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Atomic Structure Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements. 2. Compounds are composed of atoms of more than one element. The relative number of atoms of each element in a given compound is always the same. 3. Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed in chemical reactions. 2.1 Dalton 16 X + 8Y 8 X2Y 2.1 2 2.1 I. Atomic Structure A. Three Fundamental Particles 1. Electrons a. Work by Michael Faraday and George Stoney lead to the discovery of electrons b. In 1897, J.J. Thompson determined the charge to mass ratio of electrons c. In 1909, Robert Millikan determined the mass of an electron i. 9.1096 x 10-28 g Measured mass of e(1923 Nobel Prize in Physics) e- charge = -1.60 x 10-19 C Thomson’s charge/mass of e- = -1.76 x 108 C/g e- mass = 9.10 x 10-28 g 2.2 d. The electron is assigned a charge of -1 e. Cathode ray tubes - two electrodes are sealed in a glass tube with low pressure i. when high voltage is applied and rays are given off ii. these rays travel from the negative electrode ( cathode) to the positive electrode (anode) 2. Protons a. component of all atoms b. mass of the proton is 1836 times bigger than the electron i. c. 1.6726 x 10-24g assigned a charge of +1 d. Canal rays - in 1886 Eugene Goldstein observed a cathode ray tube also generates a stream of positively charged particles 3. Neutrons a. because atoms are electrically neutral - there must be an equal number of protons and electrons - but the sum of the protons and electrons do not add up to equal the mass of an atom b. in 1920 Ernest Rutherford postulated the existence of an uncharged particle c. in 1932, James Chadwick is given credit for discovering the neutron d. Heaviest of the three particles i. 1.6749 x 10-24g Rutherford’s Gold Experiment Logical conclusion atoms are mostly empty space “Honey I shrunk the KIds The modern view of the atom was developed by Ernest Rutherford (18711937). Rutherford’s Model of the Atom atomic radius ~ 100 pm = 1 x 10-10 m nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m “If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.” 2.2 Atomic number (z) of an element is the number of protons in the nucleus of each atom of that element It is the number of protons in the nucleus that determines what the element is Mass number is the total number of protons and neutrons in the nucleus of and atom Isotopes – atoms of the same element that have different mass numbers Which means that it has different number of neutrons The Atom Hydrogen Atomic number = 1 mass number = 1 Proton Electron Hydrogen has one proton, one electron and NO neutrons The Atom Hydrogen (isotope) Atomic number = 1 mass number = 2 Proton neutron Electron Hydrogen has one proton, one electron and NO neutrons The Atom Helium Proton Electron Neutron Helium has two electrons, two protons and two neutrons The Atom Lithium Protons Neutrons Electrons The Atom Beryllium Protons Neutron s Beryllium has four electrons, four protons and five neutrons. Electrons The Atom Boron Protons Electrons Neutron s Boron has five electrons, five protons and six neutrons. The Atom Carbon Protons Electrons Neutron s Carbon has six electrons, six protons and six neutrons. The Atom Nitrogen Protons Electrons Neutron s Nitrogen has seven electrons, seven protons and seven The Atom Oxygen Protons Electrons Neutron s Oxygen has eight electrons, eight protons and eight The Atom Fluorine Protons Electrons Neutron s Fluorine has nine electrons, nine protons and ten neutrons. The Atom Neon Protons Electrons Neutron s Neon has ten electrons, ten protons and ten neutrons. The Atom Sodium Protons Electrons Neutrons Sodium has eleven electrons, eleven protons and twelve Average atomic mass is the weighted average of the atomic mass’s of the naturally occurring isotopes of an element Name Hydrogen Symbol Bromine Oxygen Br atomic Number 1 35 O Na 8 11 19 Phosphorus K P 15 39.10 30.98 39 31 19 15 19 15 20 16 Magnesium Mg 12 12 12 Cl 17 24. 