Summary of Topic 5.2: Transition metal chemistry

5.2(a) derive the electron configurations of the d-block elements and their simple ions, from the Periodic Table.

Task 5.2a What rules give the electronic structures of the atoms?
What rules give the electronic structures of the ions from the electronic structures of the atoms?  Draw an electrons in boxes diagram to show the electronic structure of argon.
Draw electrons in boxes diagrams for the atoms and ions above.

5.2(b) recall the transition metals as d-block elements forming one or more stable ions which have incompletely filled d-orbitals;
The elements Sc - Zn are d-block elements.  They have atoms with electrons in d-orbitals.
The elements Ti - Cu are transition elements.  They have ions with incomplete d orbitals.       
Task 5.2b Why is iron a transition element?  Why  is CuCl  white but CuCl₂ blue?  What colour are Sc and Zn salts and why?  Why is TiO₂ white but TiCl₃ violet?

5.2(c) recall the characteristic properties of the transition elements such as chromium, iron and copper
Transition metals have higher melting points, higher boiling points and higher densities than other metals.
(i) the formation of coloured aqueous ions and other complex ions
[Cr(H₂O)₆]³⁺(aq) blue (but often seems green), dichromate Cr₂O₇^2-(aq) orange in acid, chromate [CrO₄]^2-(aq) yellow in alkali, Fe²⁺(aq) green, Fe³⁺(aq) yellow/brown,  Cu(H₂O)₆]²⁺(aq) blue. Managnate(VII) MnO₄^-(aq) purple, Tetraamminecopper(II) Cu(NH₃)₄²⁺(aq) deep blue/purple, Tetrachlorocuprate(II) CuCl₄^2-(aq) yellow, hydrated thiocyanate complex of iron (III) [Fe(SCN)(H₂O)₅]²⁺ blood red, [Co(H₂O)₆]²⁺(aq) pink, [CoCl₄]^2-(aq) blue, 
Task 5.2c(i) Match formulae above to following names: hexaaquachromium(III), tetrachlorocobaltate(II), hexaaquacopper(II), hexaaquacobalt(II).

(ii) the formation of a range of compounds in which they are present in different stable oxidation states
Transition metals show the following characteristic properties:

1. Variable oxidation states:- Transition metals have electrons of similar energy in both the 3d and 4s levels. This means that one particular element can form ions of roughly the same stability by losing different numbers of electrons. Thus, all transition metals from titanium to copper can exhibit two or more oxidation states in their compounds.

Oxidation states of some Transition Metals:

Titanium- +2, +3, +4
Vanadium- +2, +3, +4, +5
Chromium- +2, +3, +6
Manganese- +2, +3, +4, +5, +6, +7
Iron- +2, +3
Cobalt- +2, +3
Nickel- +2, +3, +4
Copper- +1, +2
When Transition Metals form positive ions they loose their electrons from the 4s sub-shell first, then the 3d sub-shell.

Task 5.2c(ii).1 Give electronic structures for the ions in the following: copper(II)oxide, Copper(I)oxide,  iron(II)sulphate, Iron(III)chloride, chromium(III)chloride.

2. Formation of complex ions:- As a lot of the transition metals have some empty spaces in their 3d-orbitals, they can receive lone pairs of electrons and form dative covalent bonds thus producing complex compounds.
Story about complex compounds at University College London

3. Coloured compounds:- When electrons move from a d-orbital (with lower energy) to another d-orbital (with higher energy), energy is taken in. This energy in the form of light is missing from the reflected light thus producing coloured compounds.

4. Catalytic properties:- As Transition Metals have variable oxidation states, they tend to have catalytic properties.  The small differences in ionisation energies make variable oxidation numbers possible.  The reversible redox reactions involving transition metal ions make lower energy routes for other reactions.
Task 5.2c(ii).2 Explain how V₂O₅ might catalyse sufur dioxide reacting with oxygen in the contact process.  Use 2V⁵⁺ + O^2- +SO₂ --> 2V⁴⁺ + SO₃ and
2V⁴⁺ +1/2O₂ --> 2V⁵⁺ +O^2-

5.2d understand the nature of the bonding in complex ions, including the aqua-ions, their shape and the cause of their colour
· stereoisomerism in such complex ions will not be examined
· an elementary treatment only is required. Students should understand that the bonding between the ligand and the metal ion is dative covalent and this causes a splitting of the D-orbitals
· colour should be related to a simple transfer of electrons between D-orbitals.

