Unit 6 - Atomic Structure

Unit 6 Atomic Structure (Ch 7,8) A/ Introduction: The Nuclear Atom Recall from Unit 1: Thomson - discovery of e- ; existence of sub-atomic particles Rutherford: Au foil experiment - the nuclear atom (small dense +ve nucleus surrounded by e- cloud) - the solar system model (electrons orbiting nucleus) Now consider atomic structure in more detail. How exactly are the e-’s arranged about the nucleus? B/ Electromagnetic Radiation and Spectra (7.1) much of what we know comes from “spectroscopy” Spectroscopy - the study of how electromagnetic radiation interacts with matter Review: light and electromagnetic radiation Freq (?) units: sec-1 or Hz Wavelength (?) units: m, cm, nm, Å Amplitude See Fig 7.1, 7.2 Fig 7.1 - Frequency and wavelength Fig 7.2 - Amplitude ?? = c c = speed of light = 2.9979 x 108 m/sec can vary ? (and ?); observe electromag. spectrum (See Fig 7.3) C/ Quantization of Energy (7.1,2) Plank - Black Body Radiation If heat solid, emits radiation (think of coals or iron in fire) As T increases: orange --> red --> white --> blue Black body radiation Max Plank - shows that to properly explain this, we require “quantization of energy” This means that atoms absorb and emit energy in specific or quantized amounts. 2. Photoelectric Effect (Einstein) - shine light on a metal and it emits e- (see Fig 7.7) Observations: as inc. intensity, inc # e- emitted but not KEelectrons as inc. freq., inc KEelectrons but not the # e- there is a minimum or threshold freq., below which no e- are emitted To explain this, Einstein viewed the incoming light as particles (photons) or packets of energy The energy of the photon is given by Ephoton = h? Notice that this model does explain the observations. But also notice that the model suggests two revolutionary ideas: The quantization of energy (again). Light is coming in packets of energy Light must (in some cases) be viewed as a particle. ie. Wave/particle duality 3. Atomic Spectra Recall: spectroscopy Emission spectroscopy - examine the emission of radiation (light) from a molecule/atom that is “excited” See Fig B7.3, 7.8 Atoms emit/absorb radiation of certain specific energies only. Rydberg equation “?-bar” = = 1/? = RH(1/n12 - 1/n22) n1 < n2 RH = 109677 cm-1 An empirical equation based on observation only (Examples later) D/ Bohr Model (7.2) applied the idea of the quantization of energy to the nuclear model of the H atom Postulates: e- travel in circular orbits Only certain orbits allowed Atoms do not radiate/absorb energy while e- stays in an orbit. When e- moves between orbits the atom absorbs/emits a photon where Ephoton = ?E = h? Eorbits = -2.18 x 10-18 (Z2/n2) J = -2.18 x 10-18/n2 for 1H, Z = 1 (atomic number) Ephoton = ?E = [-2.18 x 10-18 (1/nf2)] - [-2.18 x 10-18 (1/ni2)] = -2.18 x 10-18 (1/nf2 - 1/ni2) = h? = hc/? = hc = (-2.18 x 10-18/hc) (1/nf2 - 1/ni2) = -109677 (1/nf2 - 1/ni2) = 109677 (1/ni2 - 1/nf2) = 109677 (1/n12 - 1/n22) n1 < n2 Aha!! The Rydberg eq. Bohr model explains line spectra and the Rydberg eq.; (the n values in the Rydberg equation are orbit levels) Examples Can continue on for transitions to n = 2 (get visible lines) i>Clicker question: Transitions to the n=1 level do NOT give emission lines in the visible region. Instead they appear in the: UV region IR region microwave region radio wave region Can continue on for transitions to n = 2 (get visible lines) Cp transitions to n=1 (ultraviolet lines) etc. See Fig 7.10 Problems with the Bohr Model: Works great for 1H but…only for 1H (and other 1e- atoms) e- do not actually travel in circular orbits 3. Imposed quantization of energy on the 1H atom but why? ` E/ Wave/Particle Duality and Uncertainty (7.3) 1. de Broglie - Wave/Particle Duality Recall: light thought of classically as a wave (refraction, diffraction can be explained using wave theory) But photoelectric effect requires viewing light as particles (photons) If light can act as particles, perhaps e- can act as waves. ??