35 12 Chlorine 24.30 35.45 17 Carbon Francium C Fr 6 12.01 12 6 17 6 18 6 87 Iron Calcium Silicon Fe Ca Si 26 223 55.85 223 56 87 26 87 26 136 30 20 14 40.08 28.09 40 28 20 14 20 14 20 14 Tungsten W copper Cu 74 29 183.85 63.55 184 64 74 29 74 29 110 35 82 207.2 126.90 207 127 82 82 53 238.03 14.01 238 Sodium H Potassium Lead Iodine Uranium Nitrogen Tin Silver Manganese Aluminum Arsenic Pb I 53 U 92 N Sn Ag Mn Al As 7 50 47 25 13 33 Atomic Mass Mass number # protons # electrons # neutrons 1.01 1 1 1 0 79.90 35 35 45 80 8 16.00 16 8 8 22.99 23 11 12 11 53 92 7 7 118.71 14 119 146 7 50 50 69 107.87 54.94 108 55 47 25 47 25 61 30 27 75 13 33 13 14 42 26.98 74.92 92 125 74 33 VII Moles A. Avogadro’s number 1. 6.023 x 1023 2. this is one mole - think as though it is a dozen a. 3. One mole of anything = 6.023 x 1023 One mole of an element = atomic mass a. 1 mole of Na = 22.9898 g this is the molar mass of Na i. that is - if you take the atomic mass in grams of and element, it will = 1 mole = 6.023 x 1023 atoms How to use Avogadro’s number e.g. How many atoms are there in 10.0 moles of Mg (10.0 moles Mg) ( 6.023 x 1023atoms Mg) (1 mole of Mg) = 6.02 x 10 24 atoms of Mg e.g How many mole of Zn are in 1.2 x 1024 atoms (1.2 x 1024 atoms of Zn) ( 1mole of Zn ) (6.023 x 1023 atoms Zn) = 2.0 moles of Zn You try two How many atoms are in 3.0 moles of Cu ( 3.0 moles Cu ) ( 6.023 x 10 23 Atoms Cu 1 mole Cu ) = 1.8 x 1024 atoms of Cu How many moles of W are in 2.50 x 1025 atoms of W ( 2.50 x 1025 atoms W )( 1 mole W ) 6.023 x 10 23 Atoms of W = 41.5 moles W e.g How many grams are in 3.0 moles of Fluorine (3.0 moles F ) ( 19 g of F ) ( 1 mole of F) = 57 g of F Changing grams to atoms e.g How many atoms are in 10.0 g of Oxygen 23 Atoms of O) ( 1mole of O ) ( 6.023 x 10 (10.0 g of O ) ( 16.0 g of O) ( 1mole of O) = 3.76 x 1023 atoms of O Changing atoms to grams How many grams of potassium are in 2.30 x 1025 atoms (2.30 x 1025 atoms of K)( 1 mole of K ) ( 39.1 g of K ) ( 6.023 x 1023 atoms ) ( 1 mole of K ) = 1490 g of K Homework ( ( 25 g K )( 2.50 moles Br ( 1 mole K ) 39.1 g K )( 15.0 moles Ca 79.9 g Br 1 mole Br = .64 moles K ) = 200. g Br ) ( 6.023 x 10 atoms Ca 1 mole Ca 23 = ) 9.03 x 1024 atoms Ca ( ) ( 6.023 x 10 5.0 x 1024 atoms N ) ( atoms N 1 mole N 23 14.0 g N 1 mole N ) = 120 g N ( 2.0 g of Cu )( 1 mole Cu 64 g Cu )( 1.9 x 1022 atoms Cu ) 6.023 x 1023 atoms Cu 1 mole Cu ATOMIC STRUCTURE • properties of light • spectroscopy • quantum hypothesis • hydrogen atom • Heisenberg Uncertainty Principle • orbitals ELECTROMAGNETIC RADIATION • subatomic particles (electron, photon, etc) have both PARTICLE and WAVE properties • Light is electromagnetic radiation crossed electric and magnetic waves: Properties : Wavelength, l (nm) Frequency, n (s-1, Hz) Amplitude, A constant speed. c 3.00 x 108 m.s-1 7.1 Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1 Electromagnetic Radiation (3) • All waves have: frequency and wavelength • symbol: n (Greek letter “nu”) l (Greek “lambda”) • units: “distance” (nm) “cycles per sec” = Hertz • All radiation: l•n = c where c = velocity of light = 3.00 x 10 8 m/sec Note: Long wavelength small frequency Short wavelength high frequency increasing frequency increasing wavelength A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? l lxn=c n l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m l = 5.0 x 1012 nm Radio wave 7.1 Electromagnetic Radiation (4) Example: Red light has l = 700 nm. Calculate the frequency, n. 8 3.00 x10 m/sec 14 c = 4.29 x 10 sec = n= -7 l 7.00 