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Dalton’s Atomic Theory -each element has characteristic atoms, all of which are identical. Atoms of different elements differ in some fundamental way. Compounds are formed when atoms of different elements combine and each compound has the same relative number of atoms of each element. Reactions are reorganisations of atoms and the way they are bound together that do not alter the atoms themselves John Dalton 19th Century – atomic masses determined accurately for known elements THE NATURE OF ATOMS (p.44) -ca. 1900 J.J. Thompson created ‘cathode rays’ – a beam of negatively charged particles streaming from negative electrode. All metals seemed to contain these ‘electrons’ -proposed negative electrons were dispersed amongst cloud of postitive charge – plum pudding model of the atom: “a positive pudding with negative plums” Thompson determined charge-to-mass ratio of an electron as -1.76 x 108 C/g 1 THE NATURE OF ATOMS (p.44) -ca. 1908-1917 R.A. Millikan measured charge on a single electron by suspending charged oil droplets in an electric field -(voltage adjusted to balance droplet’s upwards electrostatic attraction with its own weight) determined electron charge = -1.6 x 10-19 C Using Thompson’s chargeto-mass ratio: ⇒ gives electron mass as: 9.1 x 10-28 g (very small!) Rutherford’s Nuclear Atom (p.47) -1906-09 Rutherford set up beam of alpha particles at thin gold foil. -Expected most particles to pass through foil with only slight deflections due to the ‘negative plums’ -Big surprise: 1 in 20,000 alpha particles were strongly deflected by > 90º or even ‘reflected’ : ⇒ atom’s positive charge was very small and concentrated nucleus 100,000 times smaller than atom! (atom is mostly empty space) 2 The Spectrum of Atomic Hydrogen (p.274) -model (theory) of structure of atom based on negative electrons orbiting a compact, positive nucleus – electrostatic attraction -spectrum of hydrogen: excited hydrogen gas becomes atomised (H-H bonds break). Energy then lost by emission of light -problem: spectrum of atomic hydrogen shows emissions at certain wavelengths only – not continuous range of wavelengths like a rainbow Wavelength, Energy and Frequency Reminder long wavelength (λ) low energy (E) low frequency (f) short wavelength (λ) high energy (E) high frequency (f) 3 4 -several series of converging lines e.g. Balmer series lies in visible region: Lines follow pattern according to: ν = R[(1/a2 ) – (1/b2 )] where a & b are whole numbers. a is characteristic of the series e.g.a = 2 for Balmer series b identifies individual line R = Rydberg constant 3.3 x 1015 Hz -photon of light emitted as atom loses energy (when electron falls from higher energy state to lower) E = hc = hν λ i.e. larger the energy drop, the higher the energy of the emitted photon energy (the shorter its wavelength) 5 The Bohr Atom (p. 274) Bohr: hydrogen’s electron only adopts certain orbits of particular energy (and distance from nucleus) – electron’s energy is quantised -energy of electron can be regarded as zero when proton and electron widely separated E = -hR/n2 n = whole number (quantisation) And E = lower (more negative) for small n, electrons more tightly held -deduced that electron’s orbit radius r = kn2 higher energy orbits ⇒ bigger radius r = 52.9 x 10-12 m (52 pm) for ground state hydrogen – the Bohr radius Quantisation – only particular discrete energy levels are allowed. Like climbing a ladder – you can’t stand between rungs! high energy Bohr calculated energy gaps between his allowed levels - they corresponded exactly to emission lines of excited atomic hydrogen low energy (Lyman series) 6 Lyman series (UV) electrons returning to ground state Balmer series (visible) electrons returning to n = 2 state Paschen series (IR) electrons returning to n = 3 state ∆E = hc/λ = hR[(1/nL2 ) – (1/nU2 )] We can now justifiably describe quantised energy levels as lower nL and upper nU ∆E = hc/λ = hR[(1/nL2 ) – (1/nU2 )] Q: What wavelength is the line emitted