Chapter 8

Electron Configuration and Chemical Periodicity

Wave Function (Orbital)

described by

described by

which are

spdf electronic configuration

Aufbau Rules

determined by

which involve

comprising

Core Electrons

Valence Electrons

Periodic Table

basis for

which summarizes

Periodic Properties

Hund’s Rule

OrbitalEnergy

PauliExclusion

e- filling

Quantum Numbers

Quantum Numbers

Principaln = 1,2,3,..

Angular momentum, l

Magneticml

Spin, ms

defines

defines

defines

defines

defines

Orbital size & energy

Electron spin

Orbitalorientation

Orbitalshape

Name

principal

Symbol Permitted Values Property

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

angular momentum

lintegers from 0 to n-1

orbital shape (0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.)

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

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

4-quantum numbers specify the energy and location of electrons around a nucleus (all we can know). This numbers are the framework for the “electronic structure of an atom”.

2

2

0 1

3d

QuantumNumber

AllowedValues

n

l

ml

Positive integers 1,2,3,4....

0 up to maxof n-1

-l,...0...+l

1

0

0

2

0

0

1

0 1-1

0

0

3

1

0 1-1 -1-2

1s 2s 2p 3s 3p

Possible Orbitals

Shapes or BoundrySurface Plots

Orbital Name

Schrodinger’s equation gives an exact solution for H-atom, but does not for many electron-atoms. Electron-electron repulsion in multi-electron split energy levels.

Ener

gy

Electrons willfill lowest energy orbitals first!

Hydrogen Atom Multi-electron atoms

Orbitals are “degenerate” or the same energy in Hydrogen!

Inner core electrons “shield” or “screen” outer electrons from the positive charge of the nucleus.

Screening Impacts1) alters the energy levels spacing & ordering in many-electron atoms.2) outer e- screened by inner electrons.

inner coreelectron

valence electron

4-quantum numbers specify all the information we can know the energy and location of electrons around a nucleus. Chemists call this the “electronic structure of an atom”.

Remember this diagram electronic configurations.

The “Aufbau Process” is used to generate the electronic configuration of elements filling the lowest energy orbitals sequentially.

1. Lower energy orbitals fill first (smaller n).2. Hund’s Rule-degenerate (i.e. orbitals with

the same energy) orbitals fill one at a time before electrons are paired in an orbital.

3. Pauli Exclusion Principle: No two electrons in an atom can have same 4-quantum numbers.

remember filling order using this device!

Chemists use “spdf notation” and “orbital box diagrams” to symbolize the “ground state electronic configuration” of elements.

spdf Notation

orbital box diagram

H

He

1s1

1s2

Element

electron shellprincipal quantum #

orbital typeangular quantum #

# of electronsin orbital

Arrow denotes an electron with “spin up” or “spin-down”.

Remember, no two electrons can have the same 4 quantum numbers!

Building electronic configuration using Aufbau and Hund

H

He

Li

Be

1s1

1s2

1s22s2

1s22s1

Atomic Number/Element

Orbital BoxDiagram

Full-electronicconfiguration

Condensed-electronicconfiguration

1s1

1s2

[He]2s1

[He]2s2

written with noble gas configuration

1s22s22p3

1s22s22p4

1s22s22p5

1s22s22p6

1s22s22p1

1s22s22p2

B

C

Atomic Number/Element

Orbital BoxDiagram

Full-electronicconfiguration

Condensed-electronicconfiguration

[He]2s22p1

[He]2s22p2

[He]2s22p3

[He]2s22p4

[He]2s22p5

[He]2s22p6

Paramagneticunpaired electrons

2p

Diamagneticall electrons paired

2p

• Diamagnetic atoms or ions:– All e- are paired.– Weakly repelled in a magnetic field.

• Paramagnetic atoms or ions:– Unpaired e- exist in an orbital– Attracted to an external magnetic field.

Unpaired electrons in orbitals gives rise to paramagnetism and is attracted to a magnetic field. Diamagnetic species contain all paired electrons and is “repelled” by the magnetic field.

