{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

patino_chm2041_chapter7

# patino_chm2041_chapter7 - Chapter 7 Chapter Atomic...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 7 Chapter Atomic Structure Chapter goals Chapter • Describe the properties of electromagnetic Describe radiation. radiation. • Understand the origin of light from excited Understand atoms and its relationship to atomic structure. structure. • Describe experimental evidence for waveparticle duality. • Describe the basic ideas of quantum Describe mechanics. mechanics. • Define the three quantum numbers (n, l , and and ml ) and their relationship to atomic structure. Electromagnetic Radiation Electromagnetic • light • dual nature: wave and particle • transverse wave: perpendicular oscillating transverse electric and magnetic fields electric • longitudinal wave: alternating areas of longitudinal compression and decompression. The direction of the wave is along the direction of propagation of • sound Transverse Waves Transverse • light • do not require medium for propagation Amplitude Amplitude height of wave at maximum Y Z Amplitude X Wavelength, λ ( lambda) distance traveled by wave in 1 complete oscillation; distance from the top (crest) of one wave to the top of the next wave. Y Z λ X • λ measured in m, cm, nm, Å (angstrom) (angstrom) • 1 Å = 1 × 1 0 −10 m = 1 × 1 0 −8 cm • frequency, ν (nu), measured in s−1 (hertz) (Hz): frequency, (nu) number of complete oscillations or cycles complete passing a point per unit time (s) passing • speed of propagation, speed distance traveled by ray per unit time distance in vacuum, all electromagnetic radiation travels at same rate travels c = 2.998 x 1010 cm/s (speed of light) 2.998 = 2.998 x 108 m/s 2.998 (slower in air) m c ( = ν( s−1) × λ ( m) ) ν( s What is the wavelength in nm of orange light, which has a frequency of 4.80 x 1014 s−1? which c=λ× ν c 2.998× 108 m s−1 2.998 λ = −− = −−−−−−−−−−−− = 6.25 × 10−7 m −− −−−−−−−−−−−− −1 ν 4.80 × 1014 s 1 nm nm 6.25 × 10−7 m × −−−−−−−−− = 625 nm 6.25 −−−−−−−−− 1 × 10−9 m Names to remember Names • Max Planck: quantized energy E = hν ~1900 • Albert Einstein: photoelectric effect ~1905 • Niels Bohr: 2-D version of atom • En=(-RH)(1/n2) Balmer, 1885, then Bohr, 1913 • Louis de Broglie: Wavelike properties of Louis matter ~1915 matter • Werner Heisenberg: Uncertainty Principle Werner ~1923 • Erwin Schrödinger: Schrödinger Equation Erwin ~1926 ~1926 Planck’s equation Planck’s • Planck studied black body radiation, such Planck as that of a heated body, and realized that to explain the energy spectrum he had to assume that: assume 1. An object can gain or lose energy by absorbing or emitting radiant energy in QUANTA of specific frequency (ν ) QUANTA 2. light has particle character (photons) light c • Planck’s equation is E = h × ν = h × ── Planck’s ── E = energy of one photon λ h = Planck’s constant = 6.626× 10−34 Planck’s Electromagnetic Spectrum Electromagnetic λ 0.01nm γ -rays Electromagnetic Spectrum Electromagnetic λ 0.01nm 1nm γ -rays x-rays Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 1nm γ -rays x-rays vacuum UV Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 1nm 400nm γ -rays UV x-rays vacuum UV Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 800nm 1nm 400nm γ -rays UV Vis. x-rays vacuum UV Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 800nm 1nm 400nm 25µ m γ -rays UV Vis. near x-rays infrared vacuum UV Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 800nm 1mm 1nm 400nm 25µ m γ -rays far IR UV Vis. near x-rays infrared vacuum UV Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 800nm 1mm 1nm 400nm 25µ m 100mm γ -rays far IR UV Vis. µ -waves near x-rays infrared vacuum UV Electromagnetic Spectrum Electromagnetic λ 0.01nm 200nm 800nm 1mm 1nm 400nm 25µ m 100mm γ -rays far IR UV Vis. µ -waves near x-rays infrared radio waves vacuum UV Electromagnetic Spectrum Electro Spectrum c =ν λ hν E= E = h c/λ c/ Compact