Science (Communication) and Wikipedia - Potentials and Pitfalls
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Physical Methods in Inorganic Chemistry .
Presentation · February 2014
DOI: 10.13140/RG.2.2.14093.23527/1
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Mutasem Z. Bani-Fwaz
King Khalid University
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3. Electronic Structure and Spectroscopy of TM
Complexes:
Magnetic Susceptibility
Dr.Mutasem Z. Bani Fwaz
Chem 522
UV-Visible Spectroscopy
4. Electronic Spectroscopy
Spectroscopy involving energy level transitions of the
electrons surrounding an atom or a molecule
Atoms: electrons are in
hydrogen-like orbitals
(s, p, d, f)
Molecules: electrons are in
molecular orbitals (HOMO,
LUMO, …)
(The LUMO of benzene)
(The Bohr model for nitrogen)
From
http://education.jlab.org
5. UV-Visible Spectroscopy
Definition: Spectroscopy in the optical (UV-Visible) range
involving electronic energy levels excited by
electromagnetic radiation (often valence electrons).
8. Each electronic state has its own term symbol
spin multiplicity
L = 0 S term
L = 1 P term
L = 2 D term
L = 3 F term
Within each term, there can be several degenerate microstates with different ML and MS
Use Russell Saunders Coupling to describe electron-electron repulsion
L = 0, 1, 2…total orbital angular momentum (term)
ML = -L…+L component of L (ML = S ml)
S = total spin quantum number (S = S s)
Ms = -S….+S component of S (MS = S ms)
2S+1
LJ
J = |L+S| ...|L-S|
orbital angular momentum
spin orbit coupling
When considering
symmetry use the
Mulliken symbol
How do we label states?
9. Question-1-: What is the ground state term symbol for the
following atoms 3Li, 14Si and 21Sc?
Note: ground state term symbol for lowest energy configuration.
You must know the following rearrangement to do that and
answer the question above:
ml 0
ml +1 0 -1
ml
+2 +1 0 -1 -2
ml +3 +2 +1 0 -1 -2 -3
s-orbital
p-orbitals
d-orbitals
f-orbitals
8
Dr. Mutasem Zaki Bani Fawaz
10. 3Li: 1s22s1
ml 0
21Sc: 1s22s22p63s23p63d14s2
ml +2 +1 0 -1 -2
14Si: 1s22s22p63s23p2
ml +1 0 -1
S ml = 0
S ms = + 1/2
S ml = 2
S ms = +1/2
S ms = +1/2
S ml = +1 + 0 = 1
+1/2 = +1
symbol
2S + 1 = 2(1/2) + 1 = 2
L = 0 S
term symbol 2S1/2
J = L + S.... ...
untile absolute value of
L - S
J = 1/2
2S + 1 = 2(1/2) + 1 = 2
L = 2 D
J = 5/2 and 3/2
2D5/2
2D3/2
2S1/2
2S + 1 = 2(1) + 1 = 3
L = 1 P
J = 2, 1, 0
3P2, 3P1, 3P0
3P0
most stable
most stable
most stable
Dr. Mutasem Zaki Bani Fawaz
9
11. Why Molecular term symbols are important?
As soon as ligands are coordinated to a transition-metal centre, a molecular symmetry
is imposed and the d orbital split according to the ligand field. In these cases,
molecular term symbols have to be used.
Term Components in an octahedral field
S A1g
P T1g
D T2g+ Eg
F A2g + T2g + T1g
G A1g + Eg + T2g + T1g
H Eg + 2 T1g + T2g
I A1g + A2g + Eg + T1g + 2 T2g
Similar splittings occur in a tetrahedral field, but the g labels are no longer applicable.
10
Dr. Mutasem Zaki Bani Fawaz
12. Selection rules
(determine intensities)
Laporte rule
g g forbidden (that is, d-d forbidden)
but g u allowed (that is, d-p allowed)
Spin rule
Transitions between states of different multiplicities forbidden
Transitions between states of same multiplicities allowed
These rules are relaxed by molecular vibrations, and spin-orbit coupling
11
Dr. Mutasem Zaki Bani Fawaz
15. Qualitative Explanation
Consider a d3 system complex in
octahedral environment.
The ground state configuration is:
________ A transition from the
dz2 dx2-y2 dxy to the dx2-y2, or the
dyz or dxz to the dz2
____ ____ ____ orbitals involve a
relatively
dxy dyz dxz minor change in
environment.