as an electron drops down from the third quantum level to the second (i.e. for the transition n = 3 → n = 2 ?) A: hc/λ = hR[(1/nL2 ) – (1/nU2 )] 656 nm rearrange: c/λ = R[(1/nL2 ) – (1/nU2 )] = R[(1/22 ) – (1/32 )] X UV Vis IR Rad = 0.139R 400 800 nm rearrange: λ = c/0.139R = 3 x 108/(0.139 x 3.29 x 1015) = 6.56 x 10-7 m (656 nm - in the visible range of electromagnetic spectrum ) Bohr’s model only works for one-electron atoms: H, He+, Li2+ 7 Q5: Using the Rydberg equation: ∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )] for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h = 6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023). a) Calculate the photon energy and wavelength corresponding to i) the three lowest energy lines of the Lyman series and ii) the convergence limit of the Lyman series. Lowest energy line: nL = 1, nU = 2 so [(1/nL2 ) – (1/nU2 )] = 0.75 ∆E = 1.636 x 10-18 J λ = 121.6 nm second lowest energy line: nL = 1, nU = 3 so [(1/nL2 ) – (1/nU2 )] = 0.8889 ∆E = 1.939 x 10-18 J λ = 102.6 nm third lowest energy line: nL = 1, nU = 4 so [(1/nL2 ) – (1/nU2 )] = 0.9375 ∆E = 2.045 x 10-18 J λ = 97.3 nm the convergence limit: nL = 1, nU = ∞ so [(1/nL2 ) – (1/nU2 )] = 1.0000 ∆E = 2.181 x 10-18 J λ = 91.2 nm Q5: Using the Rydberg equation: ∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )] for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h = 6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023). b) sketch the appearance of the Lyman series of lines in relation to the visible region of the electromagnetic spectrum UV UV 100 200 visible 400 IR 800 8 Q5: Using the Rydberg equation: ∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )] for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h = 6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023). c) By calculating the highest energy line of the Balmer series, determine if the two series of lines overlap Highest energy Balmer line: nL = 2, nU = ∞ so [(1/nL2 ) – (1/nU2 )] = 0.25 ∆E = 5.453 x 10-19 J λ = 364.7 nm (no overlap with Lyman series) Q5: Using the Rydberg equation: ∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )] for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h = 6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023). d) By considering the convergence limit of the Lyman series, calculate the ionisation energy (in kJ/mol) of the hydrogen atom. Why is the energy required to remove an electron from He much greater (2370 kJ/mol)? the Lyman convergence limit: ∆E = 2.181 x 10-18 J/photon ∆E = 2.181 x 10-18 x 6.022 x 1023 = 1.31 x 106 J/mol = 1310 kJ/mol 1st I.E. of He is higher due to greater nuclear charge and smaller atomic radius 9 pra ctise Q 7.30 s ,7 & 7 .32 .34 Particles as Waves: Orbitals (p.279) -1920s ‘particle-wave duality’ theory ⇒ waves behave as particles and particles behave as waves e.g. light behaves as photons (packets of energy) beams of electrons can be diffracted through crystals (wave behaviour) -electrons as waves inside atoms? -Heisenberg Uncertainty Principle (p.283) – the position and speed of a moving particle cannot be simultaneously defined Bohr model (based on classical physics) therefore not valid – cannot describe precise orbits ⇒ quantum mechanics needed 10 -describe electron in quantum mechanics ⇒ can talk only about the probability of finding the particle in a particular place (p. 283) -Schrodinger Equation (1926, p. 285) solutions are wavefunctions (ψ) called atomic orbitals – these are the quantised energy levels of a one-electron atom. ψ in ground state (n = 1) wavefunction describing the single electron of H -orbitals are region where there’s a high chance of finding the electron. The probability of finding it at any given point is given by ψ2 The Hydrogen 1s orbital (p. 288) - cross-section of ψ2 (probability distribution) 3-D plot shows highest electron density close to nucleus: -small chance of finding electron far away (0 chance at infinity) to draw ‘shape’ of orbital we must draw contour e.g. can draw