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Periodic Relationships Among
the Elements
Chapter 8
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Acknowledgement
Thanks to the McGraw-Hill Companies, Inc. for allowing classroom usage.
Electron Configurations of the Elements
Valence Electrons
• the outer electrons of an atom, the ones involved in
bonding (or chemical reaction)
• for the representative elements (main group elements), the
number of valence electrons is equal to the A Group Number
• for the transition metal elements (the B group elements), the
number of valence electrons is also equal to the B Group
Number
Example:
Which of the following electron configurations do represent similar chemical
properties of their atoms?
(i) 1s22s22p63s2
(ii) 1s22s22p3
(iii) 1s22s22p63s23p64s23d104p5
(iv) 1s22s1
(v) 1s22s22p6
(vi) 1s22s22p63s23p3
Hint: Two methods to solve this question.
(1) Traditional method: Look for the valence electrons (shown in next slide).
Atoms with same number of valence electrons are in the same group and
thus will have similar chemical properties. 10th ed. Table 7.3, p. 308; 9th
ed. Table 7.3, p. 300 and p. 318: Figure 8.2.
(2) Easier method: Add total electrons. Since they are the atoms and thus the
number of protons = number of electrons. Then find the atom symbol from
periodic table to see which are in the same group (vertical column).
ns2np6
ns2np5
ns2np4
ns2np3
ns2np2
ns2np1
d10
d5
d1
ns2
ns1
Ground State Electron Configurations of the Elements
4f
5f
8.2
Electron Configurations of Cations and Anions
Of Representative Elements
Na [Ne]3s1
Na+ [Ne]
Ca [Ar]4s2
Ca2+ [Ar]
Al
[Ne]3s23p1
Al3+
Atoms gain electrons
so that anion has a
noble-gas outer
electron configuration.
[Ne]
Atoms lose electrons so that
cation has a noble-gas outer
electron configuration.
Remember: electrons are first
removed from orbitals with the
highest principal quantum number.
H 1s1
H- 1s2 or [He]
F 1s22s22p5
F- 1s22s22p6 or [Ne]
O 1s22s22p4
O2- 1s22s22p6 or [Ne]
N 1s22s22p3
N3- 1s22s22p6 or [Ne]
8.2
-1
-2
-3
+3
+2
+1
Cations and Anions Of Representative Elements
8.2
Na+: [Ne]
Al3+: [Ne]
O2-: 1s22s22p6 or [Ne]
F-: 1s22s22p6 or [Ne]
N3-: 1s22s22p6 or [Ne]
Na+, Mg2+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne
Term: Isoelectronic: have the same number of electrons
Li+, H-: 1s2-
same electron configuration as He
8.2
Example:
1. Which of the following pairs are not isoelectronic?
(a)K+ and Cl(b) Ca2+ and S2(b) Al3+and N3(d) P3- and Sc3+
(c)Rh3+ and Ir3+
Answer is (c) because Rh and Ir are in the same group.
Isoelectronic species can not be from the same group.
2. Which of the following pairs are not isoelectronic with
Ar? Use the choices above.
Hint: This question asks two concepts: (1) isoelectronic
and (2) number of electrons in Ar. So the answers are (b)
and (c).
Electron Configurations of Cations of Transition Metals
When a cation is formed from an atom of a transition metal,
electrons are always removed first from the ns orbital and
then from the (n – 1)d orbitals.
Fe:
[Ar]4s23d6
Fe2+: [Ar]4s03d6 or [Ar]3d6
Mn:
[Ar]4s23d5
Mn2+: [Ar]4s03d5 or [Ar]3d5
Fe3+: [Ar]4s03d5 or [Ar]3d5
8.2
Example:
What is the ion with +3 charges that has the electron configuration as
[Ar]3d3?
(a) Sc 3+
(d) Rh 3+
(b) Cr 3+
(e) Ir 3+
(c) Co 3+
Hint: First, figure out the atom symbol for its atomic type. Since the cation
has 18+3 = 21 electrons and thus there are 21+3 = 24 protons in this
cation. Second, find the atom symbol from the periodic table, which
indicates that the atom is Cr and thus the +3 cation is Cr 3+.
Example:
A metal ion with a net +3 charge has five electrons in the 3d
subshell. What is this metal?
(a) Cr (b) Mn
(c) Fe (d) Co
(e) Ni
Hint: Similar to the previous question. In order to have electrons
in 3d, this ion must be from row number IV and is a transitional
metal ion. This species has +3 charges, which indicates that it
has three more protons than the electrons. According to the
question that it has five electrons in the 3d subshell, and thus
the total electrons in valence shells for its atomic type will be 8.
Note that the transition metals (with d electrons) losing its s
electrons prior to its d electrons and thus the valence electron
configuration will be 4s23d6, which is Fe.
