Chemistry 481(01) Spring 2014
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Transcript of Chemistry 481(01) Spring 2014
Chapter-1-1Chemistry 481, Spring 2015, LA Tech
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: CTH 311 Phone 257-4941
Office Hours:
M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th, F 9:30 - 11:30 a.m.
April 7 , 2015: Test 1 (Chapters 1, 2, 3)
April 30, 2015: Test 2 (Chapters 5, 6 & 7)
May 19, 2015: Test 3 (Chapters. 19 & 20)
May 19, Make Up: Comprehensive covering all Chapters
Chemistry 481(01) Spring 2015
Chapter-1-2Chemistry 481, Spring 2015, LA Tech
Origin of Elements in the Universe Scientists have long based the origin of our Universe on the Big Bang Theory. According to
this theory, our universe was simply an expanding fairly cold entity consisting of only Hydrogen
and Helium during it's incipient stages. Over the expanse of many years, and through a
continuing process of fusion and fission, our universe has come to consist of numerous chemical
elements, four terrestrial planets(Earth, Mars, Venus, and Mercury), and five giant gas
planets(Saturn, Jupiter, Neptune, Pluto, and Uranus).
Chapter-1-3Chemistry 481, Spring 2015, LA Tech
Predicted Nuclear Fusion ofLight Elements in the Young,Hot Universe
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Few minutes after big Bang
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Eight Steps in the History of the Earth1. The Big Bang2. Star Formation 3. Supernova Explosion4. Solar Nebula Condenses 5. Sun & Planetary Rings Form6. Earth Forms 7. Earth's Core Forms 8. Oceans & Atmosphere Forms
Chapter-1-6Chemistry 481, Spring 2015, LA Tech
Nuclear Burning
Chapter-1-7Chemistry 481, Spring 2015, LA Tech
Origin of the Elements: Nucleosynthesis•Elements formed in the universe's original stars were made from hydrogen gas condensing due to gravity. These young stars "burned" hydrogen in fusion reactions to produce helium and the hydrogen was depleted. Reactions such as those below built up all the heavier elements up to atomic number 26 in the periodic table.•When the stars got old they exploded in a supernova, spreading the new elements into space with high flux of neutrons to produce heavy elements by neutron capture.
Chapter-1-8Chemistry 481, Spring 2015, LA Tech
1. What are the two basic types of nuclear reactions? Give examples of each that occur during the formation of the Universe
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Cosmic Abundances
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Balancing Nuclear ReactionsTwo conditions must be met to balance nuclear reactions: 1. The sum of the masses of the reactants must equal the sum of the masses of the products. (i.e., the values of A must balance on both sides of the equation.) 2. The sum of the protons for the reactants must equal the sum of the protons for the products. (i.e., the values of Z must balance on both sides of the equation.)
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Balancing Nuclear Reactions2. Complete the following Nuclear reactions:a) Uranium – 238 decays by alpha radiation to
produce what other element?
b) Uranium – 238 decays by alpha radiation to produce what other element?
c) What element did we start out with if the result of beta decay is bismuth– 214?
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Balancing Nuclear Reactions2. Complete the following Nuclear reactions:d) What element is produced when mercury – 201 captures an inner shell electron with the production of a gamma ray to release excess energy?
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3. Predict the most likely modes of decay and the products of decay of the following nuclides:17F:
105Ag:
185Ta:
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Bonding Energy Curve
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Nuclear Binding EnergyThe binding energy of a nucleus is a measure of how
tightly its protons and neutrons are held together by the nuclear forces. The binding energy
per nucleon, the energy required to remove one neutron or proton from a nucleus, is a function
of the mass number A. (Dm) –mass defect(Dm) = Mass of Nuclide - mass of (p + n +e ) Proton mass: 1.00728 amuNeutron mass: 1.00867 amuElectron mass: 0.00055 amuMassdefect (Dm), then multiply by 931.5 MeV/amu
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4. Using the binding energy calculator, calculate the binding energy 235U if the mass of the this nuclide (isotope) is 235.0349 amu. ( P= 1.007277 amu, N= 1.008665 amu, e- =0.0005438 amu )
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5. What are theories that have been used to describe the nuclear stability?
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Stability of the Elements and Their Isotopes
P/N RatioWhy are elementsWith Z > 82 areUnstable?
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Magic Numbers• Nuclei with either numbers of protons or
neutrons equal to Z, N =2 (He), 8(O), 20 (Ca), 28(Si), 50(Sn, 82(Pb), or 126(?)(I)
• exhibit certain properties which are analogous to closed shell properties in atoms, including
• anomalously low masses, high natural abundances and high energy first excited states.
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The Kinetics of Radioactive DecayNuclear reactions follow 1st order kinetics
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6. How long would it take for a sample of 222Rn that weighs 0.750 g to decay to 0.100 g? Assume a half-life for 222Rn of 3.823 days?
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7. The skin, bones and clothing of an adult female mummy discovered in Chimney Cave, Lake Winnemucca, Nevada, were dated by radiocarbon analysis. How old is this mummy if the sample retains 73.9% of the activity of living tissue?
Chapter-1-23Chemistry 481, Spring 2015, LA Tech
Bohr model of the atom
Balmer later determined an empirical relationship that described the spectral lines for hydrogen.
DE = - 2.178 x 10-18
m-1 =
( )1
nf2
1
ni2-
nf = 2 ni = 3,4, 5, . . . Blamer series
Spectra of many other atoms can be described by
similar relationships.
Chapter-1-24Chemistry 481, Spring 2015, LA Tech
Bohr model of the atom• The Bohr model is a
‘planetary’ type model.
• Each principal quantum represents a new ‘orbit’ or layer.
• The nucleus is at the center of the model.
• RH = 2.178 x 10-18 JEn = -
En = RH
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Emission Spectrum of Hydrogen• Bohr studied the spectra produced when atoms were excited in a gas discharge tube.
He observed that each element produced its
own set of characteristic lines.
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Emission Spectrum of Hydrogen
• Line Spectrum• Energy is absorbed when an electron goes from a
lower(n) to a higher(n) • Energy is emitted when an electron goes from a
higher(n) to a lower(n) level • Energy changed is given by:DE = Ef - Ei
• or DE = -2.178 x 10-18 [1/n2f - 1/n2
i] J• DE is negative for an emission and positive for an
absorption • DE can be converted to l or 1/ l by l = hc/E.
Chapter-1-27Chemistry 481, Spring 2015, LA Tech
What is Bohr’s Atomic model? • explain emission spectrum of hydrogen atom• applied the idea of Quantization to electrons to orbits• energies of these orbits increase with the distance
from nucleus.• Energy of the electron in orbit n (En):• En = -2.178 x 10-18 J (Z2/n2)• En = -2.178 x 10-18 J 1/n2; Z=1 for H
Chapter-1-28Chemistry 481, Spring 2015, LA Tech
Bohr model of the atomBalmer later determined an empirical relationship that described the spectral lines for hydrogen.
DE = - 2.178 x 10-18
J ( )1
nf2
1
ni2-
nf = 2 ni = 3,4, 5, . . . Blamer series
Spectra of many other atoms can be described by
similar relationships.
En = -
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Paschen, Blamer and Lyman Series
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Calculation using the equation: E = -2.178 x 10-18 (1/nf
2 - 1/ni2 ) J, Calculate
the wavelength of light that can excite the electron in a ground state hydrogen atom to n = 7 energy level.
Chapter-1-31Chemistry 481, Spring 2015, LA Tech
The energy for the transition from n = 1 to n = 7:DE = -2.178 x 10-18 J [1/n2
f - 1/n2i]; nf = 7, ni = 1
DE = -2.178 x 10-18 [1/72 - 1/12] JDE = -2.178 x 10-18 [1/49 - 1/1] JDE = -2.178 x 10-18 [0.02041 - 1] JDE = -2.178 x 10-18 [-0.97959] J = 2.134 x 10-18 J (+, absorption)calculate the l using l = hc/E
6.626 x 10-34 Js x 3.00 x 108 m/s l = ---------------------- 2.13 x 10-18 J l = 9.31 x 10-8 m
Calculation using Bohr eqaution
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8. Using Bohr energy calculator, calculate the wavelength of light that can excite the electron in a ground state hydrogen atom from n = 5 to n = 3 energy level.
