1. Chapter 14 The Transition Elements and Their Chemistry Mn(VII) Cr(VI)
V(V)
Chapter 14
The Transition Elements and Their Chemistry

2. 14.1 Properties of the Transition Elements
14.2 The Inner Transition Elements
14.3 Highlights of Selected Transition Metals
14.4 Coordination Compounds
14.5 Theoretical Basis for the Bonding and Properties of Complexes

3. The transition elements (d block) and inner transition elements (f block) in the periodic table.

4. The Period 4 transition metals

5. The Period 4 transition metals

6. The Period 4 transition metals

7. Orbital Occupancy of the Period 4 Transition Metals

8. Sample Problem 14.1
Writing Electron Configurations of Transition
Metal Atoms and Ions
PROBLEM:
Write condensed electron configurations for the following: (a) Zr; (b) V3+; (c) Mo3+. (Assume that elements in higher periods behave like those in Period 4.)
PLAN:
The general configuration is [noble gas] ns2(n-1)dx. Recall that in ions the ns electrons are lost first.
SOLUTION:
(a) Zr is the second element in the 4d series: [Kr]5s24d2.
(b) V is the thired element in the 3d series: [Ar]4s23d3. In forming V3+, three electrons are lost (two 4s and one 3d), so V3+ is a d2 ion: [Ar]3d10.
(c) Mo lies below Cr in Group 6B(6), so we expect the same except in configuration as for Cr. Thus, Mo is [Kr]5s14d5. In forming the ion, Mo loses the one 5s and two of the 4d electrons to become a 4d3 ion: [Kr]4d3.

9. Horizontal trends in key atomic properties of the Period 4 elements.

10. Vertical trends in key properties within the transition elements

11. Aqueous oxoanions of transition elements.
One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation states.
Mn(II)
Mn(VI)
Mn(VII)
V(V)
Cr(VI)
Mn(VII)

12. Oxidation states and d-orbital occupancy of the period 4 transition metals

13. Standard electrode potentials of period 4 M 2+ ions

14. Colors of representative compounds of the Period 4 transition metals.
sodium chromate
nickel(II) nitrate hexahydrate
potassium ferricyanide
zinc sulfate heptahydrate
titanium oxide
scandium oxide
manganese(II) chloride tetrahydrate
copper(II) sulfate pentahydrate
vanadyl sulfate dihydrate
cobalt(II) chloride hexahydrate

15. Some properties of group 6B(6) elements

16. There are 6 unpaired e- in Sm.
Sample Problem 14.2
Finding the Number of Unpaired Electrons
PROBLEM:
The alloy SmCo5 forms a permanent magent because both samarium and cobalt have unpaired electrons. How many unpaired electrons are in the Sm atom (Z = 62)?
PLAN:
Write the condensed configuration of Sm and, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram.
SOLUTION:
Sm is the eighth element after Xe. Two electrons go into the 6s sublevel and the remaining six electrons into the 4f (which fills before the 5d).
Sm is [Xe]6s24f6 6s 4f 5d
There are 6 unpaired e- in Sm.

17. The bright colors of chromium (VI) compounds.

18. Some oxidation states of manganese
Orbital Occupancy
*Most common states in bold face.

19. Steps in producing a black-and-white negative.

20. Components of a coordination compound.
models
wedge diagrams
chemical formulas

21. Structures of Complex Ions:
Coordination Numbers, Geometries, and Ligands
Coordination Number - the number of ligand atoms that are bonded directly to the central metal ion. The coordination number is specific for a given metal ion in a particular oxidation state and compound.
Geometry - the geometry (shape) of a complex ion depends on the coordination number and nature of the metal ion.
Donor atoms per ligand - molecules and/or anions with one or more donor atoms that each donate a lone pair of electrons to the metal ion to form a covalent bond.

22. Coordination numbers and shapes of some complex ions

23. Some common ligands in coordination compounds

24. Names of some neutral and anionic ligands
Names of some metal ions in complex anions

25. Formulas and Names of Coordination Compounds
Rules for writing formulas:
1. The cation is written before the anion.
2. The charge of the cation(s) is balanced by the charge of the anion(s).
3. In the complex ion, neutral ligands are written before anionic ligands, and the formula for the whole ion is placed in brackets.

26. Formulas and Names of Coordination Compounds
continued
Rules for naming complexes:
1. The cation is named before the anion.
2. Within the complex ion, the ligands are named, in alphabetical order, before the metal ion.
3. Neutral ligands generally have the molecule name, but there are a few exceptions. Anionic ligands drop the -ide and add -o after the root name.
4. A numerical prefix indicates the number of ligands of a particular type.
5. The oxidation state of the central metal ion is given by a Roman numeral (in parentheses).
6. If the complex ion is an anion we drop the ending of the metal name and add -ate.

