Download Chapter 6. Electronic Structure of Atoms.

Chapter 6. Electronic Structure
of Atoms.
6.1 The wave nature of Light.
Much of our present understanding of the electronic
structure of atoms and molecules comes from light
absorbed or emitted by substances. Light has a wave
nature with wavelength and frequency
Frequency (n) measures how many
wavelengths pass a point per second:
4 x l ÷ 1 s = 4 s-1 = 4 Hz
1s
Light of different wavelengths and
frequencies:
medium
wavelength
short wavelength
high frequency
long wavelength
low frequency
Relationship between speed (c),
frequency (ν), and wavelength (λ) of light:
For light, we have:
c = νλ
where c is the speed of light (3 x 108
m/s), ν is the frequency, and λ is the
wavelength. Frequency is expressed in
cps (cycles per second), also called Hz
(‘Herz’). 1 Hz = 1 s-1 or 1/s.
What is the wavelength, in m, of radiowaves
transmitted by the local radio station WHQR at
91.3 MHz? (speed of light = 3 x 105 km/s)
91.3 MHz = 91.3 MHz x 106 Hz = 91.3 x 106 Hz
1 MHz
c = νλ, so λ = c/ν
λ
1 Hz =
1/s
= 3 x 105 km x 1000 m x ___1___ s
s
km
91.3 x 106 1
= 3.29 m
The electromagnetic spectrum:
Red Orange Yellow Green Blue Indigo Violet
wavelength (l)
frequency (n)
energy (E)
6.2. Quantized Energy and Photons:
Classical physics cannot explain:
1) black-body radiation
2) the emission of electrons from metal
surfaces on which light shines (the
photo-electric effect)
3) the emission of light from
electronically excited gas-atoms.
Black-body radiation:
In physics, a black body is an
object that absorbs all
electromagnetic radiation that
falls onto it. In the 19th
century physicists could not
explain why the frequency of
light emitted by a black-body
went up as the temperature
went up. They expected that
the frequency should remain
the same and the intensity
should just increase. We can
see this in practice, because
objects become red-hot, and
Iron heated in a furnace for
then blue and finally white hot. making wrought iron.
Blackbody radiation at different
temperatures:
Planck’s Quantum theory:
Planck solved the black-body
problem by proposing that
light was emitted in packets
that he called quanta, and
that the energy of a quantum
of light was related to its
frequency (v) by:
E
=
hν
where h is Planck’s constant,
which has a value of 6.626 x
10-34 J-s. (Dimensional
analysis: J-s x s-1 = J).
Max Planck
What is the energy of a photon of
electromagnetic radiation
that has a frequency of 400 kHz?
E  hn
1
s
1000 Hz
 6.63 10 Js  400 kHz 

kHz
Hz
34
= 2.65 x 10-28 J
Energy of a photon: Calculate energy of
light of one photon with a wavelength
of 589 nm.
We have c = νλ, so ν = c/λ
v = 3.00 x 108 m/s
589 x 10-9 m
=
5.09 x 1014 s-1.
E = 6.626 x 10-34 J-s x 5.09 x 1014 s-1
= 3.37 x 10-19 J
The photoelectric effect:
electromagnetic
radiation
If energy of radiation
is high enough, an
electron is excited to
jump to the positive
terminal
+
positive
terminal
power
source
evacuated
chamber
e
metal
surface
current
indicator
+ -
The Photoelectric effect and Photons.
In 1905 Albert Einstein
used Planck’s quantum
theory to explain the
photoelectric effect. For
each metal there is a
minimum frequency of
light below which no
electrons are emitted. He
assumed light came in
packets called quanta,
and that these need a
minimum energy E
before they could
dislodge an electron.
Albert Einstein
6.3. Line Spectra and the Bohr Model:
When a gas is placed in very low
concentration in a vacuum, and excited, light
at only a few specific wavelengths is
produced. This is called a spectrum. Why
this would be could not be explained.
line spectrum of the
H-atom in the visible
region, obtained from
moist H2 gas in a
tube with an AC
current, due to Hatoms formed by
break-up of H2
molecules.
400
700 nm
Fraunhofer (1815) lines in the sun:
The sun appears to produce a continuous spectrum,
but this is only because it has so many different
nuclei being excited. Careful examination shows that
it is a superposition of many different spectra. Even
a ‘white’ fluorescent lamp has spectral lines:
Emission-line spectrum from a standard fluorescent light
fixture. This shows the sharp, bright emission lines of
Mercury plus broad bands from the organic phosphors
coating the inside of the tube.
The Rhydberg Equation:
Sodium vapor, or H atoms, give characteristic
lines. It had been discovered by Rydberg that the
lines in the H-atom spectrum could be predicted
by an empirical equation:
1/λ =
(RH)
1 - 1
n12 n22
Where λ is the wavelength of the line in the Hatom spectrum, RH is the Rydberg constant
(1.096776 x 107 m-1), and n1 and n2 are positive
integers with n2 > n1.
