Topic 15 Transition metals

Question 1

Where are the d block elements?

Answer 1

They sit in the middle of the periodic table
Some of these elements are transition metals.
The ones you mainly need to know are the top row of the block

Question 2

What is a transition element?

Answer 2

It is a d block element that can form at least 1 stable ion with a partially filled subshell

The d sub shell can hold up to 10 electrons.
For period 4 elements, only 8 of them are transition metals. Scandium and zinc aren’t transition elements. This is because they don’t form a stable ion with a partially filled d sub shell

In the configuration, you can see that they fill up singly first then double up. This is because electrons repel each other

Chromium and copper behave differently. An electron from the 4s orbital moves into the 3d orbital to create a more stable half full or full 3d sub shell

Question 3

Scandium and zinc

Answer 3

Scandium and zinc aren’t transition metals

Scandium forms only one stable ion of Sc3+
Sc3+ has an empty d subshell. As it is not partially filled it is not a transition element

Zinc forms only 1 stable ion of Zn2+
Zn2+ has a full d - subshell. As it is not partially filled it is not a transition element

Question 4

What is the specific way transition metals lose electrons

Answer 4

They lose from the 4S subshell first then 3D

Question 5

What are the properties of transition metals?

Answer 5

Variable oxidation state, form coloured ions in solution and are good catalysts

Transition metals have variable oxidation states. This is because the electrons sit in 4s and 3d energy levels which are very close. As a result, electrons are gained and lost using a similar amount of energy when they form the ions seen in the right of the table

They also form coloured ions in solution.

Question 6

What are the colours of the transition metals when dissolved in water?

Answer 6

V2+ is violet
V3+ is green
VO^2+ is blue
VO2 + is yellow

Cr3+ is green / violet - Violet when surrounded by 6H2O. They are normally substituted so look green
Cr2O7 2- is orange

Fe2+ is pale green
Fe3+ is yellow

Co2+ is pink

Cu2+ is blue

Don’t need to know Ti, Mn or Ni

Question 7

Complex ions

Answer 7

Another property of transition metals is their ability to form complex ions

A complex ion is where a central transition metal ion is surrounded by ligands bonded by dative covalent bonds

You have a central metal atom. It is surrounded by ligands. These have at least 1 lone pair of electrons where they are used to form a dative covalent bond with the metal. A square bracket shows the full complex and the overall charge of the complex sits outside of this

Question 8

What is a ligand?

Answer 8

It is an ion, atom or molecule that has at least one lone pair of electrons. They can be monodentate, bidentate or polydentate.

Question 9

What are ligands with only one lone pair called and examples?

Answer 9

Ligands which only have 1 LP are called monodentate ligands. H2O NH3 CL- OH-

Question 10

What are ligands with 2 lone pairs called and examples?

Answer 10

Ligands which have 2 LP’s pf electrons are called bidentate ligands. Ethanedioate. Ethane-1,2-diamine.

Question 11

What are ligands with more than 1 coordinate bond called? examples

Answer 11

Ligands which have more than 1 coordinate bond are called multidentate ligands. EDTA4- is an example of a multidentate ligand. It can form 6 coordinate bonds with the central metal ion.

Question 12

What is a coordination number?

Answer 12

The shape is dependent on the size of the ligands and the coordination number

The coordination number is the number of coordinate bonds in a complex. IT IS NOT THE NUMBER OF LIGANDS

Question 13

How many ligands can you fit around a central metal ion?

Answer 13

Some ligands are small and you can fit 6 of them around a central metal ion. H2O NH3 OH-

Some ligands are larger and you can only fit 4 of them around the central metal ion - CL-.

Ethanedioate and Ethane -1,2-diamine are larger still. Normally you have 3 of these around a central metal ion

Question 14

What do complexes with a coordination number of 6 form?

Answer 14

They form octahedral shapes

Example would be CO[(H2O)6]2+ or CO[(NH3)6]2+

All bond angles in an octahedral complex are 90 degrees

Question 15

What do complexes with a coordination number of 4 form?