16. Qualitative Explanation
Since the promotion of an electron from
the t2g set of orbitals to the eg set can
involve differing changes in environment,
several peaks will be seen in the
spectrum.
17. 16
Dr. Mutasem Zaki Bani Fawaz
ml
+2 +1 0 -1 -2
d-orbitals
ML = 2 +1 +0 = 3
or L = 3
MS = 1/2 + 1/2 + 1/2 = 3/2 = S
or S = 1
2S + 1 = 2 x 3/2 + 1 = 4
4F
18. Magnetic Susceptibility
• Magnetic Properties reveal numbers of unpaired electrons
• Hund’s Rule requires maximum number of unpaired electrons in
degenerate orbitals
• Magnetic Descriptions
– Diamagnetic = all paired electrons = slightly repelled by magnetic
field
– Paramagnetic = unpaired electrons = strongly attracted by magnetic
field
• Magnetic properties are determined by one of several experimental
methods
not
17
Dr. Mutasem Zaki Bani Fawaz
19. Molar Magnetic Susceptibility = cM = cm3/mol = data from an experimental
determination
Magnetic Moment = Calculated value from cM and theoretical treatment of
magnetism. m = Bohr magnetons
Theoretical basis of magnetism
– Electron Spin: a spinning charged particle would generate a spin
magnetic moment = mS (not really spinning but a property of an
electron)
– mS = + ½ or – ½ only because the electron can either “spin” clockwise
or counter clockwise (up or down arrows)
– S = Spin Quantum Number = S mS
» Example: S = 3(+ ½ ) = 3/2
» Example: S = +1/2
18
Dr. Mutasem Zaki Bani Fawaz
)
1
(
2
S
S
s
m
)
2
(
n
n
s
m
20. Question-3-: : calculate mS for high spin Fe3+
» Determine d-electron count by using the periodic table
For iron (3+) and counting the capacity of d-orbitals by
electrons : 3d5
» Arrange the d-electrons in the 5d-orbitals as high spin
» Apply mS = g[S(S+1)]½ = 2[5/2(7/2)]½ = 5.92 Bohr
magnetons
Or [n(n+2)] ½ = [5(7)] ½ = (35) ½ = 5.92
19
Dr. Mutasem Zaki Bani Fawaz
Question and answer
21. The spin-only formula applies reasonably well to metal ions from the
first row of transition metals: (units = μB,, Bohr-magnetons)
Metal ion dn configuration μeff(spin only) μeff (experimently)
Ca2+, Sc3+ d0 0 0
Ti3+ d1 1.73 1.7-1.8
V3+ d2 2.83 2.8-3.1
V2+, Cr3+ d3 3.87 3.7-3.9
Cr2+, Mn3+ d4 4.90 4.8-4.9
Mn2+, Fe3+ d5 5.92 5.7-6.0
Fe2+, Co3+ d6 4.90 5.0-5.6
Co2+ d7 3.87 4.3-5.2
Ni2+ d8 2.83 2.9-3.9
Cu2+ d9 1.73 1.9-2.1
Zn2+, Ga3+ d10 0 0
Magnetic properties
23. 22
Dr. Mutasem Zaki Bani Fawaz
Homework-1-.
Q1) The [Ni(pyridine)4(H2O)2]Br2 has d-d absorption
bands at 27,000, 16,500, and 10,150 cm-1. No symmetry
splitting is observed, explain your answer?
Q2) Calculate the spin-only magnetic moment for
[Ni(H2O)6]Cl2.
24. 23
Dr. Mutasem Zaki Bani Fawaz
Q3) Indicate the oxidation number of the
central atom and the term symbol for the
ground state in each of the following
complexes.
a) [Pt(CN)4]2-
b) [Cr(NH3)6](NO3)3
c) Na2[CdCl4]
d) [Co(H2O)5Cl]2+
e) K3[V(C2O4)3]
32. Ligands
Common Organic Ligands
B) Binding Modes Ligands and Nomenclature
1) Bridging is possible with organometallic ligands
2e(0)
-
-
4e (-2)
6e(-3)
4e(-1)
6e(-2)
6e(-1)
6e(0)
2e(0)
2e(0) or 4e(0)
2e(-1)
4e(-1)
2e(-1)
4e(0)
CO, N2, PR3, H2, NH3
All neutral 2e(0)
CN-, Cl-, Br-, I-, H-
2e(-1)
h1-C6H5 2e(-1)
h6-
h3-
h5-
h4-
-2
8
Dr. Mutasem Zaki Bani Fawaz
33. Examples of Electron Counting
1) Cr(CO)6
– Total charge on ligands = 0, so charge on Cr = 0, so Cr = d6
– 6 CO ligands x 2 electrons each = 12 electrons
– Total of 18 electrons
2) (h5-C5H5)Fe(CO)2Cl
– Total charge on ligands = 2-, so Fe2+ = d6
– (h5-C5H5
- = 6) + (2CO x 2 = 4) + (Cl- = 2) = 12 electrons
– Total of 18 electrons
3) Charged complex: [Mn(CO)6]+
– Total ligand charge = 0, so Mn+ = d6
– 12 electrons from 6 CO ligands gives a total of 18 electrons
4) M—M Bond: (CO)5Mn—Mn(CO)5
– Each bond between metals counts 1 electron per metal: Mn—Mn = 1 e-
– Total ligand charge = 0, so Mn0 = d7
– 5 CO ligands per metal = 10 electrons for a total of 18 electrons per Mn
9
Dr. Mutasem Zaki Bani Fawaz
34. Why we need MO diagram for
diatomic molecules
Dr. Mutasem Zaki Bani Fawaz 10
As bond order increase the bond energy increase and bond distance or
length decrease and IR frequency increase
HOMO:highest occupied molecular orbital
LUMO:lowest unoccupied molecular orbital
43. Infrared Spectroscopy
1) Number of Bands is determined by group theory
2) Position of IR Bands
a) Electron Density determines Wavenumbers
Cr(CO)6 n = 2000 cm-1 [V(CO)6]- n = 1858 cm-1 [Mn(CO)6]+ n = 2095 cm-1
45. Dr. Mutasem Zaki Bani Fawaz 21
Q1) For each of the following pairs of metal complexes,
circle the one that should have the highest average carbonyl
IR stretching frequency. What does this tell you about the
relative electron density on the metal centre (electron-rich or deficient)?
Briefly discuss your reasoning for each case.
a) [Rh(CO)4]1- or [Fe(CO)4]2-
b)(PF3)3Mo(CO)3 or (PCl3)3Mo(CO)3
Homework-2-.
46. Dr. Mutasem Zaki Bani Fawaz 22
Q2) The IR spectrum below shows signals due to CO stretching frequency, draw
out suitable structure which fit the molecular formula Fe2(CO)9, Briefly discuss
your answer?
47. Dr. Mutasem Zaki Bani Fawaz 23
Q3) Explain the differences in the IR CO stretching frequencies of
the following two complexes. Fe(CO)5 2025 and 2000 cm-1 and
Fe(CO)3(PPh3)2 1944, 1886 and 1881 cm-1.
48. Dr. Mutasem Zaki Bani Fawaz 24
Q4) (h5-Cp)2Co2(CO)3 has IR bands at 1965 cm-1 and 1812 cm-1 but (h5-
Cp)2Co2(CO)2 has only a single band at 1792 cm-1. Provide structures for both
molecules that are consistent with the data and the 18 e- rule. Show clearly the
number of M-M bonds in each structure.
49. Dr. Mutasem Zaki Bani Fawaz 25
Q5) Consider the compound (h5-Cp)2Co2(CO)2 has IR band at 2000 cm-1.
Provide structure for molecule that is consistent with the data and the 18 e- rule.
Show clearly the number of M-M bonds.
51. Isotope
patterns
63Cu: 69.1%
65Cu: 30.9%
35Cl: 75.5%
37Cl: 24.5%
63Cu35Cl
63Cu37Cl
65Cu35Cl
65Cu37Cl
Abundance of first isotope peak M: .691 x .755 = .521
For next isotope peak: M+2: .691 x .245 = .169
.309 x .755 = .233} .403
For next isotope peak: M+4: .309 x .245 = .076
52. Fragmentation
1. The charge is likely to remain on Metal-containing fragment
MLn
M Ln-1
+ + L
L+ + M Ln-1 Less likely
2. Rearrangement involving hydrogen migration are frequent
(M => Si, Ge, Sn, Pb…)
R
R
M
H R
R
M H
C
H2 CH2
R
R
M O
H
CH3
R
R
M H
MeHC O
53. Rearrangement process
“McLafferty” type of rearrangement involving metal instead of H
O
MR3
+
CH2
C
H2
C
H2
R
O
MR3
+
C
H2
R
CH2
CH2
+
This rearrangement depends on readiness of metal to
become pentacoordinate (using it’s d-orbitals)
Observed in: Keto-organotin: RCO (CH2)3 Sn (CH3)3
54. Influence of operating conditions
Electron deficient Alkyls and Aryls groups gives information about
molecular complexity of the vapor.
