Chemistry Rules!'06 |
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Trends and Patterns - Transition Elements |
A transition metal is a d-block element that has at least one ion with a partially filled d-orbital.
Note, this deliberately excludes zinc, which is a d-block element from being considered a transition metal.
(2) Electron configurations :For the elements:
Sc | 1s22s22p63s23p6 | 4s23d1 |
Ti | 4s23d2 | |
V | 4s23d3 | |
Cr | 4s13d5 | |
Mn | 4s23d5 | |
Fe | 4s23d6 | |
Co | 4s23d7 | |
Ni | 4s23d8 | |
Cu | 4s13d10 |
Note that the electron configurations for chromium and copper do not follow the pattern. With a d5 or d10 configuration every d orbital is either half-full or totally full and this gives an exceptionally stable, symmetrical arrangement. This allows for only one electron to be in the 4s orbital.
For the electron configurations of ions the 4s electrons are removed first and then any 3d electrons necessary are removed.
e.g. | Sc2+ | 1s22s22p63s23p6 | 3d1 |
Ti2+ | 3d2 | ||
Fe3+ | 3d5 | ||
Cr3+ | 3d3 | ||
Mn2+ | 3d5 |
The majority of transition metals can form ions in more than one oxidation state, i.e. a variable number of electrons can be removed to form compounds. This is why it is very important to include roman numerals in the compound's name to indicate the oxidation state of the transition element.
e.g. | Fe2(SO4)3 | Iron(III) sulphate (or ferric sulphate) |
CuCl | Copper(I) chloride (or cuprous chloride) | |
KMnO4 | Potassium Manganate(VII) (or potassium permanganate) | |
V2O5 | Vanadium(V) oxide (or vanadium pentoxide) |
Most of the transition metal ions form compounds that are brightly coloured, e.g. copper(II) carbonate is green, iron(III) oxide is red, sodium dichromate(VI) is orange and potassium manganate(VII) is purple.
Ions with a d5 or d10 configuration, e.g. manganese(II) and copper(I) ions, generally give colourless or very weakly coloured compounds.
(5) Catalytic behavior :Many transition metals are useful as catalysts for industrial reactions, either as the metal or in compounds.
e.g. | Fe | in the Haber process (manufacture of ammonia) |
Ni | in the hydrogenation of fats/alkenes | |
V2O5 | in the Contact process (manufacture of sulphuric acid) | |
Pt | in the production of nitric acid |
A complex ion is formed when an aqueous transition metal ion interacts with other small molecules or ions, known as ligands, in the solution. There are a variety of common ligands, e.g. H2O, NH3, Cl- and -CN.
A ligand might be charged or neutral but must have at least one lone pair of electrons to form dative bonds with the metal ion.
e.g. hexaaquacopper(II) ion : | ![]() |
or | [Cu(H2O)6]2+ |
Just as with the reactivity series of metals, ligands have a variety of strengths. That it is to say they have variable abilities to bind to the transition metal ion.
A strong ligand will displace a weaker ligand from the complex ion, forming a new complex ion.
The relative strengths, in order of increasing strength, of some ligands are:
Cl- < H2O < NH3 < -CN
By Le Chatelier's principle a high concentration of a weaker ligand can displace a stronger ligand as the position of equilibrium moves to the right to reduce the concentration of the weaker ligand in the solution.
This ligand strength series can be seen by the reactions of copper(II) aqueous ions.
When concentrated hydrochloric acid, HCl, is added to Cu2+(aq) the solution turns green, forming the tetrachlorocuprate(II) ion, [CuCl4]2- :
This is an equilibrium reaction and diluting the solution will result in the original complex being formed.
When (dilute) ammonia, NH3, solution is added to Cu2+(aq) ions the solution turns a dark violet, forming the tetraamminediaquacopper(II) ion, [Cu(NH3)4(H2O)2]2+ :
Again, this is an equilibrium reaction and diluting the solution will result in the original complex being formed. However, since the concentration of HCl needs to be very high for the colour change to occur it shows that the Cl- ligand is weaker than water. As the concentration of NH3 only needs to be weak, the NH3 ligand is shown to be stronger than water.
Another example of ligand exchange is :
Copper(II), iron(II) and iron(III) aqueous ions all form distinctive coloured precipitates when sodium hydroxide(aq) is added to samples of them :
Cu2+(aq)
+ 2-OH(aq) Cu(OH)2(s)
Fe2+(aq)
+ 2-OH(aq)
Fe(OH)2(s)
Fe3+(aq)
+ 3-OH(aq)
Fe(OH)3(s)
Iron(II) ions can be oxidised by acidified manganate(VII) ions to give iron(III) ions. When this is reaction is conducted as part of a titration then the concentration of the iron(II) solution can be calculated.
The end point of the titration comes when one additional drop of manganate(VII) ions do not react with any iron(II) ions and the solution takes on a purple colour.
Equation :MnO4-(aq)
+ 8H+(aq) + 5Fe2+(aq) 5Fe3+(aq)
+ Mn2+(aq) + 4H2O(l)
If 10.20 cm3 of a 0.020 mol dm-3 solution of acidified potassium manganate reacts completely with 25.00 cm3 of a solution of iron(II) sulphate, what is the concentration of the iron(II) sulphate?
stage 1 - calculate the amount of manganate ions used -
amount
of manganate
|
= concentration x volume |
= 0.020 mol dm-3 x 0.0102 dm3 | |
= 0.000204 mol |
From the chemical equation above the ratio of MnO4-:Fe2+ is 1:5.
amount
of iron(II)
|
= 5 x concentration of manganate |
= 5 x 0.000204 mol | |
= 0.001020 mol |
concentration
of iron(II)
|
= amount / volume |
= 0.001020 mol / 0.025 dm3 | |
= 0.0408 mol dm-3 |
written by Dr Richard Clarkson : © Saturday, 1 November 1997
updated : Sunday, 30th October, 2005
mail to: chemistryrules@rjclarkson.demon.co.uk
created with the aid of ChemWindow®5.1
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