Answer 15

They form a tetrahedral shape and square planar shapes
Examples are [CuCl4]2-. 109.5 degrees in this tetrahedral shape

A specific example of square planar is the anti cancer drug cis - platen. Pt[(NH3)2(Cl)2]. Bond angles are always 90 degrees

Question 16

How to calc the overall charge of a complex shape?

Answer 16

Complexes have a charge which is the same as its total oxidation state.

Total oxidation state = total oxidation state - total oxidation state of ligands

Question 17

What is haem in haemoglobin?

Answer 17

It is a multidentate ligand that is found in the molecule haemoglobin.

It is a protein that is used to transport oxygen around the body in blood.
The structure is octahedral. 4 of the nitrogens come from 1 multidentate ligand called haem. One of the coordinate bonds comes from a large protein called globein. The final coordinate bond comes from either an oxygen or water molecule

Oxygen substitutes the water ligand where the oxygen concentration is high to form oxyhaemoglobin. This is transported around the body. Oxyhaemoglobin then gives up oxygen to a place where it is needed. Water takes the place and haemoglobin returns back to the lungs to start the process again

Question 18

How is haemoglobin and CO2 connected?

Answer 18

Carbon monoxide is a poisonous gas that causes headaches, unconsciousness and even death. It is known as the silent killer

If co is inhaled the water ligand is replaced with a carbon monoxide ligand/ Unfortunately carbon monoxide bonds strongly so it’s not readily replaced y oxygen or water. This means that oxygen can’t be transported and leads to oxygen starvation in organs. Hence why carbon monoxide is poisonous

Question 19

What is optical isomerism?

Answer 19

Complex ions show optical isomerism

Complexes are optical isomers when they are non superimposable mirror images

Octahedral complexes with 3 bidentate ligands show optical isomers like the ones shown to the left

Question 20

What is cis - trans isomerism in complex ions

Answer 20

Octahedral complexes with 4 ligands of the same type and 2 ligands of diff type display cis-trans isomerism.
If the 2 different ligands are opposite each other you have a trans isomer. If the 2 diff ligands are adjacent to each other you have a cis isomer

Square planar complexes with 2 ligands of the same type and 2 ligands of a different type display cis-trans isomerism. If the 2 ligands are opposite each other - you have a trans isomer but if the 2 ligands are adjacent we have a cis isomer.

Question 21

What is d orbital splitting?

Answer 21

The D subshell is split into 2 when ligands bond with the central metal ion.

For example in terms of arrows, 22111 will split into 11 above and 221 below

When electrons absorb light energy some move from the lowest energy level (ground state) to higher energy level orbitals (excited state)

In order for this to happen, the energy from the light must equal the change in energy

Question 22

What is the size of the energy gap dependent on?

Answer 22

The central metal ion and its oxidation state
The type of ligand
The coordination number

Question 23

How do coloured complexes work?

Answer 23

Some frequencies of visible light are absorbed by transition metal complexes. The frequencies absorbed depends on the size of the energy gap

Frequency increases from one side at red to the other side of the visible spectrum at violet/purple. Frequencies which aren’t absorbed are reflected or transmitted

[Cu(H2O)6]2+ absorbs frequencies that produce red light. The complementary colour is light blue (cyan) so we observe this colour.

For those complexes where we have a full or empty 3d subshell no electrons can migrate to the higher energy level. This means we see these complexes as colourless or white.

Question 24

What are complementary colours?

Answer 24

White light is made of the colours we see above. When this hits a transition metal solution, only one frequency is absorbed. Any frequency which is not absorbed are reflected or transmitted.

The combination of all these frequencies create a complementary colour that we observe. Essentially if we mix the complementary colour and the absorbed colour we get white light.

