Chemistry 481(01) Spring 2016 Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu 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 5, 2016: Test 1 (Chapters 1, 2, 3, 4) April 28, 2016: Test 2 (Chapters (6 & 7) May 17, 2016: Test 3 (Chapters. 19 & 20) May 18, Make Up: Comprehensive covering all Chapters Chapter-1-1
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-1-2
Predicted Nuclear Fusion of Light Elements in the Young, Hot Universe Chapter-1-3
Few minutes after big Bang Chapter-1-4
Eight Steps in the History of the Earth 1. The Big Bang 2. Star Formation 3. Supernova Explosion 4. Solar Nebula Condenses 5. Sun & Planetary Rings Form 6. Earth Forms 7. Earth's Core Forms 8. Oceans & Atmosphere Forms Chapter-1-5
Nuclear Burning Chapter-1-6
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 super nova, spreading the new elements into space with high flux of neutrons to produce heavy elements by neutron capture. Chapter-1-7
1. What are the two basic types of nuclear reactions? Give examples of each that occur during the formation of the Universe Chapter-1-8
Cosmic Abundances Chapter-1-9
Balancing Nuclear Reactions Two 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.) Chapter-1-10
Balancing Nuclear Reactions 2. 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? Chapter-1-11
Balancing Nuclear Reactions 2. 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? Chapter-1-12
3. Predict the most likely modes of decay and the products of decay of the following nuclides: 17 F: 105 Ag: 185 Ta: Chapter-1-13
Bonding Energy Curve Chapter-1-14
Nuclear Binding Energy The 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 amu Neutron mass: 1.00867 amu Electron mass: 0.00055 amu Massdefect (Dm), then multiply by 931.5 MeV/amu Chapter-1-15
4. Using the binding energy calculator, calculate the binding energy 235 U if the mass of the this nuclide (isotope) is 235.0349 amu. ( P= 1.007277 amu, N= 1.008665 amu, e - =0.0005438 amu ) Chapter-1-16
5. What are theories that have been used to describe the nuclear stability? Chapter-1-17
Stability of the Elements and Their Isotopes P/N Ratio Why are elements With Z > 82 are Unstable? Chapter-1-18
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. Chapter-1-19
The Kinetics of Radioactive Decay Nuclear reactions follow 1 st order kinetics Chapter-1-20
6. How long would it take for a sample of 222 Rn that weighs 0.750 g to decay to 0.100 g? Assume a halflife for 222 Rn of 3.823 days? Chapter-1-21
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-1-22
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 1 ) n 2 - f n 2 i n f = 2 ni = 3,4, 5,... Blamer series Spectra of many other atoms can be described by similar relationships. Chapter-1-23
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. R H = 2.178 x 10-18 J E n = R H Z 2 n 2 E n = - m e Z2 e 4 8h 2 2 n 2 Chapter-1-24
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. Chapter-1-25
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 = E f - E i or DE = -2.178 x 10-18 [1/n 2 f - 1/n 2 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-1-26
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 (E n ): E n = -2.178 x 10-18 J (Z 2 /n 2 ) E n = -2.178 x 10-18 J 1/n 2 ; Z=1 for H Chapter-1-27
Bohr model of the atom Balmer later determined an empirical relationship that described the spectral lines for hydrogen. E n = - m e Z2 e 4 8h 2 2 n 2 DE = - 2.178 x 10-18 ( 1 1 ) J n 2 - f n 2 i nf = 2 ni = 3,4, 5,... Blamer series Spectra of many other atoms can be described by similar relationships. Chapter-1-28
Paschen, Blamer and Lyman Series Chapter-1-29
