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Chapter 6 Lecture notes

Acknowledgements: Some of the images are adopted from BLB textbook, the only purpose of which is to enhance student learning.

When atoms react, electrons interact. Electrons in an atom are arranged in a certain way, giving electronic structure. All the electrons in an atom are not exactly the same. To learn more about the electrons in an atom we use help from quantum theory. In quantum mechanics laws of physics and complicated mathematical equations are applied. We will deal with the simpler forms of the applications.

6.1  Wave nature of light

To understand the electrons and their energies we need to take a step back and understand electromagnetic radiation. All types of electromagnetic radiation move through vacuum at a speed of 3.00 x 108 m/s, the speed of light. They all have wave-like characteristics. A wave has crests and toughs. 

 Wavelength, λ

 

 Frequency, υ

 

υλ = c
υ ↑ λ ↓
, inverse proportionality

υ = frequency, λ = wavelength
c = speed of light, 3 x 108 m/s

The electromagnetic spectrum spans over 1024 s–1 frequency or 10–12 to 104 meter λ, out of which the visible range is 400–750 nm.

 

Problem 7. a) What is the frequency of radiation that has a wavelength of 0.452 pm?

 

 

 

b) What is the wavelength of radiation that has a frequency of 2.55 x 1016s–1?

 

 

 

c) Would the radiations in part (a) or part (b) be visible to the human eye?

d) What distance does electromagnetic radiation travel in 7.50 ms?

 

 

 

6.2    Quantized energy and photon

 Along with wave like characteristics light also posses particle behavior which can explain:
  1.  Emission of light from hot objects, black body radiation
  2.  Emission of electrons from metal surfaces on which light shines, photoelectric effect
  3.  Emission of light from electronically excited gaseous atoms, emission spectra

In 1900 Max Planck in his “quantum theory” related the energy of an absorbed or an emitted electromagnetic radiation with its frequency (wavelength). The energy is discrete or quantized (fixed amount).

E = hυ              h = Planck’s const., 6.626 x 10–34 J.s

According to his theory energy is emitted or absorbed in hυ , 2hυ, 3hυ, and so forth. [3hυ means three quanta.]

 

 

Problem 13. a) Calculate the smallest increment of energy (a quantum) that can be emitted or absorbed at a wavelength of 812 nm.

 

 

 

b) Calculate the energy of a photon of frequency 2.72 x 1013s–1.

 

 

 

c) What wavelength of radiation has photons of energy 7.84 x 10–18 J? In what portion of the electromagnetic spectrum would this radiation be found?

 

 

 

Problem 19. A diode laser emits at a wavelength of 987 nm. All of its output energy is absorbed in a detector that measures a total energy of 0.52 J over a period of 32 s. How many photons per second are being emitted by the laser?

 

 

 

 

 

In 1905 Albert Einstein used Planck’s quantum theory to explain the photoelectric effect of light. Radiant energy striking a shiny metal surface is a stream of tiny packets, called photons that behave like tiny particles. Each photon must have an energy proportional to the frequency of light: E = hυ. Thus radiant energy is quantized.

            Energy of photon, E = hυ

When a photon strikes the metal, the energy may be transferred to an electron in the metal. If the amount of energy is enough to overcome the attractive forces in the atom, the electron is ejected; the excess energy appears as kinetic energy of emitted electron.

 

6.3    Line spectra and Bohr model

Line spectra – Electromagnetic radiation when emitted generate line spectra. Rydberg equation relates between the lines of a spectra and wavelength (or frequency) when electrons are emitted.

 

RH = Rydberg const. 2.18 x 10–18 J or 1.096776 x 107 m–1
n1 = lower line, n2 = higher line

Bohr model of atom

According to the Bohr model an atom is a micro solar system. Orbit in fixed paths with fixed energy (allowed values). Energy is absorbed when electrons jump up to the higher orbit and emitted when they jump down to a lower orbit.

En = (–2.18 x 10–18 J)(1/n2)                  n = levels, quantum numbers

For a one-electron system (e.g. H atom) n1, is the general state (lowest energy level), n2, n3, etc. are the excited states. The values of n are integral and up to infinity.

            E = (–2.18 x 10–18 J)(1/∞2) = 0

This means the electron is the farthest away from the nucleus. About here it goes out of the nuclear attraction producing an ion. Therefore, the highest energy of an electron is zero joule. All other levels possess negative energy. When an electron jumps down from a higher level to a lower level.

