Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes
description
Transcript of Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes
Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes
H Advanced ChemistryUnit 3
Objectives #1-3 Atomic Theory
*review of electromagnetic radiation characteristics:
(diagrams)
Examples of Electromagnetic Radiation
Objectives #1-3 Atomic Theory
frequency, wavelength, energyfrequency vs. wavelength (inverse
relationship)frequency vs. energy (direct relationship)wavelength vs. energy (inverse
relationship)c=fl (c = speed of light in m/s, f =
frequency in Hz (1/s), l = wavelength in m)
Max Planck (1858-1947)
Objectives #1-3 Atomic Theory
E = hf or hc/l h = Planck’s Constant (energy for waves)
6.626 X 10-34 Js
Albert Einstein (1879-1955)
Objectives #1-3 Atomic Theory
E = mc2 (energy for particles)*Wave particle-dualityMatter has wave and particle
characteristics; acts as particle when interacting with matter; acts as wave when traveling through space (clip)
Louis de Broglie (1892-1987)
Derivation of de Broglie’s Equation:Ewaves = Eparticles hc / l = mc2
l (hc / l) = (mc2)lh = mcll = h / mv
(examples)
#1 Calculate the frequency of light having a wavelength of 6.50 X 102 nm.
c = fl (convert nm to meters)6.50 X 102 nm X 1m / 1 X 109 nm = 6.50 X 10-7 mf = (3.00 X 108 m/s / 6.5 X 10-7 m) = 4.62 X 1014 Hz
#2 Calculate the energy of the blue color of wavelength 4.50 X 102 nm emitted by an atom of copper.
E = hc / l = (6.626 X 10-34 Js) (3.00 X 108 m/s) /
4.5 X 10-7 m = 4.42 X 10-19 J
#3 Calculate the wavelength for an electron having a mass of 9.11 X 10-31 kg and traveling at a speed of 1.0 X 107 m/s. Calculate the wavelength for a ball having a mass of .10 kg and traveling at a speed of 35 m/s.
l = h/mv = (6.626 X 10-34 Js) / (9.11 X 10-31 kg) (1.0 X 107 m/s) 7.27 X 10-11 m
l = h / mv = (6.626 X 10-34 Js) / (.10 kg) (35 m/s) = 1.9 X 10-34 m
Objectives #1-3 Atomic Theory
*Work Function (Photoelectric Effect)Φ = hfo
Φ = work functionminimum energy required to remove
electron from surface of metalfo = threshold frequencyminimum frequency required to remove
electrons from surface of metal(clip, diagram and examples to follow)
Photoelectric Effect (Albert Einstein Nobel Prize 1921)
#1 A gold strip is irradiated with radiation of a frequency of
6.9 X 1012 Hz. Calculate the energy of this radiation and determine if it is sufficient to cause electrons to be released from the metal. The work function of gold is 7.7 X 10-19 J.
E = hf = 6.626 X 10-34 Js X 6.9 X 1012 Hz = 4.57 X 10-21 J; no electron ejection
#2 The ionization energy of gold is 890.1 kJ/mole of electrons. Calculate the threshold
frequency required to cause the photoelectric effect and eject an electron.
Φ = hfo
(convert I.E. into energy per electron)890.1 kJ/mole X 1000 J / 1 kJ X 1 mole electrons / 6.02 X 1023 electrons= 1.479 X 10-18 Jfo = 1.47 X 10-18 J / 6.626 X 10 -34 Js = 2.232 X 1015 Hz
Niels Bohr (1892-1962)
Objectives #1-3 Atomic Theory
*Bohr’s Equation:E = -2.178 X 10-18 J (z2/n2) OR ∆E = -
2.178 X 10-18 J (z2) X (1/n2final –
1/n2initial)
used for: determining energy changes when electrons change energy levels
for hydrogen; z = 1(clip and examples)
#1 Calculate the energy required to excite an electron from energy level 1 to energy level 3.
ΔE = -2.178 X 10-18 J (1/9 – 1/1) = -2.178 X 10-18 J (-.8889) = 1.936 X 10-18 J
#2 Calculate the energy required to remove an electron from a hydrogen atom.
