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Chemistry for Engineering
TRENDS IN THE PERIODIC TABLE
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LEARNING OUTCOMES
By the end of this session, the student should be able to:
1. Identify the trends of properties of elements in theperiodic table.
2. Arrange different elements according to the size of theiratomic and ionic radii.
3. Arrange elements according to their ionization energies.
4. Arrange elements according to their electronegativity.
5. Determine the strength and type of a bond.
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Trends in
periodictable
Atomic radius
Ionic radius
Ionization energyelectronegativity
Type of bond
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THE ATOMIC RADIUS
The electron cloud surrounding the nucleus
does not have a sharp boundary.
The atomic size is considered as the volume
containing about 90 % of the total electrondensity around the nucleus.
The atomic radius is one-half the distance
between the two nuclei in two adjacent metal
atoms or in a diatomic molecule.
Examples: For Cuoin copper metal, r=0.128 nm
For Cl2, r= 0.099 nm.
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ATOMIC RADII (in pm) OF REPRESENTATIVE ELEMENTS
(same trend for ionic radii).
1. In the same period, the effective nuclear charge increases from left to right the
added valence electron at each step is more strongly attracted by the nucleus than the
one before, so the radii decrease from left to right.
2. In the same group, the size increases with increasing number of electron orbtits
radius increases from top to bottom.
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ATOMION
cation
Removing 1 or more e-from anatom reduces electron-electron
repulsion butthe nuclear charge(p+) remains the same, so the e-
cloud shrinks
the cation is smaller than its
corresponding atom.
Na+(0.095 nm) vs. Nao (0.186 nm)
anion
Gaining 1 or more e-
increases the radius, because
the nuclear charge (p+)
remains the same but the
repulsion resulting from the
additional e
-
enlarges thedomain of the electron cloud.
The anion is larger than its
corresponding atom
Cl-(0.181 nm) vs , Clo(0.099nm)
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Arrange the following sets of atoms and ions in order ofincreasing size using the periodic table
1. Mg, Al, Ca.
2. S, Cl, S2-
3. Fe, Fe2+
, Fe3+
1. Comparison with reference to Mg:Al is to the right to Mg, thus smaller
than Mg. Ca is below Mg, thus larger. The order is Al < Mg < Ca.
2. Comparison with reference to S atom: Cl is to the right to S, thus is
smaller. The S2-
anion is larger than the S atom.
The order is Cl
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IONIZATION ENERGY (IE)
Ionization energy is the minimum
energy (in kJ/mol) required to removean electron from a gaseous atom in its
ground state.
X + energy X++ 1 e-
The 1st ionization energy (endothermic
process = requires E) is the E change for
the removal of the outermost e- from a
gaseous atom to form a +1 ion. The
more difficult it is to remove e-, the
larger the ionization energy.
2ndand 3rdIE (to remove a 2ndand a 3rd
e-) are certainly more difficult due to
the increasing attraction exerted by the
nucleus at every loss of 1 e- after the
other. 819-10-2014
Why IE decreases as we go
down a group?
A greater separation
between the e- and the
nucleus = weaker attraction.
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ELECTRON DISTRIBUTION IN AN ATOM
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nucleus
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THE ELECTRONIC CONFIGURATION OF
ELEMENTS
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(only outer shell level of electrons is represented)
THE CHEMICAL PROPERTIES OF ANY ATOM
ARE DETERMINED BY THE CONFIGURATION
OF THE ATOMS VALENCE ELECTRONS.
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EXCEPTIONS TO TREND OF IE (1stexception)
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between Group 2A and 3A elements in the same period
Ex: Be vs. B and Mg vs. Al.
IE of 3A < 2A in the same
period.
because they all have a single e-
in the outermost p subshell ( ns2
np1 ), which is well shielded by
the inner e-and the ns2e-.
less energy is needed to
remove a single p e- than to
remove an s e- from the sameenergy level.
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EXCEPTIONS TO TREND OF IE (2ndexception)
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between Groups 5A and 6A elements in the same period
ex: N vs. O and P vs. S
IE of 6A < 5A in the same period.
In 5A ( ns2 np3 ), the p e- are in 3
separate orbitals while in 6A ( ns2 np4),
the additional e- is paired with 1 of the 3p e- The proximity of two e- in the
same orbital leads to high electrostatic
repulsion, which makes it easier to
ionize an atom of the Group 6A element,
even though the nuclear charge hasincreased by one unit.
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Complete:
1. Ionization energy (IE) a.across the periodic tablefrom left to right and b.. moving down the periodic
table.
2. Comparing the trends of ionic radii and ionization energy it is
clear that there is an
c
. correlation between them.3. The d.. the atom, the more tightly its electrons are
held to the positively charged nucleus and the more difficult
they are to remove thus the e.. the ionization energy.
4. In a
f
atom (ex: group I, period 5), the electron isrelatively far from the nucleus, so the lower its g..
to remove it from the atom.
Answers:
a) increases, b) decreases, c) inverse, d) smaller, e) larger, f) large, g) ionization.
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Test yourself
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Which atom with following electron configurations has the
largest first ionization energy, explain your choice.
1s22s22p6 1s22s22p63s1 1s22s22p63s2
The atom with the largest value of IE is 1s22s22p6 (this is a Ne),because it is found at the right end of period 2, and the IEincreases from left to right in the same period of the periodictable.
As for the other 2 configurations including 3s electrons they willbe of lower IE because it is easier to loose e- from an incompleteshell.
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Test yourself
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ELECTRONEGATIVITY (ELECTRON AFFINITY)
Electron affinity is the energy change that occurs when an e- is accepted by
an atom in the gaseous state to form an anion.X(g)+ e
- X-(g)Electronegativity increases from left to right in the periodic table and vary
little within a given group. The halogens (Group 7A) have the highest
electronegativity.
In a covalent bond, the more electronegative the atom, the more it attracts
the shared e- towards it resulting in unfair sharing polarizationof the bond
with the most electronegative acquiring apartial negative charge -- vs. a +
on the least electronegative.
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IONIC BOND by full e- transfer
from the least to the most
electronegative (i.e. large
difference in electronegativity)
COVALENT BOND between
identical atoms, i.e. no
difference in electronegativity
(NON POLARIZED)COVALENT BOND (POLARIZED,
i.e. some difference inelectronegativity)
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Order the following bonds according to bond polarity
H-H, S-H, Cl-H, O-H, and F-H
(2.1) (2.1) (2.5)(2.1) (3.0)(2.1) (3.5)(2.1) (4.0)(2.1)
0 0.4 0.9 1.4 1.9
Thus the order is
H-H < S-H < Cl-H < O-H < F-HNon polar covalent bondPolar covalent bond
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Test yourself
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REFERENCES
1. Chemistry, 10thed., Raymond Chang, ISBN 978-0-
07-017264-7, McGraw Hill. Chapter 8.
2. Lecture 2 by Prof. Rasha Elnashar, GUC, WS 2013.
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