Chapter-1-33Chemistry 481, Spring 2015, LA Tech
Wave theory of the electron• 1924: De Broglie suggested that electrons
have wave properties to account for why their energy was quantized.
• He reasoned that the electron in the hydrogen atom was fixed in the space around the nucleus.
• He felt that the electron would best be represented as a standing wave.
• As a standing wave, each electron’s path must equal a whole number times the wavelength.
Chapter-1-34Chemistry 481, Spring 2015, LA Tech
De Broglie proposed that all particles have a wavelength as related by:
l = wavelength, metersh = Plank’s constantm = mass, kgv = frequency, m/s
De Broglie waves
l =h
mv
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Constructively Interfered 2D-Wave
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destructively Interfered 2D-Wave
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Two-dimensional wave - Vibrations on a Drumskin One circular node
(at the drumskin's edge)
Two circular nodes
(one at the drumskin's edge plus
one more)
Three circular nodes
(one at the drumskin's edge plus
two more)
One transverse node
(plus a circular one at the
drumskin's edge)
Two transverse nodes
(plus one at the drumskin's
edge)
Chapter-1-38Chemistry 481, Spring 2015, LA Tech
What is a wave-mechanical model?• motions of a vibrating string shows one dimensional motion.• Energy of the vibrating string is quantized• Energy of the waves increased with the nodes. • Nodes are places were string is stationary. • Number of nodes gives the quantum number. One
dimensional motion gives one quantum number.Vibrating String : y = sin(npx/l)d2y/dx2 = -(n2p2/l2)sin(npx/l) = -(n2p2/l2)y
Chapter-1-39Chemistry 481, Spring 2015, LA Tech
Quantum model of the atom• Schrödinger developed an equation to
describe the behavior and energies of electrons in atoms.
• His equation ( Wave function y) is similar to one used to describe electromagnetic waves. Each electron can be described in terms of Wave function yits quantum numbers. yn, l, ml, ms),
• y2is proportional probablity of finding the electron in a given volume. Max Born Interpretation: y2= atomic orbital
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Schrödinger Equation
y = wave function
E = total energy
V = potential energy
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Schrödinger Equation
y = wave function E = total energy V = potential energy
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Schrödinger Equation in Polar Coordinates
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Polar Coordinates
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Quantum Model of atom• Electrons travel in three dimensions• Four quantum numbers are needed• three to describe, x, y, z, and four for the spin• four quantum numbers
describe an orbital currently used to explain the arrangement, bonding and spectra of atoms.
Chapter-1-45Chemistry 481, Spring 2015, LA Tech
Four Quantum Numbers of the Atom• n value could be
1, 2, 3, 4, 5, 6. 7. . . etc.• l values depend on n value: can have
0 . . . (n - 1) values
• ml values depends on l value: can have -l . , 0 . . . +l values of ml
• ms values should always be -1/2 or +1/2
Chapter-1-46Chemistry 481, Spring 2015, LA Tech
Solutions to Shrődinger Equation
Series of allowed discrete y values:
yn, l, ml, ms
n = 1,2,3,4,5,6,7..etc.
En = -
Chapter-1-47Chemistry 481, Spring 2015, LA Tech
Components of yMathematical expression of hydrogen like orbitals
in polar coordinates:
yn, l, ml, ms (r,,) = R n, l, (r) Y l, ml, (,) R n, l, (r ) = Radial Wave Function Y l, ml, (,) =Angular Wave Function [R n, l (r )]2 or 4pr2R2 = Radial Distribution
Function or Pnl(r).
Chapter-1-48Chemistry 481, Spring 2015, LA Tech
Radial Distribution Function, Pnl(r).This is defined as the probability that an electron in
the orbital with quantum numbers n and l will be found at a distance r from the nucleus. It is related to the radial wave function by the following relationship:
; normalized by
Chapter-1-49Chemistry 481, Spring 2015, LA Tech
9. Describe the Schrödinger equation and the breaking up of wave function, y into radial and
angular component of a wave function and explain the general rule used to find the number
of radial and angular nodes of a wave function.
