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Topic 5 –Transition Elements
Revision Notes
1) Introduction
- The d-block is the ten short columns in the middle of the Periodic Table. All of the elements in the d-block have their highest energy (outermost) electron in a d sub- shell
- Transition elements have 3 characteristic properties: they form coloured ions, they have more than one oxidation state and they act as catalysts
- To be a transition metal and display these properties, there must be at least one ion that has an incomplete d sub-shell
- This definition rules out Sc (whose only ion, Sc3+, is 3d^0 ) and Zn (whose only ion, Zn2+, is 3d^10 ). Sc and Zn are d-block elements but not transition metals
- Ions of transition elements undergo three types of reaction: o Precipitation o Ligand substitution o Redox
2) Electron arrangements
- The order in which electron orbitals are filled is as follows: 1s 2s 2p 3s 3p 4s 3d 4p
- There are 2 exceptions to the pattern: chromium and copper
- Cr is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5 and Cu is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^10
- These electronic structures are more stable than the alternative structures that follow the pattern
- When transition metals, like iron and copper, form ions they lose their 4s electrons before their 3d electrons
- Fe2+^ is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^6 (4s^0 is optional)
- Cu2+^ is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^9
3) Illustration of Properties
a) Variable oxidation state
- The table below indicates the oxidation states of transition elements in compounds
Source: http://commons.wikimedia.org/wiki/File:Transition_metal_oxidation_states.png
- Ti has oxidation state +4 in TiO 2
- V has oxidation state +5 in VO 3 -^ and +4 in VO2+
- Cr has oxidation state +6 in K 2 Cr 2 O 7 and +3 in CrCl 3
- Mn has oxidation state +4 in MnO 2 and is +7 in KMnO 4
- Fe is +2 in FeSO 4 and +3 in FeCl 3. Co and Ni follow the same pattern
- Cu is +1 in CuI and +2 in CuSO 4
b) Formation of coloured ions
- The dichromate(VI) ion, Cr 2 O 7 2-, is orange and the chromate(VI) ion, CrO 4 - , is yellow
- The manganate(VII) ion, MnO 4 - , is purple
- Cu2+(aq) is blue, Fe2+(aq) is pale green, Fe3+(aq) is yellow-brown, Co2+(aq) is pink
c) Ability to act as catalysts
- Transition elements and their compounds can act as catalysts
- Fe is a heterogeneous catalyst in the Haber process, N 2 + 3H 2 2NH 3
- Ni is the catalyst for hydrogenation of alkenes e.g. CH 2 =CH 2 + H 2 CH 3 CH 3
- V 2 O 5 is the catalyst in the Contact Process, 2SO 2 + O 2 2SO 3 , which is one of the steps in the manufacture of sulphuric acid
4) Ligands and complex ions
a) Definitions
- A complex consists of a central metal ion bonded to one or more ligands e.g. [Fe(H 2 O) 6 ]2+, [CuCl 4 ]2-
- Ligands have lone pairs of electrons e.g. H 2 O, Cl-, NH 3 , SCN-
- Ligands form co-ordinate bonds with the metal ion. The ligand is an electron pair donor and the metal ion accepts an electron pair
b) Shapes
- With 6 ligands bonded to a metal ion, the complex is octahedral in shape with a bond angle of 90°
- With 4 ligands bonded to a metal ion, the complex is usually tetrahedral in shape with a bond angle of 109.5°. The exceptions are complexes of nickel and platinum which are square planar with a bond angle of 90°
c) Co-ordination number
- The co-ordination number of a complex is the number of lone pairs bonded to the central ion (not the number of ligands)
- Unidentate ligands bond through one lone pair e.g. H 2 O, NH 3 , Cl-, SCN-, CN-
