Ch 8 Atomic Electron Configuration and Chemical Periodicity

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1 Ch 8 Atomic Electron Configuration and Chemical Periodicity Pauli Exclusion principle No two electrons in an atom have the same set of quantum numbers. If two electrons occupy the same orbital, they must have opposite spin. Hund s rule When several orbitalsof equal energy are available, as in a given sublevel, electrons enter with parallel spins. Electron configuration of atoms Orbital diagrams of atoms Electron configuration of atoms Shows the numbers of electrons which are indicated by a superscript in each sublevel, describing the arrangement of electron in atom) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 Quantum number set: n = 4, l = 2, m i = varies, m s = ± ½ 4d 5 Number of electron assigned to the designated orbital Orbital type (l) Electron Shell (n) Orbital diagrams of atoms Diagram shows how the electrons are distributed among orbitals. Each orbital is represented by parentheses (), or box or circle and the electrons are shown by arrows written as or depends on spin. 9F 1s 2s 2p 1

2 SAMPLE PROBLEM Determining Electron Configuration Using the periodic table, give the full and condensed electrons configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements: (a) potassium (K: Z = 19) (b) molybdenum (Mo: Z = 42) (c) lead (Pb: Z = 82) Use the atomic number for the number of electrons and the periodic table for the order of filling for electron orbitals. Condensed configurations consist of the preceding noble gas and outer electrons. (a) for K (Z = 19) full configuration condensed configuration partial orbital diagram 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Also called spdfnotation [Ar] 4s 1 (is also called abbreviated configuration-starts with the preceding noble gas) There are 18 inner electrons and one valence electron. 4s 1 3d 4p SAMPLE PROBLEM continued (b) for Mo (Z = 42) full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 condensed configuration partial orbital diagram [Kr] 5s 1 4d 5 There are 36 inner electrons and 6 valence electrons. 5s 1 4d 5 5p (c) for Pb(Z = 82) comprehensive reading full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 condensed configuration [Xe] 6s 2 4f 14 5d 10 6p 2 partial orbital diagram 6s 2 6p 2 There are 78 inner electrons and 4 valence electrons. 2

3 SAMPLE PROBLEM Determining Quantum Numbers from Orbital Diagrams Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom. Use the orbital diagram to find the third and eighth electrons. 9 F 1s 2s 2p The third electron is in the 2s orbital. Its quantum numbers are n = 2 l = 0 m l = 0 m s = + or -1/2 The eighth electron is in a 2p orbital. Its quantum numbers are n = 2 l = 1 m l = -1, 0, or +1 m s = + or -1/2 SAMPLE PROBLEM Determining Electron Configuration of ions Using the periodic table, give the full and condensed electrons configurations: (a) Iron (K +, Z = 19) (b) bromide (Br - Z=35) Use the atomic number for the number of electrons the periodic table for the order of filling for electron orbitals. Condensed configurations consist of the preceding noble gas and outer electrons. (a) for K + (Z = 19), full configuration condensed configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 [Ar] 4s 0 (is also called abbreviated configuration-starts with the preceding noble gas) partial orbital diagram 3s 2 3p6 There are 18 inner electrons and 0 valence electrons. 3

4 SAMPLE PROBLEM continued (b) for Br - (Z = 35, 35+1 electrons, n= 2, s, p, d and orbitalsare occupied) full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 This electron configuration is the same as the noble gas Kr, krypton. Partial Orbital diagram 4p 6 4s 2 3d 10 SAMPLE PROBLEM Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic (the unpaired electron display paramagnetism, it is attracted by the external field). (a) Mn 2+ (Z = 25) (b) Hg 2+ (Z = 80) Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration hasunpaired electrons, it is paramagnetic. (a) Mn 2+ (Z = 25) Mn([Ar]4s 2 3d 5 ) Mn 2+ ([Ar] 3d 5 ) + 2e - paramagnetic (c) Hg 2+ (Z = 80) Hg([Xe]6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e - not paramagnetic (is diamagnetic) 4

5 Apparatus for measuring the magnetic behavior of a sample. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Diamagnetic materials are solids with all paired electron resulting in no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron orbits under the influence of an external magnetic field. (copper, silver, and gold). Paramagnetic materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron orbits caused by the external magnetic field. (magnesium, molybdenum, lithium, and tantalum). Take-home message Factors Affecting Atomic Orbital Energies The Effect of Nuclear Charge (Z effective ) Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions. The Effect of Electron Repulsions (Shielding) Additional electron in the same orbital An additional electron raises the orbital energy through electron-electron repulsions. Additional electrons in inner orbitals Inner electrons shield outer electrons more effectively than do electrons in the same sublevel. 5

6 Take-home message Condensed ground-state electron configurations in the first three periods. Take-home message 6

7 Take-home message Take-home message 7

8 Atomic ionic radii of the main-group and transition elements. Atomic radius-can be defined and measured, assuming the atom is spherical. 1. Atomic radii decrease across the period from left to right. 2. Atomic radii increase down a group from top to bottom. 3. The positive ions are always smaller than atoms of the same element. 4. The negative ions are always larger than atoms of the same element. SAMPLE PROBLEM Ranking Elements by Atomic Size Using only the periodic table, rank each set of main group elements in order of decreasing atomic size: (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb Elements in the same group increase in size and you go down; elements decrease in size as you go across a period. (a) Sr > Ca > Mg These elements are in Group 2A(2). (b) K > Ca > Ga These elements are in Period 4. (c) Rb> Br > Kr (d) Rb> Sr > Ca Rb has a higher energy level and is far to the left. Br is to the left of Kr. Ca is one energy level smaller than Rband Sr. Rbis to the left of Sr. 8