The colour directly opposite the colour observed is what is absorbed.
From top clockwise. Red Yellow Green Cyan Blue Magenta

Question 25

Transition metals and redox potentials

Answer 25

Redox potentials tell us how easily an ion is reduced which is the same as electrode potentials
The least stable ions have the largest redox potential and are more likely to be reduced

Zn2+ + 2e- <—-> Zn (-0.76)
Cu2+ + 2e- <—–> Cu (+0.34)
So Cu2+ is less stable

There may be a difference in redox potentials to the standard values seen in a data book. It is dependent on the environment the ions are in

Question 26

Vanadium chemistry

Answer 26

Ion Oxidation state Reduction half equation Reduction potential and then colour

V^2+ has an oxidation state pf +2. The reduction half equation is V^2+ + 2e- <===> V. The reduction potential is -1.18. The colour is violet

V^3+ has an oxidation state of +3. The reduction half equation is V^3+ + e- <===> V^2+. The reduction potential is -0.26

VO^2+ has an oxidation state of +4. The reduction half equation is VO^2+ + 2H+ + e- <====> V^3+ + H2O. The reduction potential is +0.34. The colour is blue

VO2 + The oxidation state is +5. The reduction half equation is VO2+ + 2H+ + e- <====> VO^2+ H2O. The reduction potential is +1 and the colour is yellow

1) VO2+ is reduced to VO^2+ and there is a colour change from yellow to blue.
2VO2+ + Zn + 4H+ —-> 2VO^2+ + Zn2+ + 2H2O

2) Then, VO^2+ is reduced to V3+ and there is a colour change of blue to green.
2VO^2+ + Zn + 4H+ —-> 2V^3+ + Zn2+ + 2H2O

3) Finally, V^3+ is reduced to V^2+ and the colour change is green to violet.
2V^3+ + Zn —> 2V^2+ + Zn2+

Last reduction (not shown here) won’t occur as our E theta value will be negative

Question 27

Chromium chemistry`

Answer 27

Cr2O7^2- has an oxidation state of +6 and it is orange. Also known as dichromate (VI) ions and they are good oxidising agents.

CrO4^2- has an oxidation state of + 6 and the colour is yellow. These are also known as chromate (VI ions, they are also good oxidising agents)

Cr^3+ has an oxidation state of green. This is the most stable ion, when surrounded by 6 water ligands a violet solution is formed. However, Cl- impurities replace the water ligands to make it look green.

Cr2+ has an oxidation state of +2 and is blue

Chromium ions can undergo oxidation and reduction too.

Here dichromate ions are reduced using zinc as a catalyst
Cr2O7^2- + 3Zn + 14H+ —-> 2Cr^3+ + 3Zn^2+ + 7H2O
Cr goes from +6 to +3. Zn goes from 0 to +2. Chromium will go from orange to green.

Zinc can be used to reduce Cr^3+ ions further to Cr^2+. However, Cr^2+ is very unstable and is readily oxidised back to Cr^3+ from oxygen in the air.
2Cr^3+ + Zn —-> 2Cr^3+ + Zn2+
Chromium goes from green to blue. Reduced from +3 to +2. Zinc goes from 0 to +2.

Cr^3+ is oxidised using hydrogen peroxide in alkaline solution. Yellow chromate (VI) is produced.
2Cr^3+ + 10OH- + 3H2O —-> 2CrO4^2- + 8H2O
Chromium ion goes from +3 to +6
Chromate ions (VI) ions (CrO4^2-) exist in equilibrium with dichromate ions (Cr2O4^2-)

Adding acid to yellow chromate ion solution creates orange dichromate ion solution. This exists in equilibrium so is a reversible reaction.
2CrO4^2- + 2H+ —–> Cr2O4^2- + H2O
+6 to to +6. It goes from yellow to orange.

Question 28

How is chromium hydroxide amphoteric?

Answer 28

Chromium hydroxide (Cr(H2O)3(OH)3) is amphoteric. When we add a base it acts as an acid and donates H+ ions to react with the OH- and it dissolves.
Cr(H2O)3(OH)3 + OH- —–> [Cr(OH)6]3- + 3H2O
When we add an acid it acts as a base by accepting the H+ ions and it dissolves.
Cr(H2O)3(OH)3 + 3H+ ——> [Cr(H2O)6]3+

The reactions shown above are not ligand exchange reactions. They are an example of acid - base reactions. The products are chemically changed through the addition or removal of H+ ions.
Be careful if we add excess ammonia to solid Cr(H2O)3(OH)3 we do get a ligand exchange reaction. Purple Cr(NH3)6^3+ solution is formed

Cr(H2O)3(OH)3 + 6NH3—-> [Cr(NH3)6]3+ + 3H2O + 3OH-
It goes from green grey solid to purple

Question 29

How can we hydrolyse metal aqua ions to form insoluble chromium hydroxides?