            ΔE = Ehi – Elo = hc/ λ

            hc/λ = RH[1/nlo2 – 1/nhi2]           c/λ = υ

            υ = RH/h[1/nlo2 – 1/nhi2]

Bohr model works well with hydrogen atom where there is only one electron. Energy calculations start to deviate from Bohr model in a multi-electron system, due to the repulsion energy between them. Quantum mechanics model helps clean the calculations by using Schrödinger equation.

Problem 25.  Is energy emitted or absorbed when the following electronic transitions occur in hydrogen?

a) from n = 4 to n = 2

b) from an orbit of radius 2.12 Å to one of radius 8.48 Å

c) an electron adds to the H+ ion and ends up in the n = 3 shell.

Problem 27. Using Equation 6.5, calculate the energy of an electron in the hydrogen atom when n = 2, and when n = 6. Calculate the wavelength of the radiation released when an electron moves from n = 6 to n = 2. Is this line in the visible region of the electromagnetic spectrum? If so, what color is it?

 

 

 

 

 

 

Skip 6.4

6.5  Quantum mechanics and atomic orbitals

The solution of Schrödinger eqn. generates set numbers called quantum numbers. These numbers describe the probability of finding an electron x distance away from the nucleus. Electron density is another way of expressing probability. All together these quantum numbers describe the distance, shape, orientations of the electron clouds for an allowed energy state.

  Schrödinger equation in one dimension:
After solving Schrödinger equation the one electron of hydrogen atom show its probability of finding by the color shade which gives an appearance of an electron cloud.

Orbitals and quantum numbers

Bohr introduced a single quantum number, n., to describe an orbit. The quantum mechanical model uses three quantum numbers, n, l and, ml to describe an orbital.

1. Principal quantum number, n, primary energy level

n = 1, 2, 3 . . . etc . . . . ∞  all integral values.

Increases in n values means orbitals are larger, electrons spend more time away from the nucleus, these electrons are less tightly bound to the nucleus. This quantum number provides the principal amount of energy of the level.

2. Azimuthal quantum number, l, sublevel

This quantum number defines the shape of the electron cloud (orbital). The value of l is from 0 to n–1.

 

 

3. Magnetic quantum number, ml

This quantum number describes the orientation of the orbital (electron cloud) in space. 
The allowed values range from l, l-1, …, 0,… –
l.

 

 

 

Observations:

1. Shell number “n: will have n subshells.  Relationship between n and l.
Shell, n Subshell, l  Subshell designation
1 0 1s
2 0
1
2s
2p
     
2. Relationship between l and ml: each subshell contains (2l + 1) orbitals.
Subshell, l Orbital, ml  
0 0 (1)
0 1
0
1
(3)
2 2
1
0
1
2
(5)
     
3. Total number of electrons in shell number is “n2”.
n l ml (= n2) electrons (= 2n2)
1 0 (s) 0 one 2
2 0 (s)
1 (p)
0 one
1, 0, 1 three
8

6.6  Representation of orbitals

The “s” orbitals

These are the lowest energy orbitals in any shell.  They are spherical.

The “p” orbitals

The electron density in these orbitals is concentrated into two regions on either side of the nucleus, giving a “two lobed” dumbbells.

 

The “d” orbitalsThese orbitals look like “4-leaf clovers”

Problem 41. a) For n = 4, what are the possible values of l?

b) For l = 2, what are the possible values of ml?

Problem 43. Give the numerical values of n and l corresponding to each of the following designations:

n

l

a)

3p

 

 

b)

2s

 

 

c)

4f

 

 

d)

5d

 

 

Problem 45. Which of the following represent impossible combinations of n and l. Explain why.

Explanation

a)

1p

 

b)

4s

 

c)

5f

 

d)

2d

 

Problem 47. Sketch the shape and orientation of the following types of orbitals:

a)  s                                             b)  pz                                            c)  dxy.

 

6.7   Many-electron atoms

In atoms where there are more than one electron, there exists the repulsion between the electrons. This causes a split in the sublevels, s becomes the lowest energy, p is next, d is next, and f is the last (so far). However, the different orbitals of the same sublevel (p, d, or f) possess the same energy and thus called “degenerate” orbitals.

Electron spin and Pauli Exclusion Principle

In a many-electron system scientists found a pair of lines for every line spectrum. This generated the discovery of “electron spin” – an electron behaves like a tiny sphere spinning on its own axis. During the spinning they generate tiny electromagnetic fields. The two electrons in any orbital must spin the opposite direction, so they do not repel each other. From here came the term “spin magnetic quantum number, ms. The two possible values of ms in any orbital are +½ and –½, indicating the opposed spins.

