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Hund’s Rule. For an atom in its ground-state configuration, all unpaired electrons have the same spin orientation. Therefore electrons tend to occupy all free orbitals and not pair up, so that their spins all add up to produce a general vector for the atom. Fig. 8.5.
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Hund’s Rule • For an atom in its ground-state configuration, all unpaired electrons have the same spin orientation. • Therefore electrons tend to occupy all free orbitals and not pair up, so that their spins all add up to produce a general vector for the atom.
Orbital Occupancy for the First 10 Elements, H through Ne Fig. 8.6
1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz Orbital Box Diagrams - II : B Ne B (5 e-) 1s2 2s2 2p1 C (6 e-) 1s2 2s2 2p2 N (7 e-) 1s2 2s2 2p3 O (8 e-) 1s2 2s2 2p4 F (9 e-) 1s2 2s2 2p5 Ne (10 e-) 1s2 2s2 2p6
Quantum Numbers Noble Gases Electron OrbitalsNumber of ElectronsElement 1s2 2 He 1s2 2s22p6 10 Ne 1s2 2s22p6 3s23p6 18 Ar 1s2 2s22p6 3s23p6 4s23d104p6 36 Kr 1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 54 Xe 1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 6s24f14 5d106p6 86 Rn 1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 6s24f145d106p6 7s25f146d10?
The Periodic Table of the Elements Electronic Structure H He Li Be B C N O F Ne Ar Na Mg Al Si P S Cl Kr K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Xe Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Rn Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Fr Ra Ac Rf Ha Sg Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr “ S” Orbitals “ P” Orbitals “ f ” Orbitals “ d” Orbitals
Figure 8.12: A periodic table illustrating the building-up order.
Figure 8.7: A mnemonic diagram for the building-up order (diagonal rule).
Valence and Core Electrons • Valence Electrons - Those electrons outside of a closed electron shell. These electrons take part in chemical reactions. • Core Electrons - The electrons in the closed shells. They cannot take part in chemical reactions. • Sodium 11 electrons • Valence electrons [Ne] 3s 1 --- one • Core electrons 1s 2 2s 2 2p 6 --- Ten • Chlorine 17 electrons • Valence electrons [Ne] 3s 2 3p 5---- seven • Core 2 2s 2 2p 6 ---- Ten
Condensed Ground-State Electron Configurations in the First Three Periods
Orbital Box Diagrams - III Na Ar Atomic Number Orbital Box Condensed Electron Element Diagrams(3s&3p) Configuration 11 Na [He] 3s1 12 Mg [He] 3s2 13 Al [He] 3s23p1 14 Si [He] 3s23p2 15 P [He] 3s23p3 16 S [He] 3s23p4 17 Cl [He] 3s23p5 18 Ar [He] 3s23p6 3s 3px 3py 3pz 3s 3px 3py 3pz 3s 3px 3py 3pz 3s 3px 3py 3pz 3s 3px 3py 3pz 3s 3px 3py 3pz 3px 3py 3pz 3s
Figure 8.17: Representation of atomic radii (covalent radii) of the main-group elements.
Electronic Configuration Ions • Na 1s 2 2s 2 2p 6 3s 1 Na+ 1s 2 2s 2 2p 6 • Mg 1s 2 2s 2 2p 6 3s 2 Mg+2 1s 2 2s 2 2p6 • Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al+3 1s 2 2s 2 2p 6 • O 1s 2 2s 2 2p 4 O- 2 1s 2 2s 2 2p 6 • F 1s 2 2s 2 2p 5 F- 1 1s 2 2s 2 2p 6 • N 1s 2 2s 2 2p 3 N- 3 1s 2 2s 2 2p 6
Isoelectronic Atoms and Ions • H- 1 { He } Li+ Be+2 • N- 3 O- 2 F-{ Ne } Na+ Mg+2 Al+3 • P- 3 S- 2 Cl- { Ar } K+ Ca+2 Sc+3 Ti+4 • As- 3 Se- 2 Br- { Kr } Rb+ Sr+2 Y+3 Zr+4 • Sb- 3 Te- 2 I- { Xe } Cs+ Ba+2 La+3 Hf+4
Pseudo - Noble Gas Electron Configurations Elements in groups 3A, 4A, and 5A can form cations by losing enough electrons to leave a “pseudo noble gas” configuration. By losing electrons and leaving a filled d orbital, which is quite stable! Sn [Kr] 5s24d105p2 Sn4+ [Kr] 4d10 + 4 e - Sn [Kr] 5s24d105p2 Sn2+ [Kr] 5s24d10 + 2 e - Pb [Xe] 4f145d106s26p2 Pb+2 [Xe] 4f145d106s2 + 2 e- Pb [Xe] 4f145d106s26p2 Pb+4 [Xe] 4f145d10 + 4 e- As [Ar] 3d104s24p3 As3+ [Ar] 3d104s2 + 3 e- As [Ar] 3d104s24p3 As5+ [Ar] 3d10 + 5 e- Sb [Kr] 4d105s25p3 Sb3+ [Kr] 4d105s2 + 3 e- Sb [Kr] 4d105s25p3 Sb5+ [Kr] 4d10 + 5 e-
Ranking Ions According to Size Problem: Rank each set of Ions in order of increasing size. a) K+, Rb+, Na+ b) Na+, O2-, F - c) Fe+2, Fe+3 Plan: We find the position of each element in the periodic table and apply the ideas of size: i) size increases down a group, ii) size decreases across a period but increases from cation to anion. iii) size decreases with increasing positive (or decreasing negative) charge in an isoelectronic series. iv) cations of the same element decreases in size as the charge increases. Solution: a) since K+, Rb+, and Na+ are from the same group (1A), they increase in size down the group:Na+ < K+ < Rb+ b)the ions Na+, O2-, and F- are isoelectronic. O2- has lower Zeff than F-, so it is larger. Na+ is a cation, and has the highest Zeff, so it is smaller: Na+ < F- < O2- c) Fe+2 has a lower charge than Fe+3, so it is larger: Fe+3 < Fe+2
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Ranking Ions According to Size Problem: Rank each set of Ions in order of increasing size. a) K+, Rb+, Na+ b) Na+, O2-, F - c) Fe+2, Fe+3 Plan: We find the position of each element in the periodic table and apply the ideas of size: i) size increases down a group, ii) size decreases across a period but increases from cation to anion. iii) size decreases with increasing positive (or decreasing negative) charge in an isoelectronic series. iv) cations of the same element decreases in size as the charge increases. Solution: a) since K+, Rb+, and Na+ are from the same group (1A), they increase in size down the group:Na+ < K+ < Rb+ b)the ions Na+, O2-, and F- are isoelectronic. O2- has lower Zeff than F-, so it is larger. Na+ is a cation, and has the highest Zeff, so it is smaller: Na+ < F- < O2- c) Fe+2 has a lower charge than Fe+3, so it is larger: Fe+3 < Fe+2
Periodicity of First Ionization Energy (IE1) Like Figure 8-18
Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution:
Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution: The largest jump in IE occurs after IE3 so the element has 3 valence electrons thus it is Aluminum ( Al, Z=13), its electron configuration is : 1s2 2s2 2p6 3s2 3p1
Ranking Elements by First Ionization Energy Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasingIE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar Plan: Find their relative positions in the periodic table and apply trends! Solution:
Ranking Elements by First Ionization Energy Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasingIE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar Plan: Find their relative positions in the periodic table and apply trends! Solution: a) Rn, Ar,Ne These elements are all noble gases and their IE decreases as you go down the group.