Answer 29

We can do this by adding a base (NH3 or OH- ions)

[Cr(H2O)6]3+ + 3OH- —–> [Cr(H2O)3(OH)3] + 3H20

[Cr(H2O)6]3+ + 3NH3 —–> [Cr(H2O)3(OH)3] + 3NH4+

It goes from green to a green grey precipitate

Question 30

How are chromium complexes made?

Answer 30

Most complexes can be made in a beaker, however some complexes such as chromium ethanoate Cr2(CH3COO)4(H2O)2 require a more complex method

First acidified sodium dichromate ions (orange) is reduced using zinc. It forms -
1) a green solution containing Cr3+ ions then
2) A blue solution containing Cr2+ ions

Cr2O7^2- + 3Zn + 14H+ —-> 2Cr^3+ + 3Zn^2+ + 7H2O. Orange to green.
2Cr^3+ + Zn —-> 2Cr^2+ + Zn^2+. Green to blue.
Remember - Cr^2+ ions are easily oxidised back to Cr^3+ so the whole experiment must be done under inert conditions. Normally a nitrogen atmosphere is used

Secondly, Sodium ethanoate reacts with Cr^2+ to form a red precipitate of chromium ethanoate.
2Cr^2+ + 4CH3COO- + 2H2O —-> [Cr2(CH3COO)4(H2O)2]. This is blue to red

Question 31

Transition metal substitution reactions

Answer 31

A colour change can exist when ligands in a complex exchange/substitute. These substitution reactions show ligands of a similar size being exchanged.

[Co(H2O)6]2+ + 6NH3 —-> [Co(NH3)6]2+ + 6H2O.
The colour goes from pink to straw. The shape remains octahedral. Both ammonia and water are similar size ligands and same charges so the complex shape remains octahedral.

[Cu(H2O)6]2+ + 4NH3 —–> [Cu(NH3)4(H2O)2]2+ + 4H2O. The colour goes from blue to dark blue. Shape remains octahedral. Here, substitution is partial. This reaction occurs when [Cu(H2O)6]2+ reacts with EXCESS ammonia

These next substitutions show ligands of a different size being exchanged. Cl- is a larger ligand than H2O or NH3. Only 4 CL- ion ligands can fit around so we have a change in shape and coordination number. Here are some examplesc

[Co(H2O)6]2+ + 4Cl- —–> [CoCl4]2- + 6H2O.
This goes from pink and octahedral to blue and tetrahedral.
[Cu(H2O)6]2+ + 4Cl- —–> [CuCl4]2- + 6H2O.
This goes from Blue and octahedral to Yellow and tetrahedral
[Fe(H2O)6]3+ + 4Cl- —–> [FeCl4]- + 6H2O.
This stays yellow but the shape changes from octahedral to tetrahedral

Question 32

Different ligands can form different strength bonds to the metal ion

Answer 32

In this example, the CN- ions form stronger bonds than the H2O molecules. This means this reaction is not easily reversed. The new complex formed is more stable.
[Fe(H2O)6]3+ + 6CN- —-> [Fe(CN)6]3- + 6H2O

Multidentate ligand can form complexes that are more stable than monodentate ligands. In this example, NH2CH2CH2NH2 is a bidentate ligand. This reaction is hard to reverse.
[Cu(H2O)6]2+ + 3NH2CH2CH2NH2 —–> [Cu(NH2CH2CH2NH2)3]2+ + 6H2O

Question 33

Entropy and stable complexes

Answer 33

Entropy increasing forms a more stable complex. In a ligand reaction, bonds are broken in the original complex and new ones are formed to make the new complex. Quite often the energy needed to break the bonds is similar as the energy released when new ones are formed. The enthalpy change is small like the reaction below

[Cu(H2O)6]2+ + 3NH2CH2CH2NH2 —–> [Cu(NH2CH2CH2NH2)3]2+ + 6H2O
6 coordinate bonds are breaking and then 6 are forming. This reaction is not considered reversible as the [Cu(NH2CH2CH2NH2)3]2+ product is much more stable than [Cu(H2O)6]2+. This is despite the fact that enthalpy change is small 8

The increase in stability is known as the chelate effect. When we substitute monodentate ligands with bidentate and multidentate ligands we create a solution with more particles in it. This means we increase in entropy. REACTIONS THAT HAVE AN INCREASE IN ENTROPY ARE MORE LIKELY TO OCCUR.