x 10 m • Wave nature of light is shown by classical wave properties such as • interference • diffraction Quantization of Energy (2) Energy of radiation is proportional to frequency. E = h•n where h = Planck’s constant = 6.6262 x 10 -34 J•s Light with large l (small n) has a small E. Light with a short l (large n) has a large E. When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. E=hxn E=hxc/l E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) E = 1.29 x 10 -15 J 7.2 Atomic Line Spectra • Bohr’s greatest contribution to science was in building a simple model of the atom. • It was based on understanding the SHARP LINE SPECTRA of excited atoms. Niels Bohr (1885-1962) (Nobel Prize, 1922) Line Spectra of Excited Atoms • Excited atoms emit light of only certain wavelengths • The wavelengths of emitted light depend on the element. H Hg Ne 7.3 Atomic Spectra and Bohr Model One view of atomic structure in early 20th century was that an electron (e-) traveled about the nucleus in an orbit. + Electron orbit 1. Classically any orbit should be possible and so is any energy. Atomic Spectra and Bohr Model (2) • Bohr said classical view is wrong. • Need a new theory — now called QUANTUM or WAVE MECHANICS. • e- can only exist in certain discrete orbits — called stationary states. • e- is restricted to QUANTIZED energy states. Energy of state = - C/n2 where C is a CONSTANT n = QUANTUM NUMBER, n = 1, 2, 3, 4, .... E = hn E = hn 7.3 Chemistry in Action: Element from the Sun In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines Mystery element was named Helium In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). Atomic Spectra and Bohr Model (3) Energy of quantized state = - C/n2 • Only orbits where n = integral number are permitted. • Radius of allowed orbitals = n2 x (0.0529 nm) • Results can be used to explain atomic spectra. Atomic Spectra and Bohr Model (4) If e-’s are in quantized energy states, then DE of states can have only certain values. This explains sharp line spectra. E = -C (1/22) H atom n=2 07m07an1.mov E = -C (1/12) n=1 Calculate DE for e- in H “falling” from n = 2 to n = 1 (higher to lower energy) . Energy Atomic Spectra and Bohr Model (5) n=2 n=1 DE = Efinal - Einitial = -C[(1/12) - (1/2 2)] = -(3/4)C • (-ve sign for DE indicates emission (+ve for absorption) • since energy (wavelength, frequency) of light can only be +ve it is best to consider such calculations as DE = Eupper - Elower C has been found from experiment. It is now called R, the Rydberg constant. R = 1312 kJ/mol or 3.29 x 1015 Hz so, E of emitted light = (3/4)R = 2.47 x 1015 Hz and l = c/n = 121.6 nm (in ULTRAVIOLET region) This is exactly in agreement with experiment! Hydrogen atom spectra High E Short l High n Low E Long l Low n Visible lines in H atom spectrum are called the BALMER series. 6 5 4 Energy 3 2 1 En = -1312 n2 Ultra Violet Lyman Visible Balmer Infrared Paschen n Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. Ephoton = DE = RH( 1 n2i 1 n2f ) Ephoton = 2.18 x 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J Ephoton = h x c / l l = h x c / Ephoton l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l = 1280 nm 7.3 Because you have only one H atom, the R constant R = 1312 kJ/mol Must be changed to per atom. That means that you must divide R by 6.023 x 1023 This gives you 2.18 x 10-18 J ni = 3 ni = 3 ni = 2 nf = 2 Ephoton = DE = Ef - Ei 1 Ef = -RH ( 2 nf 1 Ei = -RH ( 2 ni 1 DE = RH( 2 ni ) ) 1 n2f ) nnf f==11 7.3 From Bohr model to Quantum mechanics Bohr’s theory was a great accomplishment and radically changed our view of matter. But problems existed with Bohr theory — – theory only successful for the H atom. – introduced quantum idea artificially. • So, we go on to QUANTUM or WAVE MECHANICS Quantum or