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Chemistry 1A: Chapter 8 Page |1 Chapter 8: Periodic Properties of the Elements Homework: Read Chapter 8. Work out sample/practice exercises Suggested Chapter 8 Problems: 43, 45, 51, 55, 59, 63, 67, 71, 75, 79, 83, 99 Check for the MasteringChemistry.com assignment and complete before due date The Periodic Table: 1869 Dmitri Mendeleev (Russia) and Lothar Meyer (Germany) classified known elements by organizing similar physical and chemical properties. Mendeleev’s periodic table was an attempt to organize the known data at the time in a way that made sense. Elements were arranged by increasing atomic mass and grouped together by chemical reactivity. Where problems with mass positions occurred ( Te and I), he re-ordered by other properties. Several holes led to predictions of elements and their properties that were not yet discovered “eka-aluminum” (Ga) and “eka-silicon”(Ge). 1913 *Henry Moseley improved the periodic table by ordering the elements by increasing atomic number. More “holes” were found, which led to the discovery of more elements and the family of noble gases. The periodic table gives a great amount of information in an organized manner. Vertical columns are called groups or families. If you are aware of the properties of a couple elements in a group, you can make a good guess at the properties of the other elements in the same group. Periods are the horizontal rows in the periodic table. Many patterns can be seen or predicted following periods and groups. Electron Configurations and Orbital Diagrams: Chemistry 1A: Chapter 8 Page |2 A. Energy Levels: i. In the Bohr Atom (one electron systems) there is a one-to-one correspondence between an orbit and its energy level (En = -B/n2). Example: the 3s, 3p, 3d orbitals in Hydrogen all have the same energy. ii. In the Quantum Mechanical version of the atom, the energy level of multielectron atoms depend on both the size (1, 2, 3, 4…) and shape (s, p, d, f). B. Ground State: Filling orbitals of an atom where electrons go into the lowest energy orbitals first. i. Pauli Exclusion Principle, no two electrons in an atom may have all four quantum numbers alike. ii. Hund's Rule, when filling degenerate energy levels, each orbital fills one electron, spin unpaired, before any orbital fills with two electrons. C. Aufbau Principle: Electronic Configuration by energy i. 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s2 5f146d107p6 ii. Isoelectronic Same Electronic Configuration Chemistry 1A: Chapter 8 Page |3 Watch for exceptions: Electron Configurations (complete/short) Core electrons Valence electrons Pseudocore electrons Ground state Excited state By size By energy Pauli’s rule Hund’s Rule Paramagnetic Diamagnetic Quantum numbers (m, l ml, ms) Chemistry 1A: Chapter 8 Page |4 Quantum Numbers: A. Each electron in an atom is defined by four Quantum Numbers. The principle quantum number, n, is related to the size (90% probability that the electron is within a given radius) Name Symbol Values Significance principle n 1, 2, 3, ... size azimuthal 0, 1, ...(n-1) shape magnetic ml 0, 1, 2,.... orientation spin ms ½ electron spin B. The values of the azimuthal quantum number, , determines the shape of the orbital. In the designation of an orbital, this quantum number is represented by a letter. Value Letter Shape 0 s spherical 1 p dumbbell 2 d four-lobed 3 f eight-lobed C. The magnetic quantum number, ml, identifies the three dimensional orientation in space. For an s orbital it is 0, for p it can be -1, 0, +1, d has 5 orientations, f has 7 Three quantum numbers are required to specify an orbital: principle, azimuthal and magnetic. An orbital is a place in an atom to hold electrons. An orbital may contain a maximum of two electrons D. The fourth quantum number is the electron spin, ms. Values are ½. Stern-Gerlach experiment split a beam of silver atoms (47 electrons-odd number-paramagnetic) in two by a magnetic field. As electrons spin they generate a magnetic field. About have half the electrons point “north” (spin up) and others point “south” (spin down) Chemistry 1A: Chapter 8 Page |5 Example 1: Predict the ground state short electron configuration for each. Identify electrons