27. sodium hexafluoroaluminate
Sample Problem 14.3
Writing Names and Formulas of Coordination
Compounds
PROBLEM:
(a) What is the systematic name of Na3[AlF6]?
(b) What is the systematic name of [Co(en)2Cl2]NO3?
(c) What is the formula of tetraaminebromochloroplatinum(IV) chloride?
(d) What is the formula of hexaaminecobalt(III) tetrachloro-ferrate(III)?
PLAN:
Use the rules presented and
SOLUTION:
(a) The complex ion is [AlF6]3-.
Six (hexa-) fluorines (fluoro-) are the ligands - hexafluoro
Aluminum is the central metal atom - aluminate
Aluminum has only the +3 ion so we don’t need Roman numerals.
sodium hexafluoroaluminate

28. dichlorobis(ethylenediamine)cobalt(III) nitrate
Sample Problem 14.3
Writing Names and Formulas of Coordination
Compounds
continued
(b) There are two ligands, chlorine and ethylenediamine -
dichloro, bis(ethylenediamine)
The complex is the cation and we have to use Roman numerals for the cobalt oxidation state since it has more than one - (III)
The anion, nitrate, is named last.
dichlorobis(ethylenediamine)cobalt(III) nitrate
tetraaminebromochloroplatinum(IV) chloride
(c)
4 NH3
Br-
Cl-
Pt4+
Cl-
[Pt(NH3)4BrCl]Cl2
(d)
hexaaminecobalt(III) tetrachloro-ferrate(III)
6 NH3
Co3+
4 Cl-
Fe3+
[Co(NH3)6][Cl4Fe]3

29. Some coordination compounds of cobalt studied by werner

30. Constitutional (structural) isomers
Important types of isomerism in coordination compounds.
ISOMERS
Same chemical formula, but different properties
Constitutional (structural) isomers
Stereoisomers
Atoms connected differently
Different spatial arrangement
Coordination isomers
Ligand and counter-ion exchange
Linkage isomers
Different donor atom
Geometric (cis-trans) isomers (diastereomers)
Different arrangement around metal ion
Optical isomers (enantiomers)
Nonsuperimposable mirror images

31. Linkage isomers

32. Geometric (cis-trans) isomerism.

33. Optical isomerism in an octahedral complex ion.

34. Sample Problem 14.4
Determining the Type of Stereoisomerism
PROBLEM:
Draw all stereoisomers for each of the following and state the type of isomerism:
(a) [Pt(NH3)2Br2]
(b) [Cr(en)3]3+ (en = H2NCH2CH2NH2)
PLAN:
Determine the geometry around each metal ion and the nature of the ligands. Place the ligands in as many different positions as possible. Look for cis-trans and optical isomers.
SOLUTION:
(a) Pt(II) forms a square planar complex and there are two pair of monodentate ligands - NH3 and Br.
These are geometric isomers; they are not optical isomers since they are superimposable on their mirror images.
trans
cis

35. Sample Problem 14.5
Determining the Type of Stereoisomerism
continued
(b) Ethylenediamine is a bidentate ligand. Cr3+ is hexacoordinated and will form an octahedral geometry.
Since all of the ligands are identical, there will be no geometric isomerism possible.
The mirror images are nonsuperimposable and are therefore optical isomers.

36. Hybrid orbitals and bonding in the octahedral [Cr(NH3)6]3+ ion.

37. Hybrid orbitals and bonding in the square planar [Ni(CN)4]2- ion.

38. Hybrid orbitals and bonding in the tetrahedral [Zn(OH)4]2- ion.

39. An artist’s wheel.

40. Relation between absorbed and observed colors

41. The five d-orbitals in an octahedral field of ligands.

42. Splitting of d-orbital energies by an octahedral field of ligands.
D is the splitting energy

43. The effect of ligand on splitting energy.

44. The color of [Ti(H2O)6]3+.

45. Effects of the metal oxidation state and of ligand identity on color.
[V(H2O)6]3+
[V(H2O)6]2+
[Cr(NH3)6]3+
[Cr(NH3)5Cl ]2+

46. The spectrochemical series.
For a given ligand, the color depends on the oxidation state of the metal ion.
For a given metal ion, the color depends on the ligand.
I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO
WEAKER FIELD
STRONGER FIELD
LARGER D
SMALLER D
LONGER 
SHORTER 

47. [Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+
Sample Problem 14.6
Ranking Crystal Field Splitting Energies for
Complex Ions of a Given Metal
PROBLEM:
Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3- in terms of the relative value of D and of the energy of visible light absorbed.
PLAN:
The oxidation state of Ti is 3+ in all of the complexes so we are looking at the crystal field strength of the ligands. The stronger the ligand the greater the splitting and the higher the energy of the light absorbed.
SOLUTION:
The field strength according to is CN- > NH3 > H2O. So the relative values of D and energy of light absorbed will be
[Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+

48. High-spin and low-spin complex ions of Mn2+.

49. Orbital occupancy for high- and low-spin complexes
of d4 through d7 metal ions.
high spin: weak-field ligand
low spin: strong-field ligand
high spin: weak-field ligand
low spin: strong-field ligand

50. Sample Problem 14.7
Identifying Complex Ions as High Spin or Low Spin
PROBLEM:
Iron (II) forms an essential complex in hemoglobin. For each of the two octahedral complex ions [Fe(H2O)6]2+ and [Fe(CN)6]4-, draw an orbital splitting diagram, predict the number of unpaired electrons, and identify the ion as low or high spin.
PLAN:
The electron configuration of Fe2+ gives us information that the iron has 6d electrons. The two ligands have field strengths shown in
Draw the orbital box diagrams, splitting the d orbitals into eg and t2g. Add the electrons noting that a weak-field ligand gives the maximum number of unpaired electrons and a high-spin complex and vice-versa.
[Fe(CN)6]4-
[Fe(H2O)6]2+
SOLUTION:
potential energy
4 unpaired e-- (high spin) eg eg
no unpaired e-- (low spin)
t2g
t2g

51. Splitting of d-orbital energies by a tetrahedral field and a square planar field of ligands.

52. Hemoglobin and the octahedral complex in heme.

53. Some transition metal trace elements in humans

54. The tetrahedral Zn2+ complex in carbonic anhydrase.