disk players use lasers that emit red light with a wavelength of 685 nm. What is the energy of one photon of this light? What is the energy of one mole of photons of that red light? of λ , nm → λ , m → ν , s−1 → E, J/photon → E, J/mole nm 10−9 m 10 × −−−−−−− −−−−−−− nm c ν = −− −− λ E = hν × Avogadro’s number 10−9 m 10 685 nm × −−−−−−− = 6.85 × 10−7 m −−−−−−− 1 nm c 2.998× 108 m s−1 2.998 ν = −− = −−−−−−−−−−−− = 4.38 × 1014 s−1 −− −−−−−−−−−−−− λ 6.85 × 10−7 m E = hν = (6.626× 10−34 J⋅ s/photon)× 4.38 × 1014 4.38 s−1 = 2.90× 10−19 J/photon = (2.90× 10−19 J/photon)× 6.022× 1023 photons/mol = 1.75 × 105 J/mol The Photoelectric Effect The • Light can strike the surface of some metals Light causing electrons to be ejected. causing • It demonstrates the particle nature of light. The Photoelectric Effect The • What are some practical uses of the What photoelectric effect? photoelectric • Electronic door openers • Light switches for street lights • Exposure meters for cameras • Albert Einstein explained the effect – Explanation involved light having particlelike behavior. The minimum energy needed to eject the e− is E = h × ν (Planck’s equation) It is also called ‘threshold’ energy. It – Einstein won the 1921 Nobel Prize in Physics Einstein for this work. for Prob.: An energy of 2.0× 102 kJ/mol is required to cause a Cs Prob.: atom on a metal surface to loose an electron. Calculate the longest possible λ of light that can ionize a Cs atom. From the value of energy we calculate the frequency ( ν ) and, with this From and, we calculate lambda (λ ). we Firstly, we need to calculate the energy in J per atom; it is given in kJ Firstly, per mol of atoms... per kJ 1000 J 1 mol kJ 2 2.0× 10 ── x ───── x ───────────── = 3.3× 10−19 Joule per atom ── ───── ───────────── mol kJ 6.022 × 1023 atoms E=h× ν E 3.3 × 10−19 Joule ν = ── = ───────────── = 5.0× 1014 s-1 ── ───────────── h 6.626 × 10−34 J s Now, speed of light, c = λ ν Now, c 2.998× 108 m s-1 2.998 λ = ── = ───────────= 6.0× 10−7 m ── ─────────── ν 5.0× 1014 s-1 5.0 1 nm nm 6.0× 10 m × ──────── = 600 nm ──────── 1× 10−9 m −7 (Visible light) Prob. : A switch works by the photoelectric effect. The metal you wish to use for your device requires 6.7× 10−19 metal J/atom to remove an electron. Will the switch work if the J/atom light falling on the metal has a λ = 540 nm or greater? Why? The energy of photon is The calculated with Planck’s Equation calculated c E = h × ν = h × ── If calculated E ≥ 6.7× 10−19 J, 6.7 λ the switch will work. the 1× 10−9 m 540 nm × ────── = 5.40× 10−7 m ────── nm nm 2.998× 108 m s−1 2.998 E = 6.626× 10−34 J⋅ s × ────────── = 3.68× 10−19 J 6.626 ────────── 5.40× 10−7 m 5.40 The switch won’t open, because E < 6.7× 10−19 J. λ has to 6.7 be less than 540 nm. Atomic Line Spectra and the Bohr Atom Atomic (Niels Bohr, 1885-1962) (Niels 1885-1962 • An emission spectrum is formed by an electric emission current passing through a gas in a vacuum tube (at very low pressure) which causes the gas to emit light. very – Sometimes called a bright line spectrum. line Atomic Line Spectra and the Bohr Atom Atom • The Rydberg The equation is an empirical equation that relates the wavelengths of the lines in the hydrogen spectrum. Lines are due to transitions n2 ──── upper level n1 ──── lower level n , n = 1, 2, 3, 4, 5,… 1 1 1 = R 2 − 2 n λ n2 1 R is the Rydberg constant R = 1.097 × 107 m -1 n1 < n 2 n’ s refer to the numbers of the energy levels in the emission spectrum of hydrogen Example 5-8. What is the wavelength in angstroms of light emitted when the hydrogen atom’s energy changes from n = 4 to n = 2? changes n 2 = 4 and n1 = 2 1 1 1 = R 2 − 2 n λ n2 1 1 1 7 -1 1 = 1.097 × 10 m 2 − 2 λ 4 2 1 λ =1.097 ×107 m -1 ( 0.250 − 0.0625) 1 λ =1.097 ×10 7 m -1 ( 0.1875) 1 λ = 2.057 ×106 m -1 1 1Å λ = ─ ─ ─ ─ ─ ─ ─ ─ ─ = 4.862 × 10−7 m × ─ ─ ─ ─ ─ = 4862 Å 2.057× 106 m−1 10−10 m 2.057 That corresponds to the green line in H spectrum Atomic Line Spectra and the Bohr