Unit 5 Chemistry- Transition Metals


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  • Created by: Hannah
  • Created on: 23-06-11 10:04


Transition metal- A metal that can form one or more stable ions with a partially filled d-subshell 

Transition metals are found in the d-Block. The electronic configurations of transition metals cause their properties- the incomplete D shell. 

A d-orbital can fit 10 electrons in. So a transition metal must form at least one ion that has between 1 and 9 electrons in the d-orbital. All the period 4 d-block elements are transition metals...

Apart from Scandium and Zinc. 

When doing configurations remember:

1. The 4s shell fills up before 3d

2. Cr prefers to have 1 electron in each orbital of 3d and 1 in 4s shell as its more stable. 

3. Cu prefers to have a full 3d shell and 1 electron in the 4s shell- more stable

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Scandium and Zinc aren't transition metals:

Scandium only forms one ion, Sc 3+, which has an empty d-subshell. Scandium has the electronic configuration [Ar] 3d1 4s2, and when it loses 3 electrons to form Sc 3+, it ends up with the configuration [Ar]

Zinc only forms one ion, Zn 2+, which has a full d-subshell. Zinc has the configuration [Ar] 3d 10 4s2. When it forms Zn 2+, it loses two electrons, both from the 4s subshell. This means it keeps its full 3d subshell. 

When ions are formed, the s electrons are removed first. 

Transition metals form positive ions. 

When this happens, the s electrons are removed first, then the d electrons. 

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Transition elements are all typical metals and have similar physical properties:

1. They all have a high density

2. They all have high melting and high boiling points

3. Their ionic radii are more or less the same

Transition metals have special chemical properties:

1. Form complex ions

2. Form coloured ions

3. Good catalysts

4. Exist in variable oxidation states

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Common coloured ions and oxidation states

The colours refer to the aqueous ions:

Oxidation state +2:  V 2+ (violet) , Mn 2+ (pale pink), Fe 2+ (pale green), Co 2+ (pink), Ni 2+ (green), Cu2+ (blue)

Oxidation state +3:  V 3+ (green), Cr 3+ (green/ violet), Fe 3+ (yellow)

Oxidation state +4:  VO 2+ (blue)

Oxidation state +5:   VO2 + (yellow)

Oxidation state +6:   Cr2O7 2- (orange)

Oxidation state +7:   MnO4- (purple)

These elements show variable oxidation states because the energy levels of the 4s and the 3d subshells are very close to one another. So different numbers of electrons can be gained or lost using fairly similar amounts of energy. 

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Complex ions

A complex is a metal surrounded by coordinately bonded ligands.

1. A coordinate bond (or dative covalent bond) is a covalent bond in which both electrons in the shared pair come from the same atoms. In a complex, they come from ligands.

2. So, a ligand is an atom, ion  or molecule that donates a pair of electrons to a central metal ion. 

3. The coordination number is the number of coordinate bonds that are formed with the central metal ion. 

4. The usual coordination numbers are 6 and 4. If the ligands are small, like H2O or NH3, 6 can fit around the central metal ion. If the ligands are larger, like Cl-, only 4 can fit around the central metal ion. 

6 coordinate bonds mean an octahedral shape

4 coordinate bonds usually mean a tetrahedral shape, but in a few complexes e.g. cisplatin, 4 coordinate bonds form a square planar shape

Some silver complexes have 2 coordinate bonds and form a linear shape.

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More on complex ions:

The overall charge on the complex ion is its total oxidation state. Its put just outside the square brackets. 

The oxidation state of the metal ion =

the total oxidation state- the sum of the oxidation states of the ligands

A ligand must have at least one lone pair of electrons, or it won't have anything to use to form a coordinate bond. 

1. Ligands with one bond are called unidentate e.g. H2O:, :NH3, :Cl-

2. Ligands with more than one lone pair are called multidentate e.g. EDTA 4- has six lone pairs (hexadentate), so it can form 6 coordinate bonds with a metal ion. 

3. Bidentate ligands are multidentate ligands with two lone pairs e.g. ethane-1,2-diamine, :NH2CH2CH2:NH2. Bidentate ligands can each form two coordinate bonds with  metal ion. 