Use EDTA4- as an example. You will see the number of particles increase and so entropy increases too. It’s difficult to reverse these reactions as this would mean a decrease in entropy.
[Cr(H2O)6]3+ + EDTA4- —–> [Cr(EDTA)]- + 6H2O

Question 34

Examples of where oxidation state changes but the ligands and the coordinate number stays the same.

Answer 34

[Fe(H2O)6]2+ —–> [Fe(H2O)6]3+
Pale green to yellow. Stays an octahedral

[V(H2O)6]2+ —–> [V(H2O)6]3+
Violet to green. Stay octahedral

Question 35

Complex ion solution colours:

Answer 35

CU2+:
In aqueous solution:
[Cu(H2O)6]2+ or Cu2+. It will be blue
Add some OH- or NH3 to aqueous solution:
Cu(OH)2(H2O)4. It will be a pale blue precipitate.
Add excess OH- to the precipitate:
Insoluble in excess NaOH (No change)
Add excess NH3 to the precipitate:
[Cu(NH3)4(H2O)2]2+. Dark blue solution. Partial ligand substitution

Fe2+:
In aqueous solution:
[Fe(H2O)6]2+ or Fe2+. It will be pale green
Add some OH- or NH3 to aqueous solution:
Fe(OH)2(H2O)4. It will be a dirty green precipitate.
Add excess OH- to the precipitate:
Insoluble in excess NaOH (no change)
Add excess NH3 to the precipitate:
Insoluble in excess NH3

Fe3+:
In aqueous solution:
[Fe(H2O)6]3+ or Fe3+. Yellow
Add some OH- or NH3 to aqueous solution:
Fe(OH)3(H2O)3. Orange precipitate
Add excess OH- to the precipitate:
Insoluble in excess NaOH (No change)
Add excess NH3 to the precipitate:
Insoluble in excess NH3 (No change)

Co2+:
In aqueous solution:
[Co(H2O)6]2+ or Co2+. Pale pink.
Add some OH- or NH3 to aqueous solution:
Co(OH)2(H2O)4. Blue precipitate (Turns brown after a while)
Add excess OH- to the precipitate:
Insoluble in excess NaOH(No change)
Add excess NH3 to the precipitate:
Co(NH3)6^2+ (Turns brown after a while)

Question 36

Complex ion solution equations:

Answer 36

Cu2+:
Add some OH- or NH3 to the aqueous solution:
[Cu(H2O)6]2+ + 2OH- —-> [Cu(OH)2(H2O)4] + 2H2O
[Cu(H2O)6]2+ + 2NH3 —-> [Cu(OH)2(H2O)4] 2NH4+
1. Add excess OH- to the precipitate - Insoluble in excess NaOH (No Change)
2. Add excess NH3 to the precipitate
2. [Cu(OH)2(H2O)4] forms [Cu(NH3)4(H2O)2]2+

Fe2+:
Add some OH- or NH3 to the aqueous solution:
[Fe(H2O)6]2+ + 2OH- —-> [Fe(OH)2(H2O)4] + 2H2O
[Fe(H2O)6]2+ + 2NH3 —-> [Fe(OH)2(H2O)4] 2NH4+
1. Add excess OH- to the precipitate - Insoluble in excess NaOH (No Change)
2. Add excess NH3 to the precipitate
Insoluble

Fe3+:
Add some OH- or NH3 to the aqueous solution:
[Fe(H2O)6]3+ + 3OH- —-> [Fe(OH)3(H2O)3] + 3H2O
[Fe(H2O)6]3+ + 3NH3 —-> [Fe(OH)3(H2O)3] 3NH4+
1. Add excess OH- to the precipitate - Insoluble in excess NaOH (No Change)
2. Add excess NH3 to the precipitate
Insoluble in excess NH3

