CHM2C1-B Physical Spectroscopy
Electronic spectra of atoms (and—————diatomics)• Essential readings:– Atkins, Elements of Physical Chemistry, Chapter 13
Exercises 13.1-11, 13.22-25, 13.28-31 (page 322-323)– Housecroft & Sharpe, Inorganic Chemistry, Chapters 1.4-1.9, 20.6
Exercises (pages 9,14,21,22), Problem 20.12 (page 486)with particular emphasis on Atomic Term Symbols (energy levels) andElectronic Transitions in atoms (isolated, under ligand-fields).
• Materials that overlap with other modules will be treated only briefly.
• Electronic excitations in diatomic molecules will not be treated.
• The lecture slides can be downloaded fromhttp://www.chem.bham.ac.uk/labs/ando/chm2c1Slide.pdf•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

1= RHλ11−n21 n22



RH=Rydberg constant=109677cm−1
Lyman (n1=1), Balmer (n1=2),
Paschen (n1=3), Brackett (n1=4),...

Fig 13.1/P: The Spectrum of Atomic Hydrogen
Emitted light from Hydrogen discharge lamp
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Fig 13.2/P:
The Energy Levels of Hydrogen AtomFor Hydrogenic Atoms H, He+, Li2+, · · ·(i.e., one-electron atoms)
Z2Z2
En = −hcR 2 ∝ − 2nn
The constant R depends on the atomic mass
mN via the reduced mass µ:
me mN
R ∝ µ,µ=me + mN
However, because me  mN ,µ ' me and R ' RH
Self-test 13.1/P: The shortest wavelengthtransition in the Paschen series in hydrogenoccurs at 821 nm; at what wavelength doesin occur in Li2+?
•First •Prev •Next •Last •Go Back •Full Screen •Close •Quits, p, d Atomic Orbitals (Cartesian representation)•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitShells (n), Subshells (l), Orbitals (ml )Fig 13.3/P

Self-test 13.3/P: How many orbitals are there in a shell with n = 5 ?
Try: Self-study exercises/Inorg Chem, p.9,14•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitSelf-study exercises/Ip.9, left col.
1. If ml has values of -1, 0, +1, write down the corresponding value of l.
2. If l has values 0, 1, 2 and 3, deduce the corresponding value of n.
3. For n=1, what are the allowed values of l and ml ?
p9, right col.
1. Write down the possible types of atomic orbital for n=4.
2. Which atomic orbital has values of n=4 and l=2?
3. Give the three quantum numbers that describe a 2s atomic orbital.
p.14
1. Write down two possible sets of quantum numbers to describe an electron in a 3s atomic orbital.
2. If an electron has the quantum numbers n=2, l=1, ml =1 and ms=+ 21
which type of atomic orbital is it occupying?
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Fig 13.11/P: Appearance of the Spectrum of Atomic Hydrogen.
(Note: Hydrogenic atoms ⇒ Orbitalenergies depend only on n.)
Selection Rules for Hydrogenic Atoms∆l = ±1,(∆ml = 0, ±1)s ↔ p ↔ d ↔ f ···The principal quantum number n canchange by any amount.
Example 13.1/P: To what orbitals
may a 4d electron make spectroscopictransitions?
Self-test 13.5/P: To what orbitals
may a 4s electron make spectroscopictransitions?