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s orbitals
R n, l, (r) only no Y l, ml, (,)
s-Atomic Orbitals
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2s orbital
2s-Atomic Orbital: Probability distribution ψ2 for the 2s orbital
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2s
3s
s-Atomic orbitals
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p-Atomic orbitals
2p
3p
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Nodes in the yTotal nodes = n -1Angular nodes = lRadial nodes = n -1- lEg 4d orbital: Total nodes = 4 -1 = 3Angular nodes = l = 2Radial nodes = n -1- l = 4-1-2 = 1
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10. Consider the following radial probability density-distribution plot and respond to the associated questions.a) How many radial nodes are there?
b) If the total number of nodes is 3, what type of orbital is involved?
c) Which orbital would it be if there were one more node?
Chapter-1-56Chemistry 481, Spring 2015, LA Tech
.
Rnl(r) Pnl(r) n l
1s
1s
1 0
2s
2s
2 0
2p
2p
2 1
3s
3s
3 0
3p
3p
3 1
3d
3d
3 2
Radial wavefunctions, Rnl(r), and the radial distribution functions, Pnl(r)
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d-orbitals(dxy, dxz, dyz, dz
2 , and dx
2-y
2
orbitals)
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f-orbitals( 4fy3 , 4fx3 , 4fz3 , 4fxz2y2 , 4fyz2x2 , 4fzx2y2 , and 4fxyz orbitals)
Chapter-1-59Chemistry 481, Spring 2015, LA Tech
Screening (shielding) constant (σ)Screening (shielding) constant (σ) for each electron is calculated based on: the principle quantum number orbital type and penetration and of all other electrons in an atom.σ gives Zeff . Zeff = Z - σ; Z is the atomic number.
Chapter-1-60Chemistry 481, Spring 2015, LA Tech
Effective nuclear charge (Zeff)
Zeff is the nuclear charge felt by an electron in a multielectron atom:
a) Each electron in an atom has different Zeff.b) Each Zeff is less than atomic number (Z) since
electrons screen each other from the nucleus.c) Zeff depends on the n and l quantum number of an
electron.d) Zeff Depends on orbital type the electron is in: Zeff
of 4s > 4p > 4d > 4f.
Chapter-1-61Chemistry 481, Spring 2015, LA Tech
Radial Wave Funtions
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Radial Distribution Functions, Penetration and Shielding
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Penetration & Shielding of an Electron in Multi-electron Atom
Penetration of an electron:• Greater the penetration there is more chance of
electrons being located close to the nucleus.• Comparing s, p, d, or f orbitals within same shell (or
principle QN), penetration of an electrons are in the order: s > p > d > f
Shielding power of an electron:• Shields of other electrons depends penetration and the
orbital type. Shielding power of electrons in orbitals of that same shell are: s > p > d > f
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Slater Calculation of (Zeff)
Chapter-1-65Chemistry 481, Spring 2015, LA Tech
Slater Calculation of (Zeff)
Chapter-1-66Chemistry 481, Spring 2015, LA Tech
11. Cu: (1s2)(2s2, 2p6) (3s2,3p6) (3d10) (4s1) : there are two possible scenarios for forming Cu+ ion- ionizing 3d10 electron or 4s1. Using Slater’s Rules show which one of the electrons 4s or 3d would come out easily.If the electron is in a d or f-orbital: All electrons in groups higher than the electron in question contribute zero to s.
Each electron in the same group contributes 0.35 to s.
All those in groups to the left contribute 1.0 to s (n-3) (n-2) (n-1) (n-1)Cu: (1s2)(2s2, 2p6) (3s2,3p6) (3d10) (4s1) s(4s1) = ( 10x1 ) ( 8x 0.85)(1X10) = 26.8 Zeff = 29 – 26.6 = 2.4Cu: (1s2)(2s2, 2p6) (3s2,3p6) (3d10) (4s1) s(3d1) = ( 18x1 ) ( 9x 0.35) (0) = 21.15 Zeff = 29 – 21.15 = 7.85
Chapter-1-67Chemistry 481, Spring 2015, LA Tech
Effective nuclear charge (Zeff) of Atomic Orbitals vs. Z (atomic number)
En = -
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How do you get the electronic configuration of an atom?