Ethyl Lithium: produce following ions:
Li6Et5
+
Li5Et4
+
Li4Et3
+
Li3Et2
+
Li2Et+
Li+
All these can be derive from the
hexamer Li6Et6
In Electron deficient compounds of group II and III (Be and Al),
monomeric molecules are associated by weak bridging bonds
58. 9
% of Na = 1 x 23
84
X 100 = 27.38 %
% of H =
1 x 1
84
X 100 = 1.19 %
% of C =
1 x 12
84
X 100 = 14.28 %
% of O = 3 x 16
84
X 100 = 57.14 %
Review
Q1: What are mass percents of Na, H, C, and O in NaHCO3?
Soln: Molecular mass of NaHCO3 =
(1 x Na) + (1 x H) + (1 x C) + (3 x O)
(1 x 23) + (1 x 1) + (1 x 12) + (3 x 16) = 84 g/mol
59. Fe2O3. 3/2 H2O
The molecular mass of limonite
= (2 x Fe) + (3 x O) + 3/2 (H2O)
= (2 x 56) + (3 x 16) + 3/2 (18) = 186.7
g/mole
Q2:
60. Simplest Formula [Empirical Formula] from Chemical analysis
Q3: A 25.0 g sample of an orange compound contains 6.64 g of
potassium (K), 8.84 g of chromium (Cr) and 9.52 g of oxygen (O).
Find the empirical formula. [Ar of K = 39.19 & Cr = 52 & O = 16]
Answer: no of moles of (K) = 6.64 / 39.19 = 0.17 mole
no. of moles of (Cr) = 8.84 / 52 = 0.17 mole
no. of moles of (O) = 9.52 / 16 = 0.595 mole
To know the empirical formula divided by the smallest no. of moles
i.e. 0.170
No. of (K) atoms in the simplest formula = 0.170 / 0.170 = 1 (K)
No. of (Cr) atoms in the simplest formula = 0.170 / 0.170 = 1 (Cr)
No. of (O) atoms in the simplest formula = 0.595 / 0.170 = 3.5 (O)
The simplest formula KCrO3.5
The empirical formula is K2Cr2O7
61. Q4) Nicotine is a natural product found in tobacco leaves. It can act as a stimulant or a
depressant. It is 74.03% C, 8.70% H, and 17.27% N. Its molecular weight is 162.24 g/mol.
a. What is the empirical formula of nicotine?
b. What is the molecular formula of nicotine?
62. NMR
Nuclear Magnetic Resonance
NMR for Organometallic compounds
Electron Paramagnetic Resonance
Electron Spin Resonance
Basic principles of
Mössbauer spectroscopy
65. The number of components into which a signal is split is 2nI+1, where I is the spin quantum number and
n is the number of other nuclei interacting with the nucleus.
For proton, I = 1/2
Neighbor group has one proton
Neighbor group has two protons
Neighbor group has three protons
Two neighbor groups have one proton each
66. classification of the nuclei
• I = 1/2, 100%abundance 1H, 31P, 19F, 103Rh
• I = 1/2, low abundance 13C, 15N, 29Si, 77Se,
109Ag, 119Sn, 125Te,
183W, 195Pt, 199Hg
• I > 1/2, 100% abundance 14N, 27Al, 51V, 59Co
• I > 1/2, low abundance 11B, 121Sb, 193Ir
71. Pi bonds do not count toward this bond limit, but J may be too small to observe
C
C
Ha
Hd
C
Hc
Hb •Ha couples with Hb
•Ha couples with Hc
•Ha couples with Hd but J may be very small
free spacer
10
Dr. Mutasem Zaki Bani Fawaz
77. Electron Paramagnetic Resonance
Electron Spin Resonance
Mn+2
OH
. N
O
R1 R2
.