*this energy represents the ionization energy for hydrogen
ΔE = -2.178 X 10-18 J (-1) =2.178 X 10-18 J
Johannes Rydberg (1854-1919)
Objectives #1-3 Atomic Theory
*Rydberg Equation:1/λ = 1/91 nm (1/nL
2 – 1/nH2)
*used for: determining wavelength of photons released when electrons change energy levels
(Examples)
#1 Calculate the wavelength of light released when an electron drops from the following states:
2 to 1:1 / l = 1 /91 nm (1/1 – 1/4) = .75/91 nm l = 91nm/.75 = 121.33 nm
4 to 1:1 / l = 1 /91 nm (1/1 – 1/16) = .9375/91 nm l = 91nm/.9375 = 97.07 nm
6 to 1:1 / l = 1 /91 nm (1/1 – 1/36) = .9722/91 nm l = 91nm/.9722 = 93.60 nm
*relationships of answers: the greater the energy difference, the smaller the wavelength
Erwin Schrodinger (1887-1961)(clip)
Objectives #4-5 The Quantum Numbers and Quantum States
*Review of Quantum Theory:1. Quantum NumbersA. Principle (n)*energy level of shell of electron*n = 1,2,3…..*(old system) n = K, L, M, ….*indicates the number of sublevels in
energy level
Illustration of Principle Quantum Number
Objectives #4-5 The Quantum Numbers and Quantum States
B. Orbital (l)*indicates orbital shape*l = 0, n-1*s, p, d, f
Illustration of Orbital Quantum Number / Orbital Shapes
Objectives #4-5 The Quantum Numbers and Quantum States
C. Magnetic (ml)*indicates orientation of orbital in
space*ml = 0, +/-l *the number of ml values indicate the
number of orbitals within sublevel
Illustration of Magnetic Quantum Number
Objectives #4-5 The Quantum Numbers and Quantum States
D. Spin (ms)*indicates spin of electron*+1/2 or -1/2*allows for up to 2 electrons per orbital*s 2 electrons p 6 electrons d 10 electrons f 14 electrons
Illustration of Spin Quantum Number
Objectives #4-5 The Quantum Numbers and Quantum States
*Quantum Number Sets for Electrons in Atoms:
Illustration of Quantum States
Objectives #4-5 The Quantum Numbers and Quantum States
(examples of quantum number states problems)
Objectives #7-9 Electron Configurations of Ions / Orbital Filling and Periodic Trends
*valence electrons and occasionally the electrons contained within the d sublevel are involved in chemical bonding
*atoms tend to lose or gain electrons in such a way to complete octets (s2p6) or to form similarly stable arrangements called pseudo noble-gas configurations
(clip and examples)
Objectives #7-9 Electron Configurations of Ions / Orbital Filling and Periodic Trends
*Orbital Filling and Periodic Trends1. Ionization Energy (from period 2)Group 1 Li 5.4 evGroup 2 Be 9.3 ev (spike)Group 15 N 14.5 ev (spike)Group 17 F 17.4 ev (spike)Group 18 Ne 21.6 ev (spike)
Trends in Ionization Energy
The effective nuclear charge on a particular electron in an atom is less than the actual nuclear charge of the atom due to the screening or shielding effect of the inner core electrons. For example, the effective nuclear charge on the 3s electrons in magnesium is less than the 2p electrons because there is less nuclear charge acting on the 3s electron. Since the 2p electrons are attracted more strongly, they are more stable, and have less energy than the 3s electron
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
*ionic bonds involve the transfer of valence electrons from a metal to a nonmetal
*the tendency for a metal to lose electrons depends on its ionization energy and the tendency of a nonmetal to gain electrons depends on its electron affinity
*the loss of an electron requires a gain of energy and is therefore an endothermic process
example: Na + energy › Na+1 + e-
*the gain of an electron releases energy and is therefore an exothermic process
example: Cl + e- › Cl-1 + energy
Formation of Sodium Chloride
Formation of Crystal Lattice
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
*combinations of elements with low ionization energies and high electron affinities will cause an extremely exothermic reaction and generally be the most stable*example: Na(s) + Cl2(g) › NaCl(s) + energy
*the energy produced when the ionic bond forms is referred to as the lattice energy; this energy is also equal to the energy required to break apart the ionic bond
*chemical bonding not only involves a rearrangement of electrons but it also involves changes in energy (clip)
Illustration of Born-Haber Cycle