• Use periodic table• Periodic table is divided into orbital blocks• Each period:• represents a shell or n • Start writing electron configuration• Using following order1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d…
(building up (Auf Bau) principle:)
Chapter-1-69Chemistry 481, Spring 2015, LA Tech
ELECTRONIC CONFIGURATION OF MANY-ELECTRON ATOMS
• AUFBAU (GER. BUILDING UP) PRINCIPLE• PAULI EXCLUSION PRINCIPLE• HUND’S RULE
Chapter-1-70Chemistry 481, Spring 2015, LA Tech
Tl
5d10
6s2
6p1
Hg
4f14
5d10
6s2
Au
4f14
5d10
6s1
Hf
4f14
5d2
6s2
Lu
4f14
5d1
6s2
Li
2s1
Na
3s1
Cs
6s1
Rb
5s1
K
4s1
Fr
7s1
Pt
4f14
5d9
6s1
Ir
4f14
5d7
6s2
Os
4f14
5d6
6s2
Re
4f14
5d5
6s2
W
4f14
5d4
6s2
Ta
4f14
5d3
6s2
H
1s1
He
1s2
Rn
5d10
6s2
6p6
At
5d10
6s2
6p5
Po
5d10
6s2
6p4
Bi
5d10
6s2
6p3
Pb
5d10
6s2
6p2
Cd
4d10
5s2
Ag
4d10
5s1
Zr
4d2
5s2
Y
4d1
5s2
Pd
4d10
Rh
4d8
5s1
Ru
4d7
5s1
Tc
4d5
5s2
Mo
4d5
5s1
Nb
4d3
5s2
Lr
6d1
7s2
Ba
6s2
Be
2s2
Mg
3s2
Sr
5s2
Ca
4s2
Ra
7s2
Zn
3d10
4s2
Cu
3d10
4s1
Ti
3d2
4s2
Sc
3d1
4s2
Ni
3d8
4s2
Co
3d7
4s2
Fe
3d6
4s2
Mn
3d5
4s2
Cr
3d5
4s1
V
3d3
4s2 In
4d10
5s2
5p1
Xe
4d10
5s2
5p6
I
4d10
5s2
5p5
Te
4d10
5s2
5p4
Sb
4d10
5s2
5p3
Sn
4d10
5s2
5p2
Ga
3d10
4s2
4p1
Kr
3d10
4s2
4p6
Br
3d10
4s2
4p5
Se
3d10
4s2
4p4
As
3d10
4s2
4p3
Ge
3d10
4s2
4p2
Al
3s2
3p1
Ar
3s2
3p6
Cl
3s2
3p5
S
3s2
3p4
P
3s2
3p3
Si
3s2
3p2
B
2s2
2p1
Ne
2s2
2p6
F
2s2
2p5
O
2s2
2p4
N
2s2
2p3
C
2s2
2p2
Gd
4f7
5d1
6s2
Cm
5f7
6d1
7s2
Tb
4f9
6s2Bk
5f9
7s2
Sm
4f6
6s2Pu
5f6
7s2
Eu
4f7
6s2Am
5f7
7s2
Nd
4f4
6s2
U
5f3
6d1
7s2
Pm
4f5
6s2
Np
5f4
6d1
7s2
Ce
4f1
5d1
6s2
Th
6f2
7s2
Pr
4f3
6s2
Pa
5f2
6d1
7s2
Yb
4f14
6s2No
5f14
7s2
La
5d1
6s2Ac
6d1
7s2
Er
4f12
6s2Fm
5f12
7s2
Tm
4f13
6s2Md
5f13
7s2
Dy
4f10
6s2Cf
5f10
7s2
Ho
4f11
6s2Es
5f11
7s2
Electronic Confn
Chapter-1-71Chemistry 481, Spring 2015, LA Tech
Using the periodic table• To write the ground-state electron configuration of
an element:• Starting with hydrogen, go through the elements in
order of increasing atomic number• As you move across a period
• Add electrons to the ns orbital as you pass through groups IA (1) and IIA (2).