region within which there is 90% chance of finding electron – region is spherical so hydrogen’s occupied orbital is spherical: 11 orbitals are defined by three quantum numbers: n (shell number) l (determines orbital shape) ml (determines orbital orientation) Naming atomic orbitals (p.279) - there are n2 orbitals per shell i.e. one orbital for n = 1 four orbitals for n = 2 nine orbitals for n = 3 n l ml n l ml 12 2s orbital – higher in energy than 1s – positive and negative regions of wavefunction 1s 2s For 2s orbital there is a nodal surface where ψ = 0 and changes from +ve to -ve. n.b. probability (ψ2) always positive. 2p orbitals (p. 289) for the other three orbitals in n = 2 shell l = 1. these feature a nodal plane (ψ changes sign at the nucleus) – there are two lobes - these are called p orbitals 13 p orbitals are drawn like: or -the three p orbitals within shells n = 2, n = 3, n = 4 etc… are at right angles to each other e.g. 2px, 2py, 2pz but within each shell all are same energy (degenerate) in one electron atoms. name indicates the shell, shape of orbital and the orientation of its lobes…. e.g. 2py indicates radius of orbital. Determined by principal quantum number n indicates shape (symmetry) of orbital. Determined by quantum number l indicates orientation of orbital. Determined by quantum number ml - other orbitals occupied by electron in excited states 14 3d orbitals (p. 289) in n = 3 shell there are nine orbitals): one s, three p, and five d orbitals in this and higher energy shells l can equal 2 giving rise to d orbitals which have four lobes: Occupancy of d orbitals is important in explaining the colourful compounds and complexes formed by transition metals e.g. Fe and Cu Electron Spin and ms (p.292) Each orbital can hold two electrons. These must be of opposite spin (Pauli Exclusion Principle). Fourth quantum number – electron spin ms = +½ or -½ Pauli Exclusion Principle – each electron in any atom has a unique set of four quantum numbers -an electron’s set of quantum numbers is like a unique address within the atom - tells us shell no., type of orbital, orientation of orbital, electron spin 15 Shielding (p.295) Schrödinger equation very powerful but can’t be solved for manyelectron atoms due to repulsions between electrons -In many-electron atoms, each electron doesn’t feel full nuclear charge – they are shielded by other electrons -e.g. in He, Z = +2, but electrons repel each other so: ‘effective nuclear charge’ Zeff = +1.7 (hard to measure) -electrons in same orbital exert a small shielding effect on each other Shielding (p.295) radial distribution function = 4πr2ψ2 -electrons in inner orbitals shield outer electrons very effectively - (largely cancelling out nuclear charge) e.g. in carbon (2p2), the 2px electron is shielded slightly by 2py electron but much more by 1s electrons which lie closer to nucleus (see right) Zeff = +1.3 for Li 2s electron despite nuclear charge of +3 -We can still apply Schrödinger equation if we consider each electron in turn moving in a field resulting from a nuclear charge of +Zeff rather than +Z ⇒ Can then describe many-electron atoms using hydrogen-like orbitals 16 Building Up Principle (p.298) -to explain electron arrangement in all elements: -add successive electrons to hydrogen to form heavier elements -orbitals fill from lowest energies to highest – the Aufbau principle -in many-electron atoms, orbitals within each shell no longer degenerate e.g. 2p orbitals higher in energy than 2s orbitals - for element of atomic no. Z, we add Z electrons to fill its orbitals - Hydrogen has a half-full 1s orbital - we write its electron configuration as 1s1 He has a full 1s orbital – 1s2 Li is 1s2 2s1 or because of its He core we can write [He] 2s1 Similarly, Be is [He] 2s2 17 -Only outermost shell contains valence electrons – those available to take part in reactions -At boron (Z = 5) one 2p orbital is half occupied -For C there is a choice – sixth electron can either pair up with fifth or else go into another p orbital Hund’s