Electron Configuration and

Chemical Periodicity

GENERAL CHEMISTRY

LECTURE 8

Electron configuration is the distribution of electrons of an atom (or molecule) in atomic orbital (or molecular orbital).

a. How do we place electrons in an orbital?

b. How do we arrange the orbitals in an atom?

Writing Electronic Configuration

3d - ___ ___ ___ ___ ___

zyx p p p

222 zyxxzyzxy ddddd

1s2

2p6

3d10

A. How do we place electrons in an orbital?

Only a maximum of 2 electrons per orbital

For degenerate orbitals, place the electrons singly before pairing (Hund’s Rule)

1s - ___

2p - ___ ___ ___

Writing Electronic Configuration

* However, this scheme is not always accurate * The arrangement of orbitals is based on calculated energies from Hψ = Eψ

B. How do we arrange the orbitals in an atom?

Orbitals are arranged in increasing energy

↑n value, ↑energy

For same n value: s < p < d < f

Writing Electronic Configuration

Order for filling energy sublevels with electrons

Order for filling energy sublevels with electrons Aufbau principle

Electrons occupy orbitals in a way that minimizes the energy of the atom.

Electron fills up the lowest energy orbital available

Ground-state electronic configuration

Electron Configuration and

Chemical Periodicity

GENERAL CHEMISTRY

LECTURE 8

1. Shorthand notation:

n l # of electrons in the sublevel

as s, p, d, f

2. Orbital diagram:

C (Z=6):

C (Z=6):

Writing Electronic Configuration

1s 2 2s 2 2p 2

Example: Write the electronic configuration of the following elements:

O (Z=8): 1s 2 2s 2 2p 4

1s 2s 2p

1.

Writing Electronic Configuration

Example: Write the electronic configuration of the following elements:

Writing Electronic Configuration

Al (Z=13): 1s 2 2s 2 2p 6 2. 3s 2 3p 1

1s 2s 2p 3s 3p

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

F (Z=9)

1s 2s 2p

Third electron: n = l = ml = ms= 2 0 0 + 1

2

Determining Quantum Numbers from Orbital Diagrams

Eighth electron: n = l = ml = ms= 2 1 -1 - 1

2

-1 +1 0

3. Condensed electron configuration

Al (Z=13): 1s 2 2s 2 2p 6 3s 2 3p 1

[Ne] 3s 2 3p 1

S (Z=16):

[Ne] 3s 2 3p 4

1.

Example: Write the condensed electronic configuration of the following :

1s 2 2s 2 2p 6 3s 2 3p 4

Writing Electronic Configuration

Mo (Z = 42): 2. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5

[Kr] 5s14d5

Elements with d-orbitals

The ns sublevel fills before the (n-1)d sublevel

How about the transition metals?

Writing Electronic Configuration

* There is special stability associated with half-filled orbitals

Cr (Z=24): 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2

V (Z=23): 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3

3d 4

1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5

1s 2s 2p 3s 3p 4s 3d

[Ar] 4s 1 3d 5

Cr (Z=24):

Cr (Z=24):

Electronic Configuration of Some Transition Metals

Cu (Z=29): 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2

Ni (Z=28): 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 8

3d 9

1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10

* There is special stability associated with completely filled orbitals

1s 2s 2p 3s 3p 4s 3d

[Ar] 4s 1 3d 10

Cu (Z=29):

Cu (Z=29):

Electronic Configuration of Some Transition Metals

Exercise 1 Determining Electron Configuration

1. Give the full and condensed electron configurations of the following elements.

(a) Mo (Z = 42) (b) Pb (Z = 82)

(a) Mo (Z = 42)

[Kr] 5s14d5 condensed configuration:

full configuration: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5

Exercise 1 Determining Electron Configuration

(b) Pb (Z = 82)

[Xe] 6s24f145d106p2

full configuration: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10

condensed configuration:

5p 6 6s2 4f 14 5d 10 6p 2

1. Write the expanded electronic configuration of silver (Z=47) 2 pts

2. Write the condensed electron configuration of bromine. (Z=35) 2 pts

3. Write the quantum numbers for the last entering electron of Mg. 2 pts

Bonus: Write the quantum numbers for the last entering electron of Mg2+. 2 pts

Quiz

The Periodic table Alkali Metals

Alkaline Earths

Transition Metals

Halogens

Noble Gases

Lanthanides and Actinides

Main Group

Main Group

Group (Columns) Period (Rows)

The Periodic Table

The period number is the n value of the highest energy level.

The Periodic Table

The period number is the n value of the highest energy level.

Ni (Z=28): 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 8

Period number? 4

S (Z=16): 1s 2 2s 2 2p 6 3s 2 3p 4

Period number? 3

Group (Columns) Period (Rows)

The Periodic Table

The group number is equal to the number of outer or valence electrons.

The Periodic Table

Elements in the same group have the same chemical properties

The group number is equal to the number of outer or valence electrons.

The valence electrons are those electrons in the highest energy level and they are involved in forming compounds.

N (Z=7) 1s 2 2s 2 2p 3

Example:

Group 5

P (Z=15) 1s 2 2s 2 2p 6 3s 2 3p 3

The inner electrons are those electrons in the lower energy levels.

The relation between orbital filling and the periodic table.

Mg (Z =12): [Ne] 3s 2

[Kr] 5s 2 4d 3 Nb (Z = 41):

[Ar] 4s 2 3d 10 Ga (Z = 31): 4p1

The Periodic Trends

All physical and chemical properties of the elements is based on the electron configurations of their atoms.

