Download Descriptive Chemistry for Midterm Exam #1

Chemistry 1051
Descriptive Chemistry
Intersession
Material on Midterm #1
Note: This handout is meant to highlight important facts and reactions. Exam
questions can be based on anything in the assigned text pages. You must read
each section carefully.
Coordination Chemistry
PHHM (9th Edition): Section 24-1 pp 1002-1004, p1007 (writing formulas, pt #6 only)
PHH (8th Edition): Section, pp 986-988, pp 991 (writing formulas, pt #6 only)
Questions:
1.
Define the following terms: coordination compound, complex ion, ligand, Lewis
acid, Lewis base, coordination sphere, coordination number, oxidation state
(number) of central atom.
2.
Distinguish between free and ligand ions in a coordination compound in terms of
their reactivity in aqueous solution. Where are free and ligand ions located in the
compound formula?
3.
Determine the number of moles of AgBr which will be formed by reaction of one
mole of the following compounds with an excess of AgNO3(aq).
(a) [CoBr(NH3)5]Br2 (b) [Co(en)3]Br3
4.
(c) [CoBr2(NH3)5]Br.
Determine the coordination number and oxidation number of the central metal
atom in each of the following:
(a) [CrBr2(NH3)4]Br (b) K4[Co(C2O4)3] (c) [Al(OH)(H2O)5]SO4 (d) [Cu(CN)4]2⎯
5.
Write the formula for
(a)
(b)
(c)
(d)
A complex ion having Ni2+ as central ion and one Cl⎯ and 5 NH3 molecules
as ligands.
A complex ion having Fe3+ as central ion and five CN⎯ and one SCN⎯ as
ligands.
A complex ion having Cr3+ as central ion and 4 NH3 molecules and two Br⎯
ions as ligands.
A complex of Fe3+ and oxalate C2O42⎯ with a coordination number of six.
6.
Write the Lewis structure for the bidentate ligands oxalate (C2O42⎯) and ethylene
diamine (H2NCH2CH2NH2). What makes these species bidentate?
7.
What kind of bond is formed when a ligand reacts with a metal cation?
Answers to selected questions above
2.
Free ions are present to balance the charge on the complex ion. They are
attracted to the complex ion by cation-anion forces and separate from the complex ion
when the compound dissolves. They are fairly reactive vs. ligand ions which remain
bonded to the metal center in solution through strong covalent bonding. Ligand ions
are found within the square brackets and free ions on the outside.
3.
4.
5.
6.
7.
(a) 2 moles (b) 3 moles (c) one mole
(a) CN = 6, XCr = +3 (b) CN = 6, XCo = +2 (c) CN = 6, XAl = +3 (d) CN = 4, XCu = +2
(a) [NiCl(NH3)5]+ (b) [Fe(CN)5SCN]3⎯ (c) [CrBr2(NH3)4]+ (d) [Fe(C2O4)3]3⎯
bidentate means 2 bonding lone pairs per ligand
a coordinate covalent bond
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Text Questions on Coordination Chemistry
PHHM (9th Edition) –
page 1004 Practice examples 24.1A, 24.2B
Chapter 24 exercises: 2 (no naming), 5, 64a, 66(a) and (b)
PHH (8th Edition) –
page 988 Practice examples 25-1A, 25-1B
page 1018 Problems 1(a), 3(a), 4(a, b) 5(a, b, c, d, f) No naming
Transition elements
d- and f - block elements
PHHM (9th Edition) Chapter 23: pp 964 -969
PHH (8th Edition) Chapter 24: pp 948-954
General properties:
high melting points
good electrical conductors
moderate to extreme hardness
Transition series
Period
Highest energy subshell
first
4
3d
second
5
4d
third
6
5d
lanthanides
6*
4f
7*
5f
actinides
*found under main body of Table
Important elements: Cr, Mn, Fe, Co, Ni, Cu, Zn, Ag, Au, Cd, Hg
Properties of the first transition series Sc → Zn
1.
Electron configurations: All have an argon core.
Two 4s electrons except Cr and Cu which have one.
A number of 3d electrons ranging from one in Sc to
10 in Cu and Zn.
2.
Atomic (metallic) radii:
3.
Oxidation states: Except zinc, each exhibits 2 or more oxidation states. The
maximum oxidation state equals the group number up to
group 7. e.g. Oxidation of Cr (Group 6) is +6 in CrO42⎯.
Little variation. This is because as you proceed
across the series from left to right, the increase in
nuclear charge is balanced by a corresponding
increase in the number of inner shell 3d electrons.
The effective nuclear charge on the 4s valence
electrons remains nearly constant.
To the right of Group 7, the loss of a large number of d
electrons becomes increasingly energetically unfavourable
due to the increased nuclear charge and only the lower
oxidation states are commonly encountered.
Compounds with metals in lower oxidation states are
primarily ionic and those with the metal in higher oxidation
states are primarily covalent. Transition elements are
commonly involved in redox reactions. The term common
oxidation state refers to the oxidation state of a metal
commonly found in aqueous solution.
NOTE:
Balancing redox reactions in acid and base is required. Review the rules.
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Reaction with acid:
All except Cu displace H2 from an acid.
e.g. Fe(s) + 2 HCl(aq) → FeCl2(aq) + H2(g)
Ferromagnetism:
Atoms of Fe, Co, Ni form domains, a region containing a large number of atoms whose
magnetic fields are all directed in the same way. In a magnetic field, the normally
randomly oriented domains line up to produce a strong resultant magnetic field. When
the external magnetic field is removed, the domains remain aligned and the metal has
become a permanent magnet. Ferromagnetic atoms must have both unpaired electrons
and interatomic distances just the right size to allow ordering of atoms into domains.
Catalytic activity:
Many transition metals and their oxides act as heterogeneous catalysts.
e.g.
