inorganic - transition metals

Question 1

what is a transition metal

Answer 1

metal that form at least one stable ion with an incomplete d-shell of electrons

Question 2

what is a coordinate bond

Answer 2

one atom donates a pair of electrons to an electron-deficient atom to form a covalent bond

Question 3

what is a ligand

Answer 3

molecule or ion that forms a covalent bond with a transition metal by donating a pair of electrons - must have a lone pair

Question 4

what is a complex

Answer 4

central metal atom or ion surrounded by ligands

Question 5

what is the co-ordination number

Answer 5

number of co-ordiante bonds to the central metal atom or ion

Question 6

how do metals form complex ions

Answer 6

+ve charge on metal ion attracts -ve charge of ligand
-use lone pair to form a co-ordinate bond

Question 7

what are some comman ligands

Answer 7

Water H2O

Ammonia NH3

Chloride Cl–

Cyanide CN–

Hydroxide OH–

Ethanedioate (ox) –COO–COO–C2O42–

1,2-diaminoethane (en) H2NCH2CH2NH2

Question 8

what are monodentate ligands

Answer 8

Monodentate ligands can form only one dative bond to the central metal ion
Examples of monodentate ligands are:
Water (H2O) molecules
Ammonia (NH3) molecules
Chloride (Cl–) ions
Cyanide (CN–) ions

Question 9

what are bidentate ligands

Answer 9

Bidentate ligands can each form two dative bonds to the central metal ion
This is because each ligand contains two atoms with lone pairs of electrons
Examples of bidentate ligands are:
1,2-diaminoethane (H2NCH2CH2NH2) which is also written as ‘en’
Ethanedioate ion (C2O42- ) which is sometimes written as ‘ox’

Question 10

what are multidentate ligands

Answer 10

Some ligands contain more than two atoms with lone pairs of electrons
These ligands can form more than two dative bonds to the and are said to be multidentate ligands
An example of a multidentate ligand is EDTA4-, which is a hexadentate ligand as it forms 6 dative covalent bonds to the central metal ion

Question 11

what are the general properties of transition elements

Answer 11

Variable oxidation states
Form complex ions
Form coloured compounds
Behave as catalysts

Question 12

how do transition metals have variable oxidation states

Answer 12

Like other metals on the periodic table, the transition elements will lose electrons to form positively charged ions
However, unlike other metals, transition elements can form more than one positive ion
They are said to have variable oxidation states
Because of this, Roman numerals are used to indicate the oxidation state on the metal ion
For example, the metal sodium (Na) will only form Na+ ions (no Roman numerals are needed, as the ion formed by Na will always have an oxidation state of +1)
The transition metal iron (Fe) can form Fe2+ (Fe(II)) and Fe3+ (Fe(III)) ions

Question 13

why do transtion metals form complex ions

Answer 13

Due to the different oxidation states of the central metal ions, a different number and wide variety of ligands can form bonds with the transition element
For example, the chromium(III) ion can form [Cr(NH3)6]3+, [Cr(OH)6]3- and [Cr(H2O)6]3+ complex ions

Question 14

what colours do transition metals form in coloured compounds

Answer 14

Another characteristic property of transition elements is that their compounds are often coloured
For example, the colour of the [Cr(OH)6]3- complex (where oxidation state of Cr is +3) is dark green
Whereas the colour of the [Cr(NH3)6]3+ complex (oxidation state of Cr is still +3) is purple

Question 15

how do transition metals act as catalysts

Answer 15

Since transition elements can have variable oxidation states, they make excellent catalysts
During catalysis, the transition element can change to various oxidation states by gaining electrons or donating electrons from reagents within the reaction
Substances can also be adsorbed onto their surface and activated in the process

Question 16

complex ions with water and ammonia

Answer 16

Water and ammonia molecules are examples of neutral ligands
Both ligands contain a lone pair of electrons which can be used to form a dative covalent bond with the central metal ion
In water, this is the lone pair on the oxygen atom
In ammonia, it is the lone pair on the nitrogen atom
Since water and ammonia are small ligands, 6 of them can usually fit around a central metal ion, each donating a lone pair of electrons, forming 6 dative bonds
Since there are 6 dative bonds, the coordination number for the complex is 6
The overall charge of a complex is the sum of the charge on the central metal ion, and the charges on each of the ligands
A complex with cobalt(II) or chromium(II) as a central metal ion, and water or ammonia molecules as ligands, will have an overall charge of 2+
The central metal ion has a 2+ charge and the ligands are neutral