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- Haemoglobin is a protein found in blood that helps transport oxygen around the body

- Haemoglobin contains Fe2+ ions, which are hexa-coordinated, six lone pairs are donated to them to form six coordinate bonds

- Four of the lone pairs come from the nitrogen atoms, which form a circle around the Fe 2+. This part of the molecule is called haem

- The molecule that the four nitrogen atoms are part of is a multidentate ligand. 

- A protein called globin and either an oxygen or a water molecule also bind to the Fe2+ ion to form an octahedral structure.

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Coloured ions

Normally the 3d orbitals of transition element ions all have the same energy. 

But when ligands come along and bond to the ions, some of the orbitals are given more energy than others. This splits the 3d orbitals into two different energy levels.

Electrons tend to occupy the lower orbitals (the ground state). To jump up to the higher orbitals (excited states), they need energy equal to the energy gap, delta E. They get this from visible light. 

The energy absorbed when electrons jump up can be worked out by:

Delta E = h x v  where v = frequency of light absorbed and h = Planck's constant

The amount of energy needed to make electrons jump depends upon the central metal ion and its oxidation state, the ligands and the coordination number, as these affect the size of the energy gap

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Colours of compounds

When visible light hits a transition metal ion, some frequencies are absorbed when electrons jump up to the higher orbitals. The frequencies absorbed depend on the size of the energy gap.

The rest of the frequencies are reflected. These reflected frequencies combine to make the complement of the colour of the absorbed frequencies- this is the colour you see. 

E.g. [Cu(H2O)6] 2+ ions absorb yellow light. The remaining frequencies combine to produce the complementary colour - in this case it's blue. so [Cu(H2O)6] 2+ in solution appears blue. 

If there are no 3d electrons or the 3d subshell is full, then no electrons will jump, so no energy will be absorbed. If there's no energy absorbed, the compound will look white of colourless. 

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Transition metals can be identified by their colou

The colour of a complex can be altered by any of the factors that can affect the size of the energy gap:

1. Changes in oxidation state:

Complex: [Fe(H2O)6] 2+ (aq) ---> [Fe(H2O)6] 3+ (aq)

Oxidation state: +2 ---> +3

Colour: pale green ---> yellow

e.g. 2

Complex: [V(H2O)6] 2+ (aq) ---> [V(H2O)6] 3+ (aq)

Oxidation state: +2 ---> +3 

Colour: violet ---> green 

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2. Changes in coordination number- this always involves a change of ligand too

Complex: [Cu(H2O)6] 2+ + 4Cl- ---> [CuCl4]2- + 6H2O

Coordination number: 6 ---> 4

Colour: blue ---> yellow

3. Changes in ligand- this can cause a colour change even if the oxidation state and coordination number remain the same

Complex: [Co(H2O)6] 2+ + 6NH3 ---> [Co(NH3)6] 2+ + 6H2O

Oxidation state: +2 ---> +2

Colour: Pink ---> straw coloured

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Can be used to determine the concentration of a solution by measuring how much light it absorbs.

1. White light is shone through a filter, which is chosen to only let the colour of light through that is absorbed by the sample

2. The light then passes through the sample to a colorimeter which calculates how much light was absorbed by the sample

3. The more concentrated a coloured solution is, the more light it will absorb. So you can use the measurement to work out the concentration of a solution of transition metals. 

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Variable oxidation states


Oxidation state = +6, Formula = Cr2O7 2- (aq), Colour = orange

Oxidation state = +6, Formula = CrO4 2- (aq), Colour = yellow

Oxidation state = +3, Formula = Cr 3+ (aq), Colour = green (violet)

Oxidation state = +2, Formula = Cr 2+ (aq), Colour = blue

1. When an alkali (OH-) is added to aqueous dichromate (VI) ions, the orange colour turns yellow because aqueous chromate (VI) ions form: Cr2O7 2- (aq) + OH- (aq) ---> 2CrO4 2- (aq) + H+ (aq) orange ---> yellow

2. When an acid (H+ ions) is added to aqueous chromate (VI) ions, the yellow colour turns orange because aqueous dichromate (VI) ions form: 2CrO4 2- (aq) + H+ (aq) ---> Cr2O7 2- (aq) + OH- (aq) ... yellow --> orange

3. They're in equilibrium: Cr2O7 2- (aq) + H2O (l) <--> 2CrO4 2- (aq) + 2H+ (aq)

Position of equilibrium depends on pH, le chatelier's. If H+ is added, eq. shifts left so orange ions formed, if OH-, eq. shifts right and the yellow ions formed.