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The building-up (Aufbau) principle (Sec 13.11/P)
1. The order of occupation of orbitals is
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, ...
2. Each orbital may accommodate up to two electrons (Pauli principle)
3. Electrons occupy different orbitals of a given subshell before doubly
occupying any one of them.
4. In its ground state, an atom adopts a configuration with the greatest
number of unpaird electrons (Hund’s rule)
Self-test 13.6/P:
Predict the ground-state electron configuration of sulfur.
Self-test 13.7/P:
Predict the electron configuration of a Cu2+ ion and an S2− ion.
Exercise 13.25/P:
Give the electron configuration of the ground states of the first 18 elementsin the periodic table.
•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitSelf-study exercises/I
p.21, left col.
1.——————————————————————————————
Where, in the above argument, is the Pauli exclusion principle applied?
2. Will the three electrons in the P 3p atomic orbitals possess the same ordifferent values of the spin quantum number?
p.21, right col.
1. Values of Z for Li, Na, K and Rb are 3, 11, 19 and 37 respectively. Write
down their ground state configuration and comment on the result.
2. How are the ground state electronic configurations of O, S and Se (Z=8,16, 34 respectively) alike? Give another element related in the same way.
p.22
1. Show that the electronic configuration 1s22s22p1 for B corresponds toeach electron having a unique set of four quantum numbers.
2. The ground state of N is 1s22s22p3. Show that each electron in the 2p
level possesses a unique set of four quantum numbers.
•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitTerm symbols (Sec 13.16/P, Box 20.5/I)(2S+1)LJ• L = total orbital angular momentum quantum number
L 0 1 2 3 ...
S P D F ...
Two electrons with l1, l2 giveL = l1 + l2, l1 + l2 − 1, · · · , |l1 − l2|• S = total spin angular momentum quantum number
S = s1 + s2, s1 + s2 − 1, · · · , |s1 − s2|(But it’s easier to count the number of up and down spins)
2S + 1 = spin multiplicity (singlet, double, triplet, ...)
• J = total angular momentum quantum number
J = L + S, L + S − 1, · · · , |L − S|
Self-test 13.8/P
What terms and levels can arise from the configuration 4p13d1?
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Spin-orbit CouplingFig 13.17/P: The magnetic interaction responsible for spin-orbit coupling.
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Hund’s Rule for the Lowest Energy Term of a given configuration
1. Highest Multiplicity 2S + 1
2. Highest value of L
 Lowest value of J when less than half-filled
3. or Highest value of J when more than half-filled
Example: ...4p1 3d1 configuration(Self-test 13.8/P)⇒ 1F3, 1D2, 1P1, 3F4,3,2, 3D3,2,1, 3P2,1,0
1. ⇒ 3F4,3,2, 3D3,2,1, 3P2,1,0
2. ⇒ 3F4,3,2
3. ⇒ 3F2
In fact, it’s much easier to find the lowest term only:1. two unpaired electrons ⇒ highest S = 1 ⇒ 2S + 1 = 3
2. one in p and one in d ⇒ highest L = 1 + 2 = 3 ⇒ 3F
3. L = 3, S = 1 ⇒ J = 4, 3, 2; less than half-filled ⇒ 3F2•First •Prev •Next •Last •Go Back •Full Screen •Close •QuitExercise 13.31/P
The ground configuration of a Ti2+ ion is [Ar]3d2.
What is the term of lowest energy and which level of that term lies lowest?
1. two unpaired electrons ⇒ highest S = 1 ⇒ 2S + 1 = 3
2. two in d in parallel spin ⇒ highest L = 1 + 2 = 3 ⇒ 3F
3. L = 3, S = 1 ⇒ J = 4, 3, 2; less than half-filled ⇒ 3F2Now, try:CNO
Cr3+Mn3+Fe3+[He]2s2 2p2[He]2s2 2p3[He]2s2 2p4[Ar]3d3[Ar]3d4[Ar]3d5

3P0
4
S3/2
3P2
4
F3/2
5D06S5/2•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

Selection Rules for Many-electron Atoms• ∆S = 0• ∆L = 0, ±1,• ∆l = ±1 for the electron excited in the transition• ∆J = 0, ±1 but J = 0 → J = 0 is forbidden
The rules on J are exact. Those on l, L, S presume that these quantum
numbers are well-defined.
Example: Na atom
Na [Ne]ns1 ⇒ 2S1/2
Na [Ne]np1 ⇒ 2P3/2, 2P1/2
Na [Ne]nd1 ⇒ 2D5/2, 2D3/2
Observed lines include:
5S → 3P 616.07, 615.42 nm
3P → 3S 589.00, 589.59 nm
4D → 3P 568.82, 568.26 nm•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

Ligand-field Splitting
Atom 2L + 1 −→ OctahedralS1A1g
P
3T1gD5Eg + T2g
F7A2g + T2g + T1g

Fig 20.16/I: d1 ion in an Octahedral fielddegeneracies
132+31+3+3Fig 20.18/I: Orgel diagram for d2,d3, d7, d8 ions.

Try: Problem 20.12/I (p.486)•First •Prev •Next •Last •Go Back •Full Screen •Close •Quit

Colours from d-electrons

KMnF3

Y3Al5O12 : Cr

Dy3Al5O12

Al2O3 : CrRbNiF3BaAl12O19 : Fe

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