Free Radicals
Ascorbate
Hydroxyl
Tocopherol
etc.
Transition Metals
All that have
unpaired electrons
Spin Labels
R1 & R2 chosen
to provide
specificity
78. CH3
n=3 I=1/2
[2 X 3 X (1/2)+1] = 4
CH3
the 4 lines 1:3:3:1
ms = 1/2, ml = 3/2
17
Dr. Mutasem Zaki Bani Fawaz
79. H n=1 I=1/2
[2 X 1 X (1/2)+1] = 2 lines
Examples:
AlH3
-
Al n=1 I=5/2
H n=3 I=1/2
[2 X 1 X (5/2)+1][2 X 3 X (1/2)+1]
[6] X [4] = 24 lines
C6H6
- radical anion the 7 lines
1:6:15:20:15:6:1
18
Dr. Mutasem Zaki Bani Fawaz
82. Appearance of Mössbauer spectra
Depending on the local environments of the Fe atoms and the magnetic
properties, Mössbauer spectra of iron oxides can consist of a singlet, a
doublet, or a sextet.
Symmetric charge
No magnetic field
Asymmetric charge
No magnetic field
Symmetric or asymmetric charge
Magnetic field (internal or external)
Δ Bhf
δ
Isomer
shift
Quadrupole
splitting
Magnetic
hyperfine
field
88. Strengths and weaknesses of 57Fe
Mössbauer spectroscopy
• Sensitive only to 57Fe
(no matrix effects)
• Sensitive to oxidation state
• Allows distinction of magnetic
phases
• Very sensitive towards
magnetic phases
• Non-destructive
• Resolution limited by
uncertainty principle
• Sensitive only to 57Fe
(“sees” only 57Fe)
• Coordination ? to ±
• Paramagnetic phase data
often ambiguous
• Diamagnetic element
substitution & relaxation
• Slow
• If possible, use other
techniques as well
Strengths Weaknesses
Often a combination of Mössbauer spectroscopy with
other techniques can help solve problems that cannot
be resolved using Mössbauer spectroscopy alone.
90. Introduction To:
Solid State Chemistry
Dr. Mutasem Z. Bani Fwaz
فواز بني زكي معتصم الدكتور
الكيمياء قسم
-
خالد الملك جامعه
Dr.Mutasem Zaki Bani-Fawaz 1
91. SOLIDS
can be divided into two catagories.
Crystalline
Amorphous
Solids with rigid a highly regular
arrangement
Solids with considerable disorder in their
structures
Dr.Mutasem Zaki Bani-Fawaz 2
92. Dr.Mutasem Zaki Bani-Fawaz 3
Crystalline B
Amorphous A
Time
Tempreture Tempreture
Time
Heating curve temperature vs time A for
amorphous and B for crystalline
94. Crystals are made of infinite number of unit cells
Dr.Mutasem Zaki Bani-Fawaz 5
Crystal lattice
Unit Cell
95. = between b and c
= between c and a
= between a and b
A parallelepiped is generated by the vectors
a , b and c
Dr.Mutasem Zaki Bani-Fawaz 6
97. Lattices
• In 1848, Auguste Bravais demonstrated that
in a 3-dimensional system there are fourteen
possible lattices
• A Bravais lattice is an infinite array of
discrete points with identical environment
• seven crystal systems + four lattice centering
types = 14 Bravais lattices
• Lattices are characterized by translation
symmetry
Auguste Bravais
(1811-1863)
Dr.Mutasem Zaki Bani-Fawaz 8
98. Primitive ( P ) Body Centered ( I )
Face Centered ( F ) C-Centered (C )
LATTICE TYPES
Dr.Mutasem Zaki Bani-Fawaz 9
100. Seven systems divide into 14 Bravais
lattices :
1. Triclinic P
2. Monoclinic P, C
3. Orthorhombic P, C, I, F
4. Trigonal P
5. Hexagonal P
6. Tetragonal P, I
7. Cubic P, I, F
Dr.Mutasem Zaki Bani-Fawaz 11
101. The crystal lattice: one atomic basis
• The basis can also just consist of
one atom.
