Chapter 8 Electron Configuration and Chemical Periodicity

8-1 Dr. Wolf’s CHM 101

Chapter 8Electron Configuration and

Chemical Periodicity


Electron Configuration and Chemical Periodicity

5.1 Development of the Periodic Table

5.2 Characteristics of Many-Electron Atoms

5.3 The Quantum-Mechanical Model and the Periodic Table

5.4 Trends in Some Key Periodic Atomic Properties

5.5 The Connection Between Atomic Structure and Chemical Reactivity


Mendeleev’s Periodic Law

Arranged the 65 known elements by atomic mass and by recurrence of various physical and chemical properties.

The Periodic Table today is very similar but arranged according to atomic number (number of protons).

The arrangement led to families of elements with similar properties and at the time allowed for the prediction and properties of elements yet to be discovered.


Mendeleev’s Predicted Properties of Germanium (“eka Silicon”) and Its Actual Properties

Table 8.1

PropertyPredicted Properties of eka Silicon(E)

Actual Properties of Germanium (Ge)

atomic massappearancedensitymolar volumespecific heat capacityoxide formulaoxide densitysulfide formula and solubility

chloride formula (boiling point)

chloride densityelement preparation

72amugray metal5.5g/cm3

13cm3/mol0.31J/g*KEO2

4.7g/cm3

ES2; insoluble in H2O; soluble in aqueous (NH4)2SECl4; (<1000C)

1.9g/cm3

reduction of K2EF6 with sodium

72.61amugray metal5.32g/cm3

13.65cm3/mol0.32J/g*KGeO2

4.23g/cm3

GeS2; insoluble in H2O; soluble in aqueous (NH4)2SGeCl4; (840C)

1.844g/cm3

reduction of K2GeF6 with sodium


Remember from Chapter 4- Quantum Numbers and Atomic Orbitals

An atomic orbital is specified by three quantum numbers.n the principal quantum number - a positive integer

l the angular momentum quantum number - an integer from 0 to n-1

ml the magnetic moment quantum number - an integer from -l to +l

The three quantum numbers are actually giving the energy of the electron in the orbital and a fourth q.n. is needed to describe a property of electrons called spin. The spin can be clockwise or counterclockwise.

The spin q.n., ms can be + ½ or - ½ .

The Pauli Exclusion Principle - No two electrons in the same atom can have the same four q.n. Since the first three q.n. define the orbital, this means only two electrons can be in the same orbital and they must have opposite spins.


Table 8.2 Summary of Quantum Numbers of Electrons in Atoms

Name Symbol Permitted Values Property

principal n positive integers(1,2,3,…) orbital energy (size)

angular momentum

l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.)

magnetic mlintegers from -l to 0 to +l orbital orientation

spin ms+1/2 or -1/2 direction of e- spin


Factors Affecting Atomic Orbital Energies

Additional electron in the same orbital (makes less stable)

An additional electron raises the orbital energy through electron-electron repulsions.

Additional electrons in inner orbitals (makes outer orbital less stable)

Inner electrons shield outer electrons more effectively than do electrons in the same sublevel.

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)


The effect of another electron

in the same orbital


The effect of other electrons in inner

orbitals


The effect of orbital shape


Illustrating Orbital Occupancies

The electron configuration

n l# of electrons in the sublevel

as s,p,d,f

The orbital diagram

Order for filling energy sublevels with electrons


filled, spin-paired

half-filled

empty

A vertical orbital diagram

for the Li ground state


SAMPLE PROBLEM 8.1 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= +1/2

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

n = l = ml = ms=

2 0 0

2 1 -1 -1/2


Orbital occupancy for the first 10 elements, H through Ne.


Hund’s rule


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




A periodic table of partial ground-state electron configurations


The relation between orbital filling and the periodic table


General pattern for filling the sublevels


SAMPLE PROBLEM 8.2 Determining Electron Configuration

PLAN:

SOLUTION:

PROBLEM: Using the periodic table on the inside cover of the text (not Figure 8.12 or Table 8.4), 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)

1s22s22p63s23p64s1

[Ar] 4s1

4s1

condensed configuration

partial orbital diagram

full configuration

There are 18 inner electrons.


SAMPLE PROBLEM 8.2

continued

(b) for Mo (Z = 42)

1s22s22p63s23p64s23d104p65s14d5

[Kr] 5s14d5

(c) for Pb (Z = 82)

[Xe] 6s24f145d106p2

condensed configurationpartial orbital diagram

full configuration

5s1 4d5

condensed configuration

partial orbital diagram

full configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2

There are 36 inner electrons and 6 valence electrons.

6s2 6p2

There are 78 inner electrons and 4 valence electrons.


Defining metallic and covalent radii

Knowing the Cl radius and the C-Cl bond length, the C radius can be determined.


Atomic radii of the main-group and transition

elements.

Trends in the Periodic Table


Periodicity of atomic radius



SAMPLE PROBLEM 8.3 Ranking Elements by Atomic Size

PLAN:

SOLUTION:

PROBLEM: Using only the periodic table (not Figure 8.15)m 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 decrease in size as you go up; 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 Rb has a higher energy level and is far to the left. Br is to the left of Kr.

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


Periodicity of first ionization energy (IE1)


Energy required to remove one outermost electron.


First ionization energies of the

main-group elements



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 increases as you proceed up 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; Rb is below K.

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



The first three ionization energies of beryllium

(in MJ/mol)


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.


Electron affinities of the main-group elementsTrends in the

Periodic Table

Electron Affinity:

Energy change to add one electron.

In most cases, EA negative (energy released becauseelectron attracted to nucleus


Trends in three atomic properties


Trends in metallic behavior


Main-group ions and the noble gas configurations

Trends in the

Periodic Table

Properties of Monatomic Ions


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) can lose their np or ns and 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-


Magnetic Properties of Transition Metal Ions

A species with unpaired electrons exhibits paramagnetism. It is attracted by an external magnetic field.

Species with all paired e’s, not attracted........diamagnetic


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]4s23d6) Cr3+ ([Ar] 3d5) + 3e- paramagnetic

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

not paramagnetic (is diamagnetic)


Ionic vs. atomic radius


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 therefore 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 higher the + charge, the smaller the ion.


End of Chapter 8

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Chapter 8 Electron Configuration and Chemical Periodicity