Co2+:
Add some OH- or NH3 to the aqueous solution:
[Co(H2O)6]2+ + 2OH- —-> [Co(OH)2(H2O)4] + 2H2O
[Co(H2O)6]2+ + 2NH3 —-> [Co(OH)2(H2O)4] 2NH4+
1. Add excess OH- to the precipitate - Insoluble in excess NaOH (No Change)
2. Add excess NH3 to the precipitate
2. [Co(OH)2(H2O)4] forms [Co(NH3)6 2+

Question 37

Types of catalysts:

Answer 37

Heterogeneous catalysts:
This is a catalyst in a different phase from the reactants. Different state. By increasing the SA of the heterogeneous catalyst will increase the rate of reaction. More particles can react with the catalyst at the same time

Homogeneous:
This is a catalyst that is in the same phase as the reactants. Generally, homogeneous catalysts are aqueous in aqueous reactants.
Homogeneous catalysts forms intermediate species by reactants combining with the catalysts which react to form products. The catalyst is reformed again.

Question 38

Catalysis and the contact process

Answer 38

As transition metals have variable oxidation states they are good catalysts by receiving and losing some electrons in the d orbitals to speed up reactions

The contact process uses vanadium to make sulfuric acid.
In the manufacture of sulfuric acid, we use V2O5 as a catalyst. It is used to catalyse SO2 to SO3. This is a heterogeneous catalyst.
1) V2O5 oxidises SO2 to SO3 and itself is reduced to V2O4.
V2O5 + SO2 —-> V2O4 + SO3
2) V2O4 is oxidised by oxygen to reform V2O5

Catalysts are used to make product faster and can be used to lower the temp required for a reaction. This saves energy and money!. Better for the environment
V2O4 + 1/2O2 ——> V2O5

Question 39

Heterogeneous catalyst can be poisoned by impurities

Answer 39

Impurities can bind to the surface of a catalyst and block the active site for reactants to adsorb.
When an impurity blocks a site we call this poisoning. Catalytic poisoning reduces the SA of the catalyst for the reactants to add to. This slows down the reaction.

For the Haber process:
The hydrogen is made from methane. Methane contains sulfur impurities. Any sulfur that is not removed will adsorb to the surface forming iron sulfide. The catalyst is less efficient.

A poisoned catalyst means less product is made, the catalyst needs to be replaced or cleaned more often, increased cost of the chemical process.

Question 40

How do heterogeneous catalysts work

Answer 40

Reaction occurs on solid heterogeneous catalyst. They bond with the surface of the catalyst. This is called adsorption.
The bonds in the reactants weaken and break to form radicals (atoms/molecules with unpaired electrons). The radicals react with each other to make new substances. The new molecules are then released to the surface of the catalyst in a process called desorption

Question 41

Homogeneous catalysts energy profile

Answer 41

It looks like 2 peaks and a dip in between. In this dip, intermediates are formed. Homogeneous catalysts forms intermediate species by reactants combining with the catalyst which react to form products. The catalyst is reformed again

The catalyst lowers the AE required for the reaction to occur. Remember, the catalyst is always re-formed and is never used up

Question 42

What is autocatalysis?

Answer 42

It is another form of homogeneous catalysis where the product catalyses the reaction. It catalyses itself.

Example is where Mn2+ is the catalyst in a reaction between C2O42- and MnO4-.
Mn2+ is a product and a catalyst. This means that as the reaction proceeds, the amount of products increases and so does the rate of reaction.

The reaction uncatalysed is very slow. We are trying to react 2 negatively charged ions together which repel. As a result, this reaction has a high AE.

2MnO4- + 16H+ + 5C2O42- —-> 2Mn2+ + 8H2O + 10CO2.

First the Mn2+ catalyses the reaction converting MnO4- into Mn3+
MnO4- + 8H+ + 4Mn2+ —–> 5Mn3+ + 4H2O

Secondly, the Mn3+ formed reacts with C2O42- and Mn2+ ions are reformed.
2Mn3+ + C2O42- —–> 2Mn2+ + 2CO2