Wave Mechanics • Light has both wave & particle properties • de Broglie (1924) proposed that all moving objects have wave properties. • For light: E = hn = hc / l L. de Broglie (1892-1987) • For particles: E = mc2 Therefore, mc = h / l (Einstein) and for particles (mass)x(velocity) = h / l l for particles is called the de Broglie wavelength What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? l = h/mv h in J•s m in kg v in (m/s) l = 6.63 x 10-34 j sec / (2.5 x 10-3 kg x 15.6 m/sec) l = 1.7 x 10-32 m = 1.7 x 10-23 nm 7.4 Quantum or Wave Mechanics Schrodinger applied idea of ebehaving as a wave to the problem of electrons in atoms. E. Schrodinger 1887-1961 Uncertainty Principle W. Heisenberg 1901-1976 Problem of defining nature of electrons in atoms solved by W. Heisenberg. Cannot simultaneously define the position and momentum (= m•v) of an electron. Dx. Dp = h At best we can describe the position and velocity of an electron by a PROBABILITY DISTRIBUTION, QUANTUM NUMBERS The shape, size, and energy of each orbital is a function of 3 quantum numbers which describe the location of an electron within an atom or ion n (principal) ---> energy level l (orbital) ---> shape of orbital ml (magnetic) ---> designates a particular suborbital The fourth quantum number is not derived from the wave function s (spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½) Orbital Quantum Numbers An atomic orbital is defined by 3 quantum numbers: – n l ml Electrons are arranged in shells and subshells of ORBITALS . n shell l subshell ml designates an orbital within a subshell Where 90% of the e- density is found for the 1s orbital e- density (1s orbital) falls off rapidly as distance from nucleus increases 7.6 l = 0 (s orbitals) l = 1 (p orbitals) 7.6 Schrodinger Wave Equation Y = fn(n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus n=1 n=2 n=3 7.6 How many 2p orbitals are there in an atom? n=2 If l = 1, then ml = -1, 0, or +1 2p 3 orbitals l=1 How many electrons can be placed in the 3d subshell? n=3 3d l=2 If l = 2, then ml = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e7.6 Quantum Numbers Symbol Values Description n (major) 1, 2, 3, .. Orbital size and energy = -R(1/n2) l (angular) 0, 1, 2, .. n-1 Orbital shape type (subshell) ml (magnetic) -l..0..+l Orbital orientation in space Total # of orbitals in lth subshell = 2 l + 1 Shells and Subshells For n = 1, l = 0 and ml = 0 There is only one subshell and that subshell has a single orbital (ml has a single value ---> 1 orbital) This subshell is labeled s (“ess”) and we call this orbital 1s Each shell has 1 orbital labeled s. It is SPHERICAL in shape. Types of Orbitals (l) s orbital p orbital d orbital Schrodinger Wave Equation Y = fn(n, l, ml, ms) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 l=0 l=1 l=2 l=3 s orbital p orbital d orbital f orbital Shape of the “volume” of space that the e- occupies 7.6 s Orbitals All s orbitals are spherical in shape. p Orbitals For n = 2, l = 0 and 1 There are 2 types of orbitals — 2 subshells For l = 0 ml = 0 this is a s subshell For l = 1 ml = -1, 0, +1 this is a p subshell with 3 orbitals Typical p orbital planar node When l = 1, there is a PLANAR NODE through the nucleus. p Orbitals • The three p orbitals lie 90o apart in space d Orbitals For n = 3, what are the values of l? l = 0, 1, 2 and so there are 3 subshells in the shell. For l = 0, ml = 0 s subshell with single orbital For l = 1, ml = -1, 0, +1 p subshell with 3 orbitals For l = 2, ml = -2, -1, 0, +1, +2 d subshell with 5 orbitals l = 2 (d orbitals) 7.6 Boundary surfaces for all orbitals of the n = 1, n = 2 and n = 3 shells n= 3d 3 2 There are n2 orbitals in the nth SHELL 1 f Orbital The last quantum number is the spin quantum num Electrons act as though they spin