9 SAMPLE PROBLEM Ranking Ions by Size Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2-, Cl - (c) Au +, Au 3+ Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. (a) Sr 2+ > Ca 2+ > Mg 2+ These are members of the same Group (2A/2) and therefore decrease in size going up the group. (b) S 2- > Cl - > K + (c) Au + > Au 3+ The ions are isoelectronic; S 2- has the smallest Z eff and therefore is the largest while K + is a cation with a large Z eff and is the smallest. The higher the + charge, the smaller the ion. First ionization energies of the main-group elements. ionization energy is the measure of how difficult to remove an electron from the gaseous atom. Energy must be absorbed to cause ionization, so the ionization energy is always positive (endothermic). The first ionization energy is the energy change to remove the outer-most electron for gaseous atom to form a +1 ion. Ionization energy increases across the periodic table from left to right. Ionization energy decreases down from top to bottom in the periodic table. Ionization energy has unit of kj. 9

10 SAMPLE PROBLEM Ranking Elements by First Ionization Energy Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE 1 : (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs IE decreases as you proceed down in a group; IE increases as you go across a period. (a) He > Ar > Kr (b) Te > Sb > Sn (c) Ca > K > Rb (d) Xe> I > Cs Group 8A(18) -IE decreases down a group. Period 5 elements -IE increases across a period. Ca is to the right of K; Rbis below K. I is to the left of Xe; Cs is further to the left and down one period. Electron affinities (EA) of the main-group elements. Electron affinity the energy change accompanying the addition of 1 mol of electrons to 1 mol of gaseous atoms or ions. The first electron affinity accompanying the formation of 1 mol of monovalent (1-) gaseous anions. Atom (g) + e - ion - E = EA 1 (EA1 is usually negative since the energy is released in most of cases.) Reactive nonmetals the elements in 6A (16) and 7A (17) has more negative EA and forms negative ions. Reactive metals the elements in 1A (1) and 2A (2) has slightly negative EA and forms positive ions. Noble gas the elements in 8A (18) has slightly positive EA and do not tend to lose or gain electrons. Only larger atoms (Kr, Xeand Rn) can form compounds. 10

11 Electronegativity (EN) of the elements. The reverse of electronegativity, the ability of an atom to lose electrons, is known as electropositivity Electronegativity measures the ability of an atom to attract to itself the electrons pair forming a covalent bond. The greater the electronegativity of an atom, the greater its attraction for electrons. Electronegativity increases from left to right in the same period. Electronegativity decreases from top to bottom in the same group. Atoms with similar electronegativities will share an electron with each other to form a covalent bond. If the atoms have significant different electronegativities, electrons will be permanently transferred to one atom to form ionic bond. Electronegativity (EN) of elements H 2.20 H 2 Li 0.98 Be 1.57 B 2.04 C 2.55 N 3.04 O 3.44 F 3.98 N 3 Na 0.93 Mg 1 Al 1.61 Si 1.90 P 2.19 S 2.58 Cl 3.16 A 4 K 0.82 Ca 1.00 Sc 6 Ti 1.54 V 1.63 Cr 1.66 Mn 1.55 Fe 1.83 Co 1.88 Ni 1.91 Cu 1.90 Zn 1.65 Ga 1.81 Ge 2.01 As 2.18 Se 2.55 Br 2.96 K 3 5 Rb 0.82 Sr 0.95 Y 1.22 Zr 3 Nb 1.6 Mo 2.16 Tc 1.9 Ru 2.2 Rh 2.28 Pd 2.20 Ag 1.93 Cd 1.69 In 1.78 Sn 1.96 Sb 2.05 Te 2.1 I 2.66 X 2 6 Cs 0.79 Ba 0.89 * Hf Ta 1.5 W 2.36 Re 1.9 Os 2.2 Ir 2.20 Pt 2.28 Au 2.54 Hg 2.00 Tl 1.62 Pb 2.33 Bi 2.02 Po 2.0 At 2.2 R 7 Fr 0.7 Ra 0.9 ** Rf Db Sg Bh Hs Mt Ds Rg Uub Uut Uuq Uup Uuh Uus U Lant hani des * La 1.1 Ce 1.12 Pr 1.13 Nd 1.14 Pm 1.13 Sm 1.17 Eu 1.2 Gd 1.2 Tb 1.1 Dy 1.22 Ho 1.23 Er 1.24 Tm 1.25 Yb 1.1 Lu 1.27 Acti nide s ** Ac 1.1 Th Pa 1.5 U 8 Np 6 Pu 1.28 Am 1.13 Cm 1.28 Bk Cf Es Fm Md No Lr 11

B I N G O B I N G O. Hf Cd Na Nb Lr. I Fl Fr Mo Si. Ho Bi Ce Eu Ac. Md Co P Pa Tc. Uut Rh K N. Sb At Md H. Bh Cm H Bi Es. Mo Uus Lu P F.

B I N G O B I N G O. Hf Cd Na Nb Lr. I Fl Fr Mo Si. Ho Bi Ce Eu Ac. Md Co P Pa Tc. Uut Rh K N. Sb At Md H. Bh Cm H Bi Es. Mo Uus Lu P F. Hf Cd Na Nb Lr Ho Bi Ce u Ac I Fl Fr Mo i Md Co P Pa Tc Uut Rh K N Dy Cl N Am b At Md H Y Bh Cm H Bi s Mo Uus Lu P F Cu Ar Ag Mg K Thomas Jefferson National Accelerator Facility - Office of cience ducation

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