.37  10–19J, then one mole of these photons will supply
This is the magnitude of enthalpies of reactions (Section 5.4), so radiation can break chemical bonds, producing
what are called photochemical reactions.
SAMPLE EXERCISE 6.3 continued
PRACTICE EXERCISE
(a) A laser emits light with a frequency of 4.69  1014 s–1. What is the energy of one photon of the radiation from
this laser? (b) If the laser emits a pulse of energy containing 5.0  1017 photons of this radiation, what is the total
energy of that pulse? (c) If the laser emits 1.3  10–2 J of energy during a pulse, how many photons are emitted
during the pulse?
Answers: (a) 3.11  10–19 J, (b) 0.16 J, (c) 4.2  1016 photons
The Nature of Energy
Another mystery
involved the emission
spectra observed from
energy emitted by
atoms and molecules.
The Nature of Energy
• One does not observe a
continuous spectrum, as
one gets from a white
light source.
• Only a line spectrum of
discrete wavelengths is
observed.
Line Emission Spectrum of Hydrogen Atoms
Bohr’s Model of
the Atom (1913)
http://media.pearsoncmg.com/ph/esm/es
m_brown_chemistry_10/ch06/FlameTest
sforMetals.html
1. e- can only have specific
(quantized) energy
values
2. light is emitted as emoves from one energy
level to a lower energy
level
En = -RH (
1
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
The Bohr Ring Atom
• He didn’t know why but only certain
energies were allowed.
• He called these allowed energies energy
levels.
• Putting Energy into the atom moved the
electron away from the nucleus.
• From ground state to excited state.
• When it returns to ground state it gives off
light of a certain energy.
The Nature of Energy
•
Niels Bohr adopted Planck’s
assumption and explained
these phenomena in this way:
1. Electrons in an atom can only
occupy certain orbits
(corresponding to certain
energies).
The Nature of Energy
•
Niels Bohr adopted Planck’s
assumption and explained
these phenomena in this way:
2. Electrons in permitted orbits
have specific, “allowed”
energies; these energies will not
be radiated from the atom.
The Nature of Energy
•
Niels Bohr adopted Planck’s
assumption and explained
these phenomena in this way:
3. Energy is only absorbed or
emitted in such a way as to
move an electron from one
“allowed” energy state to
another; the energy is defined by
E = hn
The Nature of Energy
The energy absorbed or emitted from
the process of electron promotion or
demotion can be calculated by the
equation:
E = −RH (
1
1
- 2
nf2
ni
)
where RH is the Rydberg
constant, 2.18  10−18 J, and ni
and nf are the initial and final
energy levels of the electron.
ni = 3
ni = 3
ni = 2
nf = 2
nnf f==11
Ephoton = E = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
E = RH( 2
ni
)
)
1
n2f
)
We are worried about the change
• When the electron moves from one energy
level to another.
 E = Efinal - Einitial
 E = -2.178 x 10-18 J Z2 (1/ nf2 - 1/ ni2)
SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom
Using the picture below, predict which of the following electronic transitions produces the spectral line having
the longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3.
E=
hc
l
PRACTICE EXERCISE
Indicate whether each of the following electronic transitions emits energy or requires the absorption of energy:
(a) n = 3 to n = 1; (b) n = 2 to n = 4 .
SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom
Using Figure 6.13, predict which of the following electronic transitions produces the spectral line having the
longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3.
Solution The wavelength increases as frequency decreases (l = c/n . Hence the longest wavelength will be
associated with the lowest frequency. According to Planck’s equation, E = hn, the lowest frequency is associated
with the lowest energy. In Figure 6.13 the shortest vertical line represents the smallest energy change. Thus, the
n = 4 to n = 3 transition produces the longest wavelength (lowest frequency) line.
PRACTICE EXERCISE
Indicate whether each of the following electronic transitions emits energy or requires the absorption of energy:
(a) n = 3 to n = 1; (b) n = 2 to n = 4 .
Answers: (a) emits energy, (b) requires absorption of energy
Calculate the wavelength (in nm) of a photon
emitted by a hydrogen atom when its electron
drops from the n = 5 state to the n = 3 state.
Ephoton = E = RH(
1
n2i
1
n2f
)
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = E = -1.55 x 10-19 J
Ephoton = hc / l
l = hc / Ephoton
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm
The Wave Nature of Matter
• Louis de Broglie stated that if light can have
material properties, matter should exhibit wave
properties.
• He demonstrated that the relationship between
mass and wavelength was
h
l = mv
The Quantum Mechanical Model
• A totally new approach.
• De Broglie said matter could be like a wave.
• De Broglie said they were like standing
waves.
• The vibrations of a stringed instrument.