Calculation using the equation: E = -2.178 x 10-18 (1/n 2 f - 1/n 2 i ) J, Calculate the wavelength of light that can excite the electron in a ground state hydrogen atom to n = 7 energy level. Chapter-1-30
Calculation using Bohr eqaution The energy for the transition from n = 1 to n = 7: DE = -2.178 x 10-18 J [1/n 2 f - 1/n 2 i]; n f = 7, n i = 1 DE = -2.178 x 10-18 [1/7 2-1/1 2 ] J DE = -2.178 x 10-18 [1/49-1/1] J DE = -2.178 x 10-18 [0.02041-1] J DE = -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 10 8 m/s l = ---------------------- 2.13 x 10-18 J l = 9.31 x 10-8 m Chapter-1-31
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-1-32
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-1-33
De Broglie waves De Broglie proposed that all particles have a wavelength as related by: l = h mv l = wavelength, meters h = Plank s constant m = mass, kg v = frequency, m/s Chapter-1-34
Constructively Interfered 2D-Wave Chapter-1-35
destructively Interfered 2D-Wave Chapter-1-36
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-1-37
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) d 2 y/dx 2 = -(n 2 p 2 /l 2 )sin(npx/l) = -(n 2 p 2 /l 2 )y Chapter-1-38
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 y its quantum numbers. y n, l, m l, m s ), y 2 is proportional probablity of finding the electron in a given volume. Max Born Interpretation: y 2 = atomic orbital Chapter-1-39
Schrödinger Equation y = wave function E = total energy V = potential energy Chapter-1-40
Schrödinger Equation y = wave function E = total energy V = potential energy Chapter-1-41
Schrödinger Equation in Polar Coordinates Chapter-1-42
Polar Coordinates Chapter-1-43
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-1-44
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 m l values depends on l value: can have -l., 0... +l values of m l m s values should always be -1/2 or +1/2 Chapter-1-45
Solutions to Shrődinger Equation Series of allowed discrete y values: y n, l, ml, ms n = 1,2,3,4,5,6,7..etc. E n = - m e Z2 e 4 8h 2 2 n 2 Chapter-1-46
Components of y Mathematical expression of hydrogen like orbitals in polar coordinates: y n, l, ml, ms (r,, ) = R n, l, (r) Y l, ml, (, ) R n, l, (r ) Y l, ml, (, ) = Radial Wave Function =Angular Wave Function [R n, l (r )] 2 or 4pr 2 R 2 = Radial Distribution Function or P nl (r). Chapter-1-47
Radial Distribution Function, P nl (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-1-48
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. Chapter-1-49
s-atomic Orbitals R n, l, (r) only no Y l, ml, (, ) s orbitals Chapter-1-50
2s-Atomic Orbital: Probability distribution ψ 2 for the 2s orbital 2s orbital Chapter-1-51
s-atomic orbitals 2s 3s Chapter-1-52
p-atomic orbitals 2p 3p Chapter-1-53
Nodes in the y Total nodes = n -1 Angular nodes = l Radial nodes = n -1- l Eg 4d orbital: Total nodes = 4-1 = 3 Angular nodes = l = 2 Radial nodes = n -1- l = 4-1-2 = 1 Chapter-1-54
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-1-55
. Radial wavefunctions, R nl (r), and the radial distribution functions, P nl (r) R nl (r) P nl (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 Chapter-1-56
d-orbitals (d xy, d xz, d yz, d z2, and d x2 - y 2 orbitals) Chapter-1-57
f-orbitals ( 4fy3, 4fx3, 4fz3, 4fxz2y2, 4fyz2x2, 4fzx2y2, and 4fxyz orbitals) Chapter-1-58
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 Z eff. Z eff = Z - σ; Z is the atomic number. Chapter-1-59
Effective nuclear charge (Z eff ) Z eff is the nuclear charge felt by an electron in a multielectron atom: a) Each electron in an atom has different Z eff. b) Each Z eff is less than atomic number (Z) since electrons screen each other from the nucleus. c) Z eff depends on the n and l quantum number of an electron. d) Z eff Depends on orbital type the electron is in: Z eff of 4s > 4p > 4d > 4f. Chapter-1-60