core, [valence], pseudocore. Draw the orbital diagram for the outershell electrons, identify if species is diamagnetic or paramagnetic, write out the quantum numbers for each of the outershell electrons. Ag Ag+1 Ga Ga+3 Fe Fe+2 Fe+3 Sn Sn+2 Sn+4 P P-3 P+3 P+5 S S-2 Ce Ce +2 Ce+3 Ce+4 Chemistry 1A: Chapter 8 Page |6 Coulombs Law: F = kQ1Q2/d2 Since Energy is force times distance, E = kQ1Q2/d Coulombs Law describes attractions and repulsions between charged particles. Attraction is stronger as atomic sizes decrease and charge differences increase. Effective Nuclear Charge: Negatively charged electrons are attracted to the positively charged nucleus and repelled by other electrons in the atom. The force of attraction depends on the magnitude of the net nuclear charge acting on an electron and the average distance between the nucleus and the electron. We can estimate the net attraction and average environment of a single electron through Zeff (effective nuclear charge), which is always smaller than the total charge of the nucleus. Zeff = Ztotal - Sscreening constant S is generally close to the # of core electrons, (i.e. for Na, 10 core electrons, 1 valence electron. The Zeff = 11-10 = +1 for the valence electron 3s1) Chemistry 1A: Chapter 8 Page |7 Penetration: • The closer an electron is to the nucleus, the more attractions it experiences. • The degree of penetration is related to the orbitals radial distribution function • The radial distribution function shows that the 2s orbital penetrates more than the 2p • The weaker penetration of the 2p sublevel means that electrons in the 2p sublevel are more shielded from the attractive force of the nucleus • The deeper penetration of the 2s electrons means electrons in the 2s sublevel experience a greater attractive force to the nucleus and are not shielded as effectively • Penetration causes the energies of sublevels in the same principal level to not be degenerate (2s and 2p are different energies) • In the 4th and 5th principle levels, the effects of penetration cause the s orbital to be lower in energy than d orbitals of the previous principal level (4s is lower than 3d) • The energy separations between one set of orbitals and the next become smaller beyond the 4s so the ordering can vary among elements causing variations (exceptions) in the electron configurations of the transition metals and their ions Periodic Trends: Properties of the elements follow a periodic pattern: Same column have similar properties and in a period a pattern repeats. This is explained with the quantum-mechanical model because the number of valence electrons and types of orbitals they occupy are periodic. Chemistry 1A: Chapter 8 Page |8 Size (atomic and ionic radii): Atomic radii increase from right to left; top to bottom of periodic table. Zeff is the effective nuclear charge. Zeff = Zactual – electron shielding This is the charge felt by the outer electrons that are shielded from the full power of the positive nuclear charge Along a period the Zeff increases left to right pulling in electrons closer to the nucleus and causing the atoms to decrease in size. Transition metals in the same d block are roughly the same size Vertically, the size of the orbitals (quantum number n) increases from top to bottom Chemistry 1A: Chapter 8 Page |9 Ionic Radii: Cations lose electrons and are therefore smaller than the original atom Anions gain electrons and are larger than the original atom Isoelectronic series (all have the same number of electrons, same electron configuration) the size increases as the charge of the nuclei decreases. (smallest Sr+2, Rb+1, Kr, Br-1, Se-2 largest) Chemistry 1A: Chapter 8 P a g e | 10 First Ionization Energy, Ei: Energy required to remove the outermost ground state electron, endothermic Ionization Energy decreases from right to left; top to bottom of periodic table. The small nonmetals require the highest ionization energy, they do not want to lose electrons Large metals have lowest ionization energy, they want to lose electrons and become positively charged cations. Minor irregularities occur a. Ei of Be is larger than B and that of Mg is larger than Al. An explanation is that Be and Mg lose an s2 electron while B and Al are losing the p1 electron. An