When writing formulae, the central atom is put first, then the negative ions and then follow any neutral molecules.  Everything is then put in square brackets and the charge added.  For example, tetraaquachlorocopper (II) would be written as [CuCl(H₂O)₄]⁺
When naming a positive complex ion, the ligands are written first in alphabetical order, then the name of the metal and then its oxidation number.  For a negative complex ion the name of the metal is changed.  Iron-ferrate, copper-cuprate, chromium-chromate, cobalt-cobalate, nickel-nickelate, zinc-zincate
The following prefixes may be needed:- di, tri, tetra, penta, hexa

For example, [CrCl(H₂O)₅]⁺² would be written as pentaaquachlorochromium (III)
and [NiCl₄]^2- is written as tetrachloronickelate(II)
Task 5.2d.1 Write the corresponding name or formulae for the following complex ions:
[Fe(H₂O)₆]²⁺, diamminesilver(I) ion, [Co(NH₃)₆]³⁺, hexacyanoferrate(III) ion, [Fe(OH)₂(H₂O)₄]⁺, tetraaquadichlorochromium(III) ion, [CuCl₃]^2-, tetrachlorocuprate(II) ion.

Complex ions (cationic or anionic) are composed of a central metal ion surrounded by a cluster of anions or molecules, called ligands (species donating electron(s)). In transition metal complexes, non-bonded pairs of electrons on the ligand form co-ordinate bonds or dative covalent bond to the central ion by donating these unshared electron pairs into vacant orbitals of the transition metal ion. The number of co-ordinate bonds from ligands to the central ion is known as the co-ordination number of the central ion.

Co-ordination number: 6, 4, 2 Shape: Octahedral [Fe(H₂O)₆]³⁺, tetrahedral [Ni(CO)₄]²⁺ ,and linear [CuCl₂]^-, respectively
     
Task2.5d.2 Show the location of the unpaired electrons which form the dative covalent bond.
Draw similar diagrams for [Ag(NH₃)₂]⁺ and [Ni(CO)₄]²⁺.
 Incomplete d orbitals in Fe³⁺ accept electrons to form the coordinate bonds.

Task 2.5d.3 Draw diagrams to show incomplete d orbitals in Cu²⁺.

The colour of a substance is due to the light that it reflects.
A black substance absorbs all light, none is reflected.
A white substance absorbs no light, it is all reflected.
A coloured substance absorbs some wavelengths (of visible light), the remainder are reflected.
Atoms can absorb energy when electrons are promoted from lower to higher energy levels.  Transitions between many orbitals require more energy than is available in visible light.  Transitions between d orbitals can happen with the energy from visible light so transition metal compounds are coloured.
d orbitals are normally of similar energy but when surrounded by ligands the orbitals are split into higher and lower energy sets.  [Ti(H₂O)₆]³⁺ is violet.  Green light is absorbed by this ion leaving red and blue light to combine and give the ion its violet colour.  The electronic transition involved is shown below.

Task5.2d.4 Draw the ground state and an excited state of the Fe²⁺ ion.  Using the diagram explain why iron(II) sulfate is green.

5.2e understand simple ligand exchange processes

Ligand exchange- This involves the replacement of one ligand by another.