Radial Wave Funtions Chapter-1-61
Radial Distribution Functions, Penetration and Shielding Chapter-1-62
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 Chapter-1-63
Slater Calculation of (Z eff ) Chapter-1-64
Slater Calculation of (Z eff ) Chapter-1-65
11. Cu: (1s 2 )(2s 2, 2p 6 ) (3s 2,3p 6 ) (3d 10 ) (4s 1 ) : there are two possible scenarios for forming Cu+ ionionizing 3d 10 electron or 4s 1. 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: (1s 2 )(2s 2, 2p 6 ) (3s 2,3p 6 ) (3d 10 ) (4s 1 ) s (4s 1 ) = ( 10x1 ) ( 8x 0.85)(1X10) = 26.8 Z eff = 29 26.6 = 2.4 Cu: (1s 2 )(2s 2, 2p 6 ) (3s 2,3p 6 ) (3d 10 ) (4s 1 ) s (3d 1 ) = ( 18x1 ) ( 9x 0.35) (0) = 21.15 Z eff = 29 21.15 = 7.85 Chapter-1-66
Effective nuclear charge (Z eff ) of Atomic Orbitals vs. Z (atomic number) E n = - m e Z eff 2 e 4 8h 2 2 n 2 Chapter-1-67
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 order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d (building up (Auf Bau) principle:) Chapter-1-68
ELECTRONIC CONFIGURATION OF MANY-ELECTRON ATOMS AUFBAU (GER. BUILDING UP) PRINCIPLE PAULI EXCLUSION PRINCIPLE HUND S RULE Chapter-1-69
Electronic Conf n Li 2s 1 Na 3s 1 Be 2s 2 Mg 3s 2 H 1s 1 He 1s 2 B 2s 2 2p 1 Al 3s 2 3p 1 C 2s 2 2p 2 Si 3s 2 3p 2 N 2s 2 2p 3 P 3s 2 3p 3 O 2s 2 2p 4 S 3s 2 3p 4 F 2s 2 2p 5 Cl 3s 2 3p 5 Ne 2s 2 2p 6 Ar 3s 2 3p 6 K 4s 1 Rb 5s 1 Cs 6s 1 Fr 7s 1 Ca 4s 2 Sr 5s 2 Ba 6s 2 Ra 7s 2 Sc 3d 1 4s 2 Y 4d 1 5s 2 Lu 4f 14 5d 1 6s 2 Lr 6d 1 7s 2 Ti 3d 2 4s 2 Zr 4d 2 5s 2 Hf 4f 14 5d 2 6s 2 La 5d 1 6s 2 Ac 6d 1 7s 2 V 3d 3 4s 2 Nb 4d 3 5s 2 Ta 4f 14 5d 3 6s 2 Ce 4f 1 5d 1 6s 2 Th 6f 2 7s 2 Cr 3d 5 4s 1 Mo 4d 5 5s 1 W 4f 14 5d 4 6s 2 Pr 4f 3 6s 2 Pa 5f 2 6d 1 7s 2 Mn 3d 5 4s 2 Tc 4d 5 5s 2 Re 4f 14 5d 5 6s 2 Nd 4f 4 6s 2 U 5f 3 6d 1 7s 2 Fe 3d 6 4s 2 Ru 4d 7 5s 1 Os 4f 14 5d 6 6s 2 Pm 4f 5 6s 2 Np 5f 4 6d 1 7s 2 Co 3d 7 4s 2 Rh 4d 8 5s 1 Ir 4f 14 5d 7 6s 2 Sm 4f 6 6s 2 Pu 5f 6 7s 2 Ni 3d 8 4s 2 Pd 4d 10 Pt 4f 14 5d 9 6s 1 Eu 4f 7 6s 2 Am 5f 7 7s 2 Cu 3d 10 4s 1 Ag 4d 10 5s 1 Au 4f 14 5d 10 6s 1 Gd 4f 7 5d 1 Cm 6s 2 5f 7 6d 1 7s 2 Zn 3d 10 4s 2 Cd 4d 10 5s 2 Hg 4f 14 5d 10 6s 2 Tb 4f 9 6s 2 Bk 5f 9 7s 2 Ga 3d 10 4s 2 4p 1 In 4d 10 5s 2 5p 1 Tl 5d 10 6s 2 6p 1 Dy 4f 10 6s 2 Cf 5f 10 7s 2 Ge 3d 10 4s 2 4p 2 Sn 4d 10 5s 2 5p 2 Pb 5d 10 6s 2 6p 2 Ho 4f 11 6s 2 Es 5f 11 7s 2 As 3d 10 4s 2 4p 3 Sb 4d 10 5s 2 5p 3 Bi 5d 10 6s 2 6p 3 Er 4f 12 6s 2 Fm 5f 12 7s 2 Se 3d 10 4s 2 4p 4 Te 4d 10 5s 2 5p 4 Po 5d 10 6s 2 6p 4 Tm 4f 13 6s 2 Md 5f 13 7s 2 Br 3d 10 4s 2 4p 5 I 4d 10 5s 2 5p 5 At 5d 10 6s 2 6p 5 Yb 4f 14 6s 2 No 5f 14 7s 2 Chapter-1-70 Kr 3d 10 4s 2 4p 6 Xe 4d 10 5s 2 5p 6 Rn 5d 10 6s 2 6p 6
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-1-71
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-1-72
Exception to Building Up Principle a) Electronic Configuration of d-block and f- block elements d 5 or d 10 and f 7 or f 14 are stable Cr :[Ar] 3d 4 4s 2 wrong Cr :[Ar] 3d 5 4s 1 correct Cu :[Ar] 3d 9 4s 2 wrong Cu :[Ar] 3d 10 4s 1 correct Chapter-1-73
Lanthanoids La 5d 1 6s 2 Ce 4f 1 5d 1 6s 2 Pr 4f 3 6s 2 Nd 4f 4 6s 2 Pm 4f 5 6s 2 Sm 4f 6 6s 2 Eu 4f 7 6s 2 Gd 4f 7 5d 1 6s 2 Tb 4f 9 6s 2 Dy 4f 10 6s 2 Ho 4f 11 6s 2 Er 4f 12 6s 2 Tm 4f 13 6s 2 Yb 4f 14 6s 2 Chapter-1-74
Actinoids Ac 6d 1 7s 2 Th 6f 2 7s 2 Pa 5f 2 6d 1 7s 2 U 5f 3 6d 1 7s 2 Np 5f 4 6d 1 7s 2 Pu 5f 6 7s 2 Am 5f 7 7s 2 Cm 5f 7 6d 1 7s 2 Bk 5f 9 7s 2 Cf 5f 10 7s 2 Es 5f 11 7s 2 Fm 5f 12 7s 2 Md 5f 13 7s 2 No 5f 14 7s 2 Chapter-1-75
Exception to Building Up Principle 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] 3d 6 4s 2 Cation, Fe 3+ :[Ar] 3d 5 Chapter-1-76
12. Give the ground state electronic configurations of following in core format. a) Mo b) Ag c) V 3+ d) Mn 2+ e) Cr 2+ f) Co 3+ g) Cr 6+ h) Gd 3+ Chapter-1-77