s electron spends more time closer to the nucleus and is therefore harder to remove. Additionally, the p electrons are shielded somewhat by the s electrons and feel a smaller Zeff. b. Ei of N is larger than O. The explanation lies in the difference between losing an electron from a half filled orbital verses a filled orbital. Oxygen’s last filled electron is 2p4 . Electrons repel each other and when electrons are forced to share space in a filled orbital they are slightly higher in energy, so it is slightly easier to remove one giving O a smaller ionization energy compared to N. Chemistry 1A: Chapter 8 Higher Ionization Energies: Energy required removing the second, third of even more electrons from an atom Larger amounts of energy are required to remove each successive electron. It is relatively easier to remove from partially filled valence shell and harder to remove from filled d shells or core electrons. Electron Affinity, Eea: Energy given away when adding an electron, exothermic: The greatest negative value (most preferred) electron affinity is for F. Small nonmetals. Ignore noble gases. -Eea generally decreases from right to left; top to bottom of periodic table. P a g e | 11 Chemistry 1A: Chapter 8 P a g e | 12 Electronegativity: (Chapter 9: page 394-395) The ability to attract electrons toward the atom Electronegativity is important in the covalent bonding or ionic transfer of electrons in molecules and compounds. (Ch 9: Lewis Structure, Valence Shell Electron Pair Repulsion, determining polarity of a molecule) Increases from left to right; bottom to top of periodic table. Small nonmetals are much better at attracting electrons. Ignore the noble gases as most do not attract electrons (except xenon which may make a few compounds such as XeOF4, XeCl2 or XeF4) http://www.green-planet-solar-energy.com/electronegativity-values.html Chemistry 1A: Chapter 8 P a g e | 13 Metallic Character: Increases from right to left; top to bottom of periodic table Octet Rule: Main group elements tend to undergo reactions that leave them with 8 outer shell electrons, obtaining the noble gas configuration. Chemistry by Group: Alkali Metals: Metallic, soft enough to cut with a knife, silver color, low melting points, malleable, conductive, reactive and must be stored under oil to prevent reaction with air and moisture. Chemistry 1A: Chapter 8 P a g e | 14 Reaction of alkali metal (M) with a halogen (X) MX Reactions: 2 M + X2 2 M + H2 6 M + N2 4 M + O2 2 M + 2 H2O Alkaline Earth Metals: Metallic, silver color, malleable, conductive, can lose 2 electrons easily causing them to be powerful reducing agents. Reactions: M + X2 (X is a halogen) M + H2 2 M + O2 M + 2 H2O Halogens: Nonmetals, diatomic molecules, high electron affinities (tendency to gain electrons), powerful oxidizing agents Reactions: (X is a halogen) 2 M + n X2 2MXn H2 + X2 Chemistry 1A: Chapter 8 P a g e | 15 Noble Gases: Unreactive nonmetals, low melting and boiling points, colorless, odorless, filled core electron configuration. He and Ne undergo no known reactions. Argon is known to form HArF. Kr and Xe may react with fluorine (XeF2, XeF4, XeF6, XeOF4) Problems: 1. Use the concepts of effective nuclear charge, shielding, and n value of the valence orbital to explain the trends in atomic radius as you (a) move across the periodic table, (b) move down the periodic table. 2. Is the order of electron removal upon ionization simply the reverse of electron addition upon filling? Why or why not. Complete the short electron configurations for thalium…Tl, Tl+1, Tl+3 3. Arrange the elements as follows: C, Mg, He, Sr, O, Fr a) Increasing metallic character b) Increasing atomic size c) Increasing Ionization energy d) Increasing electronegativity 4. Identify 2 cations and 2 anions that are isoelectronic with Xe. Place them in order of increasing atomic radii. 5. Estimate the normal melting point of Br from the given melting points of atoms in the same group… F = -219°C, Cl = -101°C, I = 114°C 6. Estimate the density of Kr from the given density of atoms in the same group at STP… Ne = 0.90 g/L, Ar= 1.78 g/L, Xe = 5.86 g/L 7. Use Coulomb’s Law to arrange the ionic compounds by expected increasing normal melting points. BaO, CsI, NaCl, Fe2O3