• Add electrons to the np orbital as you pass through Groups IIIA (13) to 0 (18).
• Add electrons to (n-1) d orbitals as you pass through IIIB (3) to IIB(12) and add electrons to (n-2) f orbitals as you pass through the f -block.
Chapter-1-72Chemistry 481, Spring 2015, LA Tech
Writing electron configurations
• Electron configurations can also be written for ions.
• Start with the ground-state configuration for the atom.
• For cations, remove a number of the outermost electrons equal to the charge.
• For anions, add a number of outermost electrons equal to the charge.
Chapter-1-73Chemistry 481, Spring 2015, LA Tech
Exception to Building Up Principle a) Electronic Configuration of d-block and f-
block elements d5 or d10 and f7 or f14 are stable Cr :[Ar] 3d4 4s2 wrong Cr :[Ar] 3d5 4s1 correct Cu :[Ar] 3d9 4s2 wrong Cu :[Ar] 3d10 4s1 correct
Chapter-1-74Chemistry 481, Spring 2015, LA Tech
Lanthanoids
Gd
4f7
5d1
6s2
Tb
4f9
6s2
Sm
4f6
6s2
Eu
4f7
6s2
Nd
4f4
6s2
Pm
4f5
6s2
Ce
4f1
5d1
6s2
Pr
4f3
6s2
Yb
4f14
6s2
La
5d1
6s2
Er
4f12
6s2
Tm
4f13
6s2
Dy
4f10
6s2
Ho
4f11
6s2
Chapter-1-75Chemistry 481, Spring 2015, LA Tech
Actinoids
Cm
5f7
6d1
7s2
Bk
5f9
7s2
Pu
5f6
7s2
Am
5f7
7s2
U
5f3
6d1
7s2
Np
5f4
6d1
7s2
Th
6f2
7s2
Pa
5f2
6d1
7s2
No
5f14
7s2
Ac
6d1
7s2
Fm
5f12
7s2
Md
5f13
7s2
Cf
5f10
7s2
Es
5f11
7s2
Chapter-1-76Chemistry 481, Spring 2015, LA Tech
Electronic Configuration of Transition Metal cations
d-block and f-block elements
d orbitals are lower in energy than s orbitals
f orbitals are lower in energy than d orbitals
E.g. Neutral atom Fe :[Ar] 3d6
4s2
Cation, Fe3+
:[Ar] 3d5
Exception to Building Up Principle
Chapter-1-77Chemistry 481, Spring 2015, LA Tech
12. Give the ground state electronic configurations of following in core format.
a) Mo
b) Ag
c) V3+
d) Mn2+
e) Cr2+
f) Co3+
g) Cr6+
h) Gd3+
Chapter-1-78Chemistry 481, Spring 2015, LA Tech
Magnetic Properties of Atoms a) Paramagnetism? attracted to magnetic field due to un-paired
electrons. b) Ferromagnetism? attracted very strongly to magnetic field due to
un-paired electrons. c) Diamagnetism? Repelled by a magnetic field due to paired
electrons.
Chapter-1-79Chemistry 481, Spring 2015, LA Tech
13. Give the ground state electronic configurations of the following in valence orbital box format and give the number of unpaired electrons.
a) Mn:
b) Co:
c) Fe2+:
d) Nd3+:
Chapter-1-80Chemistry 481, Spring 2015, LA Tech
Periodic trends• Many trends in physical and chemical properties
can be explained by electron configuration.• We’ll look at some of the more important
examples.Atomic radiiIonic radiiFirst ionization energiesElectron affinities
Chapter-1-81Chemistry 481, Spring 2015, LA Tech
• Cations have smaller radii than neutral atoms.
• Anions have larger radii than neutral atoms
• The more charge on the ion more effect on the
radii.
How does Ionic radii of elements vary?
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14. How do you measure atomic (ionic) radii (size)?
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Atomic radii of elements going down a group?