= h/mv Wave/particle duality Is light a particle or a wave? Is an electron a particle or a wave? Yes Example 7.39a But first an i>clicker question… ` i>Clicker question Cp a 95.8 kg fullback moving at 8.76 m/sec with an electron (mass = 9.109 x 10-31 kg) moving at 1/10 the speed of light (c = 3.00 x 108 m/sec). The wavelength of the fullback would be: much larger much smaller approximately the same as that of the electron. Example 7.39a 2. Heisenberg and the Uncertainty Principle ?x ?p > h/4? The more precisely we know the position, the less precisely we know the momentum (and vice versa). Example) 7.39b We have now seen three non-classical concepts: Q. of E. (E = h?) Wave/particle duality (? = h/mv) Uncertainty (?x ?p > h/4?) Why do we not observe these in the macroscopic world? - the effects are negligible in the macro. world (notice h is very small) F/ Schroedinger’s Wave Eq. (7.4) If classical mechanics doesn’t allow for these three concepts then we need a new mechanics. Schroedinger Wave eq. set up a wave eq. for a particular system (1H atom) ? = E? Solve eq. --> get ? ‘s each with an E (energy) ? is a (wave) function; ?(x,y,z) ? is also called an orbital, (not an orbit) a possible state for the e- If ? is a func., then can plug in a point (x,y,z), get the value of ? at that pt. What is the sig. of that value? None Rather ?2 a probability of finding the e- at that point ?2 is a probability map of where the e- is most likely to be We will consider later what these orbitals look like. But first… Quantum numbers each ? has three numbers associated with it # name determines allowed values n principal size n = 1,2,3… l ang. mom. shape l = n-1,n-2,…0 ml magnetic orientation ml= l,l-1,…0,-1,-2…-l Shell - orbitals with the same n value Subshell - orbitals with the same n and l value Ex) n = 1 n = 2 n = 3 i>Clicker Question: Which of the following is NOT an allowed quantum number combination? n=2; l=0; ml=0 n=3; l=1; ml=-1 n=2; l=1; ml=1 n=2; l=0; ml=-1 How many orbitals would there be in the n=3 shell? 3 b) 4 c) 5 d) 9 e) 14 What do these orbitals look like? How do they map out electron density? Recall: ?2 a prob. of finding e- at that pt. e- dot density diagram - series of “snapshots” of atom w/ e- appearing as a dot - as density of dots inc., prob. of finding e- inc. Fig 7.16A n=1, l=0 ie. a 1s orbital Notice shape: s orbitals are spherical in shape Also: Fig 7.16B ?2 vs. r plot - maps out prob. of finding e- as move out in one direction Fig 7.16C,D radial probability distribution/plot - maps out prob. that e- lies at that radius in any direction (ie. prob. that e- lies on the surface of a sphere of radius r) How big is this orbital? Recall: ?2 never reaches zero (Therefore these orbitals are “infinite” in size.) Instead use a 90% contour diagram - plot shows a contour or surface within which there is a 90% probability of finding the electron This plot also shows the shape of the orbital See Fig 7.16E i>Clicker Questions: The greatest probability of finding the electron in a 1s orbital is at the nucleus (ie. when r=0). True b) False i>Clicker Questions: As one moves out from the nucleus in one direction, the probability of finding the electron decreases. a) True b) False Can likewise consider 2s (n=2, l=0) and 3s (n=3, l=0) orbitals. Fig 7.17 - the same as 1s except for the appearance of spherical nodes Now consider when l=1 (p-orbitals) 2p orbitals Since “l” now has a different value, these orbitals must have a different shape. p orbitals have a “planar node” cutting through them Fig 7.18 Also since l=1, then ml = 1,0,-1 ie. There are three 2p orbitals, with the same shape but differing only in orientation Now consider when l=2 (d-orbitals) 3d orbitals (2 planar nodes cutting through them) Fig 7.19 i>Clicker Question: The diagram below is a ?2 vs r plot for which orbital? a) Is b) 2s c) 3s d) 2p e) 3p i>Clicker Question: The diagram below is a ?2 vs r plot for which orbital? a) 2s b) 3p c) 4p d) 3d e) 4d i>Clicker Question: The diagram below is a radial probability plot for which orbital? a) 2s b) 3p c) 4p d) 3d e) 4d G/ Electron Configurations (8.2,4) have seen various possible orbitals for the 1H atom the lowest energy orbital is the 1s (n=1, l=0, ml=0) Therefore we will place the e- in the 1s orbital But we find that there are 2 possible ways for the e- to be placed in the orbital Introduce e- “spin” (Fig 8.1) 1H atoms passing through a magnetic field are deflected Why? e- possess “spin” - Thus as a spinning charge, they produce a magnetic moment e- can spin two ways about an axis This is described by a fourth quantum number, ms For an e-, ms= +1/2 or -1/2 resulting in a magnetic moment up or down therefore we talk about “spin up” or “spin down” Now return to