- Bidentate ligands can form two bonds because they have two lone pairs e.g. 1,2- diaminoethane, H 2 NCH 2 CH 2 NH 2 , which bonds through the lone pairs on the nitrogen atoms
f) Cis-trans isomerism in complexes
- Square planar complexes, such as [Ni(NH 3 ) 2 Cl 2 ], can show cis-trans isomerism
cis-isomer trans-isomer
- Octahedral complexes, such as [Pt(NH 3 ) 4 Cl 2 ], can show cis-trans isomerism because the chlorides can be at either 90° (cis) or 180° (trans)
both diagrams show the cis isomer
both diagrams show the trans isomer
g) Optical isomerism in complexes
- Optical isomers are non-superimposable mirror images
- Optical isomers rotate plane-polarised light in opposite directions by the same number of degrees
- Examples include [Co(H 2 NCH 2 CH 2 NH 2 ) 3 ]3+^ which is abbreviated to [Co(en) 3 ]3+^ in the following diagram
Source of this diagram: http://www.tau.ac.il/chemistry/OnlineCourses/UrbakhLectures/complexes.pdf
h) Cis-platin
- Cis-platin is an anti-cancer drug
- It binds to DNA and prevents it from replicating
h) Haemoglobin
- Porphyrin is a planar tetradentate ligand
Source: http://en.wikipedia.org/wiki/Porphyrin
- When porphyrin forms four co-ordinate bonds with Fe2+, the resulting complex is called haem
- The iron achieves a co-ordination number of 6 by bonding to the protein globin and to either an oxygen or water molecule. If the sixth bond is to oxygen, the complex produced is oxyhaemoglobin. If the sixth bond is to water the complex is deoxyhaemoglobin.
- The oxygen bonds only weakly to the Fe2+. Carbon monoxide forms a much stronger bond with Fe2+^ and is therefore poisonous because it reduces the oxygen carrying capacity of the blood
4) Precipitation reactions
- When reacted with NaOH(aq), solutions of copper (II), cobalt (II), iron (II) and iron (III) salts produce precipitates of the metal hydroxide
- These reactions can be used as tests for Cu2+(aq), Co2+(aq), Fe2+(aq) and Fe3+(aq)
Cu2+(aq) + 2OH-(aq) → Cu(OH) 2 (s) blue solution → blue precipitate Co2+(aq) + 2OH-(aq) → Co(OH) 2 (s) pink solution → blue precipitate Fe2+(aq) + 2OH-(aq) → Fe(OH) 2 (s) pale green solution → green ppt Fe3+(aq) + 3OH-(aq) → Fe(OH) 3 (s) yellow solution → orange precipitate
5) Ligand substitution reactions
- In ligand substitution reactions, one ligand is displaced by a stronger ligand.
a) Copper(II) with ammonia solution
- With excess concentrated NH 3 (aq) added dropwise, two reactions occur
- Firstly, a precipitation reaction because ammonia solution contains some OH-(aq) produced by the following reaction: NH 3 (aq) + H 2 O(l) NH 4 +(aq) + OH-(aq)
[Cu(H 2 O) 6 ]2+(aq) + 2OH-(aq) → Cu(OH) 2 (s) + 6H 2 O(l) Blue solution blue precipitate
- Constructing more complicated half-equations involves 2 extra steps, namely: balancing oxygens by adding water and balancing hydrogens by adding H+ e.g. oxidation of V2+^ to VO3-
Write down formulae: V2+^ → VO 3 - Balance atoms (vanadiums): not needed Balance oxygens: V2+^ + 3H 2 O → VO 3 - Balance hydrogens: V2+^ + 3H 2 O → VO 3 -^ + 6H+ Balance charge: V2+^ + 3H 2 O → VO 3 -^ + 6H+^ + 3e-
b) Constructing overall equations
- To produce the overall equation, multiply one or both half-equations until the number of electrons is the same e.g. iron(II) and manganate(VII)
Fe2+(aq) → Fe3+(aq) + e- MnO 4 - (aq) + 8H+(aq) + 5e-^ → Mn2+(aq) + 4H 2 O(l)
- In this case, multiply the first half-equation by 5 to get 5 electrons in both.
5Fe2+(aq) → 5Fe3+(aq) + 5e-
- Now add the half-equations together, cancelling the electrons at the same time.