Dr.Mutasem Zaki Bani-Fawaz 12
102. Element Schoenflies Hermann-
Mauguin
Operation
Rotation axis Cn n n-fold rotation (360º/n)
Identity E 1 nothing
Plane of symmetry m Reflection
Center of symmetry i -1 or “1 bar” Inversion
Improper rotation axis Sn - n-fold rotation + reflection
Rotary Inversion axis - -n or “n bar” n-fold rotation + inversion
Dr.Mutasem Zaki Bani-Fawaz 13
Point Group = Collection (group) of symmetry operators that all
pass through the same point. The group must be closed, have an
identity element, and every element must have an inverse.
Space Groups = Collections of symmetry operators that are
compatible with three-dimensional crystallographic (i.e.
translational) symmetry. There are 230 space groups. Because
protein and nucleic acid molecules are chiral, there are only 65
“biological” space groups.
103. Symmetry Elements
Glide reflection (mirror plane + translation)
reflects the asymmetric unit
across a mirror and then
translates parallel to the mirror.
A glide plane changes the
handedness of figures in the
asymmetric unit. There are no
invariant points (points that map
onto themselves) under a glide
reflection.
Dr.Mutasem Zaki Bani-Fawaz 14
104. Symmetry Elements
Screw axes (rotation + translation)
Dr.Mutasem Zaki Bani-Fawaz 15
Screw Axis
A NM screw rotation operation consists of a 360/N rotation
followed by a displacement of M/N the unit cell dimension parallel to the axis.
For example a 31 screw axis parallel to the c-axis represents a 120 deg rotation
followed by a displacement of 1/3c.
106. 1. Cubic Three 4-fold axes, four 3-fold axes
and six 2-fold axes, 9m, i
2. Tetragonal One 4-fold axis and four 2-fold axes, 5m, i
3. Orthorhombic Three 2-fold axis, 3m, i
4. Hexagonal One 6-fold axis and six 2-fold axes, 7m, i
5.Rhombohedral One 3-fold axis and three 2-fold axes, 3m, i
6. Monoclinic One 2-fold axis, 1m, i
7. Triclinic Non axes or planes but only i
System Required rotation symmetry
Dr.Mutasem Zaki Bani-Fawaz 17
107. a) Monoclinic, P 21 /c b) Orthorhombic, I bca
Dr.Mutasem Zaki Bani-Fawaz 18
112. 6. Enclose in parenthesis
(2,0,0)
(0,3,0)
(0,0,1)
Miller Indices for planes
4. Take reciprocal
3. Find intercepts along axes
2. select a crystallographic
coordinate system
1. Select an origin not on the plane
5. Convert to smallest integers in the same ratio
x
y
z
2 3 1
1/2 1/3 1
: 3 2 6
: (326)
Dr.Mutasem Zaki Bani-Fawaz 23
125. Dr.Mutasem Zaki Bani-Fawaz 36
By mono single X-ray structure analysis, we can say what is
the structure we have-----------
B(Me)3 is a monomer
[Al(Me)3]2 is a dimmer
129. Dr.Mutasem Zaki Bani-Fawaz 40
P21/n is not a standard space group. There is nothing
wrong to refine the structure in P21/n it is equivalent to
P21/c. It was better to use P21/n in the past because they
used only diagonal terms for refinement ie neglecting the
correlation between parameters. That was causing
unstable refinement in P21/c. That's not the case anymore
with the full matrix.
You'll often see Dunitz, J. D. (1979). X-ray Analysis and
the Structure of Organic Molecules, p 205–206 as a
reference for this.
However, for comparison, it's nice if everybody is using
the same unit cell setting.
130. Dr.Mutasem Zaki Bani-Fawaz 41
Quiz 1
The Following Figure represent the powder X-ray diffraction
pattern of one types of materials:
Study Figure well and answer the following question:
132. Dr.Mutasem Zaki Bani-Fawaz 43
How to determine the unit cell in figure by simple calculations.
Show your calculations.
133. 44
Quiz 2
Q1) Fill the following table by lattice types (P, C, I, F, or R) and required rotation symmetry for each of the following Crystal
system:
Crystal System Bravais Lattices Required rotation symmetry
Cubic
Tetragonal
Orthorhombic
Hexagonal
Trigonal
Monclinic
Triclinic
134. Q2) Write the set of axial constraints(dimensions, angles) for each of the following crystal system:
Cubic
Tetragonal
Orthorhombic
Hexagonal
Trigonal
Monclinic
Triclinic
135. Q3) List the full meaning conveyed by the following space groups:
a) Monoclinic, P 21 /c b) Orthorhombic, I bca
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