Rule – the lowest energy configuration for an atom is to have the maximum no. of unpaired electrons all adopting parallel spin reason for maximising electron spin: mutual repulsion of electrons is minimised -Only outermost shell contains valence electrons – those available to take part in reactions -At boron (Z = 5) one 2p orbital is half occupied -For C there is a choice – sixth electron can either pair up with fifth or else go into another p orbital Hund’s Rule – the lowest energy configuration for an atom is to have the maximum no. of unpaired electrons all adopting parallel spin ⇒ carbon’s two 2p electrons have parallel spins. 18 -similarly N has three unpaired electrons at oxygen electrons begin to pair up so O has only two unpaired electrons: At Ne, n = 2 shell is full - atom has eight valence electrons - a stable octet ⇒ Ne forms no compounds After Ne third period elements (n = 3) starts to fill same as 2nd period - 3s orbital starts to fill first at sodium (Na): 19 The Building-up Principle has a surprise for us after Argon (Z = 40) 4s orbital is lower in energy than 3d so fills first. Therefore K has config [Ar] 4s1 rather than [Ar] 3d1. The Building-up Principle has a surprise for us after Argon (Z = 40) 4s orbital is lower in energy than 3d so fills first. Therefore K has config [Ar] 4s1 rather than [Ar] 3d1. Radial distribution function of 4s and 3d orbitals explains why: most probable distance lies further from nucleus for 4s BUT 4s electron density is greater than 3d close to nucleus (4s ‘penetrates’ nucleus more than 3d) penetration by 4s lowers its energy 20 filling of d orbitals (10 e) corresponds to transition metals (d-block): (p.285) Periodicity (p.50) Groups (vertical columns) and periods (horizontal rows) of periodic table are named after orbitals that are partially filled: e.g. S is in group VI (six valence electrons) [Ne] 3s2 3p4 and period 3 (valence electrons are in n = 3 shell) e.g. Ca is in group II (two valence electrons) [Ar] 4s2 and period 4 (valence electrons are in n = 4 shell) repetition of orbital filling s > p > d > s > p > d leads to periodicity (repeating chemical properties) of the elements e.g. halogens all have s2p5 configuration ⇒ similar chemistry ⇒ all form HX gases, react vigorously with metals to form MX salts - Mendeleev in Siberia ca. 1870 ⇒ periodic table 21 Atomic Number (Z) (p.49) Moseley (1913) fired electron beam (cathode ray) at elements. Emitted Xray wavelength depended on atomic number. Since no. of protons in nucleus of atom = no. of electrons in neutral atom He found way to count electrons in an atom ☺ Neutrons and Isotopes (p.49) Dalton’s original assumption that all atoms of the same element have same mass – WRONG! E.g. Sn atoms can differ in mass by over 10%! Some H atoms weighed twice as much as others - Why? Today we ‘weigh’ atoms and molecules directly using mass spectrometer Mass Spectrometry (p.86) -sample is vapourised and ionised (e.g. bombarded by electrons to give negative charge) -then accelerated towards electromagnet that deflects ions amount of deflection gives mass-to-charge (m/z) ratio e.g. most Ne atoms are about 20 times the mass of hydrogen atom but a few are 21 or 22 times as heavy – why? 22 - three different masses of Ne seen…...we say there are three isotopes of neon isotopes are atoms of the same element with differing mass isotopes contain differing numbers of neutrons – neutral particles found in all atomic nuclei except hydrogen-1 (1H) A ZX 20 10 Ne 21 10 Ne 22 10 Ne atomic mass unit (a.m.u. or Dalton) – defined as one-twelfth the mass of carbon-12 atom or 1.66 x 10-24 g -the proton and neutron both ‘weigh’ about 1 a.m.u. -differing numbers of neutrons change mass of atom but not the number electrons required to stay neutral -atoms of different isotopes have the identical chemistry – mass difference affects physical properties only e.g. 2H2O ice (solid D2O, density = 1.1 g/cm3) cubes sink in H2O water (density = 1.0 g/cm3) 23 Most elements are occur as a mixture of isotopes (fluorine is an exception): mass number (A)– the total number of nucleons (protons + neutrons) in nucleus of the atom of an isotope e.g. an isotope of nitrogen (Z = 7) with eight neutrons has a mass number of 15 an isotope of phosphorus (Z = 15) with 16 neutrons has a mass number of 31 A - mass number written above element symbol, atomic number below: Z e.g. carbon-12 (six protons, six neutrons) ⇒ 126C ….natural isotopes of Ne (Z = 10) ⇒ 2210Ne, 2110Ne, 2010Ne X Q: The most abundant isotope of tin has a mass number of 120? How many neutrons are in 120Sn nuclei? A: First find atomic number of Sn from periodic table. No. of protons = 50. 120 – 50 = 70 neutrons There are more neutrons than protons – typical of heavy atoms. 24 relative atomic mass (atomic weight) of an element is a weighted average of isotope masses of that element e.g. chlorine consists of 75% chlorine-35 and 25% chlorine-37. Its R.A.M. is 35.5 e.g. the Mr of hydrogen is 1.008 (more than 1.000 because of traces of the heavier isotopes: deuterium 2H and tritium 3H) Q: Fill in every blank box regarding the atoms or ions of the following isotopes: Species 239 No. of neutrons 4+ 1 Np [Rn]5f 3 1 mercury-201 53 2+ Cr [He] 1 2 1s 201 24 43 bismuth-211 6 34 [Kr] 83 answers at the end of this slide sequence - try without looking first!! 25 Ions (p.53) -Atoms can donate or accept electrons to form atomic ions (ionic bonding)….. -or share electrons with other atoms (covalent bonding) similarly, a molecule can also become a charged molecular ion if it gains or loses electrons e.g. NH4+, NO2-, Atoms become ions when they accept or donate electron(s) Al Na F O + + Al3+ Na+ e2e- + + 3eeFO2- Positive ions are called cations Negative ions are called anions Cation and Anion formation Postive ions are called cations Negative ions are called anions 26 Molecules (p.53) -Atoms form bonds in which electrons are shared between atoms > bonded groups form molecules e.g. H2O, PCl3, N2O the molecular formula of a substance tells us the exact number of atoms in the smallest unit of a substance e.g. H2O is the molecular formula of water a triatomic molecule with two covalent O-H bonds. -N2 is a diatomic molecule with a triple covalent bond between the two atoms N≡N 27 -most molecules are polyatomic e.g. P4 estradiol -molecules can become ions if a covalent bond is broken unevenly leaving positive and negative fragments: -e.g. CH3COOH H2SO4 SO42- CH3COO- + H+ + 2H+ Structural Formulae (p.55) the structural formula tells us which atoms are bonded to which in a molecule e.g. the structural formula of i) water is H–O–H ii) hydrogen is H–H iii) ammonia is H–N–H H Empirical Formulae (p. 55) the empirical formula tells us which atoms are present in the substance in their simplest whole-number ratios e.g. hydrogen peroxide is H–O–O–H ⇒ empirical formula is HO the empirical formula of i) ethanol (C2H5OH) is C2H6O ii) glucose (C6H12O6) is CH2O 28 The Mole (p.77) -Atoms too small to count -Atoms weigh different amounts so need way to count atoms/molecules involved in reactions -Chemists invented the mole – no. of particles in a sample that translates its atomic/molecular weight into grams 1 mole = 6.023 x 1023 particles -6.02 x 1023 = Avogadro’s number (NA) -defined as the number of atoms in 12.00 g of 12C. -mass of a mole of an element (g) = R.A.M. of element (Daltons) -1 mole Ne = 20.18 g -1 mole Cu = 63.6 g -1 mole H2 = 2.016g -1 mole O3 = 48.0g -n.b. 1 mole carbon = 12.01 g not 12.00g due to 13C and 14C -therefore 6.02 x 1023 a.m.u. = 1g -Q: A silicon chip weighs 5.68 mg. How many atoms does it contain? -A: convert to moles: 5.68 mg = 5.68 x 10-3 g R.A.M. Si = 28 g/mol no. moles = 5.68 x 10-3/28 = 2.02 x 10-4 mol no. atoms = 2.02 x 10-4 mol x 6.02 x 1023 = 1.22 x 1020 atoms 29 A: Species 239 Np4+ 6 + Li 3 He+ mercury-201 53 Cr2+ 77 Se2bismuth-211 No. of neutrons 146 3 1 121 29 No. of protons 93 3 2 80 24 No. of nucleons 239 6 3 201 53 Electron configuration [Rn]5f1 [He] 1s1 [Xe]6s25d10 [Ar]3d4 43 128 34 83 77 211 [Kr] [Xe]6s25d106p3 30