1. Chapter 6 Electronic Structure of Atoms
Lecture Presentation
Chapter 6 Electronic Structure of Atoms
James F. Kirby
Quinnipiac University
Hamden, CT

2. Electronic Structure
This chapter is about electronic structure—the arrangement and energy of electrons in atoms.
Why do we get colors?
Why do different chemicals
give us different colors? 
Ans: By emitting light of specific
wavelength in the visible region.
Extremely small particles have properties, some of them can be explained by wave nature and others by particle nature (waves & particles = wavecles)
Different elements display different colors (K, Ba, Sr salts with Al, Mg, Charcoal and sulfur fuels).
 light as a wave, refraction and diffraction are two examples

3. Electromagnetic Radiation
One of the ways that energy travels through space.
Three characteristics:
Wavelength (l)
Frequency (n)
Speed  (c)

4. 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.
Frequency (n) is the number of waves that pass through a particular point in 1 second (cycles/s = Hertz).
Speed(c) speed of light in vacuum (3.00 x 108 m/s, SI units).
=186,000 miles/s or ~300,000 km/s
A frequency of 1 hertz means that something happens once a second. Wavelength tells you the type of light. Amplitude tells you about the intensity of the light
Since all E-M waves do not vary in height or depth as they go along, we can use a point halfway as our measuring origin for amplitude.
For E-M radiations of all kinds, we substitute "c" for "V" and therefore the velocity is a constant.
Wavelength tells you the type of light. Amplitude tells you about the intensity of the light. Frequency tells about wavelength-short or long.

5. Electromagnetic Radiation
All electromagnetic radiation are different forms of light
All electromagnetic radiation travels at the same velocity: The speed of light (c) is 3.00  108 m/s (=186,000 miles/s).
c = 

6. What is the frequency of light with wavelength of 500nm?
ln = c
n = c/l
n = 3.00 x 108 m/s / 500 x 10-9m
n = 6.00 x 1014 C/s-1 or Hertz

7. ln = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m
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 n
ln = c
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

8. The Nature of Energy
The wave nature of light does not explain how an object can glow when its temperature increases.

9. The Nature of Energy—Quanta “Heated Solids Problem”
When solids are heated, they
emit electromagnetic radiation
over a wide range of wavelengths (l).
Radiant energy emitted by an
object at a certain temperature
depends on its wavelength
(hence, frequency n).
Max Planck (1900) explained it by
assuming that energy (E) is emitted or absorbed in discrete units (quantum) of size hn. E = h
where h is Planck’s constant,
6.626  10−34 J∙s.

10. The Photoelectric Effect
Einstein (1905) used quanta to explain the photoelectric effect.
Each metal has a different energy at which it ejects electrons.
No es emitted by a given metal below threshold frequency (ν0) regardless of intensity of light.
With greater than ν0 the number of es emitted increased with intensity (intensity depends on length from source, not on frequency)
The KE of es increased with increased ν
KE e-
hn = KE + W
or KE = hn - W
where W is the work function and depends how strongly electrons are held in the metal
He concluded that light has both-wave and particle nature. Photon is a “particle” of light and energy of photon is proportional to its frequency (E = h)

11. E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
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 nm.
E = hn
(n = c/ l)
E = h x c / l
E = 6.63 x (J•s) x 3.00 x 10 8 (m/s) / x 10-9 (m)
E = 1.29 x J

12. Atomic Emissions
Another mystery in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules.

13. Continuous vs. Line Spectra
For atoms and molecules, one does not observe a continuous spectrum (the “rainbow”), as one gets from a white light source.
Only a line spectrum of discrete wavelengths is observed. Each element has a unique line spectrum.
VIBGYOR

14. The Hydrogen Spectrum 1 n2
Johann Balmer (1885) discovered a simple formula relating the four lines to integers. Line spectrum – each line corresponds to a discrete wavelength:
Johannes Rydberg advanced this formula.
Neils Bohr explained why this mathematical relationship works. 1 =
( ) 1 n2 l
Johann Balmer, Swiss mathematical physicist.
Johannes Rydberg , Swedish physicist 
Neils Bohr, Danish physicist

15. The Bohr Model Atom (1913)
Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:
Electrons in an atom can only occupy certain orbits (corresponding to certain energies).
Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.
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 = h
More –ve value means more stable the electron or more tightly bound to nucleus.