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-1-78
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) Fe 2+ : d) Nd 3+ : Chapter-1-79
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 radii Ionic radii First ionization energies Electron affinities Chapter-1-80
How does Ionic radii of elements vary? 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. Chapter-1-81
14. How do you measure atomic (ionic) radii (size)? Chapter-1-82
Atomic radii of elements going down a group? Chapter-1-83
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 Z eff of 6s 2 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] 4f 1 5d 1 6s 2 Chapter-1-84
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. Chapter-1-85
Ionic radii Cations These are smaller than the atoms from which they are formed. Anions These are larger than the atoms from which there are formed.. Chapter-1-86
Isoelectronic configurations Species that have the same electron configurations. Example Each of the following has an electron configuration of 1s 2 2s 2 2p 6 O 2- F - Ne Na + Mg 2+ Al 3+ Chapter-1-87
Ionization energy First ionization energy The 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-1-88
First ionization energy 2500 He Ne 2000 First ionization energy (kj/mol) 1500 1000 Ar Kr Xe Rn 500 0 0 20 40 60 80 100 Atomic number Chapter-1-89
Changes of I.E. Across a period Chapter-1-90
16. Why is the ionization energy of P (11.00 ev) greater than S (10.36 ev)? Chapter-1-91
How does Electron Affinity vary in the periodic table? Electron Affinity depends on Z eff of the nucleus to the outermost electron in the valence shell. Going down the group Z eff for the outer most shell decrease hence the Electron Affinity also increase Going across the period Z eff for the outer most shell increase hence the Electron Affinity also decrease Chapter-1-92
Electron affinity Atomic number Chapter-1-93
Electronegativity The 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. Chapter-1-94
Electronegativity Electronegativities 4 3.5 3 Electronegativity is a periodic property. 2.5 2 1.5 1 0.5 0 20 40 60 80 100 Atomic number Chapter-1-95
17. How you define electronegativity? Chapter-1-96
Electronegativity Scales Pauling Electronegativity, c P Mulliken Electronegativity, c M The Allred-Rochow, c AR Sanderson electronegativity Allen electronegativity Chapter-1-97
Pauling Electronegativity, c P E A-A and E B-B bond-energy of homonuclear A-A & B-B diatomic molecules E A-B bond-energy of heteronuclear A-B diatomic molecule c A c B are electronegativity values of A and B Pauling comments that it is more accurate to use the geometric mean rather than the arithmetic mean Chapter-1-98
Mulliken Electronegativity, c M 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 Chapter-1-99
The Allred-Rochow, c AR The effective nuclear charge, Z eff 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, r cov. When r cov is expressed in ångströms, Chapter-1-100
Sanderson, c s 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. Allen, c A The simplest definition of electronegativity is that of Allen, bases on average energy of the valence electrons in a free atom where ε s,p are the oneelectron energies of s- and p-electrons in the free atom and n s,p are the number of s- and p-electrons in the valence shell. Chapter-1-101
18. Calculate the electronegativity (X) (X m, X ar ) for Cl. [ X m = 1/2(I+A e ); X ar = 0.744+ 0.359 Z eff /r 2 ] a) X m When E i and E ea kj per mol X m = 1.97 x 10-3 (1251 + 349 ) + 0.19 X m = 3.342 (3.54) a) X ar When r ( 0.99 Angstroms) and Z eff = 6.12 X ar = 0.359 Z eff /r 2 + 0.744 X ar = 2.98 (2.83) Chapter-1-102