This periodic table shows the two “f-block series (lanthanides 4f-block and actinides (5f) where it really is supposed to be, not removed from the table.

s-block f-block d-block p-block

Transition Metals

Lanthanides (4f-block)

Acthanides (5f-block)

Lanthanides (4f-block)

Anthanides (5f)

Transition Elements (d-block)

Representative or Main Group Elements

p-blocks-block

8.2

ns1

ns2

d1 d5

ns2 n

p1

ns2 n

p2

ns2 n

p3

ns2 n

p4

ns2 n

p5 ns2 n

p6

d10

4f

5f

Ground State Electron Configurations of the Elementsns

2

d1 d5odd

beha

vior

odd

beha

vior

4f1 4f2 4f144f10

What is the electron configuration of Mg?

What are the possible quantum numbers for the last (outermost) electron in Cl?

What is the electron configuration of Mg?

Mg 12 electrons1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons

Abbreviated as [Ne]3s2 [Ne] 1s22s22p6

What are the possible quantum numbers for the last (outermost) electron in Cl?Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons

Last electron added to 3p orbital

n = 3 l = 1 ml = -1, 0, or +1 ms = ! or -!

Using the periodic table on the inside cover of the text and 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)

(b) for Mo (Z = 42) 36 inner electrons and 6 valence electrons1s22s22p63s23p64s23d104p65s14d5

[Kr] 5s14d5

(c) for Pb (Z = 82) 78 inner electrons and 4 valence electrons.

[Xe] 6s24f145d106p2

condensedpartial orbital diagram

full configuration

condensed

partial orbital diagram

full configuration 1s22s22p63s23p64s23d104p65s24d10

5p66s24f145d106p2

6s2 6p2

5s1 4d5 5p

(a) for K (Z = 19)1s22s22p63s23p64s1

[Ar] 4s1

4s1

condensedorbital diagram

full configurationThere are 18 inner electrons.

3d 4p

Metals lose electrons so that cation has a noble-gas outer electron configuration.

Na [Ne]3s1 Na+ [Ne]

Ca [Ar]4s2 Ca2+ [Ar]

Al [Ne]3s23p1 Al3+ [Ne]

Non-metals gain electrons so that anion has a noble-gas outer electron configuration.

H 1s1 H- 1s2 or [He]

F 1s22s22p5 F- 1s22s22p6 or [Ne]

O 1s22s22p4 O2- 1s22s22p6 or [Ne]

N 1s22s22p3 N3- 1s22s22p6 or [Ne]

Metals loose electrons (oxidized) and become cations. Non-metals gain electrons (reduced) and become anions. The electronic configuration of each reflects this change in the number of electrons.

Metals and non-metal ions tend to form electronic states closest to their nearest noble gas configuration.

Na+, Al3+, F-, O2-, and N3- are all said to be “isoelectronic with Ne” as they have the same electronic configuration....all subshells are filled.

Isoelectronic species are two different elements with the same electronic configuration--but not the same nuclear configuration.

Na: [1s22s22p63s1] =====> Na+: [1s22s22p6] = [Ne]oxidation

oxidationAl: [1s22s22p63s23p1] =====> Al3+: [1s22s22p6] = [Ne]

N: [1s22s22p3] =====> N3-: [1s22s22p6] = [Ne]reduced

O: [1s22s22p4] =====> O2-: [1s22s22p6] = [Ne]reduced

F: [1s22s22p5] =====> F-: [1s22s22p6] = [Ne]reduced

When a transition-metal cation is formed from an atom of a transition metal, electrons are removed first from the ns orbital, then from the (n-1)d orbital.

Fe: [Ar]4s23d6 Fe2+: [Ar]4s03d6 or [Ar]3d6

Mn: [Ar]4s23d5 Mn2+: [Ar]4s03d5 or [Ar]3d5

Fe3+: [Ar]4s03d5 or [Ar]3d5Fe: [Ar]4s23d6

(n-1)d (ns)

Transition Metal Transition Metal Cation

Write the full electronic configuration the following ions: Sc+3, Zn+2,Co2+ and Co3+ . Distinguish if each is paramagnetic or diamagnetic.

Diamagnetic Sc3+ Paramagnetic Zn2+ , Co2+ , Co3+

Write the full electronic configuration the following ions: Sc+3, Zn+2,Co2+ and Co3+ . Distinguish if each is paramagnetic or diamagnetic.