16. The Bohr Model
The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation
E = −hcRH
( ) 1
nf2
ni2 –
where RH is the Rydberg constant,  107 m−1, and ni and nf are the initial and final energy levels of the electron.
−hcRH =
-2.18 x J

18. l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
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. i f
DE = -RH
( ) 1 n2
Ephoton =
Ephoton = x J x (1/25 - 1/9)
Ephoton = DE = x J
E = hn
Ephoton = h x c / l
E = h c / l
l = h x c / Ephoton
l = 6.63 x (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm

19. Limitations of the Bohr Model
It only works for hydrogen!
Classical physics would result in an electron falling into the positively charged nucleus. Bohr simply assumed it would not!
Circular motion is not wave-like in nature.

20. Important Ideas from the Bohr Model
Points that are incorporated into the current atomic model include the following:
Electrons exist only in certain discrete
energy levels.
2) Energy is involved in the transition of
an electron from one level to another.

21. Quantum Mechanics
Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated.
This is known as quantum mechanics.
Psi is a symbol for wave function.

22. Quantum Mechanics
The solution of Schrödinger’s wave equation is designated with a lowercase Greek psi ().
The square of the wave equation, 2, gives the electron density, or probability of where an electron is likely to be at any given time.

23. Quantum Numbers
Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies.
Each orbital describes a spatial distribution of electron density.
An orbital is described by a set of three quantum numbers.

24. Principal Quantum Number (n)
The principal quantum number, n, describes the energy level on which the orbital resides.
The values of n are integers ≥ 1.
These correspond to the values in the Bohr model.

25. Angular Momentum Quantum Number (l)
This quantum number defines the shape of the orbital.
Allowed values of l are integers ranging from 0 to n − 1.
We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.

26. Magnetic Quantum Number (ml)
The magnetic quantum number describes the three-dimensional orientation of the orbital.
Allowed values of ml are integers ranging from −l to l:
Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, and so forth.
for a given value of l
ml = -l, …., 0, …. +l
if l = 0 (s orbital), ml = s orbital
if l = 1 (p orbital), ml = -1, 0, p orbitals
if l = 2 (d orbital), ml = -2, -1, 0, +1, d orbitals
if l = 3 (f orbital), ml = , -1, 0, +1, + 2, +3 7f orbitals

27. Magnetic Quantum Number (ml)
Orbitals with the same value of n form an electron shell.
Different orbital types within a shell are subshells.

28. s Orbitals The value of l for s orbitals is 0.
They are spherical in shape.
The radius of the sphere increases with the value of n.

29. s Orbitals For an ns orbital, the number of peaks is n.
For an ns orbital, the number of nodes (where there is zero probability of finding an electron) is n – 1.
As n increases, the electron density is more spread out and there is a greater probability of finding an electron further from the nucleus.

30. p Orbitals The value of l for p orbitals is 1.
They have two lobes with a node between them.

31. d Orbitals The value of l for a d orbital is 2.
Four of the five d orbitals have four lobes; the other resembles a p orbital with a doughnut around the center.

32. f Orbitals Very complicated shapes
Seven equivalent orbitals in a sublevel
l = 3

33. Energies of Orbitals—Hydrogen
For a one-electron hydrogen atom, subshells on the same energy level have the same energy.
Chemists call them degenerate orbitals.

34. Energies of Orbitals— Many-electron Atoms
As the number of electrons increases, so does the repulsion between them.
Therefore, in atoms with more than one electron, not all orbitals on the same energy level are degenerate (equal).
Orbital sets in the same sublevel are still degenerate.
Energy levels start to overlap in energy (e.g., 4s is lower in energy than 3d.)

35. Spin Quantum Number, ms
In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy.
The “spin” of an electron describes its magnetic field, which affects its energy.
This led to the spin quantum number, ms.
The spin quantum number has only two allowed values, +½ and –½.
Two possible spinning motions of an electron result in two different magnetic fields.