Question 17

complexes with hydroxide and chloride ions

Answer 17

Hydroxide and chloride ions are examples of negatively charged ligands
Both ligands contain a lone pair of electrons which can be used to form a dative covalent bond with the central metal ion
Hydroxide ligands are small, so 6 of them can fit around a central metal ion and the complex formed will have a coordination number of 6
Chloride ligands are large ligands, so only 4 of them will fit around a central metal ion
Complexes with 4 chloride ligands will have a coordination number of 4
A complex with cobalt(II) or copper(II) as a central metal ion and chloride ions as ligands, will have an overall charge of 2-
The central metal ion has a charge of 2+
Each chloride ligand has a charge of 1-
There are 4 chloride ligands in the complex, so the overall negative charge is 4-
The overall positive charge is 2+
Therefore, the overall charge of the complex is 2-
A complex with chromium(III) as a central metal ion and hydroxide ions as ligands, will have an overall charge of 3-
The central metal ion has a charge of 3+
Each hydroxide ligand has a charge of 1-
There are 6 hydroxide ligands in the complex, so the overall negative charge is 6-
The overall positive charge is 3+

Question 18

what are the shapes of the complex ions

Answer 18

-linear
-tetrahedral
-square planar
-octohedral

Question 19

what are linear complex ions

Answer 19

Central metal atoms or ions with two coordinate bonds form linear complexes
The bond angles in these complexes are 180o
The most common examples are a copper (I) ion, (Cu+), or a silver (I) ion, (Ag+), as the central metal ion with two coordinate bonds formed to two ammonia ligands

Question 20

what are tetrahedral complex ions

Answer 20

When there are four coordinate bonds the complexes often have a tetrahedral shape
Complexes with four chloride ions most commonly adopt this geometry
Chloride ligands are large, so only four will fit around the central metal ion
The bond angles in tetrahedral complexes are 109.5o

Question 21

what are square planar complex ions

Answer 21

Sometimes, complexes with four coordinate bonds may adopt a square planar geometry instead of a tetrahedral one
Cyanide ions (CN-) are the most common ligands to adopt this geometry
An example of a square planar complex is cisplatin
The bond angles in a square planar complex are 90o

Question 22

what is a octahedral complex

Answer 22

Octahedral complexes are formed when a central metal atom or ion forms six coordinate bonds
This could be six coordinate bonds with six small, monodentate ligands
Examples of such ligands are water and ammonia molecules and hydroxide and thiocyanate ions
It could be six coordinate bonds with three bidentate ligands
Each bidentate ligand will form two coordinate bonds, meaning six coordinate bonds in total
Examples of these ligands are 1,2-diaminoethane and the ethanedioate ion
It could be six coordinate bonds with one multidentate ligand
The multidentate ligand, for example EDTA4-, forms all six coordinate bonds
The bond angles in an octahedral complex are 90o

Question 23

how can transition element complexes exhibit stereoisomerism

Answer 23

Even though transition element complexes do not have a double bond, they can still have geometrical isomers
Square planar and octahedral complexes with two pairs of different ligands exhibit cis-trans isomerism (this is a special case of E-Z isomerism)

Question 24

how can complex ions exhibit optical isomerism

Answer 24

Octahedral complexes with bidentate ligands also have optical isomers
This means that the two forms are non-superimposable mirror images of each other
They have no plane of symmetry, and one image cannot be placed directly on top of the other
The optical isomers only differ in their ability to rotate the plane of polarised light in opposite directions

Question 25

why are complex ions coloured

Answer 25

-in transition metal compounds the five d orbitals do not all have the same energy - some electrons are in the ground state and excited state
-gap in energy between the d orbitals corresponds to the energy of the UV/visible light
-electrons absorb the UV/visible light to provide the energy to promote/ excite the electrons to the higher energy level
-the colours seen are the ones that are not absorbed and are reflected back
-the remaining colours are absorbed due to the transtition metal ions ability to excite the electrons to higher orbitals

Question 26

what factors affect the colour of the compound

Answer 26

-the identity of the metal
-oxidation state of the metal
-identity of the ligands
-co-ordination number
if any of these factors are changed then the size of the enrgy gap between the higher and lower d orbital chnages - so the frequency of the light absorbed changes and so the colour of the light seen changes

Question 27

what is the equation to calculate the gap of energy in the d orbital

Answer 27

delta E = hf = hc/wavelength of light

h=plancks constant
c=velocity of light

Question 28

what is Plancks constant

Answer 28

6.63 x10-34 J

Question 29

what is the value of the velocity of light

Answer 29

3.00 x108

Question 30

what can be used to measure absorbance of uv/visible light

Answer 30

uv/visible light spectrometer

Question 31

how does a uv/visible light spectrometer work

Answer 31

-light is passed through the complex and the frequences of uv/vis light passing through is detected - those that do not pass are absorbed

Question 32

how is the conc of a complex measured using the light spectrometer

Answer 32

-the more concentrated the solution the more light is absorbed and thus they can be used to measure conc of the solution

Question 33

how can colorimeters find the conc of a complex solution

Answer 33

-a colour of light is chosen that the compounds absorb
-the strenth of the absorption of a range of solutions of known conc is measured and a graph/calibration curve is produced
-the conc of a soltuion of unknown conc can be found measuring the absorption and using the graph to read off

Question 34

what is the wavelength of a substance with a high frequency

Answer 34

low wavelength

Question 35

what is the wavelength of a substance with a low frequency

Answer 35

high wavelength

Question 36

what are the two types of catalyst

Answer 36

heterogeneous
homogeneous

Question 37

what is a heterogeneous catalyst

Answer 37

in a different physical state (phase) from the reactants
eg
vanadium(V) oxide, V2O5, is a solid catalyst used in the Contact process for making sulfuric acid
Another example is the use of solid iron, Fe, in the Haber process for making ammonia

Question 38

how does a heterogeneous catalyst work

Answer 38

-one of the reactants is adsorbed into the surface and forms bonds with the atoms
-into places known as active sites
-an effective catalyst forms many bonds
-they then either oxidise or reduce another element by changing their oxidation state

Question 39

what are some of the problems with heterogeneous catalysts

Answer 39

-very expensive
In order to minimise the cost and maximise the efficiency of the catalyst the following measures can be taken:
Increasing the surface area of the catalyst
Coating an inert surface medium with the catalyst to avoid using large amounts of the catalyst
-catalyst poisonings - so it lowers the efficiency

Question 40

how do heterogeneous catalysts increase the speed of the reaction

Answer 40

-adsorption increases the conc of the reactants - brings them closer to the gas phase increase the collisions
-weaken the bonds
-put in the favourable orientation

Question 41

what are homogeneous catalysts

Answer 41

in the same phase as the reactants
-reactions involve intermediate species formed from a reactant and the catalyst and which then reacts further and regenerates the catalyst

Question 42

how does an acid catalysis esterification

Answer 42

-acid catalysts protonates one of the reactants
-species produced then reacts with anther reactant to give the products and reforms the catalyst

Question 43

how do transition metals act as catalysts

Answer 43

Transition element ions can adopt more than one stable oxidation state
This means that they can accept and lose electrons easily to go from one oxidation state to another
They can therefore catalyse redox reactions, by acting as both oxidising agents and reducing agents

Question 44

how does the iron ion act as a catalyst

Answer 44

For example, iron (Fe) is often used as a catalyst due to its ability to form Fe(II) and Fe(III) ions, acting as an oxidising agent and a reducing agent
When Fe(II) acts as a reducing agent, it will reduce another species and become oxidised itself
Fe2+ → Fe3+ + e-

The Fe3+ formed in the catalytic cycle, can then also act as an oxidising agent by oxidising another species and getting reduced itself to reform the Fe2+ ion
Fe3+ + e- → Fe2+

Iron(II) ions catalyse the reaction between iodide ions, I-, and peroxodisulfate ions, S2O82-
The overall reaction is quite slow because the repulsion of two negative ions coming together hinders the reaction
S2O82- + 2I- → I2 + 2SO42-

The reaction is quite slow even though it is energetically favourable. Both ions are negatively charged, so they are unlikely to make successful collisions with one another. However, if iron(II) ions are added to the reaction, the rate is much quicker. Starch is often added to this reaction, which will form a blue-black colour showing the formation of iodine
The addition of iron(II) ions reduces the peroxodisulfate to sulfate ions and produces iron(III) in the process
S2O82- + 2Fe2+ → 2SO42- + 2Fe3+

The iron(III) ions will oxidise iodide ions to iodine and then are reduced once again to iron(II)
2I- + 2Fe3+ → I2 + 2Fe2+

Question 45

why are reactions involving the catalyst often quite slow

Answer 45

two negative charges repelling one and other

Question 46

what is autocatalysis

Answer 46

Autocatalysis the term used to describe a reaction which is speeded up by a product which acts as a catalyst for the reaction
If you plot a rate graph of concentration versus time it has an usual shape
The gradient becomes steeper during the course of the reaction which tells you the rate is speeding up, not slowing down over time as the reactants become used up

Question 47

example of autocatalysis

Answer 47

An example of an autocatalysed reaction takes place between manganate(VII) ions and oxalate (ethandioate) ions
You can see that one of the products is manganese(II) ions - this is the catalyst
As more manganese(II) is formed the reaction speeds up
Like to the role of iron(II) in the previous section, manganese(II) ions take part in a redox cycle between two different oxidation states (+2 → +3 → +2)
4Mn2+ (aq) + MnO4– (aq) + 8H+ (aq) → 5Mn3+ (aq) + 4H2O (aq)