C2H4(g) + H2(g)
Ni
⎯⎯→
C2H6(g)
catalyst
Compound colour:
A coloured species absorbs certain frequencies of visible light and reflects the rest. The
color of transition metal species is imparted by electronic transitions involving partly
occupied d-orbitals. Repulsions between ligand electrons and the d-electrons of a
metal cation removes some of the degeneracy of the d-orbitals. An electron can absorb
a visible photon and move to a higher energy d-orbital from a lower energy d-orbital..
The energy gap between the d-orbitals is affected by the type of ligand. This affects the
frequencies of absorbed and reflected visible light. The compounds and complexes of a
given transition metal have different colors depending on which ligands are present.
Transition vs. Main Group Elements:
Special properties such as complex ion formation, coloured species, multiple oxidation
states, magnetic properties and catalytic properties are due to the importance of the
involvement of d-orbitals in bonding which is much less frequent or non-existent with the
main group elements.
Questions on transition elements:
PHHM (9th Edition) – Chapter 23: 3, 5, 7, 13, 14
PHH (8th Edition) – Chapter 24: 17, 19, 21, 25, 26
Compare the bonding in MnO and Mn2O7 in terms of ionic vs. covalent character.
Explain the term domain as it applies to ferromagnetism.
What two properties does an element need to be ferromagnetic?
Explain what happens to a sample of Fe, Co or Ni in a magnetic field.
Explain why only lower oxidation states are commonly encountered for metals from
Groups 8−12 (8B, 1B, 2B).
Which first transition series element does not react with acids to yield H2(g)?
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The Halogens
production, properties and uses
PHHM (9th Edition): Chapter 22, pp 920-928
PHH (8th Edition): Section 23.2, pp 908-912
Properties
(a)
(b)
(c)
(d)
Exist as non-polar, diatomic molecules, X2
Relatively low boiling and melting points which increase with size from F2 to I2
Reactivity increases from I2 to F2 with smaller size and increasing
electronegativity.
F2 is the most powerful oxidizing agent.
Production
Electrolysis:
A method used widely in industry for the production of important chemicals, especially
metals and gases.
A strong electric current is passed through a concentrated aqueous solution of a salt or
a melt (liquid salt) using a power supply connected to two electrodes called the anode
and cathode. Oxidation occurs at the anode and reduction of the cathode. The overall
balanced equation is the sum of the oxidation and reduction half-reactions balancing
electrons lost and gained.
Fluorine:
F2: by electrolysis of HF
2 HF(in KHF2) → H2(g) + F2(g)
Chlorine:
Cl2: by electrolysis of concentrated, aqueous NaCl
2 NaCl(aq) + 2 H2O(l) → H2(g) + Cl2(g) + 2 NaOH(aq)
NOTE:
H2O is a lot easier to reduce than Na+. Hence it is reduced at the cathode.
oxidation of anode:
reduction of cathode:
Bromine:
2 Cl⎯(aq) → Cl2(g) + 2 e⎯
2 H2O(l) + 2 e⎯ → H2(g) + 2 OH⎯(aq)
Br2: by displacement from seawater or brine
Cl2(g) + 2 Br⎯(aq)
Iodine:
1.
⎯pH
⎯=3.5
⎯→
Br2(l) + 2Cl⎯(aq)
I2: by a two step process starting with NaIO3
Reduction with hydrogen sulfite ion.
IO3⎯(aq) + 3 HSO3⎯(aq) → I⎯(aq) + 3 SO42⎯(aq) + 3 H+(aq)
2.
Reverse disproportionation in acidic solution.
5 I⎯(aq) + IO3⎯(aq) + 6 H+(aq) → 3 I2(s) + 3 H2O(l)
Disproportionation: A redox reaction in which a reactant containing an element
in an intermediate oxidation state (e.g. Cu+) undergoes both
oxidation and reduction to form a product in a higher
oxidation state (e.g. Cu2+) and one in a lower oxidation state
(e.g. Cu)
e.g.
2 Cu+(aq) → Cu(s) + Cu2+(aq)
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Reverse disproportionation:
A redox reaction in which reactants containing
different oxidation states of one element
combine to form a product of intermediate
oxidation state. When iodine is produced by
reverse disproportionation in acidic solution
(equation 2 above), the oxidation states of
I⎯(XI = −1) and IO3⎯ (XI = +5) combine to form
I2 (XI = 0)
Some uses of the halogens:
General:
all are used to make halogenated organic compounds
e.g. polyvinyl chloride (PVC)
Additional: Cl2 is used as bleach for paper and textiles and as disinfectant for drinking
water and in the production of chlorinated organic compounds.
Br2 is used to make brominated organic compounds such as those used in
fire retardants. AgBr is the principal light-sensitive agent in photography.
F2 was used to make now-banned chlorofluorocarbons (CFC’s) which
were used as propellants in spray cans. One of its most important
compounds is the polymer teflon (polytetrafluoroethylene).
I2 is used to make pharmaceuticals.
Questions on halogens:
PHHM (9th Edition): Chapter 22: 7, 9, 11, 15
PHH (8th Edition): Chapter 23: 22, 23, 29, 30
1.
Balance the redox reactions responsible for the production of iodine in acidic
medium.
(a)
(b)
IO3⎯(aq) + HSO3⎯(aq) → I⎯(aq) + SO42⎯(aq)
I⎯(aq) + IO3 (aq)⎯ → I2 (s)
2.
Give one use for chlorine, bromine, fluorine and iodine.
3.
Explain the terms disproportionation and reverse disproportionation.
Extractive Metallurgy
PHHM (9th Edition): Section 23-2 pp 969-970
PHH (8th Edition): Section 24-2 pp 954-956
Developed by Dr. Chris Flinn
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