36. Orbital Summary
principal quantum number, n: related to size and energy of the orbital
angular momentum quantum number, l:related to shape of the orbital
magnetic quantum number, ml : related to orientation of the orbital
spin quantum number, ms : related to electron spin orientation
Allowed values:
• n = 1,2,3 …,n
• l = 0,1,2,…,(n-1)
• ml = (-l, …, 0. …+l)
• ms = +1/2, -1/2
Orbital Energy & Size increases with n: At higher n, the electron is farther away from the nucleus...this is a higher energy configuration. There is a larger area of space where you are likely to find an electron at higher E’s.
Orbital shape is the same no matter the value of n: 3s looks like 1s, except it’s bigger and has more nodes. Same for p, d, f, etc.
Number of nodes in an orbital goes as n – 1: 1s has zero nodes, 2s has one node, 3s has two nodes... 2px, 2py, 2pz each have one node, 3px, 3py, 3pz each have two nodes the 3d orbitals each have two nodes, 4d have three, etc.
(Note that number of nodes indicates relative energy)
All atoms have all orbitals…but in an unexcited atom, only those closest
to the nucleus will be occupied by electrons

37. Pauli Exclusion Principle
No two electrons in the same atom can have exactly the same energy.
Therefore, no two electrons in the same atom can have identical sets of quantum numbers.
This means that every electron in an atom must differ by at least one of the four quantum number values: n, l, ml, and ms.
If you have two electrons in an orbital, ms can
be +½ & -½ , but not +½ & +½ and -½ & -½

38. Electron Configurations
The way electrons are distributed in an atom is called its electron configuration.
Three rules:
1.Electrons fill orbitals first starting with lowest energy orbitals and moving upwards (Aufbau principle)
2. No more than two electrons can be placed in each orbital and no two electrons can fill one orbital with the same spin (Pauli exclusion principle)
(no two electrons in an atom can have the same four quantum numbers.)
3. For degenerate (equal energy) orbitals, electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule). The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins)

39. Orbital Quiz • The shape of a given type of orbital changes as n
increases.
• The number of types of orbitals (subshells) in a given energy level is the same as the value of n.
• The hydrogen atom has a 3s orbital.
• The number of lobes on a p-orbital increases as n increases. That is, a 3p orbital has more lobes than a 2p orbital.
• The electron path is indicated by the surface of
the orbital
F, T, T, F, F

40. Electron Configurations
The most stable organization is the lowest possible energy, called the ground state.
Each component consists of
a number denoting the energy level;
a letter denoting the type of orbital;
a superscript denoting the number of electrons in those orbitals.
4p5
4p5
4p5

41. Orbital Diagrams Each box in the diagram represents one orbital.
Half-arrows represent the electrons.
The direction of the arrow represents the relative spin of the electron.

42. Hund’s Rule
“For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”
This means that, for a set of orbitals in the same sublevel, there must be one electron in each orbital before pairing and the electrons have the same spin, as much as possible.

43. Condensed Electron Configurations
Elements in the same group of the periodic table have the same number of electrons in the outer most shell. These are called valence electrons.
The filled inner shell electrons are called core electrons. These include completely filled d or f sublevels.
We write a shortened version of an electron configuration using brackets around a noble gas symbol and listing only valence electrons.

44. What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
= 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
core electrons
valence electrons
What are the possible quantum numbers for the last (outermost) electron in Cl?
Cl 17 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5
= 17 electrons
Last electron added to 3p orbital
n = 3
l = 1(p)
ml = -1, 0, +1
ms = ½ or -½

45. Periodic Table We fill orbitals in increasing order of energy.
Different blocks on the periodic table correspond to different types of orbitals: s = blue, p = pink (s and p are representative elements); d = orange (transition elements); f = tan (lanthanides and actinides, or inner transition elements)

46. Order of orbitals (filling) in multi-electron atom
Look for n+l value and if it is same for two, the one with lower n value comes first (lower energy)
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

47. Some Anomalies
Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.

48. Chromium as an Anomaly
For instance, the electron configuration for chromium is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4.
This occurs because the 4s and 3d orbitals are very close in energy.
These anomalies occur in f-block atoms with f and d orbitals, as well.

49. The reason for this irregularity is that a slight stability is associated with the half-filled (d5) and completely filled (d10) subshells.

50. How many unpaired electrons are present in Al,N,Si,S?
Al (13): 1s22s22p63s23p1 1
N (7): 1s22s22p3 3
Si (14): 1s22s22p63s23p2 2
S (16): 1s22s22p63s23p4 2