Atom Atom • An absorption spectrum is formed by shining a An absorption beam of white light through a sample of gas. beam – Absorption spectra indicate the Absorption wavelengths of light that have been wavelengths absorbed. absorbed Every element has a unique spectrum. Thus we can use spectra to identify elements. elements. This can be done in the lab, stars, This fireworks, etc. fireworks, Bohr Model of the Atom Bohr • planetary model • considers only the particle nature of the considers electron electron • p+ & n packed tightly in ‘tiny’ nucleus • electrons traveling in circular paths, electrons orbits, in space surrounding nucleus orbits, • size, energy, and e– capacity of orbits increase as does distance from nucleus (orbital radius) (orbital • orbits quantized (only certain levels exist) orbits quantized e– e– e– P+ n e– e– e– e– Energy Levels E n=6 n=5 n=4 n=3 n=2 n=1 n=1 -1 -1 E ∝ ── r2 -1 E ∝ ── n2 E n=6 n=5 n=4 n=3 n=2 n=1 n=1 Exciting the electron from ground level (n = 1) to upper levels (n > 1) Energy is absorbed E n=6 n=5 n=4 n=3 n=2 Decay of the electron from upper levels to lower levels: Energy is emitted Emission of Photons hν One photon photon per per transition transition n=1 n=1 E n=6 n=5 n=4 n=3 n=2 n=1 n=1 Balmer Series, nf = 2, for hydrogen. There are other series. hν Calculating E Difference Between two Levels Calculating A school teacher was the first to find this! school Johann Balmer Johann 1 1 ∀ ∆ E= |Efinal − Einitial| = RH(── − ──) E= n f 2 n i2 18 • RH= 2.18 x 10--18 J/atom = 1312 kJ/mol J/atom • ni and nf = principal quantum numbers of and the initial and final states: n f < ni the • 1,2,3,4…. Problem: Calculate ∆ E and λ for the violet line and of Balmer series of H. ninitial = 6 nfinal = 2 of 1 1 18 ∀ ∆ E= RH(── − ──) RH= 2.18 x 10--18 J/atom E= (── J/atom nf2 ni2 1 1 18 ∆ E= 2.18 x 10--18 J(── − ──) = 4.84 x 10−19 J 22 62 ∆ E = hν ν = c/λ Then, ∆ E = hc/λ hc 6.626× 10−34 J⋅ s × 2.998x108 ms−1 hc λ = ── = ───────────────────── ∆E 4.84 x 10−19 J 4.84 λ = 4.104 x 10−7 m × (1 Å/10−10m) = 4104 Å = 410.4 nm 4.104 Å/10 m) 410.4 Bohr Model of the Atom Bohr • Bohr’s theory correctly explains the H Bohr’s emission spectrum and those of hydrogenemission like ions (He+, Li2+ … 1e− species) • The theory fails for all other elements The because it is not an adequate theory: it doesn’t take into account the fact that the (very small) electron can be thought as electron having wave behavior. having The Wave Nature of the Electron The • In 1925 Louis de In Broglie published his Ph.D. dissertation. dissertation. – A crucial element crucial of his dissertation is that electrons have wave-like properties. properties. – The electron The wavelengths are described by the de Broglie relationship. relationship. h λ= mv h = Planck’ s constant m = mass of particle v = velocity of particle The Wave-Particle Duality of the Electron Electron • Consequently, we now know that Consequently, electrons (in fact - all particles) have both a particle and a wave like character. particle • This wave-particle duality is a This fundamental property of submicroscopic particles (not for macroscopic ones.) particles The Wave-Particle Duality of the Electron The • Example: Determine the wavelength, in m and Å, Example: of an electron, with mass 9.11 x 10-31 kg, having a of velocity of 5.65 x 107 m/s. velocity h = 6.626 x 10−34 Js = 6.626 x 10−34 kg m2/s h 6.626× 10−34 kg m2s−1 kg 6.626 λ = ── = ──────────────────── mv 9.11× 10−31kg × 5.65x107 ms−1 mv −11 λ = 1.29 × 10−11 m Good: Good: 1Å within within −11 λ = 1.29 × 10−11 m ────── = 0.129 Å atomic atomic 10−10 m dimensions 10 dimensions The Wave-Particle Duality of the Electron The • Example: Determine the wavelength, in m, of a Example: 0.22 caliber bullet, with mass 3.89 x 10-3 kg, 0.22 having a velocity of 395 m/s, ~ 1300 ft/s. having h = 6.626 x 10−34 Js = 6.626 x 10−34 kg m2/s 6.626 h 6.626× 10−34 kg m2s−1 kg 6.626 λ = ── = ──────────────── mv 3.89× 10−3kg × 395 