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Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes. H Advanced Chemistry Unit 3. Objectives #1-3 Atomic Theory. *review of electromagnetic radiation characteristics: (diagrams). Examples of Electromagnetic Radiation. Objectives #1-3 Atomic Theory.
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Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes H Advanced Chemistry Unit 3
Objectives #1-3 Atomic Theory *review of electromagnetic radiation characteristics: (diagrams)
Objectives #1-3 Atomic Theory frequency, wavelength, energy frequency 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)
Objectives #1-3 Atomic Theory E = hf or hc/l h = Planck’s Constant (energy for waves) 6.626 X 10-34 Js
Objectives #1-3 Atomic Theory E = mc2 (energy for particles) *Wave particle-duality Matter has wave and particle characteristics; acts as particle when interacting with matter; acts as wave when traveling through space (clip)
Derivation of de Broglie’s Equation: Ewaves = Eparticles hc / l = mc2 l (hc / l) = (mc2)l h = mcl l = 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 m f = (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. • = 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 function minimum energy required to remove electron from surface of metal fo = threshold frequency minimum frequency required to remove electrons from surface of metal (clip, diagram and examples to follow)
#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 J fo = 1.47 X 10-18 J / 6.626 X 10 -34 Js = 2.232 X 1015 Hz
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
Objectives #1-3 Atomic Theory *Rydberg Equation: 1/λ = 1/91 nm (1/nL2 – 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
Objectives #4-5 The Quantum Numbers and Quantum States *Review of Quantum Theory: • Quantum Numbers • 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
Objectives #4-5 The Quantum Numbers and Quantum States • Orbital (l) *indicates orbital shape *l = 0, n-1 *s, p, d, f
Objectives #4-5 The Quantum Numbers and Quantum States • Magnetic (ml) *indicates orientation of orbital in space *ml = 0, +/-l *the number of ml values indicate the number of orbitals within sublevel
Objectives #4-5 The Quantum Numbers and Quantum States • 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
Objectives #4-5 The Quantum Numbers and Quantum States *Quantum Number Sets for Electrons in Atoms:
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 Trends • Ionization Energy (from period 2) Group 1 Li 5.4 ev Group 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)
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 lessthan 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
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)
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 kJ where ∆Hof 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: