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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / A. Transition Metals
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS
A. Transition Metals
1. Electron Configuration
a) General trend
The principal quantum number of the d orbitals is one less than the principal quantum number
of the s orbital of the period. As you move from left to right in the first and second row of the
transition metals, the general trend is the s orbital is filled first with 2 electrons, and then the d
orbitals are filled.
e.g. 23V is the third transition element in period 4, the last part of its electron configuration is
...4s23d3.
b) Half-filled stabilization and filled stabilization
There are a few exceptions to the general trend above. Due to the closeness in energy of the
ns and n-1d orbitals, it is energetically more favorable when the d orbitals are half-filled or
completely filled and the ns orbital has one or no electrons.
e.g.
should be ... 4s23d4, but is actually ... 4s13d5
2
9
1
10
29Cu should be ... 4s 3d , but is actually ... 4s 3d
2
4
1
5
42Mo should be ... 5s 4d , but is actually ... 5s 3d
2
8
10
46Pd should be ... 5s 4d , but is actually ... 4d
24Cr
There are other exceptions as well. Your instructor will tell you which ones you should know.
2. Transition Metal Complexes
a) Transition metals and transition metal ions are Lewis acids (i.e. electron pair
acceptors).
i) Molecules or negative ions with an unshared pair(s) of electrons are Lewis bases
(i.e. electron pair donors).
ii) Transition Metal Complex: consist of a central transition metal or ion
surrounded by 2, 4 or 6 electron pair donors.
iii) The electron pair donors are called ligands.
iv) The ligands form coordinate covalent bonds with the metal or the metal ion.
Example(1): Ag+ forms strong coordinate covalent bonds with 2 NH3 molecule: [Ag(NH3)2] +
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / A. Transition Metals
v) No ion – simple, polyatomic, or complex – can exist without a counter balancing
charged partner ion.
vi) Together the complex ion and its partner make a coordination compound.
Example(2): If the negative partner ion in the previous example is Cl‾, the coordination compound is
[Ag(NH3)2]Cl.
b) Some common ligands
Neutral: H2O, NH3, CO, NH2CH2CH2NH2, (ethylenediamine – en)
Charged: F‾, Cl‾, OH‾, CN‾, SCN‾, EDTA
c) Determining the charge on the transition metal.
Example(3): Determine the charge on the transition metal in the coordination compounds below:
[Ag(NH3)2]Cl
[Co(CN)3(H2O)3]Cl
[Pt(NH3)4(OH)2]SO4
Example(4): Determine the charge on the transition metal in the complex ions below:
[Fe(CN)6]‾4
[Ru(Br)2(H2O)4]+
[Pt(Cl)(NH3)3]+
d) Structure of transition metal complexes
The geometric structure of a of transition metal complex depends on both the metal or metal
ion and the ligand(s). The most common structures are:
i) six ligands – octahedral
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / A. Transition Metals
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ii) four ligands – square planar
iii) four ligands – tetrahedral
e) Some special ligands
i) ethylenediamine (en) NH2CH2CH2NH2
Since en has two N atoms, each with a lone pair of electrons, it is able to bond to the transition
metal at two sites. It is classified as a bidentate ( two-toothed) ligand.
ii) diethylenetriamine (dien) NH2CH2CH2NHCH2CH2NH2
Since dien has three N atoms, each with a lone pair of electrons, it is able to bond to the
transition metal at three sites. It is classified as a tridentate ( three-toothed) ligand.
iii) EDTA (ethylenediaminetetraacetate ion) is a hexadentate (six-toothed) ligand.
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / A. Transition Metals
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f) Isomers
Like organic compounds, transition metal complexes can have many isomers if the ligands are
not all identical. e.g.
g) Colors
Almost all complexes are highly colored. The color is due to the absorption of one color of light
from white light, the rest of the light is not absorbed. The unabsorbed light appears not as
white any longer, but as the complementary color of the absorbed color.
The light absorbed causes an electron to make a transition to a slightly higher energy level.
The d orbitals of an atom or ion usually all have the same energy. However, in a complex,
some of the d orbitals point directly at the ligands, and some point between the ligands. Those
pointing at the ligands have a higher energy than those pointing between the ligands. Thus,
there is a splitting in the energy of the d orbitals. The color is the result of an electron
absorbing light and going from a low energy d orbital to a higher energy d orbital.
JUST FOR FUN. There are many biochemical complexes. The Fe+2 in hemoglobin is
complexed on five sides with the part of the molecule called the heme. The sixth side bonds to
the O2 molecule on its way to the cells, and with the CO2 molecule on its way back to the
lungs. CO is a much stronger ligand than O2 or CO2. Thus if you inhale CO, it complexes with
the iron, and will not let go. So, the iron can no longer carry either O2 or CO2.
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group VII–Halogens
B. Group VII, the Halogens
1. Physical Properties
a) Formula
F2
Cl2
Br2
I2
b) Physical
c) Melting
d) Boiling
e) Bond Energy
Phase @ RT Point ºC
Point ºC
kJ/mol
gas
gas
liquid
solid
–220
–101
–7
114
–188
–34
59
185
158
243
193
151
Example(1): What intermolecular force accounts for the variation in the melting and boiling points of
the halogens.
2. Oxidizing and Reducing Agent Strength
F2(aq) + 2e ‫ ⇄־‬2F‾(aq)
+2.87
Cl2(aq) + 2e‫ ⇄ ־‬2Cl‾(aq)
+1.36
Br2(aq) + 2e‫ ⇄ ־‬2Br‾(aq)
+1.07
I2(aq)
+ 2e‫ ⇄ ־‬2I‾(aq)
+0.54
Example(2): Which is the better oxidizing agent Cl2 or Br2?
Example(3): Which is the better reducing agent, Br‾ or I‾?
3. Preparation of the Halogens
None of the halogens are found in their elemental form in nature. This is due to their high
reactivity. In general the halogens are prepared by oxidation of the anions.
a) Chemical oxidation.
Any oxidizing agent above the halogen in the Electrochemical Series (table of reduction
potentials) can oxidize the halide ion to its elemental form.
Example(4): Which oxidizing agent cannot produce Br2 from Br‾: Cl2, MnO4‾(acid solution), or Cu+2?
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group VII–Halogens
Example(5): F2 cannot be produced by chemical oxidation. Explain why.
b) Electrolysis
All of the halogens can be produced by the electrolysis of the anions. Recall (from your study
of electrochemistry) that molten halide salts can be electrolyzed. However, not all aqueous
solutions of halide salts can be electrolyzed.
Example(6): Under standard conditions, which halogen(s) cannot be produced by the electrolysis of an
aqueous solution of the halide ion?
4. Reaction with Main Group Metals
a) Group I: 2M + X2 → 2MX
Example(7): Write the balanced equation showing the reaction of K with Br2.
b) Group II: M + X2 → MX2
Example(8): Write the balanced equation showing the reaction of Mg with Cl2.
c) Group III: Al
Ga
In
Tl
+
+
+
+
X2
X2
X2
X2
→
→
→
→
AlX3, Al2X6
GaX3, Ga2X6
InX3, In2X6
TlX
d) Group IV Ge + X2 → GeX4
Sn + X2 → SnX4
Pb + X2 → PbX2
d) Group V Sb + X2 → SbX3 or SbX5
Bi + X2 → BiX3
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group VII–Halogens
5. Reactions With Nonmetals
a) With Group VI
i) O2 + X2 → OF2, Cl2O, Br2O
Example(9): Why is the formula of the oxygen–fluorine compound written as OF2 and not F2O?
Example(10): Why doesn’t the compound I2O form? (Hind: it has to do with size.)
ii) S + F2 → SF2, SF4, SF6, S2F2
S + Cl2 → SCl2, SCl4, S2Cl2
S + Br2 → S2Br2
Example(11): Do the dot formula of S2X2. (Hint: it is like a peroxide, the 2 S atoms are bonded with a
single bond.)
iii) Se and Te: form compounds similar to those of S with X2.
b) With Group V
i) N2 + X2 → NX3
ii) P + X2 → PX3, PX5
iii) As + X2 → AsX3, AsX5
6. Hydrogen Halides
a) Synthesis by direct reaction with H2
F2 + H2 → HF
Cl2 + H2 → HCl
100%, Explosive
100%
Br2 + H2 ⇄ HBr
~50%
I2
+ H2 ⇄ HI
~10%
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group VII–Halogens
b) Synthesis by halide salt plus an acid at a high temperature
CaF + H2SO4 → 2HF(g) + CaSO4(s)
NaF + H2SO4 → HF(g) + NaHSO4
H2SO4 is not used to make HBr or HI, since H2SO4 is an oxidizing agent and will oxidize the
ions at high temperatures. H3PO4 is used with Br‾ and I‾.
NaBr + H3PO4 → HBr + NaH2PO4
Example(12): Write the molecular equation showing the reaction between NaI and H3PO4
c) Properties of the hydrogen halides
HF
HCl
HBr
HI
Boiling Point
°C
20
–85
–67
–35
Acid Strength
weak
strong
stronger
strongest
Bond Energy
kJ/mol
565
421
366
299
Example(13): What causes the boiling point of HF to be so much higher than the other hydrogen
halides?