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
*the formation of an ionic compound; such as the following reaction:
Na(s) + 1/2Cl(2)(g) › NaCl(s) + ∆Hof =
-410.9 kJwhere ∆Ho
f refers to the standard heat of formation which is the energy change involved when a compound is formed from its elements, involves a series of energy changing steps known as the Born-Haber cycle
*these steps are as follows:
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
1. Sublimation or Vaporization of nongaseous reaction components:
Here: Na(s) › Na(g) 108 kJ which represents the energy of sublimation or vaporization (an endothermic process)
2. Breaking the bonds of any gaseous components:Here 1/2Cl2(g) › Cl(g) 122 kJ which represents the
dissociation energy (an endothermic process)(now that all reactants are gaseous, ions must be
formed)
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
3. Formation of the positive ion:Here: Na(g) › Na+1
(g) + e- 496 kJ which represents the ionization energy (an endothermic process)
4. Formation of the negative ion:Here: Cl(g) + e- › Cl(g)
-1 -349 kJ which represents the electron affinity energy (an exothermic process)
5. Formation of the ionic compound by combining the two ions formed together:
Here: Na+1(g) + Cl(g)
-1 › NaCl(s) -788 kJwhich represents the lattice energy (an exothermic process)*the overall energy change, ∆Ho
f, is equal to the sum of all these changes:
∆Hof = ∆Ho
fNa + ∆HofCl + IE Na - EA Cl - ∆Hlattice
= -411 kJ
Example II Show all of the energy changing steps and determine the heat of formation for LiF given the following:
Sublimation energy for Li = 161 kJIonization energy for Li = 520 kJBond dissociation energy for F2 = 77 kJ
per ½ moleElectron affinity of fluorine = -328 kJLattice energy for LiF = -1047 kJ
Li(s) > Li(g) +161 kJLi(g) > Li+1 +520 kJ1/2F2(g) > F(g) +77 kJF(g) + e-1 > F-1
(g) -328 kJLi+1
(g) + F-1 (g)
> LiF(s) -1047 kJ--------------------------------------------
-617 kJ
Example III
Mg(s) > Mg(g) +147 kJMg(g) > Mg+1 (g) +738 kJMg(g)
+1 > Mg+2 (g) +1451 kJ
Cl2(g) > 2Cl(g) +242 kJ2Cl(g) + 2e-1 > 2Cl-1(g) -698 kJMg+2
(g) + 2Cl-1 > MgCl2(s) -2326 kJ--------------------------------------------
-446 kJ
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
*Relationship of lattice energy and ionic chargeConsider the following lattice energy data from the above
example problems:NaCl 788 kJLiF 1030 kJMgCl2 2326 kJKCl 701 kJ**strength of ionic bonds:KCl ‹ NaCl ‹ LiF ‹ MgCl2**size of ions:LiF ‹ MgCl2 ‹ NaCl ‹ KCl**charge of ions:Na +1, Cl -1 Li +1, F -1 K +1, Cl -1 Mg +2, Cl-1
Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy
**formula: Eel. = KQ1Q2/d where “K” is a constant of electrical charge, where “Q1” and “Q2” are the charges of the ions involved, where “d” distance separating the ions
As the magnitude of the charges in an ionic compound increases, the lattice energy increases (affects lattice energy the most)
As the size of the ions involved decrease, the lattice energy increases
(examples)
Examples:
Arrange the following ionic compounds in order of increasing lattice energy:
NaF CsI CaOCsI, NaF, CaO
Arrange the following ionic compounds in order of increasing lattice energy:
AgCl, CuO, CrN
Objectives #13-14 Phase Changes and Phase Diagrams
Objectives #13-14 Phase Changes and Phase Diagrams
*Important Parts of a Heating/Cooling Curve(see curve in lecture guide)A. Specific heat of solid added (endothermic)B. Specific heat of liquid added (endothermic)C. Specific heat of gas added (endothermic)D. Melting (heat of fusion added) (endothermic)E. Freezing (heat of solidification released) (exothermic)F. Boiling (heat of vaporization added) (endothermic)G. Condensing (heat of condensation released)
(exothermic)(graph interpretation practice)
Objectives #13-14 Phase Changes and Phase Diagrams
*where the graph is increasing or decreasing, specific heat is being added or subtracted (which results in the temperature changing)
*where the graph is not changing, a phase change is occurring and these is no change in the temperature of the substance
*key equations:to change temperature: Q =mc∆tto change phase:∆H = moles of material X molar heat of phase
change(examples)
Example I:Calculate the energy change involved
to convert 1 mole of ice at -25oC to steam at 125oC.
(analysis of steps involved)
Example I
Step I: -25C > 0CQ = (18.0 g) (2.09J/gC) (25C) = 941 J = .941 kJStep II: meltΔH = (1 mole) (6.01 kJ/mol) = 6.01 kJStep III: 0C > 100C Q = (18.0 g) (4.18 J/gC) (100C) = 7524 J = 7.52 kJ
Example I
Step IV: boil ΔH = (1 mole) (40.67 kJ/mole) = 40.67 kJStep V: 100C > 125C Q = (18.0 g) (1.84 J/gC) (25C) = 828 J = .828 kJ Total = 56.0 kJ
Example II
Step I: 140C > 100CQ = (100.0 g) (1.84J/gC) (40C) = -7360 J = -7.360 kJStep II: condensationΔH = (100.0 g/18.0 g) (40.67 kJ/mol) = -225.9 kJStep III: 100C > 0C Q = (100.0 g) (4.18 J/gC) (100C) = -41800 J = -41.8 kJ
Example II
Step IV: freeze ΔH = (100.0 g/18.0 g) (6.01 kJ/mole) = -33.4 kJStep V: 0C > -30C Q = (100.0 g) (2.09 J/gC) (30C) = -6270 J = -6.27 kJ Total = -315 kJ
Objectives #13-14 Phase Changes and Phase Diagrams
*Interpreting Phase Diagrams*a phase diagram allows one to
determine the phase that a substance is in at a given temperature and pressure
*the phase diagram only shows one substance in its various phases
*a typical phase diagram:(see diagram in lecture guide and next
slide)
Objectives #13-14 Phase Changes and Phase Diagrams
Objectives #13-14 Phase Changes and Phase Diagrams
*Interpreting Phase Diagrams*a phase diagram allows one to
determine the phase that a substance is in at a given temperature and pressure
*the phase diagram only shows one substance in its various phases
*a typical phase diagram:(see diagram in lecture guide)
Objectives #13-14 Phase Changes and Phase Diagrams
*the boundaries between different phase regions represent areas of equilibrium in which the two phase changes are occurring at the same rate; for example at the liquid – gas boundary, molecules of gaseous vapor are moving into the liquid phase while molecules of liquid are moving into the gaseous phase
*if a point on the diagram does not fall on any line, only one phase is present
Objectives #13-14 Phase Changes and Phase Diagrams
*the following lines on the graph represent phase change boundaries:Line Segment Phase Change
BoundaryA-B Liquid – gas
(vaporization ↔ condensation
A-C Solid – gas (sublimation ↔ deposition)
A-D Solid – liquid (melting ↔ freezing)
Objectives #13-14 Phase Changes and Phase Diagrams
Critical Point the endpoint of the vapor-pressure curve; beyond this point of critical temperature and critical pressure, the liquid and gas phases can not be distinguished from each other
Normal Boiling Point the location on the vapor-pressure curve where the vapor pressure is 1 atm
Normal Melting Point the location on the solid-liquid curve where the melting (freezing) point is at 1 atm
Triple Point the point where all 3 curves intersect and all three phases are in equilibrium
Objectives #13-14 Phase Changes and Phase Diagram
(practice worksheets)
*some general relationships and observations to note:
*if the solid-liquid line curves to the right with increasing pressure, then the melting point is also increasing (this is the norm)
*if the solid-liquid line curves to the left with increasing pressure, then the melting point is decreasing (this is not the norm; water follows this pattern)
Illustration of Phase Diagram for Water
Objectives #15-17 Structure of Solids, Properties and Applications
*solids come in two general types: crystalline or amorphous*crystalline solids contain particles arranged in a well-defined
pattern called a crystal lattice with flat faces and definite angles; examples include NaCl or diamonds
*amorphous solids lack any well defined structure; examples include wax, rubber, or glass
