General Chemistry - Valdosta State University

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Atomic Electronic
Configurations and
Chemical Periodicity
Chapter 8
Chapter 8
1
Electron Spin
Effective Nuclear Charge
Effective nuclear charge - The charge experienced by an
electron in a many-electron atom.
Chapter 8
2
Electron Spin
Effective Nuclear Charge
- Electrons are attracted to the nucleus, but repelled by
the electrons that screen it from the nuclear charge.
- The nuclear charge experienced by an electron
depends on its distance from the nucleus and the
number of core electrons.
- As the average number of screening electrons (S)
increases, the effective nuclear charge (zeff) decreases.
- As the distance from the nucleus increases, S increases
and zeff decreases.
zeff = z - S
Chapter 8
3
Electron Spin
Pauli Exclusion Principle
Pauli’s Exclusion Principle - no two electrons can have
the same set of 4 quantum numbers.
- Therefore, two electrons in the same orbital must
have opposite spins.
-One electron has a spin of +½, the other –½.
Chapter 8
4
Electron Spin
Magnetism
Diamagnetic – Property of a substance that is repelled
by a magnetic field.
Paramagnetic – Property of a substance that is attracted
by a magnetic field.
Whether a substance is diamagnetic or paramagnetic
depends on its electron configuration.
- Atoms without “paired” spins result in a
paramagnetic substance.
Chapter 8
5
Electron Assignment
Orbitals and Quantum Numbers
- Orbitals can be ranked in terms of energy to yield an
orbital energy diagram.
- As n increases, note that the spacing between energy
levels becomes smaller.
- As shielding increases the energy of an orbital
increases.
Chapter 8
6
Electron Assignment
Chapter 8
7
Electron Configuration
Electron configurations tells us in which orbitals the
electrons for an element are located.
Three rules:
- Electrons fill orbitals starting with lowest n and
moving upwards.
- no two electrons can fill one orbital with the same
spin (Pauli Exclusion Principle).
- for degenerate orbitals, electrons fill each orbital
singly before any orbital gets a second electron
(Hund’s rule).
Chapter 8
8
Electron Configuration
Chapter 8
9
Electron Configuration
- There is a shorthand way of writing electron
configurations
- Write the core electrons corresponding to the filled
Noble gas in square brackets.
- Write the valence electrons explicitly.
Example, P: 1s22s22p63s23p3
but Ne is 1s22s22p6
Therefore, P: [Ne]3s23p3.
Chapter 8
10
Electron Configuration of Ions
Cations:
To form a cation, an electron from the highest
principle quantum number and angular momentum
quantum number is removed.
Mg: [ 1s22s22p63s2]  Mg2+: [1s22s22p6] + 2eAnions:
To form an anion, an electron is added to the
highest principle quantum number and angular
momentum quantum number.
Cl: [1s22s22p63s23p5] + 1e-  Cl-: [1s22s22p63s23p6]
Chapter 8
11
Atomic Properties and Periodic Trends
Chapter 8
12
Atomic Properties and Periodic Trends
Electron Shells in Atoms
- Elements in the same column have the same electron
configuration.
Consider: Ne: 1s2 2s22p6
Ar: 1s2 2s22p6 3s23p6
Both elements have the same electron configuration:
[Element]ns2np6
-Therefore, the elements in the periodic table should
exhibit regular variations in there physical properties.
Chapter 8
13
Atomic Properties and Periodic Trends
Atomic Size
- Atomic size varies consistently through the periodic
table.
- As we move down a group, the atoms become
larger.
- As we move across (left to right) a period, atoms
become smaller.
- There are two factors at work:
- principal quantum number, n
- the effective nuclear charge, zeff
Chapter 8
14
Atomic Properties and Periodic Trends
Atomic Size
- As the principle quantum number increases, the
distance of the outermost electron from the nucleus
becomes larger. Hence, the atomic radius increases.
- As we move across the periodic table, there is an
increased attraction between the nucleus and the
outermost electrons. This attraction causes the atomic
radius to decrease.
Chapter 8
15
Atomic Properties and Periodic Trends
Atomic Size
Chapter 8
16
Atomic Properties and Periodic Trends
Ionization Energy
First Ionization Energy - The first ionization energy, I1,
is the amount of energy required to remove an
electron from a gaseous atom:
Na(g)  Na+(g) + e- The larger ionization energy, the more difficult it is to
remove the electron.
- There is a sharp increase in ionization energy when a
core electron is removed.
Chapter 8
17
Atomic Properties and Periodic Trends
Ionization Energy
Chapter 8
18
Atomic Properties and Periodic Trends
Ionization Energy
- Ionization energy decreases down a group.
- As the atom gets bigger, it becomes easier to remove
an electron from the most spatially extended orbital.
- Ionization energy generally increases across a period.
- Two exceptions: removing the first p electron and
removing the fourth p electron have a lower
energies.
- This indicates that half-filled and completely filled
subshells are more stable.
Chapter 8
19
Atomic Properties and Periodic Trends
Ionization Energy
The s electrons are more effective at shielding than p
electrons. Therefore, forming the s2p0 becomes more
favorable.
When the p subshell has four electrons, one orbital has
two electrons. When this electron is removed, the
resulting s2p3 configuration is more stable than the
starting s2p4 configuration (the final state has a much
lower electron-electron repulsion).
Chapter 8
20
Atomic Properties and Periodic Trends
Ionization Energy
Chapter 8
21
Atomic Properties and Periodic Trends
Electron Affinity
Electron affinity – The energy required to add an
electron to an atom in the gaseous state:
Cl(g) + e-  Cl-(g)
- Electron affinity can either be exothermic (as the
above example) or endothermic.
Chapter 8
22
Atomic Properties and Periodic Trends
Electron Affinity
Chapter 8
23
Atomic Properties and Periodic Trends
Ion Sizes
- Cations are smaller than the parent atom.
- Anions are larger than the parent atom.
Chapter 8
24
Periodic Trends and Chemical Properties
- Since the representative (main group) elements have
the same valence electron configuration, their chemical
properties should be similar.
Example:
2 Li(s) + Cl2(g)  2 LiCl(s)
2 Na(s) + Cl2(g)  2 NaCl(s)
2 K(s) + Cl2(g)  2 KCl(s)
Chapter 8
25
Homework
4, 20, 22, 32, 34, 44
Chapter 8
26
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