A) What is the electron configuration of Mg and Mg2+?

B) What are the possible quantum numbers for the last (outermost) electron in Cl?

C) Is ground state F paramagenetic or diamagnetic?

What is the spdf and condensed electron configuration of Mg and Mg2+ ? Mg 12 electrons

What are the possible quantum numbers for the last (outermost) electron in Cl?

Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electronsLast electron added to 3p orbitaln = 3 l = 1 ml = -1, 0, or +1 ms = ! or -!

C) Is ground state F paramagenetic or diamagnetic?

Mg 1s22s22p63s2 [Ne]3s2

Mg2+ 1s22s22p63s0 [Ne]3s0 = [Ne]

9F1s 2s 2p

Unpaired electron = PARAMAGNETIC

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.

SOLUTION:

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 (is diamagnetic)

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)

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.

(c) Hg2+(Z = 80)

Identify n and l quantum numbers for each of the following.

What neutral element has the following orbital-filling diagram?

fourth shellthird shell

3p 4dz2

Gallium = Ga

Using spdf notation write the complete electron configuration of O, Cl, Ti, Zn?

What are the possible quantum numbers for the last (outermost) electron in Cl?

Using condensed spdf notation what is the

electronic configuration of Br and Br -

Periodicity in the chemical reactivity of elements occurs because of periodicity in the electronic structure of valence electrons!

1-electron outer s-orbital

2-electrons outer d-orbital

5-electrons outer p-orbital

We must know these properties and be able to apply them!

Amount of energy to remove 1 mole e- from 1 mole of gaseous atoms or element

Amount of energy to add 1 mole e- to 1 mole of gaseous atoms or element

Here’s a different view on one periodic table! Know these properties cold (see homework and examples)

1) Inner core electrons : electrons residing in the lower n shells of an element--located closer to the nucleus.

2. Outer core or VALENCE e- : total number of e- in the highest n-value shell.

The number of valence e- is given by the Group Number in the periodic table for Group A representative elements.

Electrons in elements are categorized either as inner-core electrons or outer valence-electrons.

n =2

n =3

n =4

n =5

n =6

Inner core electrons “shield” or “screen” outer electrons from the positive charge of the nucleus.

Screening Impacts1) alters the energy levels spacing & ordering in many-electron atoms.2) outer e- screened by inner electrons.

inner coreelectron

valence electron

Effective nuclear charge (Zeff) is the electrostatic force felt by the outer valence electrons taking into “shielding” by internal core electrons.

To a good approximation: effective nuclear charge, Zeff, is given by:

Zeff = Z – core e-

# of inner core non-valence electrons

# protonsin atom

Effective Nuclear charge

Larger Zeff means more “pull” or electrostatic force between nucleus and electrons.

8.2

ns1

ns2

d1 d5

ns2 n

p1 ns

2 np2

ns2 n

p3 ns

2 np4

ns2 n

p5

ns2 n

p6

d10

4f

5f

ns2

d1 d5

The number of valence electrons can be found by counting the # of e- in the outer most shell (main-group elements only)

Configuration Element Z (p+)Core

ElectronsValence

Electrons ZeffectiveRadius(pm)

[Ne]3s1 Na 11 10 1 1 186

[Ne]3s2 Mg 12 10 2 2 160

[Ne]3s23p1 Al 13 10 3 3 143

[Ne]3s23p2 Si 14 10 4 4 132

[Ne]3s23p3 P 15 10 5 5 128

[Ne]3s23p4 S 16 10 6 6 127

[Ne]3s23p5 Cl 17 10 7 7 99

[Ne]3s23p6 Ar 18 10 8 8 98

[Ar]4s1 K 19 18 1 1 227

[Ar]4s2 Ca 20 18 2 2 197

[Ar]4s23d1 Sc 21 18 3 3 135

Zeff = Z – core e-****

The effective nuclear charge (Zeff) (“pull” on valence electrons) increases across a period and upward in a group! Know this trend and others follow!

Zeff Increases

incr

easi

ng Z

eff

Atomic radii decrease across a Period because the effective nuclear increases. Down a group the atomic radi get larger!