The Uncertainty Principle
• Heisenberg showed that the more precisely
the momentum of a particle is known, the
less precisely is its position known (flashlight) :
h
(x) (mv) 
4
• In many cases, our uncertainty of the
whereabouts of an electron is greater than
the size of the atom itself!
What’s possible?
• You can only have a standing wave if you
have complete waves.
• There are only certain allowed waves.
• In the atom there are certain allowed waves
called electrons.
• 1925 Erwin Schroedinger described the wave
function of the electron.
• Much math but what is important are the
solution.
There is a limit to what we can
know
• We can’t know how the electron is moving
or how it gets from one energy level to
another.
• The Heisenberg Uncertainty Principle.
• There is a limit to how well we can know
both the position and the momentum of an
object.
Probability
Distance from nucleus
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
l = h/mu
h in J•s m in kg u in (m/s)
l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
l = 1.7 x 10-32 m = 1.7 x 10-23 nm
SAMPLE EXERCISE 6.5 Matter Waves
What is the wavelength of an electron moving with a speed of
5.97  106 m/s? (The mass of the electron is 9.11  10–28 g.)
SAMPLE EXERCISE 6.5 Matter Waves
What is the wavelength of an electron moving with a speed of 5.97  106 m/s? (The mass of the electron is 9.11
 10–28 g.)
Solution
Analyze: We are given the mass, m, and velocity, n, of the electron, and we must calculate its de Broglie
wavelength, l
Plan: The wavelength of a moving particle is given by Equation 6.8, so l is calculated by inserting the known
quantities h, m, and n. In doing so, however, we must pay attention to units.
Solve: Using the value of Planck’s constant,
and recalling that
we have the following:
SAMPLE EXERCISE 6.5 continued
Comment: By comparing this value with the wavelengths of electromagnetic
radiation shown in Figure 6.4, we see that the wavelength of this electron is about the
same as that of X rays.
PRACTICE EXERCISE
Calculate the velocity of a neutron whose de Broglie wavelength is 500 pm. The mass of a
neutron is given in the table on the back inside cover of the text to be 1.67  10–24 g.
Chemistry in Action: Element from the Sun
In 1868, Pierre Janssen detected a new dark line in the solar
emission spectrum that did not match known emission lines
Mystery element was named Helium
In 1895, William Ramsey discovered helium in a mineral of
uranium (from alpha decay).
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eWave function (Y) describes:
1. energy of e- with a given Y
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
7.5
Quantum Numbers
• There are many solutions to Schroedinger’s
equation
• Each solution can be described with
quantum numbers that describe some aspect
of the solution.
• Principal quantum number (n) size and
energy of of an orbital.
• Has integer values >0
QUANTUM NUMBERS
The shape, size, and energy of each orbital is a function
of 3 quantum numbers which describe the location of
an electron within an atom or ion
n (principal)
---> energy level
l (orbital) ---> shape of orbital
ml (magnetic) ---> designates a particular
suborbital
The fourth quantum number is not derived from the
wave function
s (spin)
---> spin of the electron
(clockwise or counterclockwise: ½ or – ½)
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
n=3
Where 90% of the
e- density is found
for the 1s orbital
e- density (1s orbital) falls off rapidly
as distance from nucleus increases
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Angular (or azimuthal) momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
shape of the orbital.
integer values from 0 to n-1
l = 0 is called s
l = 1 is called p
l =2 is called d
l =3 is called f
l =4 is called g
Shape of the “volume” of space that the e- occupies
Types of Orbitals (l)
s orbital
p orbital
d orbital
l = 0 (s orbitals)
l = 1 (p orbitals)
p Orbitals
this is a p sublevel
with 3 orbitals
These are called x, y, and z
3py
orbital
There is a PLANAR
NODE thru the
nucleus, which is
an area of zero
probability of
finding an electron
p Orbitals
• The three p orbitals lie 90o apart in
space
d orbitals
f Orbitals
For l = 3,
orbitals
f sublevel with 7
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
ml = -1
ml = -2
ml = 0
ml = -1
ml = 0
ml = 1
ml = 1
ml = 2
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
ms = -½
SAMPLE EXERCISE 6.6 Subshells of the Hydrogen Atom
(a)
•
•
Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n = 4.
Give the label for each of these subshells.
How many orbitals are in each of these subshells?
•
Analyze and Plan : We are given the value of the principal quantum number, n.
We need to determine the allowed values of l and ml for this given value of n and
then count the number of orbitals in each subshell.
• Solution :
– a. There are four subshells in the fourth shell, corresponding to the four
possible values of l (0, 1, 2, and 3).