It took nearly thirty years to explain this simple
equation.
Example:
Calculate the wavelength in nm of the band in the spectrum of
atomic hydrogen that occurs with n1 = 2 and n2 = 3.
1/λ
=
( R H)
1 - 1
n12 n22
410 434 486
λ
= 1.096776 x 107 m-1 x (1/(22) – 1/(32))
= 1.096776 x 107 m-1 x (1/4 – 1/9)
= 1/(1.096776 x 107 m-1 x (0.13889))
= 6.56 x 10-7 m = 6.56 x 10-7 m x 1 nm
10-9 m
= 656 nm
656 nm
The Bohr Model of the Atom.
According to classical physics, an
electron moving in a circular orbit as
in the Rutherford model should
continuously radiate away energy and
spiral into the nucleus. Bohr got
around this problem with three
postulates:
1. Only orbits with definite radii are
permitted
2. An electron in a specific orbit has
an allowed energy
3. Energy is emitted or absorbed only
when it changes from one state to
another.
(1885-1962)
Niels Bohr
6.3 Line Spectra and the Bohr Model.
Recall that there are three important properties that
classical physics cannot explain:
1) Black-body radiation. Why does the wavelength of light
emitted shorten as the temperature of the light emitter
goes up? This is explained by Planck’s relationship (before
this time the relationship between the energy of light and
its frequency was not understood). E = hν
2) The photoelectric effect. Why does each metal have a
minimum frequency of light required to excite electrons
from its surface? Light of lower frequency, no matter how
intense, excites no electrons. This is answered by Einstein
using Planck’s equation. Each photon must have a
minimum energy to excite an electron.
3) The existence of line spectra. Why do atoms in e.g.
Na vapor or H-atoms, emit light as line spectra? Why
do the lines in the H-atom spectrum fit the Rhydberg
equation?
1/λ
=
(RH)
1 - 1
n12
n22
Bohr calculated the value of RH in the
Rydberg equation from the physics of motion
and interacting charges, and found that he
could reproduce it exactly.
The energies of transitions in the H-atom are
given by:
ΔE = hν = hc/λ = (-2.18 x 10-18J)(1/nf2 – 1/ni2)
The ni is the principal quantum number of
the initial state, and nf of the final state. Thus
transitions can occur when nf is greater than
ni, i.e. energy is absorbed, or where nf is less
than ni, when light is emitted.
Niels Bohr was the first to offer an explanation for
line spectra:
In the Bohr
model of the
H-atom, the
orbits closest
to the nucleus
(n = 1) are of
lowest energy.
As the energy
increases,
n increases.
The line spectra
correspond to
transitions
between these
orbits.
n=1
n=2
n=3
n=4
n=5
n=6
electron orbits
nucleus
Bohr’s Model of the Hydrogen Atom
Energy levels in the Bohr model of the Hatom
Bohr’s Model of the Hydrogen Atom
absorption of
a photon as
electron is
excited to
higher energy
(n = 3) state
n=6
n=5
n=4
n=3
Energy
n=2
hv
absorption of a photon
n=1
Ground State
nucleus
e
Bohr’s Model of the Hydrogen Atom
n=6
n=5
n=4
e
n=3
n=1
Energy
n=2
Ground State
nucleus
“excited state”
Bohr’s Model of the Hydrogen Atom
emission of
a photon as
electron
returns to
ground state
n=6
n=5
n=4
n=3
n=1
Energy
n=2
hv
Ground State
nucleus
e
ground state
Transitions corresponding to
the Balmer series
n=6 → n=2
n=4 → n=2
n=5 → n=2
n=3 → n=2
Figure 6.13.
Problem: using Fig. 6.13, predict which of the
following transitions produces the longest
wavelength. n = 2 to n = 1, n = 3 to n = 2, or n
= 4 to n = 3? Ans: n = 4 to n = 3.
Figure 6.13.
Problem: using Fig. 6.13, predict which of the
following transitions produces the longest
wavelength. n = 2 to n = 1, n = 3 to n = 2, or n
= 4 to n = 3? Ans: n = 4 to n = 3.
The energies of individual levels in the
hydrogen atom
Note: One can calculate the energy of
any level in the H-atom from the Bohr
model. For a level with quantum
number n, the energy is given by:
En = - RH x (1/n2)
Limitations of the Bohr model.
The Bohr model only works for atoms/ions with a
single electron (e.g. H or He+). It cannot account for
the more numerous lines in multi-electron atoms
such as Na or He. It does, however, make two
important postulates:
1) electrons exist only in discrete energy levels, and
2) energy is involved in moving an electron from one
level to another. Energy is absorbed as photons to
excite the electron from one level to a higher energy
level, and is emitted as photons in dropping to a
lower energy level.