*the crystal lattice of a crystalline solid, which is a three dimensional array showing the location of individual particles, is actually made up of many repeating individual parts called the unit cell; for example the repeating pattern on wall paper
*the simplest common type of unit cell is the cubic unit cell where all sides are equal in length and consist of all 90o angles
Objectives #15-17 Structure of Solids, Properties and Applications
*the three types of cubic unit cells are:primitive or simple cubic – lattice points
only occur at the corners of unit cellbody-centered cubic – lattice points
occur at the corners and in the centerface-centered cubic – lattice points
occur at the corners and faces
Types of Cubic Unit Cells
Objectives #15-17 Structure of Solids, Properties and Applications
*except for the atom in the center of the body-centered cubic unit cell, all of the atoms located at the lattice points are actually shared to various degrees by other unit cells; in order to determine the net number of atoms in a unit cell and thus its chemical formula, one must know the fraction of an atom that occurs in each position of the unit cell as follows:
Objectives #15-17 Structure of Solids, Properties and Applications
Position in Unit Cell Fraction in Unit CellCenter 1Face ½Edge ¼Corner 1/8
Example of Face Centered Unit Cell in Sodium Chloride
Objectives #15-17 Structure of Solids, Properties and Applications
(examples)Example #1: Determine the number of sodium
and chloride ions in the NaCl unit cell using the diagram on the screen as a guide.
For Na: 12 edges X 1/4 each = 3 Na ions 1 ion NaFor Cl: 6 faces X ½ each = 3 Cl ions 8 corners X 1/8 each = 1 Cl ion4 Na ions to 4 Cl ions; NaCl
Example of Body Centered Unit Cell in Cesium Chloride
Example #2 The element iron crystallizes in a form called alpha iron, which has a body centered unit cell. How many iron atoms are in the unit cell?
1 Fe atom at center8 corners X 1/8 each = 1 Fe atomresults in 2 iron atoms
X-Ray Crystallography
*the layers of atoms in the crystal lattice acts as an effective diffraction grating that can be used to scatter a beam of x-rays
*the diffraction, or scattering, of the x-rays produces a characteristic pattern of light and dark areas on a x-ray detector
*by examining the areas of light and dark and measuring the angles of deflection, the original crystal structure of the material can be deduced
*this analytical technique has been used to determine the structure of DNA and other molecular crystals
Instrumentation(clip)
Objectives #15-17 Structure of Solids, Properties and Applications
*Types of Bonding in Solids and their Influence on Properties
Type of Solid
Form of Unit Particle
Forces Between Particles
Properties Examples
Molecular Atoms or molecules
London, DD, HB
Fairly soft, low melting points, poor conductors
Gases, sugars, dry ice
Covalent-Network
Atoms connected in a network of covalent bonds
Covalent bonds
Very hard, high melting points, poor conduction
Diamond, quartz, SiO2
Ionic Positive and negative ions
Electrostatic attractions
Hard and brittle, high melting point, poor conductor
Salts such as NaCl
Objectives #15-17 Structure of Solids, Properties and Applications
*Types of Bonding in Solids and their Influence on Properties
Metallic Atoms Metallic bonds
Soft to very hard, low to high melting point, malleable and ductile, excellent conductor
Metallic elements such as copper
Dry Ice – Example of Molecular Solid
Diamond – Example of Covalent Network Solid
Sodium Chloride – Example of Ionic Solid
Copper – Example of Metallic Solid