– b. These subshells are labeled 4s, 4p, 4d, and 4f. The number given in the
designation of a subshell is the principal quantum number, n; the letter
designates the value of the angular (or azimuthal) quantum number, l: for
l = 0, s;
l = 1, p;
l = 2, d;
l = 3, f;
– c. There is one 4s orbital (when l = 0, there is only one possible value of
ml: 0). There are three 4p orbitals (when l = 1, there are three possible
values of ml : 1, 0, -1). There are five 4d orbitals (when l = 2, there are five
allowed values of ml : 2, 1, 0, -1, -2). There are seven 4f orbitals (when l =
3, there are seven permitted values of ml : 3, 2, 1, 0, -1, -2, -3)
PRACTICE EXERCISE 6.6
(a) What is the designation (or label) for the subshells with n = 5 and l = 1?
(b) How many orbitals are in this subshell?
(c) Indicate the values of ml for each of these orbitals.
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Y.
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
diamagnetic--not magnetic [magnetism dies]; in fact they are slightly
repelled by a magnetic field. Occurs when all electrons are PAIRED.
paramagnetic--attracted to a magnetic field; lose their magnetism when
removed from the magnetic field; HAS ONE OR MORE UNPAIRED
ELECTRONS
ferromagnetic--retain magnetism upon introduction to, then removal
from a magnetic field
What about ferromagnetic?
clusters of atoms have their unpaired electrons
aligned within a cluster, clusters are more or less
aligned and substance acts as a magnet. Don't
drop it!!
1.When all of the domains, represented by these
arrows are aligned, it behaves as a magnet.
2.This is what happens if you drop it: The
domains go in all different directions and it no
longer operates as a magnet.
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
How many electrons can an orbital hold?
If n, l, and ml are fixed, then ms = ½ or - ½
Y = (n, l, ml, ½) or Y = (n, l, ml, -½)
An orbital can hold 2 electrons
How many 2p orbitals are there in an atom?
n=2
If l = 1, then ml = -1, 0, or +1
2p
3 orbitals
l=1
How many electrons can be placed in the 3d
subshell?
n=3
3d
l=2
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals which can hold a total of 10 e-
Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=1 l = 0
n=3 l = 1
n=2 l = 1
“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
Be
Li
B5
C
3
64electrons
electrons
22s
222s
22p
12 1
BBe
Li1s1s
1s
2s
H
He12electron
electrons
He
H 1s
1s12
The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
Ne97
C
N
O
F
6
810
electrons
electrons
electrons
22s
222p
22p
5
246
3
Ne
C
N
O
F 1s
1s222s
Why are d and f orbitals always
in lower energy levels?
• d and f orbitals require LARGE amounts of
energy
• It’s better (lower in energy) to skip a
sublevel that requires a large amount of
energy (d and f orbtials) for one in a higher
level but lower energy
This is the reason we learn to read electron
configurations using your periodic table.
Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
H
1s1
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Cl 17 electrons
1s22s22p63s23p5
1s < 2s < 2p < 3s < 3p < 4s
2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n=3
l=1
ml = -1, 0, or +1
ms = ½ or -½
Outermost subshell being filled with electrons
Exceptions to the Aufbau Principle
• Remember d and f orbitals require LARGE
amounts of energy
• If we can’t fill these sublevels, then the next best
thing is to be HALF full (one electron in each
orbital in the sublevel)
• There are many exceptions, but the most
common ones are
d4 and d9
For the purposes of this class, we are going to
assume that ALL atoms (or ions) that end in d4
or d9 are exceptions to the rule. This may or
may not be true, it just depends on the atom.
Exceptions to the Aufbau Principle
d4 is one electron short of being HALF full
In order to become more stable (require less
energy), one of the closest s electrons will
actually go into the d, making it d5 instead of d4.
For example: Cr would be [Ar] 4s2 3d4, but since
this ends exactly with a d4 it is an exception to
the rule. Thus, Cr should be [Ar] 4s1 3d5.
Procedure: Find the closest s orbital. Steal one
electron from it, and add it to the d.
Try These!
• Write the shorthand notation
for:
Cu
[Ar] 4s1 3d10
1 4f14 5d5
[Xe]
6s
W
1 4f14 5d10
[Xe]
6s
Au
Keep an Eye On Those Ions!
• Electrons are lost or gained like they
always are with ions… negative ions
have gained electrons, positive ions
have lost electrons
• The electrons that are lost or gained
should be added/removed from the
highest energy level (not the highest
orbital in energy!)
Keep an Eye On Those Ions!
• Tin
Atom: [Kr] 5s2 4d10 5p2
Sn+4 ion: [Kr] 4d10
Sn+2 ion: [Kr] 5s2 4d10
Note that the electrons came out of
the highest energy level, not the
highest energy orbital!