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Chapter 8

Electron Configuration and Chemical Periodicity

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Electron Configuration and Chemical Periodicity

8.1 Development of the Periodic Table

8.2 Characteristics of Many-Electron Atoms

8.3 The Quantum-Mechanical Model and the Periodic Table

8.4 Trends in Three Key Atomic Properties

8.5 Atomic Structure and Chemical Reactivity

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The “Fourth” Quantum Number Three quantum numbers came out of Schroedinger’s work, but he was dealing with single electron atoms and ions.

Discrepancies became apparent upon examination of multi-electron species

Stern and Gerlach: fire a beam of H atoms thru a magnet. Unexpected result is that the incomingbeam is split into TWO separate beams, each containing HALF the H atoms. The orientation of the magnetic moment of the electron must also be quantized!

Explain the results by introducing a fourth quantum number: the spin quantum number (ms; either +1/2 (α) or -1/2 (β))

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Observing the effect of electron spin.

The Stern-Gerlach experiment.

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Factors Affecting Atomic Orbital Energies

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. Shielding by inner electrons greatly lowers the Zeff felt by outer electrons.

Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions.

The Effect of Nuclear Charge (Zeffective)

The Effect of Electron Repulsions (Shielding)

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Figure 8.2 Penetration and orbital energy.

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Illustrating Orbital Occupancies

The electron configuration

n l # of electrons in the sublevel

as s,p,d,f

The orbital diagram (box or circle)

Figure 8.3

Order for filling energy sublevels with electrons.

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dark - filled, spin-paired

light - half-filled

no color-empty

An orbital diagram for the Li ground state.

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Determining Quantum Numbers from Orbital Diagrams

PLAN:

SOLUTION:

Use the orbital diagram to find the third and eighth electrons.

PROBLEM: Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom.

9F

1s 2s 2p

The third electron is in the 2s orbital. Its quantum numbers are:

n = l = ml = ms= 2 0 0 + or -

The eighth electron is in a 2p orbital. Its quantum numbers are:

2 n = l = ml = ms= -1, 0, or +1 1 + or -

1

2

1

2

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Figure 8.4 Condensed ground-state electron configurations in the first three periods.

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A periodic table of partial ground-state electron configurations.

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Orbital filling and the periodic table.

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Determining Electron Configuration

PLAN:

SOLUTION:

PROBLEM: Using the periodic table on the inside cover of the text (not Figure 8.5 or Table 8.3), give the full and condensed electron 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)

1s22s22p63s23p64s1

[Ar] 4s1 condensed configuration

partial orbital diagram

full configuration

There are 18 inner electrons.

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SAMPLE PROBLEM 8.2

continued

(b) for Mo (Z = 42)

1s22s22p63s23p64s23d104p65s14d5 [Kr] 5s14d5

(c) for Pb (Z = 82)

[Xe] 6s24f145d106p2

condensed configuration partial orbital diagram

full configuration

condensed configuration

partial orbital diagram

full configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2

There are 36 inner electrons and 6 valence electrons.

There are 78 inner electrons and 4 valence electrons.

Determining Electron Configuration

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Figure 8.7 Defining metallic and covalent radii.

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Figure 8.8

Atomic radii of the main-group and transition

elements.

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Figure 8.9 Periodicity of atomic radius.

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SAMPLE PROBLEM 8.3 Ranking Elements by Atomic Size

PLAN:

SOLUTION:

PROBLEM: Using only the periodic table (not Figure 8.8), 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 as you go down; elements decrease in size as you go across a period.

(a) Sr > Ca > Mg These elements are in Group 2A(2), and size decreases up the group.

(b) K > Ca > Ga These elements are in Period 4, and size decreases across a period.

(c) Rb > Br > Kr Rb has a higher energy level and is far to the left. Br is to the left of Kr in Period 4.

(d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr in the same period.

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Figure 8.10 Periodicity of first ionization energy (IE1).

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Figure 8.11 First ionization energies of the main-group elements.

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Figure 8.12 The first three ionization energies of beryllium (in MJ/mol).

For more data on sequential ionization energies of the elements, go to http://www.webelements.com.

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SAMPLE PROBLEM 8.4 Ranking Elements by First Ionization Energy

PLAN:

SOLUTION:

PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1:

(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

These three elements are all in Group 8A(18), IE decreases down a group.

These are all in Period 5, IE increases across a period.

Ca is to the right of K; Rb is below K.

I is to the left of Xe; Cs is furtther to the left and down one period.

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SAMPLE PROBLEM 8.5 Identifying an Element from Successive Ionization Energies

PLAN:

SOLUTION:

PROBLEM: Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration:

IE1 IE2 IE3 IE4 IE5 IE6

1012 1903 2910 4956 6278 22,230

Look for a large increase in energy which indicates that all of the valence electrons have been removed.

The largest increase occurs after IE5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s23p3 and the element must be phosphorous, P (Z = 15).

The complete electron configuration is 1s22s22p63s23p3.

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Figure 8.13 Electron affinities of the main-group elements.

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Figure 8.14 Trends in three atomic properties.

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Figure 8.15 Trends in metallic behavior.

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Figure 8.16 The trend in acid-base behavior of element oxides.

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Figure 8.17 Main-group ions and the noble gas electron configurations.

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SAMPLE PROBLEM 8.6 Writing Electron Configurations of Main-Group Ions

PLAN:

SOLUTION:

PROBLEM: Using condensed electron configurations, write reactions for the formation of the common ions of the following elements:

(a) Iodine (Z = 53) (b) Potassium (Z = 19) (c) Indium (Z = 49)

Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are usually isoelectronic with the nearest noble gas.

Metals in Groups 3A(13) to 5A(15) lose the np and ns or just the np electrons.

(a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5s24d105p5) + e- I- ([Kr]5s24d105p6)

(b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s1) K+ ([Ar]) + e-

(c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s24d105p1) In+ ([Kr]5s24d10) + e-

In ([Kr]5s24d105p1) In3+([Kr] 4d10) + 3e-

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Figure 8.18 The Period 4 crossover in sublevel energies.

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Figure 8.19

Apparatus for measuring the magnetic behavior of a sample.

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SAMPLE PROBLEM 8.7 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions

PLAN:

SOLUTION:

PROBLEM: Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic.

(a) Mn2+ (Z = 25) (b) Cr3+ (Z = 24) (c) Hg2+ (Z = 80)

Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic.

paramagnetic (a) Mn2+(Z = 25) Mn ([Ar] 4s23d5) Mn2+ ([Ar] 3d5) + 2e-

(b) Cr3+(Z = 24) Cr ([Ar] 4s13d5) Cr3+ ([Ar] 3d3) + 3e- paramagnetic

(c) Hg2+(Z = 80) Hg ([Xe] 6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e-

not paramagnetic (diamagnetic)

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Figure 8.20 Depicting ionic radius.

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Figure 8.21 Ionic vs. atomic radii.

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SAMPLE PROBLEM 8.8 Ranking Ions by Size

PLAN:

SOLUTION:

PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking:

(a) Ca2+, Sr2+, Mg2+ (b) K+, S2-, Cl- (c) Au+, Au3+

Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons.

(a) Sr2+ > Ca2+ > Mg2+

(b) S2- > Cl- > K+

These are members of the same Group 2A(2), and decrease in size going up the group.

The ions are isoelectronic; S2- has the smallest Zeff and therefore, is the largest while K+ is a cation with a large Zeff and is the smallest.

(c) Au+ > Au3+ The greater the + charge, the smaller the ion.