ms−1 mv −34 −24 λ = 4.31 × 10−34 m = 4.31 × 10−24 Å too small! It doesn’t apply to macro-objects! Quantum Mechanical Model of the Atom Quantum considers both particle and wave nature of electrons • Heisenberg and Born in 1927 developed the concept Heisenberg of the Uncertainty Principle: Uncertainty It is impossible to determine simultaneously both the position (x,y,z) and momentum (mv) of an electron (or any other small particle). any • Consequently, we must speak of the electrons’ Consequently, positions about the atom in terms of probability functions, i.e., wave equation written for each electron. functions • These probability functions are represented as orbitals These orbitals in quantum mechanics. They are the wave equations squared and plotted in 3 dimensions. squared Schrödinger’s Model of the Atom Schrödinger’s Basic Postulates of Quantum Theory 1. Atoms and molecules can exist only in certain Atoms energy states. In each energy state, the atom or molecule has a definite energy. When an atom or molecule changes its energy state, it must emit or absorb just enough energy to bring it to the new energy state (the quantum condition). energy 2. Atoms or molecules emit or absorb radiation (light) Atoms as they change their energies. The frequency of the light emitted or absorbed is related to the energy change by a simple equation. energy E = hν = hc λ Schrödinger’s Model of the Atom Schrödinger’s 3. The allowed energy states of atoms and The molecules can be described by sets of numbers called quantum numbers. quantum • Quantum numbers are the solutions of the Quantum Schrödinger, Heisenberg & Dirac equations. Schrödinger, • Four quantum numbers are necessary to describe energy states of electrons in atoms. atoms. .. Schr o dinger equation b2 ∂2Ψ ∂2Ψ ∂2Ψ − 2 2 + 2 + 2 + VΨ = EΨ 8π m ∂ x ∂ y ∂ z Orbital Orbital • region of space within which one can expect to region find an electron find • no solid boundaries • electron capacity of 2 per orbital electron per • space surrounding nucleus divided up into space large volumes called shells shells • shells subdivided into smaller volumes called shells subshells subshells • orbitals located in subshells • as shells get further from nucleus, energy, size, as and electron capacity increase and • shells, subshells, and orbitals described by shells, quantum numbers quantum Quantum Numbers Quantum • The principal quantum number has the The symbol n symbol • n = 1, 2, 3, 4, ... indicates shell 1, shell • K, L, M, N, … shells shells • as n increases, so does size, energy, and as electron capacity electron The electron’s energy depends principally energy on n . on Quantum Numbers Quantum • The angular momentum (azimuthal) quantum quantum number has the symbol . It indicates number It subshell. subshell. = 0, 1, 2, 3, 4, 5, .......(n-1) 0, from 0 to maximum (n-1) for each n from = s, p, d, f, g, h, ....... Subshells s, Subshells tells us the shape of the orbitals. tells • These orbitals are the volume around the These atom that the electrons occupy 90-95% of the time. time. This is one of the places where Heisenberg’s Uncertainty principle comes into play. Uncertainty Magnetic Quantum Number, ml Magnetic • The symbol for the magnetic quantum number is The m . It specifies the orientation of the orbital. For a given l, m = ­ , (­ + 1), (­ +2), .....0, ......., ( ­2), ( ­1), • If = 0 (or an s orbital), then m = 0. If 0. – Notice that there is only 1 value of m . Notice This implies that there is one s orbital per n This value. n ≥ 1 • If = 1 (or a p orbital), then m = -1,0,+1. – There are 3 values of m . There Thus there are three p orbitals per n value. Thus n≥ 2 Magnetic Quantum Number, m Magnetic • If = 2 (or a d orbital), then m = -2, -1, 0, +1, +2. – There are 5 values of m . There Thus there are five d orbitals per n value. n ≥ 3 Thus • If = 3 (or an f orbital), then m = -3, -2, -1, 0, +1, +2, +3. -3, – There are 7 values of m . 