6.4. The wave behavior of matter.
Louis de Broglie proposed that matter also had a wave
property. For any particle
λ = h/mv
(I have colored ‘v’ (velocity) red to
distinguish it from ‘v’ (frequency)
(E = hv, E = mv2, so hv = mv2, λ = v/v, so λ = hv/mv2)
where m is its mass, and v is its velocity. The quantity
mv is its momentum. The electron thus is not only a
particle, but also a wave. This was confirmed in that
electrons can be diffracted by crystals. Since electrons
are waves, it is inappropriate to treat them only as
particles, as in the Bohr model of the atom.
Problem:
What is the wavelength of an electron moving at a speed
of 5.97 x 106 m/s? (h = 6.63 x 10-34 J-s and 1 J = 1 Kgm2/s2, mass of electron = 9.11 x 10-28 g)
λ = h/mv
mass of electron in kg = 9.11 x 10-28 g x 1 kg
1000 g
= 9.11 x 10-31 kg
λ=
6.63 x 10-34 J-s______ x 1 kg-m2/s2
9.11 x 10-31 kg x 5.97 x 106 m/s
1J
=
1.22 x 10-10 m or 0.122 nm
Note: for an ordinary object like a golf ball m is so large that λ is too short
to measure.
The uncertainty principle.
If the electron is a wave, we cannot
state its position with any
accuracy. Werner Heisenberg
postulated the uncertainty
principle, which states that we
cannot precisely know the exact
momentum of an electron and also
its location in space.
This means that we cannot use the
Bohr model of the atom which has
the electron as a particle circling
the nucleus. We must instead
consider it as a wave.
Werner Heisenberg
(1901-1976)
6.5. Quantum mechanics and Atomic
Orbitals.
In 1926 Erwin
Schrodinger
proposed the
wave equation
which led to
quantum
mechanics. All
we can predict
is a probability
density (Ψ2)
(psi-squared)
for the electron.
Erwin Schrödinger (1887-1961)
Orbitals and Quantum numbers:
The solution to the Schrodinger waveequation leads to a set of wavefunctions that
yields 4 types of quantum numbers instead
of the single quantum number yielded by the
Bohr model.
These are:
1) The principal quantum number, n, which
has values of 1,2,3,… This corresponds to
the quantum number n in the Bohr model of
the atom.
Quantum numbers (contd.)
2) The Azimuthal quantum number l, which has values
of 0 to (n-1) for each value of n. The different values of l
correspond to orbital types as follows:
l
=
0
1
2
3
letter used
=
s
p
d
f
3) The magnetic quantum number ml, can have values
of –l through 0 to +l for each value of l.
Value of l
0
1
2
3
possible values of ml
0
-1,0,+1
-2,-1,0,+1,+2
-3,-2,-1,0,+1,+2,+3
4) The spin quantum number (ms). This can
have values of +½ or –½. This means that for
each value of ml there are two values of ms. It
is this that leads to the occupation of each
orbital by two electrons of opposite spin, i.e.
with ms = +½ or –½.
These quantum numbers lead to the shells
(different values of n) and subshells
(different values of l) that lead to our modern
understanding of chemistry. The number of
orbitals in each sub-shell (1 for s, 3 for p, 5
for d, and 7 for f sub-shells) is determined by
ml, and ms determines that only two
electrons of opposite spin can occupy each
orbital.
shells
subshells
orbitals
n
l
orbital
ml
_________________________________________________
= 0,1,..(n-1)
= -l, -(l-1),..0,..(l-1), l
1
0
1s
0
2
0
2s
0
1
2p
-1, 0, +1
3
0
3s
0
1
3p
-1, 0, +1
2
3d
-2,-1, 0, +1, +2
4
0
4s
0
1
4p
-1, 0, +1
2
4d
-2, -1, 0, +1, +2
3
4f
-3, -2, -1, 0, +1, +2, +3
__________________________________________________
Representation of orbitals:
Schrödinger’s model:
z
y
x
s-orbital
p-orbital
(1 of 3)
d-orbital
(1 of 5)
f-orbital
(1 of 7)
s-orbitals
z
1s, 2s, 3s, 4s,....
y
x
p-orbitals
z
z
y
x
2pz
z
y
y x
x
2py
2px
d-orbitals
y
z
z
y
x
dxz
x
dyz
dxy
y
z
d(x2-y2)
dz 2
x
6.7 Many-electron atoms.
Electron spin and the Pauli exclusion
principle.
The Pauli exclusion principle states that no
two electrons in an atom can have all four
quantum numbers, n, l, ml, and ms, the
same. The result of this is that each orbital
can hold a maximum of two electrons, with
ms = +½ or -½.
Failure of the Bohr model for
many-electron atoms.
For the H-atom, all the sub-shells within the
same shell have the same energy. This is
why the Bohr model of the atom works for
the H-atom. It generates only a single
quantum number, n, which predicts only the
energy of shells, but not the existence of
sub-shells. On the other hand, for a manyelectron atom, the sub-shells within a single
shell do not have the same energy. This is
the cause of the failure of the Bohr model
when applied to a multi-electron atom.
The H-atom compared to many-electron
atoms:
4p
n=4
n=4
4s
4p
4d
4f
4s
n=3
3s
3p
3d
3s
n=1
1s
H-atom. All subshells within
same shell have the same energy.
2p
n=2
energy
energy
2p
3p
n=3
n=2
2s
3d
2s
n=1
1s
Many-electron atom. Subshells
within same shell have different
energies.
Section. 6.8 Electron Configurations.
The way in which the electrons are
distributed among the various orbitals of the
atom is know as the electron configuration.
The orbitals are filled in order of increasing
energy, two electrons of opposite spin per
orbital. This is known as an orbital diagram:
Li
1s
2s
Electron configurations (contd.)
Each orbital is represented by a box. A half-arrow
pointing up represents an electron with a positive
spin (ms = +½), and arrow pointing down represents
an arrow with a negative spin (ms = -½ ). This is often
referred to as ‘spin-up’ and ‘spin-down’.
arrows pointing
in opposite
directions =
opposite spins
box represents
one orbital
1s
2s
Electrons having opposite spins are said to be spinpaired when they are in the same orbital.
Examples of electron configurations:
Electron configurations can be represented
by little boxes (one per orbital) with arrows to
indicate electron population of the orbitals.
Alternatively, these can be written out with
superscripts to show number of electrons in
each orbital:
Oxygen atom:
Phosphorus:
Sodium:
indicate no of
electrons in sub-shell
1s22s22p4
1s22s22p63s23p3
1s22s22p63s1
indicate principal quantum number
Comparison of different representations
of electron configurations:
Oxygen atom:
1s22s22p4
1s
Phosphorus:
2p
1s22s22p63s23p3
1s
Sodium:
2s
2s
2p
3s
1s22s22p63s1
1s
2s
2p
3s
3p
Hund’s rule.
Orbitals that have the same energy are known as
degenerate. Hund’s rule states that for
degenerate orbitals, the lowest energy is attained
when the number of electrons with the same spin
is maximized.
Thus with phosphorus the three electrons in the
3p sub-shell spread out to occupy it evenly, or
the four electrons in the 2p shell of oxygen do
the same:
3p
phosphorus:
oxygen:
2p
Electrons in highest
occupied sub-shell
spread out to occupy
it as evenly as possible
Condensed Electron Configurations.
In order to avoid writing out all the electrons
in an atom, the electronic configuration is
abbreviated by writing only the electrons in
the outermost occupied shell, the valence
shell. This is called a condensed electron
configuration. e.g.
Na:
Li:
P:
[Ne]3s1
[He]2s1
[Ne]3s23p3
Transition metals:
The row ended by Ar marks the beginning of
the 4th row (n = 4). In this row we encounter
for the first time transition elements, where
the 3d orbital is being filled up. Note that the
4s orbital is lower in energy than the 3d
orbitals, so that the 3d orbital does not start
to fill right away. The first two electrons, for
K and Ca, go into this orbital, so we have:
K
Ca
[Ar]4s1
[Ar]4s2
Number of electrons in the 3d orbitals of
transition metal ions:
After the 4s orbital has been filled, we
start to fill the 3d orbital in accordance
with Hund’s rule, since the 3d sub-shell
is the highest energy subshell.
e.g.
Mn:
[Ar]4s23d5
Zn:
[Ar]4s23d10
Ni:
[Ar]4s23d8
Cu:
[Ar]4s23d9
Lanthanides and Actinides.
Here we are filling up f-orbitals,
although the energies of the 4f and 5d
orbitals are very close to each other,
and the electrons can also occupy the
5d orbitals:
La:
Lu:
[Xe]6s25d1
[Xe]6s25f14
6.9. Electron configurations and the
periodic table.
Each group of elements in the periodic
table has a characteristic electronic
configuration. e.g. halogens have the
characteristic configuration ns2np5 (n =
2 = F, n = 3 = Cl, n = 4 = Br, n = 5 = I) for
the highest energy occupied shell. The
valence electrons determine the
chemical properties of the elements,
and occupy the highest energy shell.
Problem: Write the electronic configuration of Bi
In solving problems of this type use the
periodic table as your guide, The orbitals fill
up following exactly along the rows in the
periodic table with increasing Z:
1s
1s
2s
3s
2p
d-orbitals
3p
4s
3d
4p
5s
6s
4d
5d
5p
6p
Bi
Lanthanides fit in here
1s22s22p63s23p64s23d104p65s24d105p66s24f14
5d106p3