H atom Can write an electron configuration - the arrangement or distribution of the atom’s e- into orbitals … (overhead) Also He How do we put the second e- into the 1s orbital? Pauli exclusion Princ. - No two e- in the same atom can have the same 4 quantum numbers Therefore if the first e- is “spin up” the second must be “spin down” ie e- # 1 n= 1, l= 0, ml= 0, ms= 1/2 e- # 2 n= 1, l= 0, ml= 0, ms= -1/2 … (on overhead) Now continue to Li (Z=3) Pauli Excl. Princ. Also implies that we can have a max. of 2e- per orbital (since there are only two possible ms values.) Therefore we must place the third e- into higher E orbital What is that orbital? Aufbau Process - fill e- into orbitals in order of increasing E For 1H, orbital E depends on n only (overhead) But if >1 e-, then one e- can “shield” another from feeling the full extent of the attraction to the +ve nucleus (The e- “feels” a nuclear charge of Z*, instead of Z.) Not all e- within a shell shield equally well. Ex) 2s e- shield 2p e- better than 2p e- shield 2s e- Why? Because the 2s e- can lie very close to the nucleus cp’d to a 2p e-; 2s e- are said to “penetrate” very close to the nucleus (Recall the rad. prob. plot Fig 7.17B) What does this mean? The 2p e- feels less of a pull towards the nucleus; it is less stable; it is of higher energy Therefore get “Aufbau Filling Order” (subshells in order of increasing energy) 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f See Fig 8.7 (alternately consider this mnemonic diagram…) Now return to Li (Z=3) The third e- goes into the 2s orbital. 1s2 2s1 Continue… Be Z=4 1s2 2s2 B Z= 5 1s2 2s2 2p1 C Z=6 1s2 2s2 2p2 But before we consider the box diagram… Hunds Rule of Maximum Multiplicity When filling e- into orbitals of a sublevel, one does so placing the e- into separate orbitals, with parallel spins, as much as possible Now return to C… And continue… Also ions: anions- continue to add e- in the same manner cations - remove e- in the reverse order ex)…(on overhead) Valence e- : the outermost e- For elements in the s- or p-block, it is the outermost s and p e- ex) O has 6 valence e- Ca has 2 valence e- For elements in the d-block, it is the outermost d and s e- ex) Co has 9 valence e- Exceptions in e- configurations H/ Periodic Table - have seen e- configurations; can now return to the Periodic Table Recall: elements in the same group have similar chemical reactivity. Now we see why… Chemical reactivity is determined by the outermost (valence) electrons. Elements in the same group have the same valence e- configuration. Ex) O 1s2 2s2 2p4 S 1s2 2s2 2p6 3s2 3p4 I/ Atomic Properties (8.4) Physical and chemical properties are based on e- configuration. Therefore we can observe trends in properties on the Periodic Table Atomic Radius How big is an atom? Difficult to answer (since ?2 never reaches 0) Instead use radius of 90% contour --> atomic radius But also commonly use how close atoms get to each other ? radii = distance between atoms metallic radii (overhead) covalent radii Also ionic and van der Waals (non-bonding) radii i>Clicker Question: Which atom has the largest radius? a) H b) He How do atomic radii compare w/ each other? As go -->, radii decrease Why? As go -->, Z inc, Z* inc. Therefore e- are pulled in tighter. As go down, radii inc (n value of valence e- inc) 2. Ionization Energy energy required to remove outermost e- A(g) --> A+(g) + e- As go -->, IE inc (same reasoning as above) As go down, IE dec (e- is further away; less pull to nuc.) Can also have 2nd, 3rd IE A+(g) --> A2+(g) + e- Successive IE’s get larger. Why? Now pulling e- from a cation of increasingly larger +ve charge. But increases are not always constant Ex) Ca(g) --> Ca+(g) + e- Ca+(g) --> Ca2+(g) + e- Ca2+(g) --> Ca3+(g) + e- 1st IE < 2nd IE <<< 3rd IE Why? Ca2+ has a noble gas configuration (very stable). 3. Electron Affinity the opposite process A(g) + e- --> A-(g) - generally -ve (energy released; a favourable process) As go -->, EA inc. As go down, EA dec. However there are exceptions 4. Chemical Trends chemical properties also exhibit trends Ex) as go -->, IE inc, EA inc Therefore tendency to lose e- dec, to gain e- inc. (tendency to be oxidized dec,tendency to be reduced inc.) Demonstrations on video clip : 2 Na(s) + 2 H2O(l) --> 2 NaOH(aq) + H2(g) 2 Na(s) + Cl2(g) --> 2 NaCl(s) 2 H2(g) + O2(g) --> 2 H2O(l)