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Chromium ions can be oxidised and reduced

1. Dichromate (VI) ions can be reduced using a good reducing agent, such as zinc and dilute acid:

Cr2O7 2- (aq) + 14H+ (aq) + 3Zn (s) ---> 3 Zn 2+ (aq) + 2Cr 3+ (aq) + 7H2O (l)

orange ---> green 

2. Zinc will reduce Cr 3+ further to Cr 2+:

2Cr 3+ (aq) + Zn (s) --> Zn 2+ (aq) + 2 Cr 2+ (aq)

green ---> blue

But unless you use an inert atmosphere, Cr 2+ will oxidise straight back to Cr 3+

3. You can oxidise Cr 3+ to chromate (VI) ions with hydrogen peroxide,H2O2 in an alkaline solution:

2Cr 3+ (aq) + 10OH- (aq) + 3H2O2 (aq) ---> 2CrO4 2- (aq) + 8H2O (l)

green ---> yellow

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Can exist in two oxidation states: +2 as Co 2+ and +3 as Co 3+. It prefers to be in the +2 state though. 

1. Co 3+ can be made by oxidising Co2+ (aq) with hydrogen peroxide in alkaine conditions:

2Co 2+ (aq) + H2O2 (aq) ---> 2Co 3+ (aq) + 2OH- (aq)

2. You can also oxidise Co 2+ with air in ammoniacal solution. Place CO 2+ ions in an excess of aqueous ammonia, which causes [Co(NH3)6] 2+ ions to form. If these complex ions are left to stand in air, they are oxidised to [Co(NH3)6] 3+

[Co(H2O)6] 2+ (aq pink solution) --NH3--> [Co(H2O)4(OH)2] (s blue ppt.) 

---excess NH3--> [Co(NH3)6] 2+ (aq straw coloured sol.) --stand in air--> [Co(NH3)6] 3+ (aq dark brown solution)

Done this way as [Co(NH3)6] 2+ are far easier to oxidise than [Co(H2O)6] 2+

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(remember titrations using transition element ions are redox titrations)

TM catalysts:

Good catalysts because they change oxidation states by gaining or losing electrons within their d-orbitals. They transfer electrons to speed up reactions. 

Heterogeneous catalysts:

It is in a different phase from the reactants i.e. in a different physical state. E.g. in the Haber process, gases are passed over a solid iron catalyst. 

Reaction happens on the surface of the heterogenous catalyst. So, increasing the surface area of the catalyst increases the number of molecules that can react at the same time, increasing the rate of reaction.

Support mediums used to make the SA larger. E.g. catalytic converters contain a ceramic lattice coated with a thin layer of rhodium. Minimises cost. 

Rh acts as catalyst: 2CO (g) + 2NO (g) --Rh--> 2CO2 (g) + N2 (g)

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Heterogenous catalyst examples:

1. Iron in the Haber process for making ammonia:

N2 (g) + 3H2 (g) --Fe catalyst---> 2NH3 (g)

2. Vanadium (V) oxide in the Contact process for making sulfuric acid:

SO2 (g) + 1/2 O2 (g) --V2O5 catalyst---> SO3 (g)

3. Chromium (III) oxide in the manufacture of ethanol from CO:

CO (g) + 2H2 (g) ---Cr2O3 (s) catalyst----> CH3OH (g)

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Impurities can poison hetergenous catalysts:

1. During a reaction, reactants are absorbed onto active sites on the surfaces of heterogenous catalysts

2. Impurites in the reaction mixture may also bind to the catalysts surface and block reactants from being absorbed. Called catalyst poisoning

3. Catalyst poisoning increases the cost of a chemical process because lead product can be made in a certain time with an amount of energy. The catalyst may even need replacing or regenerated.

Lead poisons the catalyst in catalytic converters: The converters reduce harmful emissions from car engines. Lead can coat the surface of the catalyst in the converter, so the vehicles fitted with them can only have unleaded petrol

Sulfur poisons that iron catalyst in the Haber process. The Hydrogen in the process is produced from methane. Methane is obtained from natural gas which contains impurities, including sulfur compounds. Any sulfur that is not removed is absorbed onto the iron, forming iron sulfide, and stopping the iron being effective.

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This was really useful, thank you! 

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