Incr

easi

ng A

tom

ic R

adiu

s Decreasing Atomic Radius

Increasing n

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

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

SOLUTION:(a) Sr > Ca > Mg These elements are in Group 2A(2).

(b) K > Ca > Ga These elements are in Period 4.

(c) Rb > Br > KrRb has a higher n engery 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.

The atomic radii of cations are smaller than their ground state atoms, while anions are larger than their ground state atom (i.e. remove e- => smaller, add e- => bigger) Greater cation charge => smaller & vis versa

Cations get smaller (greater Zeff)

Anions get larger (lower Zeff)

Ranking Ions by Size

PLAN:

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.

Ranking Ions by SizeRank each set of ions in order of decreasing size, and explain your ranking:

(a) Ca2+, Sr2+, Mg2+

(b) K+, S2!, Cl!

Sr2+ > Ca2+ > Mg2+

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+ Au+ > Au3+

The higher the + charge, the smaller the ion.

Ionization energy is the minimum energy (kJ/mol) required to remove 1 mole of e- from one mole of a gaseous atom in its ground state.

Electron affinity is the energy required to add (reduce) 1 mole of e- to an atom in the gas state to form an anion. It’s a measure of an atom’s ability to “accept” an e-.

higher nuclear charge => smaller diameter => harder to remove e-

First ionization energies of the main-group elements.“How hard it is to remove an electron”

Electron affinity is largest (most negative) for chlorine and fluorine (i.e like to gain electroncs = reduced).

Easily oxidizedmetal

The increasing effective nuclear Charge (Zeff) and it’s impact on atomic radius can help us understand the trend in ionization energies of elements.

higher nuclear charge => smaller diameter => harder to remove e-

We can remove more than 1 electron from a ground state atom. It requires more energy to remove subsequent electrons.

I1 + X (g) X+(g) + e-

I2 + X (g) X2+(g) + e-

I3 + X (g) X3+(g) + e-

I1 first ionization energy

I2 second ionization energy

I3 third ionization energy

I1 < I2 < I31s2 2s2 2p3

1s2 2s2 2p1

1s2 2s2

1s2 2s1

1s2 2s2 2p2

1s2 2s2 2p4

The ionization energy increases dramatically when an core electron is removed from a non-valence shell (blue area shows big jumps in IE).

Ranking Elements by First Ionization EnergyUsing the periodic table to rank the elements in each set in order of decreasing IE1:

(a) Kr, He, Ar

(b) Sb, Te, Sn

(c) K, Ca, Rb

(d) I, Xe, Cs

Ranking Elements by First Ionization EnergyUsing the periodic table to rank the elements in each set in order of decreasing IE1:

(a) Kr, He, Ar

(b) Sb, Te, Sn

(c) K, Ca, Rb

(d) I, Xe, Cs

He > Ar > Kr

Te > Sb > Sn

Ca > K > Rb

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.

Identifying an Element from Successive Ionization Energies

PLAN:

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 number valence electrons is reflected in the periodic table for Group A elements....find the group with that number of valence electrons.

IE6

22,230

SOLUTION: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 electron configuration is: 1s22s22p63s23p3.

Identifying an Element from Successive Ionization EnergiesName the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration:

IE1 IE2 IE3 IE4 IE5

1012 1903 2910 4956 6278

Metallic behavior increases as we move down a group and from left to right on the periodic table.

--Metals have low IE

-- Tend to be oxidized to

metal ions.

--Charge is group numbers

Most metallic elements

Main Group (or representative) metals form ionic basic oxides when reacted with oxygen while non-metals form covalent acidic oxides with oxygen.

Increasing Acidity

Li2O BeO B2O3 CO2 OF2

K2O

Cs2O

Rb2O

CaO

SrO

BaO

Ga2O3

Tl2O3

In2O3

GeO2

PbO2

SnO2

SeO3

TeO3

Br2O7

I2O7

6A(16)

7A(17)

4A(14)

3A2A1A

2

4

5

6

Increasing Basicity Na2O

IonicOxides

CovalentOxides

basic acidic

Properties of Oxides Across a Period