about an axis through their ce Because there are two directions in which they can spin, the spin quantum number has 2 values +1/2,-1/2 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. number of electrons in the orbital or subshell 1s1 principal quantum number n angular momentum quantum number l Orbital diagram H 1s1 7.8 “Fill up” electrons in lowest energy orbitals (Aufbau principle) ?? Be Li B5 C 3 64electrons electrons 22s 222s 22p 12 1 BBe Li1s1s 1s 2s H He12electron electrons He H 1s 1s12 7.7 The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). Ne97 C N O F 6 810 electrons electrons electrons 22s 222p 22p 5 246 3 Ne C N O F 1s 1s222s 7.7 Energy of orbitals in a multi-electron atom Energy depends on n and l n=3 l = 2 n=3 l = 0 n=2 l = 0 n=3 l = 1 n=2 l = 1 n=1 l = 0 7.7 Exceptions to the Aufbau Principle • Remember d and f orbitals require LARGE amounts of energy • If we can’t fill these sublevels, then the next best thing is to be HALF full (one electron in each orbital in the sublevel) • There are many exceptions, but the most common ones are d4 and d9 For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d4 or d9 are exceptions to the rule. This may or may not be true, it just depends on the atom. Exceptions to the Aufbau Principle • The next most common are f1 and f8 • The electron goes into the next d orbital • Example: – La [Xe]6s2 5d1 – Gd [Xe]6s2 4f7 5d1 Exceptions to the Aufbau Principle d4 is one electron short of being HALF full In order to become more stable (require less energy), one of the closest s electrons will actually go into the d, making it d5 instead of d4. For example: Cr would be [Ar] 4s2 3d4, but since this ends exactly with a d4 it is an exception to the rule. Thus, Cr should be [Ar] 4s1 3d5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d. Why are d and f orbitals always in lower energy levels? • d and f orbitals require LARGE amounts of energy • It’s better (lower in energy) to skip a sublevel that requires a large amount of energy (d and f orbtials) for one in a higher level but lower energy This is the reason for the diagonal rule! BE SURE TO FOLLOW THE ARROWS IN ORDER! What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s22s22p63s23p5 1s < 2s < 2p < 3s < 3p < 4s 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n=3 l=1 ml = -1, 0, or +1 ms = ½ or -½ 7.8 Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei Mass Number A ZX Atomic Number 1 1H 235 92 2 1H U Element Symbol (D) 238 92 3 1H (T) U 2.3 2.3 Do You Understand Isotopes? d. Canal rays - in 1886 Eugen Goldstein observed a cathode ra tube also generates a stream of positively charged particles 14 How many protons, neutrons, and electrons are in 6 C ? 6 protons, 8 (14 - 6) neutrons, 6 electrons How many protons, neutrons, and electrons are 11 in 6 C ? 6 protons, 5 (11 - 6) neutrons, 6 electrons 2.3 Keep an Eye On Those Ions! • Tin Atom: [Kr] 5s2 4d10 5p2 Sn+4 ion: [Kr] 4d10 Sn+2 ion: [Kr] 5s2 4d10 Note that the electrons came out of the highest energy level, not the highest energy orbital! Try These! • Write the shorthand notation for: 1 3d10 [Ar] 4s Cu 1 4f14 5d5 [Xe] 6s W 1 4f14 5d10 [Xe] 6s Au Odds and ends VI. Hund’s rule - electrons must be placed in each orbital before pairing them up VII. Valance electrons - those in the outer most energy level - gives elements its chemical properties VIII. orbital notation - lines IX. Electron configuration - 1s2 2s2 2p6 X. Nobel gas notation - {Ar} 4s1 XI. Electron dot notation - only use the valance electron ELEMENTS THAT EXIST AS DIATOMIC MOLECULES Remember: BrINClHOF These elements only exist as PAIRS. Note that when they combine to make compounds, they are no longer elements so they are no longer in pairs! On page 87 do questions 1-11 Need your books today