Atomic Size (Ionic Size)

Ionization energy

Electron Affinity

Trends in Atomic Size

1. Change in n The higher the n value (energy level) the larger the atom. The n value increases down a group.

Thus atomic size increase down a group

Incr

ease

Nuclear Charge, Z

Moving across a period, each element has an increased nuclear charge and the electrons are going into the same shell (2s and 2p or 3s and 3p, etc.).

Effective Nuclear Charge, Zeff

Screen of electron charge from 10 core electrons

(-) (-)

(12+)

Inner electrons block the nuclear charge more effectively than outer electrons.

Trends in Atomic Size

2. Change in Zeff

>

SHIELDING EFFECT

Trends in Atomic Size

2. Change in Zeff The higher the effective nuclear charge (Zeff), the smaller the atom.

The Zeff increases from left to right.

Thus atomic size decreases from left to right.

Increase Screen of electron charge from 10 core electrons

(-) (-)

(12+)

Exercise 2 Ranking Elements by Atomic Size

SOLUTION:

2. 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

(a) Sr > Ca > Mg

(b) K > Ca > Ga

(c) Rb > Br > Kr

(d) Rb > Sr > Ca

Trends in Ionic Radius

Cations are smaller than the atoms from which they are formed.

Trends in Ionic Radius

Anions are larger than the atoms from which they are formed.

For isoelectronic cations, the more positive the ionic charge, the smaller the ionic radius.

For isoelectronic anions, the more negative the charge, the larger the ionic radius.

Trends in Ionic Radius

ISOELECTRONIC: Ions (or atoms) with the same electronic configuration

Trends in Ionic Radius

Example:

Ne (Z=10): 1s22s22p6

Na (Z=11) Na+ + e

1s22s22p6

Mg2+ :

1s22s22p6

1s22s22p62s1

IONIC RADIUS:

Ne > Na+ > Mg2+

Ionic vs. atomic radii.

N3- > O2- > F- > Na+ > Mg2+ > Al3+

Exercise No. 3 Ranking Ions by Size

Answer:

3. 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+

Sr2+ > Ca2+ > Mg2+

S2- > Cl- > K+

Au+ > Au3+

(a)

(b)

(c)

Ionization energies decrease as atomic radii increase.

Ionization Energy

The energy required (in kJ) for the complete removal of 1 mol of electrons from 1 mol of gaseous atoms or ions.

Atom (g) ion+ (g) + e- IE > 0

Mg+(g) → Mg2+(g) + e- IE2 = 1451 kJ

Na (g) + 496 kJ/mol Na+ (g)+ e-

Example:

First ionization energies as a function of atomic number

First ionization energies as a function of atomic number

I1 (Mg) vs. I1 (Al)

3s 3p

3s 3p

Mg

Al

I1 (P) vs. I1 (S)

3s 3p

P

S

3s 3p

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

Mg (g) → Mg+(g) + e- I1 = 738 kJ

Mg+(g) → Mg2+(g) + e- I2 = 1451 kJ

IE1 < IE2 < IE3

Exercise No. 4 Ranking Elements by First Ionization Energy

4. 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

(a) He > Ar > Kr

(b) Te > Sb > Sn

(c) Ca > K > Rb

(d) Xe > I > Cs

Answer:

F(g) + e- → F-(g) EA = -328 kJ

F(1s22s22p5) + e- → F-(1s22s22p6)

Li(g) + e- → Li-(g) EA = -59.6 kJ

Electron Affinity

Electron affinity is the amount of energy released when an electron is added to an isolated gaseous atom to form an ion with a 1- charge.

Atom (g) + e- ion- (g) EA1 <0 Example:

Electron affinities of main-group elements

Second Electron Affinities

O(g) + e- → O-(g) EA = -141 kJ

O-(g) + e- → O2-(g) EA = +744 kJ

Metals and Nonmetals

Metallic behavior

• Diamagnetic atoms or ions:

– All e- are paired.

– Weakly repelled by a magnetic field.

• Paramagnetic atoms or ions:

– Unpaired e-.

– Attracted to an external magnetic field.

Magnetic Properties

Paramagnetic

Exercise No. 5 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions

SOLUTION:

5. 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)

paramagnetic

(a) Mn2+ (Z = 25)

Mn ([Ar] 4s23d5) Mn2+ ([Ar] 3d5) + 2e-

paramagnetic

paramagnetic

(b) Cr3+(Z = 24)

paramagnetic

Cr ([Ar] 4s13d5) Cr3+ ([Ar] 3d3) + 3e-

SOLUTION:

(c) Hg2+(Z = 80)

(diamagnetic)

Hg ([Xe] 6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e-

(diamagnetic)

Exercise No. 5 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions

Trends in three atomic properties.

Periodic Properties of the Elements

6. Identify the element described in each of the following:

a. Lowest IE1 in Period 5

b. Most metallic element c. Period 4 element with filled outer orbital

d. Period 3 element whose 2- ion is isoelectronic with Ar

e. Period 4 transition element that forms 3+ diamagnetic ion.

(1 pt each)

Exercise No. 6

Summary

Write the electron configuration of atoms and ions

Predict the trends of atoms (ions) using the periodic table

Atomic Size ( and Ionic size)

Ionization Energy

Electron Affinity

Determine the magnetic properties of atoms

Diamagnetic

Paramagnetic

Review

1. For the following groups of elements select the one that has the property noted: 1.5 pts each

a. The largest atom: Mg, Mn, Mo, Ba, Bi, Br

b. The lowest first ionization energy: B, Sr, Al, Br, Mg, Pb

c. The most negative electron affinity: As, B, Cl, K, Mg, S

d. The largest unpaired electrons: F, N, S2-, Mg2+, Sc3+, Ti3+

2. Arrange the following isoelectronic compounds in order of increasing radius: Rb+, Y3+, Br-, Kr, Sr2+,Se2- 2 pts