2Mn3+ (aq) + C2O42- (aq) → 2CO2 (g) + 2Mn2+ (aq)

The manganese(II) is not present in the beginning of the reaction, but as it is formed is speeds up the reaction and is re-generated during the redox cycle
This reaction is easily followed on a colorimeter as the rate at which the purple manganate(VII) ion is consumed accelerates with time

Question 48

explain the contact process in terms of the catalysis via V2O5

Answer 48

The manufacture of sulfuric acid is a very important piece of industrial chemistry that makes use of heterogeneous catalysis
The first step of the process is roasting sulfur in air to produce sulfur dioxide
S (s) + O2 (g) → SO2 (g)

The second step is an equilibrium reaction which is catalysed by vanadium(V) oxide, V2O5,
2SO2 (g) + O2 (g) ⇌ 2SO3(g)

The vanadium(V) oxide catalyst converts sulfur dioxide into sulfur trioxide and is reduced to vanadium(IV) oxide
SO2 (g) + V2O5 (s) → V2O4 + SO3 (g)

The vanadium(V) oxide is then re-generated by reaction with oxygen, fulfilling its role as a catalyst
O2 (g) + 2V2O4 (s) → 2V2O5 (s)

This reaction shows that a variable oxidation state can also be utilised in heterogenous catalysis

Question 49

what is the ratio of MnO4- ions to Fe2+ ions

Answer 49

1:5
x moles of Mn04- by 5

Question 50

what is the ratio of MnO4- ions to C2O4 ions

Answer 50

1:2.5
so multiply by mol of Mn04- by 2.5

Question 51

what are redox titrations

Answer 51

Redox titration involves an oxidising agent being titrated against a reducing agent
Electrons are transferred from one species to another
In acid-base titrations indicators are used to show the endpoint of a reaction; however redox titrations using transition metal ions naturally change colour when changing oxidation state, so indicators are not always necessary
They are said to be ‘self-indicating’
Two common transition metal ion redox titrations are manganate(VII) and dichromate(VI) titration

Question 52

how does potassium manganate act as an oxidising agent

Answer 52

-reduced as oxidises the reactant
-add with dilute sulphuric
In acidic solutions it is reduced to the almost colourless manganese(II) ion (the ion is actually pink, but in low concentrations it is effectively colourless)
The reduction equation for the manganate(VII) ion is
MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (aq)

purple colourless

The potassium ion is a spectator ion so can be left out of the equations
By convention the potassium manganate(VII) solution is placed in the burette, so that as it reacts with the reducing agent the solution becomes colourless
At the endpoint the manganate(VII) ion becomes in excess so the first appearance of a permanent colour change marks the endpoint
The colour seen is pink; if the colour is purple then you have overshot the endpoint and there is too much manganate(VII) in excess
Analysis of iron in iron(II)sulfate tablets is typical manganate(VII) titration

Question 53

why does a titration with C2O4 need to be warmed

Answer 53

the MnO4- ion and C2O4 ion are both -ve therefore they repel and the reaction slows - so needs to be warmed to start the reaction again - takes a while for the purple manganeate to react

Question 54

A health supplement tablet contain iron(II)sulfate was analysed by titration. A tablet weighing 2.25 g was dissolved in dilute sulfuric acid and titrated against 0.100 mol dm-3 KMnO4 .The titration required 26.50 cm3 for complete reaction. Calculate the percentage by mass of iron in the table

Answer 54

Step 1: Write the balanced equation for the reaction

oxidation: Fe2+ (aq) → Fe3+ (aq) + e-

reduction: MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (l)

overall: MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

Step 2: Determine the amount of MnO4- used in the titration

moles of MnO4- = 0.0265 dm3 x 0.100 mol dm-3 = 0.00265 mol

Step 3: Determine the amount of iron in the reaction

  From the equation for the reaction we know the reacting ratio  MnO4- : Fe2+ = 1: 5

   ∴ moles of Fe2+ = 0.00265 mol MnO4- x 5 = 0.01325 mol

Step 4: Convert moles into mass of iron

Mass of iron = 0.01325 mol x 55.85 gmol-1 = 0.740 g

Step 5: find the percentage of iron in the tablet

∴ % Fe in the tablet = (0.740/ 2.25) x 100 = 32.9%

Question 55

why is copper iodide a white powder

Answer 55

Full (3)d (sub)shell or (3)d10
No (d-d) transitions possible/ cannot absorb visible/white light