Pauli exclusion principle states that “no two electrons in an atom can have the same set of four quantum numbers, n, l, ml, and ms”. An orbital can hold only two electrons, and they must have opposite spins.

6.8  Electron configurations

Arrangements of electrons in an atom in the increasing order of energy.

  • n1 < n2 < n3 . . .         1s < 2s < 3s . . .

  • 2p < 3p < 4p . . .        3d < 4d < 5d . . .

But it is not like: 1s2s2p3s3p3d4s4p4d4f . . . .

Rather it is: 1s2s2p3s3p4s3d4p5s4d5p6s4f5d6p7s . . . .

The afbau principle – gives you the right order of the orbitals with few exceptions.

f

X

X

X

4f

5f

6f

7f

d

X

X

3d

4d

5d

6d

7d

p

X

2p

3p

4p

5p

6p

7p

s

1s

2s

3s

4s

5s

6s

7s

 

1

2

3

4

5

6

7

 

 

In writing electron configurations note the following.
No more than 2 electrons per orbital.

s sublevel   1 orbital   1 x 2 = 2 e max.
p sublevel  3 orbital   3 x 2 = 6 e max.
d sublevel   5 orbital   5 x 2 = 10 e max.
f sublevel  7 orbital   7 x 2 = 14 e max.

Electron configuration describes the first three quantum numbers.

n by number
l by letter
superscripts for number of electrons in the sublevel

Hund’s rule

Hand’s rule states that “for degenerate orbitals, the lowest energy is attained when the number of electrons with same spin is maximized”. This means that you have to place one electron in each degenerate orbital before doubling (opposed spin) up. This is an extra info that is not provided by electron configuration.

C (6e):   1s22s22p2       ↑↓     ↑↓     _   _  ___
                                         1s     2s            2p

Condensed (abbreviated) electron configurations:
In this form of electron configuration the number of electrons of the nearest noble gas of lower atomic number is represented by its chemical symbol in brackets.

Li (3e):  [He]2s1
Mg (12e):  [Ne]3s2

The inner shell electrons in the bracket are core electrons. The electrons after the noble gas are outer-shell electrons, or valence electrons, which participate in chemical reactions.

 

Transition metals

These elements have “d” electrons as their outermost electrons or valence electrons.

Mn (25e):  [Ar]4s23d5            or   [Ar]   ↑↓     _   _   _   _   _

Ni (28e):  [Ar]4s23d8              or   [Ar]   ↑↓     ↑↓   ↑↓  ↑↓   _   _
                                                                      4s                    3d

 

Lanthanides and actinides

Lanthanides have 4f electron as valence electrons (element 57 – 70). These are known as rare earth elements.

            La (57e):  [Xe]6s25d1         Ce:  [Xe]6s25d14f1        Pr:  [Xe]6s24f3

Actinides have 5f electrons as valence electrons following actinium.

 

 

 

6.9  Electron configurations and the periodic table

 Anomalous electron configurations – do not follow the afbau arrangement

Examples:       Cr (25e): [Ar]4s23d5 rather than [Ar]4s23d4
                         Cu (29e): [Ar]4s13d10 rather than [Ar]4s23d9

These anomalous behaviors are due to the closeness of the 3d and 4s orbital energies. It frequently occurs when there are enough electrons to lead to precisely half-filled sets of degenerate orbitals (as in Cr) or to a completely filled d subshell (as in Cu). There are few similar cases in the f-block elements. These are minor departures from the expectation, but are not of great chemical significance.

 

Problem 55. What is the maximum number of electrons that can occupy each of the following subshells?

a) 3d                                   b) 4s                                  c) 2p                                 d) 5f?

Problem 59. Write the condensed electron configurations for the following atoms, using the appropriate noble-gas core abbreviations.

a) Cs                                                                            b) Ni

c) Se                                                                            d) Cd

e) Ac                                                                           f) Pb.

Problem 61. Draw the orbital diagrams for the valence electrons of each of the following elements, and indicate how many unpaired electrons each has.

a) S

b) Sr

c) Fe

d) Zr

e) Sb

Problem 63. Identify the specific element that corresponds to each of the following electron configurations.

a)  1s22s22p63s2

b)  [Ne]3s23p1

c)  [Ar]4s13d5

d)  [Kr]5s24d105p4

Problem 65. Describe what is wrong with the following electron configurations for atoms in their ground states?

a) 1s22s23s1

b) [Ne]2s22p3

c) [Ne]3s23d5

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