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Atomic Structure and Periodicity A. Atomic structure and periodicity 1 m=109 nm or 10-9 m 1 nm 1 J = 1 kg x m2/s2 1. Electromagnetic radiation a. a form of radiant energy b. three properties of waves - wavelength ( - Greek lambda) (shorter wavelength = higher energy) - frequency (v - Greek nu) (Hz - 1/s or s-1) - speed (c)- equal for all kinds of e-m radiation ( = 2.9979 x 108 m/s) c. v = c (wavelength and frequency are inversely related) Sample problem : Calculate the wavelength of a 101.5 mHz frequency given off by a radio station. Solution : = c/ v (101.5 mHz = 101.5 x 106 Hz or 101.5 x 106 s-1). = 2.9979 x 108 m/s / 101.5 x 106 s-1 = 2.954 m 2. Nature of matter a. Planck - energy is quantized E = nhv (n = integer, h = Planck's constant 6.626 x 10-34 Js) b. quantum - a small "packet" of energy - a discrete, yet varying amount (not all quanta are the same) c. Einstein - photons (packets of light energy) are quantized (Ephoton = hv = hc/) v = c/ d. Einstein : E = mc2 e. dual nature of light/matter - diffraction (scattering of light from a regular array of lines or points)- property of waves ; photoelectric effect- evidence of particle nature f. de Broglie's equation : = h/mv (where m = mass (in kg) and v = velocity in m/s) Sample problems : 1. Calculate the energy of a single photon and a mole of photons emitted from a vapor lamp giving off light with a wavelength of 412.0 nm. Solution : Ephoton = hv = hc/ = (6.626 x 10-34 J·s)(2.9979 x 108 m/s) / (412.0 x 10-9 m) = 4.821 x 10-19 J E mol = (4.821 x 10-19 J/e-) (6.022 x 1023 e-/mol) = 290.3 kJ 2. The first ionization energy of aluminum is 577 kJ/mol. Is light with a wavelength of 215 nm capable of ionizing an atom of aluminum in the gas phase? Solution : kJ/mol are given, first calculate J/atom : 577 kJ/ (6.022 x 1023) = 9.58 x 10-22 kJ/atom = 9.58 x 10-19 J/atom Second - calculate wavelength from energy needed and compare to given wavelength : = hc / E = (6.626 x 10-34 J·s)(2.9979 x 108 m/s) / 9.58 x 10-19 J/atom = 2.07 x 10-7 m = 207 nm Answer : No. The 207 nm wavelength is a higher energy wavelength than the 215 nm wavelength. 3. The atomic spectrum of hydrogen a. Contrast to a continuous spectrum - all wavelengths b. emission spectrum of hydrogen -pattern of light emitted from an excited hydrogen atom) - a line spectrum- indicates quantized nature of hydrogen (only certain wavelengths absorbed or emitted) 4. The Bohr model - quantum model a. energy levels available : E = -2.178 x 10-18 J(Z2/n2) - Z = nuclear charge (number of protons)(=1 for hydrogen atom) - n = integer- corresponds to energy levels - neg. sign indicates negative energy (e- is bound to nucleus therefore has lower energy than free e- where n = ) - J is joules b. ground state lowest possible energy state c. calculation of energy of quantization in hydrogen for change in energy : E = Ef - Ei (e.g. change in E when electron moves from n = 6 to n = 5) - positive sign indicates E (light) absorbed - negative sign indicates E (light) emitted d. calculation of change in electron energies E= -2.178 x 10-18 J(1/nfinal2 - 1/ninitial2) e. spectral transition diagram : ("spectral" because light is usually absorbed or emitted) f. failure of Bohr model in other elements -the Bohr model did not hold true for polyelectronic atoms 5. Quantum Mechanical Model of atom a. Heisenberg, Schrodinger, de Broglie - developed wave , or quantum mechanics - bound electron is a standing wave - only certain circumferences will allow whole numbers of half waves - Schrodinger's equation HΨ = EΨ - Ψ- wave function, or "orbital" (function of coordinates x, y and z)( does not indicate pathway of electron) - H - operator - contains mathematical terms which produce total energy of the atom - E - total energy of atom (kinetic and potential) - a specific wave function is an orbital b. Heisenberg uncertainty principle-cannot know both momentum and position of electron at a given time c. Physical meaning of wave function in light of uncertainty principle : probability of elocation - probability distribution- represents square of wave function - produces an e- density map - radial probability distribution - total probability within spherical shells - orbital - no definite size-defined as area in which 90% of total eprobability 6. Quantum numbers-describe orbitals a. Principal quantum number (n = 1, 2, 3, ...) - indicates size and energy of orbital (higher n, higher energy-less bound, less neg. energy) b. Angular momentum number (azimuthal quantum number) (l) - integral values from 0 to n-1 - indicates shape of orbital (l = 0 = s; l = 1 = p; l = 2 = d; l = 3 = f ) c. magnetic quantum number (ml) - integral values from l to - l including 0 - indicates position of orbital in space relative to other orbitals d. Spin quantum number (ms) - indicates rotation of electron - -½ or -½ Table 1 : Quantum numbers for first four energy levels n ml Orbital l (n = 1, (ml = -l …0…+l) designati (l = 0 to n2, 3…) on 1) No. of orbitals (= n2) Total number of electrons (= 2n2) (2) 1s 1 0 0 1 2 2s 2 0 0 1 2 (8) 2p 1 1, 0, -1 3* 6 3s 3 0 0 1 2 (18) 3p 1 1, 0, -1 3 6 3d 2 2, 1, 0, -1, -2 5** 10 4s 4 0 0 1 2 4p 1 1, 0, -1 3 6 (32) 4d 2 2, 1, 0, -1, -2 5 10 4f 3 3, 2, 1, 0, -1, -2, -3 7 14 * The 3 p orbitals are x, y, and z (e.g. 2px, 2py, 2pz) ** The 5 d orbitals are xz, yz, xy, x2-y2, and the z2 (e.g. 3dxz, 3dyz, 3dxy, 3dx2-y2, and the 3dz2) Sample problems : 1. Which of the following sets are incorrect? a. n = 2, l = 2, ml = -1 b. n = 3, l = 2, ml = -3 c. n = 4, l = 2, ml = -1 Answer : sets a (l cannot exceed n-1)and b (ml must be between + l and - l) 2. How many electrons can have the quantum numbers n = 5, ml = +1? Answer : 8 In n = 5, l will equal 0, 1, 2, 3, 4. For l = 1, 2, 3, 4 there will be one orbital in which ml = +1, and each orbital will contain two electrons. 7. Orbital shapes and energies a. nodal surface (nodes)- regions of zero probability of finding an eb. hydrogen atom - orbitals in energy levels (same n value) are degenerate - have the same energy 8. Electron Spin and the Pauli Principle a. Pauli exclusion principle - in a given atom, no two electrons can have the same set of four quantum numbers (i.e. -only two electrons per orbital) 9. Polyelectronic atoms a. three forces affecting distribution of electrons - kinetic energy of electrons (outward) - potential energy of attraction of nucleus (inward) - potential energy of repulsion of electrons b. electron correlation problem-since paths of electrons cannot be known exactly, repulsions cannot be calculated exactly- must make approximations c. shielding - inner electrons shield outer electrons from nuclear charge (held less tightly) d. orbitals in polyelectronic atoms are not degenerate (s < p < d < f) e. penetration effect -ex. In general, 2p electrons are closer to the nucleus and therefore have a lower energy than 2s, however, 2s electrons spend a small, but significant amount of time closer to the nucleus and are said to "penetrate" and thus are considered to have a lower energy state and are filled first d. Electron configuration problems Sample problems : 1. What is the electron configuration of neon? Answer : Neon has ten electrons. In the ground state these will be in the lowest energy levels possible giving the following : (1s22s22p6) 2. What is the electron configuration of chromium ? Answer : [Ar] 4s13d5 (note this is an exception to the normal rules) 3. What is the identity of the element with the electron configuration 1s22s22p63s13p1 in an excited state ? Answer : Magnesium - 12 e-, with one 3s1 e- moved up into the 3p1 orbital. 4. Write the electron configuration for the following orbital notation for an atom with 8 protons, identify the species (atom, ion etc) and write its formula . Answer : 1s22s22p6, oxide ion , O2- 5. How many unpaired electrons are found in nitrogen in its ground state? Answer : Three. 6. Give a set of values for the four quantum numbers of the valence electrons of phosphorus in its ground state. Answer : The valence electrons of phosphorus are the 3s23p3 electrons : Electrons n ms l ml 3s 3 0 0 +½ 3s 3 0 0 -½ 3p 3 1 -1 3p 3 1 0 3p 3 1 +1 +½ (or -½) +½ (or -½) +½ (or -½) 10. History of periodic table a. Mendeleev given most of the credit because of his emphasis on using it as a tool to predict yet unknown elements. 11. Aufbau ("building up") principle and the periodic table a. Aufbau principle - as protons are added to successive atoms of elements, so are electrons b. Hund's rule - the lowest energy configuration for an atom is the one having the most unpaired electrons allowed by the Pauli principle in degenerate orbitals c. Valence electrons - electrons in the outermost principal quantum energy level of an atom d. core electrons - "inner" electrons e. configurations of representative elements vs. transition metals and inner transition metals (lanthanides and actinides) f. exceptions to expected configurations - chromium [Ar] 4s13d5 and copper [Ar] 4s13d10 g. arrangement of periodic table- representative elements, transition metals, lanthanides, actinides, s block, p block, d block and f block h. IUPAC periodic table- numbers Groups 1-18 instead of 1A-8A and "B" designation for transition metals i. determination of electron configurations 12. Periodic trends in atomic properties a. major factors affecting trends - nuclear charge - increases across a period causing a greater attraction for electrons - shielding - additional energy levels down a group causes a decrease in the attraction of the nucleus for the outermost electrons - electron configuration - most important - groups have similar electron configurations giving them similar chemical and physical properties b. ionization energy- the energy needed to remove an electron from a gaseous atom (X(g) -->X+(g) + e-) in its ground state - first ionization energy (I1)- energy needed to remove the first electron - second ionization energy (I2) - energy needed to remove second electron - note pattern in aluminum [Ne]3s23p1 I1- removes p1electron = 580 kJ/mol I2 removes one 3s electron = 1815 kJ/mol - higher due to positive charge on + Al ion I3 removes remaining 3s electron = 2740 kJ/mol (Al2+ ) I4 removes core 2p electron = 11,600kJ/mol - core electrons come from stable noble gas configuration - periodic trend is for ionization energy to increase across a period - increased nuclear charge with no increase in shielding -exceptions-due to electron repulsion i.e. nitrogen and oxygen N: O: - easier to remove outer 2p electron from oxygen than nitrogen because of doubly occupied 2porbital -group trend is to decrease down a group - increased shielding, outer electrons are farther from nucleus c. electron affinity - the energy change associated with the addition of an electron to a gaseous atom : X(g) + e- ---> X (g) - sign of energy change- negative if exothermic, positive if endothermic - periodic trend - negative electron affinities increase across a period (more energy released) - exceptions : carbon has a negative electron affinity (becomes more stable) and nitrogen has a positive electron affinity (becomes unstable- higher energy) C: N: - an electron added to carbon goes into an empty orbital whereas an electron added to nitrogen goes into an occupied orbital resulting in electron repulsion - Group trend - electron affinities become more positive (less negative) because less energy is released due to shielding and increased distance from nucleus (less attraction for e-) Note relationship between ionization energy ( energy change for X(g) -->X+(g) + e-) and electron affinity ( energy change for X(g) + e- ---> X (g)). Since the changes are opposite their energies are opposite. Sample problem : What is the electron affinity for the ionization of the Al2+ ion (The ionization energy for Al+(g) + e- Al2+(g) = 1850 kJ/mol)? Answer : -1850 kJ/mol : Electron affinity is the energy change associated with the addition of an electron to a gaseous atom making the reaction : Al2+(g) + e- Al+(g). The electron affinity is therefore opposite (EA = -IE) and equals -1850 kJ/mol. c. Atomic radii - defined as half the distance between the nuclei of diatomic molecules of the same element (e.g. O2) - periodic trend - atomic radii decrease across a period - due to increased nuclear charge - Group trend - atomic radii increase down a group - large n value, increased shielding d. ion size - metallic ions lose electrons and become smaller due to loss of volume occupied by lost electrons and by the increased ratio of positive to negative charge - nonmetallic ions gain electrons and increase in size for reasons opposite of metallic ions - trend for metallic ions is to decrease in size (consider 3rd period metallic ions Na+, Mg2+, Al3+) - trend for nonmetallic ions is to also decrease in size (consider the 3rd period nonmetallic ions P3-, S2-, Cl-) 13. Information contained in the Periodic Table a. number of valence electrons - main determiner of chemical properties - know well b. electron configuration - know representative elements and Cr and Cu c. know : metals, nonmetals, alkali metals, alkaline earth elements, halogens, noble gases, transition metals, inner transition metals ( lanthanides and actinides) and metalloids (semimetals) d. characteristics of metals and nonmetals - metals - low ionization energies, low (less negative) electron affinities, tend to form cations - nonmetals - high ionization energies, higher (more negative) electron affinities, tend to form anions 14. The Alkali metals - Group 1A H, Li, Na, K, Rb, Cs, and Fr a. hydrogen is a nonmetal mainly because of its small size - holds electron tightly b. density increases down group - typical of all groups - mass increases at a faster rate than size c. melting points and boiling points generally decrease d. tend to lose outer e- making them reducing agents - lower ionization energy makes them stronger reducing agents thus order of reducing strength is Cs > Rb > K > Na > Li for solid metals only - in aqueous solution the reducing strength is Li > K > Na Why change in position for lithium in aqueous solution?- Lithium's small size gives it a high charge density which makes it more able to bond to water (hydration energy - energy released when a substance bonds to water in solution) and form the Li+ ion (and thus lose its valence electron) more readily than K or Na. -If lithium is a stronger reducing agent and bonds to water more strongly than K or Na, then why do experiments show that K and Na react more violently with water? - Na and K have lower melting points causing them to react with water in the liquid state giving more surface area resulting in a faster reaction. Main point of this - thermodynamics (energy change) tells us if a reaction will be spontaneous or not, kinetics (study of reaction mechanisms/rates) tells us how fast a reaction will occur.