4.05 atomic structure and electronic...

4.05 Atomic Structure and Electronic Configuration Aug ‘171

4.05 Atomic Structure and Electronic Configuration

Dr. F. O. GarcesChemistry 100Miramar College

Learning Objectives/Terminology-AufbauHund’s RulePauli Exclusion PrincipleDiamagnetism/ParamagnetismNe-Va-S-P

Arranging the electrons in an atom


Electronic ConfigurationHow are the electrons of an atom arranged in the atom?

What are shells and orbitals arrange outside the nuclei of an atom?

How is the e- arrangement liken to that of a Hotel room (Hotel del Orbital) ?

What is the importance of the valence electrons and how do these influence chemistry?


Prelude to eConfigurationPlacing electrons in orbitals to complete the electron configuration.


Shells and Orbitals


Relative Energies for Shells and Orbitals

1

2

3

4

567∞8

s p d f

Relative Energies of the orbitals


Hotel del Orbitals

s p d f

1

2

3

4

Filling Order (Aufbau Principle)

Like the hotel del Orbital, the shells of an atom are staggered and are filled based on their relative energies


Hotel del Orbitals and a pictorial view of the shell arrangements

The filling order can be memorized by the following scheme

1

2

3

4


2s• H = 1s1

• He = 1s2

• Li = 1s2 2s1

• Be = 1s2 2s2

• B = 1s2 2s2 2p1

• C = 1s2 2s2 2p2

• N = 1s2 2s2 2p3

• O = 1s2 2s2 2p4

• F = 1s2 2s2 2p5

• Ne = 1s2 2s2 2p6

1s

The First 10 elements:

2p

+ +


Electron ConfigurationHydrogen and Helium

H

HeHe = 1s2

H = 1s1

1s 2s 2p

1s 2s 2p

Pauli Exclusion Principle



1

2

3

4

567∞8

s p d f



Electron Configuration:Lithium to Beryllium

Be

+ +

=

Li+ +

=

Li = 1s2 2s1

Be = 1s2 2s21s 2s 2p

1s 2s 2p

Aufbau Process



1

2

3

4

567∞8

s p d f



Electron Configuration: Boron to Nitrogen

N

+ +

=

C

+ +

=

B

+ +

=

B = 1s2 2s2 2p1

N = 1s2 2s2 2p3

1s 2s 2p

1s 2s 2p

1s 2s 2p

C = 1s2 2s2 2p2 Hund’s Rule

Paramagnetic



1

2

3

4

567∞8

s p d f



Electron Config: Oxygen to Neon

O

+ +

=

F

+ +

=

Ne

+ +

=

F = 1s2 2s2 2p5

Ne = 1s2 2s2 2p6

1s 2s 2p

1s 2s 2p

1s 2s 2p

O = 1s2 2s2 2p4

Diamagnetic



1

2

3

4

567∞8

s p d f



Electron Configuration using the Periodic Table

When using the Periodic table to determine the electron

configuration of an atom, it is important to understand the

layout of the periodic table.

The row of a periodic table can be used to determine which

energy level, 1, 2, 3... the valence electrons are located. The

column of the periodic table helps determine how many valence

electron an atom possesses. Elements in the 1A column (family)

have one valence electron, in the IIA column, elements have two

valence electron, in the IIIA family, elements have three

valence electrons, and so on.


The elements designated in -block is where the very last electrons are found in the s-orbitals. Therefore the elements in portion of the periodic table is referred to as the s-block elements.

The elements designated in -block is where the very last electrons are found in the p-orbitals. The elements found in this portion of the periodic table is referred to as the p-block elements.

The same can be said about the -block and -block with the elements found in this portion of the periodic table referred to as the d-block (transition metals) and f-block (man-made) elements.

Electron Configuration using the Periodic Table


Electron Configuration: ne-va-s-p

e- config. for Sulfur ?

1IA

18VIIIA

1 2IIA

13IIIA

14IVA

15VA

16VIA

17VIIA

2

3 3IIIB

4IVB

5VB

6VIB

7VIIB

8 9VIIIB

10 11IB

12IIB

4

5

6

7

H1s1

Li2s1Na3s1

K4s1

Rb5s1

Cs6s1

Fr7s1

Be2s2

Mg3s2

Ca4s2

Sr5s2

Ba6s2

Ra7s2

Sc3d1

Ti3d2

V3d3

Cr4s13d5

Mn3d5

Fe3d6

Co3d7

Ni3d8

Zn3d10

Cu4s13d10

B2p1

C2p2

N2p3

O2p4

F2p5

Ne2p6

He1s2

Al3p1

Ga4p1

In5p1

Tl6p1

Si3p2

Ge4p2

Sn5p2

Pb6p2

P3p3

As4p3

Sb5p3

Bi6p3

S3p4

Se4p4

Te5p4

Po6p4

Cl3p5

Be4p5

I5p5

At6p5

Ar3p6

Kr4p6

Xe5p6

Rn6p6

Y4d1

La5d1

Ac6d1

Cd4d10

Hg5d10

Ag5s14d10

Au6s15d10

Zr4d2

Hf5d2

Db6d2

Nb4d3

Ta5d3

Jl6d3

Mo5s14d5

W6s15d5

Rf7s16d5

Tc4d5

Re5d5

Bh6d5

Ru4d6

Os5d6

Hn6d6

Rh4d7

Ir5d7

Mt6d7

Ni4d8

Ni5d8

• ne - number of electrons; the total number of electrons• va - valence electrons; the number of valence electrons• s - shell of valence electrons• p - previous noble gas


