Partial Orbital Diagram and

Condensed Configurations

Partial orbital diagram: shows only the

highest energy sublevels being filled.

Condensed electron configuration: has

the element symbol of the previous

noble gas in square brackets

1

Al (Z = 13) 1s22s22p63s23p1 ↑↓

3s 3p

↑

Al has the condensed configuration [Ne]3s23p1

Learning Check

Full Electron

Configuration

Element Condensed

Electron

Configuration

Partial

Orbital

Diagram

1s22s22p63s1 Na [Ne]3s1

1s22s22p63s2 Mg [Ne]3s2

1s22s22p63s23p4 S [Ne]3s23p4

1s22s22p63s23p5 Cl [Ne]3s23p5

2

Electron Configuration and

Group

Elements in the same

group have the same outer

electron configuration

Similar outer electron

configurations correlate

with similar chemical

behavior.

3

Period 4 Elements

Write the full and

condensed electron

configuration of

the following

elements:

◦ Ca (20)

◦ Ti (22)

◦ Mn (25)

◦ Ni (28)

◦ As (33)

◦ Kr (36)

4

Period 4 Elements

5

*Colored type indicates the sublevel to which the last electron is added.

Period 4 Elements

6

*Colored type indicates the sublevel to which the last electron is added.

Orbital Filling

and the Periodic Table

7

The order in which the orbitals are filled can be obtained directly

from the periodic table.

Categories of Electrons

Inner (core) electrons: are those an atom

has in common with the previous noble gas

and any completed transition series.

Outer electrons: are those in the highest

energy level (highest n value).

Valence electrons: are those involved in

forming compounds.

◦ For main group elements: outer electrons

◦ For transition elements: outer electrons and

any (n-1)d electrons

8

Learning Check

Give the condensed electron

configurations, partial orbital

diagrams showing valence electrons

only, and number of inner electrons

for the following elements:

◦ Potassium (K, Z=19)

◦ Technetium (Tc, Z=43)

◦ Lead (Pb, Z=82)

9

Sample Problem 8.2

SOLUTION:

(a) For K (Z = 19)

[Ar] 4s1 condensed configuration

1s22s22p63s23p64s1 full configuration

There are 18 inner electrons.

partial orbital diagram

↑

4s 4p 3d

10

Sample Problem 8.2

SOLUTION:

(b) For Tc (Z = 43)

[Kr]5s24d5 condensed configuration

1s22s22p63s23p64s23d104p65s24d5 full configuration

There are 36 inner electrons.

partial orbital diagram

↑↓

5s 5p 4d

↑ ↑ ↑ ↑ ↑

11

Sample Problem 8.2

SOLUTION:

(c) For Pb (Z = 82)

[Xe] 6s24f145d106p2 condensed configuration

1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2 full configuration

There are 78 inner electrons.

partial orbital diagram ↑↓

6s 6p

↑ ↑

12

13

Trends in the Periodic Table

1. Atomic Size

2. Ionization Energy

3. Electron Affinity

14

Defining atomic size

Atomic size: an approximation of the

space between adjacent nuclei

15 Metallic radius Covalent radius

Trends in Atomic Size

Influenced by 2 factors:

◦ Down a group, n dominates: As n increases,

the probability that the outer electrons

will be farther from the nucleus increases.

◦ Across a period, Zeff dominates: As Zeff

increases, the outer electrons are pulled

closer to the nucleus.

16

What is Zeff?

Inner core electrons “shield” outer electrons

from the positive charge of the nucleus.

Zeff (effective nuclear charge) is the force

felt by the outer valence electrons taking

into account the “shielding” effect by core

electrons.

17

18

Periodicity of Atomic Radius

19

Learning Check

Using only the periodic table, rank

the elements in each set in order of

increasing size.

◦ Se, Br, Cl

◦ I, Xe, Ba

20

Cl < Br < Se

Xe < I < Ba

Trends in Ionization Energy (IE)

IE is the energy required to remove 1

mole of electrons from 1 mole of a

gaseous atom in its ground state.

𝐼𝐸1 + 𝑋(𝑔) → 𝑋(𝑔)+ + 𝑒−

𝐼𝐸2 + 𝑋(𝑔)+ → 𝑋(𝑔)

2+ + 𝑒−

𝐼𝐸3 + 𝑋(𝑔)2+ → 𝑋(𝑔)

3+ + 𝑒−

21

IE1 < IE2 < IE3

Trends in Ionization Energy (IE)

As atomic size decreases, it takes more

energy to remove an electron.

Atoms with a low IE tend to form cations.

22

This trend is the inverse of the trend in atomic size. 23

Learning Check

Rank the elements in each of the

following sets in order of increasing

first ionization energy.

◦ Sb, Sn, I

◦ Sr, Ca, Ba

24

Sn < Sb < I

Ba < Sr < Ca

Successive Ionization Energies

The first three ionization

energies of beryllium

(Z=4).

25

Successive Ionization Energies

Drastic jump in IE occurs after outer

(valence) electrons have been

removed.

26

Identifying an Element Based on

its IE

Name the Period 3 element with the

following ionization energies (in

kJ/mol) and write its electron

configuration:

27

IE1 IE2 IE3 IE4 IE5 IE6

1012 1903 2910 4956 6278 22,230

Phosphorus, 1s22s22p63s23p3