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Lanthanoide Contration
• Filling of the 4f orbitals in the lanthanides, which occur within the third series of transition elements, causes these transition metals to be smaller than expected because the 4f orbitals are very poor nuclear shielders and Zeff of 6s2
obitals increase and the atomic radii decrease.• 3rd-series elements have nearly the same
effective nuclear charge as the 2nd-series elements, and thus, nearly the same size
Ce [Xe] 4f1
5d1
6s2
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15. Why the atomic radius of Zr (1.64) which is in 5th period is almost similar to a element, Hf (1.65) in 6th period.
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Ionic radiiCationsThese are smaller than the atoms from which they are formed.
AnionsThese are larger than the atoms from which there are formed..
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Isoelectronic configurations
Species that have the same electron configurations.ExampleEach of the following has an electron configuration of 1s2 2s2 2p6
O2- F- Ne
Na+ Mg2+ Al3+
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Ionization energy
• First ionization energyThe energy to remove one electron from a
neutral atom in the gas phase.• A(g) + first ionization energy A+(g) + e-
• This indicates how easy it is to form a cation. Metals tend to have lower first ionization energies than nonmetals.
• They prefer to become cations.
Chapter-1-89Chemistry 481, Spring 2015, LA Tech
0
500
1000
1500
2000
2500
0 20 40 60 80 100
First ionization energyHe
Ne
Ar
Kr
Xe
Rn
Firs
t io
niza
tion
ene
rgy
(kJ/m
ol)
Atomic number
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Changes of I.E. Across a period
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16. Why is the ionization energy of P (11.00 eV) greater than S (10.36 eV)?
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• Electron Affinity depends on Zeff of the nucleus to the outermost electron in the
valence shell.
• Going down the group Zeff for the outer most shell decrease hence the Electron
Affinity also increase
• Going across the period Zeff for the outer most shell increase hence the Electron
Affinity also decrease
How does Electron Affinity vary in the periodic table?
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Electron affinity
Atomic number
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ElectronegativityThe ability of an atom that is bonded to another atom or
atoms to attract electrons to itself.It is related to ionization energy and electron affinity.
It cannot be directly measured.
The values are unitless since they are relative to each other.
The values vary slightly from compound to compound but still provide useful qualitative predictions.
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Electronegativities
0.5
1
1.5
2
2.5
3
3.5
4
0 20 40 60 80 100
Elec
tron
egat
ivit
y
Atomic number
Electronegativity is a
periodic property.
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17. How you define electronegativity?
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Electronegativity Scales• Pauling Electronegativity, cP
• Mulliken Electronegativity, cM
• The Allred-Rochow, cAR
• Sanderson electronegativity
• Allen electronegativity
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Pauling Electronegativity, cP
EA-A and EB-B bond-energy of homonuclear A-A & B-B diatomic moleculesEA-B bond-energy of heteronuclear A-B diatomic moleculecA cB are electronegativity values of A and BPauling comments that it is more accurate to use the geometric mean rather than the arithmetic mean
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Mulliken Electronegativity, cM
The Mulliken electronegativity can only be calculated for an element for which the electron affinity is known• For ionization energies and electron affinities in
electronvolts
• For energies in kilojoules per mole
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The Allred-Rochow, cAR
The effective nuclear charge, Zeff experienced by valence electrons can be estimated using Slater's rules, while the surface area of an atom in a molecule can be taken to be proportional to the square of the covalent radius, rcov. When rcov is expressed in ångströms,
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Sanderson, cs
Sanderson has also noted the relationship between electronegativity and atomic size, and has proposed a method of calculation based on the reciprocal of the atomic volume.
The simplest definition of electronegativity is that of Allen, bases on average energy of the valence electrons in a free atom
Allen, cA
where εs,p are the one-electron energies of s-
and p-electrons in the free atom and ns,p are
the number of s- and p-electrons in the valence
shell.
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18. Calculate the electronegativity (X) (Xm, Xar) for Cl. [ Xm= 1/2(I+Ae); Xar= 0.744+ 0.359 Zeff/r2 ]a) Xm When Ei and Eea kJ per mol
Xm= 1.97 x 10-3(1251 + 349 ) + 0.19 Xm = 3.342 (3.54)a) Xar When r ( 0.99 Angstroms) and Zeff = 6.12
Xar= 0.359 Zeff/r2 + 0.744 Xar = 2.98 (2.83)