5Fe2+(aq) + MnO 4 - (aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H 2 O(l)
- Any species that appears on both sides of the equation needs to be cancelled e.g. H+
c) Redox titration calculations
- As with acid-base reactions in Module 2811, redox reactions can be used to determine unknown concentrations etc.
- The method involves three steps: Step 1 Calculate moles using concentration x volume/ Step 2 Use the redox equation to calculate moles of other chemical Step 3 Use moles from step 2 to calculate unknown concentration etc
Example – A student takes 25.0 cm^3 of aqueous hydrogen peroxide, H 2 O 2 , and places this in a conical flask and then adds sulphuric acid to acidify the hydrogen peroxide.
The student titrates this sample of acidified hydrogen peroxide against a solution containing 0.0200 mol dm-3^ MnO 4 - (aq) ions. For complete reaction with the acidified hydrogen peroxide, the student used 17.5 cm^3 of this solution containing MnO 4 - (aq) ions.
Calculate the concentration, in mol dm-3, of the aqueous hydrogen peroxide. 2 mol MnO 4 -^ reacts with 5 mol H 2 O 2.
Step 1 moles MnO 4 -^ = conc x vol/ = 0.02 x 17.5/ = 3.5 x 10-
Step 2 moles H 2 O 2 = 3.5 x 10-4^ x 5/ = 8.75 x 10-
Step 3 conc H 2 O 2 = moles/volume = 8.75 x 10-4/25 x 10- = 0.035 mol dm-
Source: OCR paper January 2004
d) Estimation of copper in alloys
- An alloy is a mixture of metals. Brass is an alloy of copper and zinc (about 60 to 70% copper). Bronze is an alloy of copper and tin
- The percentage of copper in brass can be estimated by reaction with iodide and titration of the iodine produced with thiosulphate, S 2 O 3 2-
- Iodide ions reduce Cu2+^ to Cu+
Cu2+^ + I-^ → Cu+^ + ½I 2
- Cu+^ then reacts with more iodide ions to form a precipitate of CuI
Cu+^ + I-^ → CuI
- Overall: Cu2+(aq) + 2I-(aq) → CuI(s) + ½I 2 (aq) White (but appears brown due to I 2 )
- Brass is dissolved in nitric acid producing copper (II) nitrate and zinc nitrate
- An excess of potassium iodide, KI(aq), is added. This precipitates CuI(s) and forms iodine, I 2
- The amount of iodine produced can be found by titration against sodium thiosulphate, Na 2 S 2 O 3
2S 2 O 3 2-^ + I 2 → S 4 O 6 2-^ + 2I-
- Near the end-point, starch solution is added which produces a blue-black colour with the remaining iodine
- The end-point is when the blue-black colour just disappears
e) Estimation of copper - example
- A solution was prepared by dissolving some copper (II) sulphate to give 250 cm^3 of aqueous solution
- 25.0 cm^3 of this solution was treated with an excess of aqueous potassium iodide, KI
2Cu2+(aq) + 4I-(aq) → 2CuI(s) + I 2 (aq)
- The iodine produced was titrated with 0.100 mol dm-3^ sodium thiosulphate
I 2 (aq) + 2S 2 O 3 2-(aq) → 2I-(aq) + S 4 O 6 2-(aq)
(a) Calculate the amount of S 2 O 3 2-^ ions in the titre
Moles S 2 O 3 2-^ = conc x vol/ = 0.100 x 22.0/ = 2.2 x 10-3^ mol
(b) Calculate the amount of I 2 produced
Moles I 2 = moles S 2 O 3 2-/ = 1.1 x 10-3^ mol
(c) Calculate the amount of Cu2+^ ions in 25.0 cm^3 of solution
Moles Cu2+^ = moles I 2 x 2 = 2.2 x 10-3^ mol
(d) Calculate the concentration of the aqueous copper (II) sulphate solution in mol dm-
Conc Cu2+^ = moles/vol = 2.2 x 10-3/(25.0/1000) = 0.088 mol dm-