E.g. [ Cu(H₂O)₄ ]²⁺_(aq) + 4Cl^- <----> [ CuCl₄ ]^2-_(aq) + 4H₂0_(l)
              blue                                     green
Draw a diagram to show the difference in splitting between d orbitals in the metal atoms in the above comlexes.
The Cl^- ligands have replaced the H₂O ligands.  The colour change is because the different ligands create a different amount of splitting between d orbitals in the copper ion.  In the above case the position of the equilibrium lies to the right.
Deprotonation seems similar but is not just an exchange.  It involves a water ligand losing a hydrogen ion (proton) to a proton acceptor such as an hydroxide ion.
[Cu(H₂O)₆]²⁺  + OH^-   -----> [Cu(OH)(H₂O)₅]⁺ + H₂O

5.2f recall the formation of hydroxide precipitates on the addition of aqueous solutions of sodium hydroxide or ammonia, and that some hydroxide precipitates react with an excess of strong alkali, and some react with an excess of ammonia; limited to Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺,Co²⁺, Ni²⁺, Cu²⁺, Zn²⁺
· students will be expected to recall the colours of the hydroxide precipitates
· the concepts of deprotonation and ligand exchange should be applied to these reactions

Deprotonation- In this process, hydrogen ions are split off from water molecules which have been acting as ligands in a complex ion.

E.g. [ Fe(H₂O)₆ ]³⁺_(aq) -----> [ Fe(OH)(H₂O)₅ ]²⁺_(aq) + H⁺_(aq)

5.2g recall the oxidation states of vanadium (+2, +3, +4, +5) in its compounds, and deduce, given E^o values, reagents for the interconversion of the metal ions, oxo anions and oxo cations of the element in these oxidation states
· students will be expected to recall the colours of the aqueous solutions of vanadium compounds in the oxidation states listed above
 

Vanadium can be reduced from the +5 states right through to the +2 state by zinc.  Each step can be predicted using E^o values.  For example for the first step (reduction from +5 to +4):
Zn²⁺(aq) + 2e^- ---> Zn(s)                                 E^o    = -0.76 V  
VO₂⁺(aq)  + 2H⁺(aq) + e- --->VO²⁺(aq)  + H₂O(l)      E^o    = +1.00 V
Zinc is the stronger reducing agent here because its E^o value is more negative than for vanadium.  So Zinc will reduce vanadium +5 to +4.
OR
Write the half equations with the most negative on top (as above).  Apply the anticlockwise rule.  So VO₂⁺(aq) is reduced to VO²⁺(aq) while Zn(s)  is oxidised to Zn²⁺(aq).
OR
The equation for the reduction of vanadium is
2VO₂⁺(aq)  + 4H⁺(aq) + Zn(s) ---> 2VO²⁺(aq)  + 2H₂O(l) + Zn²⁺(aq)
The cell diagram is:
Zn(s)|Zn²⁺(aq)|| [2VO₂⁺(aq)  + 4H⁺(aq)],[VO²⁺(aq)  + H₂O(l)]|Pt
Ecell =  E^o_R -   E^o_L    = +1.00 V - (-0.76 V)  = +1.76 V

Task 5.2g The large positive value of Ecell shows that the reaction to reduce vanadium is spontaneous.
Use the data below to show  that zinc reduces vanadium to V²⁺, that managanateVII oxidises V²⁺ back up to vanadium V and that iodide oxidises vanadium from the +2 to +3, from +3 to +4 but not from +4 to +5.
V³⁺(aq) + e^- ---> V²⁺(aq)     E^o    = - 0.26 V
VO²⁺(aq + 2H⁺(aq) +e^- ---> V³⁺(aq) + H₂O(l)   E^o    = +0.34 V
I₂(aq) + 2e^- ---> 2I^-(aq)        E^o    = +0.54 V
MnO₄^-(aq) + 8H⁺(aq) + 5e^- = Mn²⁺(aq) +4H₂O(l)  E^o = +1.51V

5.2h describe reactions for the interconversion of the oxidation states of vanadium in aqueous solution

To oxidise V²⁺ to VO₂⁺, add acidified manganate (VII) to the vanadium (II) V²⁺ -----> VO₂⁺

To reduce VO₂⁺ to V²⁺, add zinc to the dioxovanadium (V)      VO₂⁺ -----> V²⁺

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