2x +1 orbitals There Thus there are seven f orbitals per n value, n ≥ 4 Thus • Theoretically, this series continues on to g, h, i, Theoretically, etc. orbitals. etc. – Practically speaking atoms that have been Practically discovered or made up to this point in time only have electrons in s, p, d, or f orbitals in their ground state configurations. their #orbitals n2 2l + 1 ml n shell l subshell s 0 1K0 1 s 0 2L0 1 –1,0,1 1p 3 0 3 M0 s 1 –1,0,1 1p 3 2d 5 -2,-1,0,1,2 0 4N0 s 1 –1,0,1 1p 3 2d 5 -2,-1,0,1,2 3 f -3,-2,-1,0,1,2,3 7 Max #e– 1 2 28 4 6 2 6 18 9 10 2 6 16 10 32 14 Maximum two electrons per orbital Electrons Indicated by Shell and Subshell and Symbolism #electrons nl principal number 4s1 3s2 4d 12 # letter: s, p, d,.. orbital letter: s, 5p4 3p 7 4f 5 there are 4 electrons in the 5p orbitals 4f14 The Shape of Atomic Orbitals Atomic • s orbitals are spherically symmetric. A plot of the surface density as a function of the distance from the nucleus for an s orbital of a hydrogen atom • It gives the probability of finding the electron It at a given distance from the nucleus at A node (nodal node point or surface) point is a point where Ψ and the and probability probability density (Ψ 2) are density zero p orbitals orbitals • p orbital properties: – The first p orbitals appear in the n = 2 The shell. shell. • p orbitals are peanut or dumbbell shaped orbitals volumes. volumes. – They are directed along the axes of a They Cartesian coordinate system. Cartesian • There are 3 p orbitals per n level. There – The three orbitals are named px, py, pz. – They have an = 1. – m = -1,0,+1 3 values of m p Orbitals Orbitals y y z x y px x z py x pz z d orbitals orbitals • d orbital properties: – The first d orbitals appear in the n = 3 shell. • The five d orbitals have two different shapes: – 4 are clover leaf shaped. – 1 is peanut shaped with a doughnut around it. – The orbitals lie directly on the Cartesian axes The or are rotated 45o from the axes. or • There are 5 d orbitals per n level. – The five orbitals are named d xy , d yz , d xz , d x 2 - y 2 , d z 2 – They have an = 2. – m = -2,-1,0,+1,+2 5 values of m d Orbitals Orbitals y y z z z yx x dx2–y2 dxy x y z x dz2 y z x dxz dyz f orbitals orbitals • f orbital properties: – The first f orbitals appear in the n = 4 The shell. shell. • The f orbitals have the most complex The shapes. shapes. • There are seven f orbitals per n level. – The f orbitals have complicated names. – They have an = 3 – m = -3, -2, -1,0, +1,+2, +3 7 values of -3, m – The f orbitals have important effects in The f orbitals orbitals • f orbital shapes Prob: A possible excited state for the H atom has an electron in a 5d orbital. List all possible sets of quantum numbers n, l , and ml for this electron n = 5, l=2 five possible ml : -2, -1, 0, +1, +2 Then, there are five sets of (n, l , ml ) Then, (5, 2, -2) -2) (5, 2, -1) (5, -1) (5, 2, 0) (5, 2, +1) (5, +1) (5, 2, +2) (5, +2) Prob: Which of the following represent valid sets of quantum numbers? For a set that is invalid, explain briefly why it is not correct. explain a) n = 3, l = 3, ml : 0 3, b) n = 2, l = 1, ml : 0 b) No: maximum l = n-1 No: l = 0, 1, and 2, no 3 c) n = 6, l = 5, ml : −1 d) n = 4, l = 3, ml : −4 d) No: minimum value of ml : −l, that is ml : −3 that −3 Prob: What is the maximum number of orbitals that can be identified by each of the following sets of quantum numbers? When “none” is the correct answer, explain the reason. correct Answer a) n = 4, l = 3 a) Seven b) n = 5, 25 c) n = 2, l = 2 c) None Why? ml : -3, -2, -1, 0, 1, 2, 3 -3, n2 maximum l = n - 1 d) n = 3, l = 1, ml : -1 One, just that described #s. Prob: State which of the following are incorrect designations for orbitals according to the quantum theory: 3p, 4s, 2f, and 1p quantum Answer a) 3p a) correct Why? n = 3, l = 1 maximum l = 3-1=2 b) 4s correct correct c) 2f c) incorrect maximum l = 1 (l = 3 for f) d) 1p incorrect maximum l = 0 (l = 1 for p) ...
View Full Document

{[ snackBarMessage ]}