Electron Configuration: Sulfur

p - previous noble gas = Ne (10 e-)

ne - number of electrons; the total number of electrons; this equals the number of protons or atomic number = 16

Va - valence electrons; the number of valence electrons = 6

s - shell of valence electrons = 3

1IA

18VIIIA

1 2IIA

13IIIA

14IVA

15VA

16VIA

17VIIA

2

3 3IIIB

4IVB

5VB

6VIB

7VIIB

8 9VIIIB

10 11IB

12IIB

4

5

6

7

H1s1

Li2s1

Na3s1

K4s1

Rb5s1

Cs6s1

Fr7s1

Be2s2

Mg3s2

Ca4s2

Sr5s2

Ba6s2

Ra7s2

Sc3d1

Ti3d2

V3d3

Cr4s13d5

Mn3d5

Fe3d6

Co3d7

Ni3d8

Zn3d10

Cu4s13d10

B2p1

C2p2

N2p3

O2p4

F2p5

Ne2p6

He1s2

Al3p1

Ga4p1

In5p1

Tl6p1

Si3p2

Ge4p2

Sn5p2

Pb6p2

P3p3

As4p3

Sb5p3

Bi6p3

S3p4

Se4p4

Te5p4

Po6p4

Cl3p5

Be4p5

I5p5

At6p5

Ar3p6

Kr4p6

Xe5p6

Rn6p6

Y4d1

La5d1

Ac6d1

Cd4d10

Hg5d10

Ag5s14d10

Au6s15d10

Zr4d2

Hf5d2

Db6d2

Nb4d3

Ta5d3

Jl6d3

Mo5s14d5

W6s15d5

Rf7s16d5

Tc4d5

Re5d5

Bh6d5

Ru4d6

Os5d6

Hn6d6

Rh4d7

Ir5d7

Mt6d7

Ni4d8

Ni5d8

S

Ne

e- config for SulfurS = [Ne]3s23p4

• ne - number of electrons; the total number of electrons• va - valence electrons; the number of valence electrons• s - shell of valence electrons• p - previous noble gas


1

2

3

4

567∞8

Relative Energies of the orbitals and the filling order.

Electron arrangement for the Sulfur atom

The 16 electrons for sulfur occupy the shells & orbitals of sulfur from the lowest energy to the highest.

s p d f

S

16 total electrons

e- config for SulfurS = [Ne] 3s2 3p4


1

2

3

4

567∞8


Electron arrangement for the Scandium atom


s p d f

21 total electrons

e- config for ScandiumSc = [Ar]4s23d1

Sc


1

2

3

4

567∞8


Electron arrangement for the Scandium atom


s p d f

Sn

50 total electrons

e- config for TinSn = [Kr]5s24d105p2


The First 20 elements:

2s• H = 1s1

• He = 1s2

• Li = 1s2 2s1

• Be = 1s2 2s2

• B = 1s2 2s2 2p1

• C = 1s2 2s2 2p2

• N = 1s2 2s2 2p3

• O = 1s2 2s2 2p4

• F = 1s2 2s2 2p5

• Ne = 1s2 2s22p6

• Na = 1s2 2s22p63s1

• Mg = 1s2 2s22p63s2

• Al = 1s2 2s22p63s23p1

• Si = 1s2 2s22p63s23p2

• P = 1s2 2s22p63 s23p1

• S = 1s2 2s2 2p63s23p4

• Cl = 1s2 2s2 2p63s2 3p5

• Ar = 1s2 2s2 2p6 3s2 3p6

1s

2p

+ +

+ +

3p

3s


Assignment:Determine the electron configuration.

1. Write out the electron configurations for the following atoms and ions. Determine the number of unpaired electrons in the ground state.

13Al 12Mg+250Sn 15P-3

34Se-132Ge+2

2. Write the electron box diagram for the following elements. Which have identical electron configuration (isoelectronic) ?

S-2 Cl Ar Ca+2

3. What does the term: Aufbau, Pauli Exclusion and Hund’s Rule mean?

4. How does atomic radius change as one goes left to right along the periodic table? Explain*.

5. How does the ionization change as one goes down the periodic table? Explain*.

* Explain, means explain why.

4.05 atomic structure and electronic...

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