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group VI
C. Group VI
1. Physical Properties
a) Formula
b) Physical
c) Melting
d) Boiling
e)Classification
Most Common
Phase @ RT Point ºC
Point ºC
O2
S8
Se8, Sen
Ten
gas
solid
solid
solid
–210
115
220
450
–183
445
685
990
f) Allotropes: are different formulas for the same element
molecules of an element in a solid).
Nonmetal
Nonmetal
Nonmetal
Metalloid
(or different ways of packing the
i) Oxygen: O2 is diatomic oxygen, O3 is ozone.
Example(1): a) Do the dot formula of ozone.
b) Show its 2 resonance structure.
c) What is the O to O bond order?
ii) Sulfur: S8 ring, and S8 open, and S2
Example(2): a) Do the dot formula of S2
b) Why is S2 much less stable than O2?
2. Occurrence
The Group VI elements all occur free in nature, as well as in ionic and covalent compounds.
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group VI
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Example(3): Oxygen is the second most electronegative atom, but at room temperature it is almost nonreactive. What explains its low reactivity at room temperature?
3. Metal Oxides
a) Metal oxides that are soluble in water produce basic solutions. The oxide ion acts
as a very strong proton acceptor. All group I oxides are soluble, and in group II the oxides of
Ca, Sr, and Ba are soluble. The metal oxide is referred to as the anhydride of the base that it
forms in water.
Example(4): Write a molecular, ionic and net ionic equation showing the reaction of K2O and BaO with
water.
b) Insoluble metal oxides of any group will react with a strong acid to produce water.
Example(5): Write the balanced molecular equation showing the complete reaction of Al2O3 with HCl.
4. Nonmetal Oxides
a) Most nonmetal oxides react with water to from an acidic solution. The nonmetal
oxide is referred to as the anhydride of the acid that it forms in water.
SO2 + H2O ⇄ (H2SO3) ⇄ H+ + HSO3ˉ
N2O5 + H2O ⇄ (2HNO3) → 2H+ + 2NO3ˉ
SO3 + H2O ⇄ (H2SO4) → H+ + HSO4ˉ
Example(6): Write the balanced molecular equation showing the reaction of P4O10 with H2O to form
H3PO4.
Example(7): What is the anhydride of H2CO3?
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group V
D. Group V
1. Physical Properties
a) Formula
b) Physical
c) Melting
d) Boiling
e)Classification
Most Common
Phase @ RT Point ºC
Point ºC
N2
P4
As4, Asn
Sb4, Sbn
Bi
gas
solid
solid
solid
solid
–210
44
613
631
271
–196
280
(sublimes)
1750
1650
Nonmetal
Nonmetal
Metalloid
Metaloid
Metal
Example(1): What accounts for the extraordinarily low chemical reactivity of N2 at room temperature?
Example(2): a) What is the bonding arrangement in P4?
b) What is its geometric shape?
c) What accounts for the extraordinarily high chemical reactivity of P4?
Example(3): Why does the metallic character increase going down a group?
2. Oxides of Nitrogen
a) Neutral compounds
N2O
NO
NO2
N2O4
N2O3
N2O5
dinitrogen oxide (nitrous oxide), laughing gas
nitrogen oxide (nitric oxide), a free radical
nitrogen dioxide, a free radical, poisonous
dinitrogen tetroxide, a dimer of NO2
dinitrogen trioxide
dinitrogen pentoxide, decomposes to NO2 and O2
Example(4): Determine the oxidation state of each N atom in the above compounds.
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group V
Example(5): a)Do the dot formula of NO2.
b) Using the above dot formula, show how NO2 form the dimer N2O4.
b) Polyatomic ions
NO2‾ and NO3‾
Example(6): Name the two polyatomic ions of nitrogen.
c) Associated acids
HNO2 and HNO3
Example(7): What are the anhydrides of HNO2 and HNO3?
3. Oxides of Phosphorus
a) Neutral compounds
P4O6 and P4O10
b) Polyatomic ions
PO4‾3, HPO4‾2, H2PO4‾1
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VIII. DESCRIPTIVE CHEMISTRY OF SELECT ELEMENTS / B. Group V
c) Associated acid
H3PO4
d) Dehydration between phosphoric acid molecules
Example(8): Below is the structure of phosphoric acid. Draw the structure formed by the dehydration
between two phosphoric acid molecules. (Diphosphoric acid.)
O
║
H―O―P―O―H
|
O
|
H
Example(9): Draw the structure of triphosphoric acid.
JUST FOR FUN
ATP ( adenosine triphosphate) is a molecule which is able to transfer energy directly to
reactant molecules during chemical reactions within a cell. The energy is released when ATP
is hydrolyzed to ADP (adenosine diphosphate).
ATP + H2O
ADP + H3PO4 ∆G° = –30kJ/mol
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