Periodic Variation in Physical Properties
8.3
Sizes of atoms: Trends
• size increases while going down a Group
• Why?
• Because orbital size increases with increasing n
number, the size of atom increases. The highest energy
electrons can be farther away from the nucleus.
• size decreases going across Period
• Why?
• effective nuclear charge increases. The larger the
effective nuclear charge, the stronger hold of the nucleus
on electrons. The electron clouds shrink.
• Remember n does NOT increases--e- cannot be farther
from nucleus.
Ionic Radius
Cation is always smaller than atom from which it is formed. This is
because the nuclear charge remains the same but the reduced electron
repulsion resulting from removal of electrons make the electron clouds
shrink.
Anion is always larger than atom from which it is formed. This is because
the nuclear charge remains the same but electron repulsion resulting from
the additional electron enlarges the electron clouds.
8.3
•From top to bottom, both the atomic and ionic radius increase within
group.
•Across a period the anions are usually larger than cations.
• For ions derived from different groups, size comparison is meaningful
only if the ions are isoelectronic.
E.g. Na+ (Z=11) is smaller than F- (Z=9).
**For isoelectronic series (or species), the larger the effective nuclear
charge results in a smaller radius. Simply, the more positively charged
ion, smaller the size; the more negatively charged ion, the larger the
size.
Radius of tripositive ions < dipositive ions<unipositive ions
Al3+ <Mg2+ <Na+
Radius of uninegative ions < dinegative ions
O2- > F- (oxide ion is larger than fluoride ion)
Example:
Which of the following is a correct order of atomic radii?
(a) Ar < P < Na < Ca < Cs (b) Cs < Rb < K < Na < Li
(c) Na < Mg < Al < Si < P (d) Rb < Mg < C < F < He
(e) Li < H < Al < K < Ar
Answer: (a)
Example:
Which of the following is a correct order of ionic radii?
(a)Na+ < Mg2+ < Al3+ < O2- < F- < N3(b) Al3+ < Mg2+ < Na+ < F- < O2- < N3(c) F- < O2- < N3- < Na+ < Mg2+ < Al3+
(d) Al3+ < Mg2+ < Na+ < N3- < O2- < F(e) Na+ < Mg2+ < Al3+ < O2- < N3- < FAnswer: (b)
Ionization energy (I.E.) is the minimum energy (kJ/mol)
required to remove an electron from a gaseous atom in its
ground state.
I1 + X (g)
I2 + X+(g)
I3 +
X2+
(g)
X+(g) + e-
I1 first ionization energy, remove 1st electron
X2+(g) + e- I2 second ionization energy, remove 2nd electro
X3+(g)
+
rd electron
I
third
ionization
energy,
remove
3
3
e
I1 < I2 < I3
8.4
General Trend in First Ionization Energies
Increasing First Ionization Energy
Increasing First Ionization Energy
8.4
First IE:
Trends down a Group
• remember that Cs is more reactive than Na
• Easier to remove an e- from Cs than from Na
• As size increases, I.E. decreases
Trends across a Period
• remember that Cl gains rather than loses an e• Easier to remove an e- from Al than from Cl
• As size decreases, I.E. increases
Exceptions: Ionization Energy
1.Occur between Group 2A and 3A
Group 3A: ns2np1. one single electron in p orbital.
• removing the first p electron is less than expected
•Why? This p electron is shielded by inner ns2 electrons.
Less energy is needed to remove a single p electron than
to remove a pair of s electron of the same n level.
2. Occur Group 5A and 6A
5A: ns2np3 6A: ns2np4
• removing the fourth p electron is less than expected
•Why? 2nd electron in a p orbital increases the
electron-electron repulsion, which makes it easier to
ionize an atom of Group 6A.
Example:
Arrange the following in order of the increasing
first ionization energy: Na, Cl, Al, S, Cs
Hint: Ionization energy increases across a row of
the periodic table and decreases down a column
or group.
Cs < Na < Al < S < Cl
Electron affinity (EA) is the negative of the energy change
that occurs when an electron is accepted by an atom in the
gaseous state to form an anion.
X (g) + e-
X-(g)
F (g) + e-
X-(g)
DH = -328 kJ/mol
EA = +328 kJ/mol
O (g) + e-
O-(g)
DH = -141 kJ/mol
EA = +141 kJ/mol
8.5
Variation of Electron Affinity With Atomic Number (H – Ba)
Overall trend:
Becomes more negative
from left to right across the
period, and from bottom to
top in the group.
The values varied little in
the group. The EA of
metals are generally lower
than those of nonmetals.
8.5
Properties of Oxides Across a Period
basic
acidic
Across a period, oxides change from basic to amphoteric to
acidic.
Going down a group, the oxides become more basic.
8.6
Generally speaking, metal oxides and hydroxides are
basic; while nonmetal oxides are acidic.