transition metals (A2 inorganic chemistry) Flashcards
why are scandium and zinc not transition metals?
they cannot form at least one stable ion with a partially filled d-subshell
scandium forms only one stable ion, which is Sc³⁺, which has an empty d-subshell
zinc forms only one stable ion, which is Zn²⁺, which has a full d-subshell, as the 2 electrons are lost from the 4s orbital
electron configurations of transition metals
Ti - [Ar] 3d² 4s²
V - [Ar] 3d³ 4s²
Cr - [Ar] 3d⁵ 4s¹
Mn - [Ar] 3d⁵ 4s²
Fe - [Ar] 3d⁶ 4s²
Co - [Ar] 3d⁷ 4s²
Ni - [Ar] 3d⁸ 4s²
Cu - [Ar] 3d¹⁰ 4s¹
orbitals fill up singly first as they repel each other
in chromium and copper, one electron moves from the 4s orbital into the 3d sub-shell as a half-full or full 3d sub-shell gives them a more stable configuration
properties of transition metals
exhibit variable oxidation states - electrons sit in the 4s and 3d energy levels, which are very close, so as a result electrons are gained and lost using a similar amount of energy when they form ions
formation of coloured ions in solution
ability to form complex ions
act as good catalysts
colours formed by transition metal ions in solution - V²⁺, V³⁺, VO²⁺, VO₂⁺, Cr³⁺, Cr₂O₇²⁻, Mn²⁺, MnO₄⁻, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺, Cu²⁺
V²⁺ - violet
V³⁺ - green
VO²⁺ - blue
VO₂⁺ - yellow
Cr³⁺ - green/violet (violet when surrounded by 6H₂O, however these are normally substituted so usually looks green)
Cr₂O₇²⁻ - orange
Mn²⁺ - pale pink
MnO₄⁻ - purple
Fe²⁺ - pale green
Fe³⁺ - rust (orange/yellow)
Co²⁺ - pink
Ni²⁺ - green
Cu²⁺ - blue
explain how transition metals gain and lose electrons
transition metals fill the 4s orbital first and then the 3d, however they lose electrons from the 4s orbital first, this is unusual as usually atoms will lose electrons in the same order they gained them, however this can be explained by the fact that the 4s orbital has a slightly lower energy when empty, but becomes higher in energy once filled
structure of complex ions
central transition metal ions surrounded by ligands bonded by coordinate/dative covalent bonds, ions, atoms, or molecules that donate at least 1 lone pair of electrons
types of ligands and examples
monodentate e.g. H₂O, NH₃, Cl⁻ - have only one lone pair of electrons
bidentate e.g. ethanedioate (C₂O₄²⁻), ethane-1,2-diamine (NH₂CH₂CH₂NH₂) - 2 lone pairs of electrons
multidentate e.g. haem, EDTA⁴⁻ - form more than one coordinate bond, EDTA⁴⁻ can form 6 coordinate bonds with the central metal ion
complex shapes and examples, what factors affect the shape of a complex?
shape is dependent on size of ligands and coordination number (number of coordinate bonds)
some ligands are small and we can fit 6 of these around a central metal ion (H₂O, NH₃)
some ligands are larger and we can fit only 4 of these around the central metal ion (Cl⁻)
ethanedioate and ethane-1,2-diamine are even larger, we usually have 3 of these surrounding the central metal ion
complexes with a coordination number of 6 form octahedral shapes, with a 90° bond angle
complexes with a coordination number of 4 form tetrahedral shapes with a bond angle of 109.5°, or square planar (90° bond angle) in cis-platin, [Pt(NH₃)₂(Cl)₂] (aq)
complexes with a coordination number of 2 form linear shapes with 180° bond angles, specifically Tollen’s reagent [Ag(NH₃)₂]⁺ (aq)
how to work out charge of complex central metal ion
total oxidation state of metal = total oxidation state - total oxidation state of ligands
haem, haemoglobin, and carbon monoxide poisoning
in haemoglobin, an iron ion is surrounded by 5 nitrogens and one molecule of oxygen or water
4 of the nitrogens come from the multidentate ligand haem
1 of the coordinate bonds (another nitrogen) comes from a large globin protein
in the lungs, where oxygen concentration is high, oxygen substitutes the water ligand in haemoglobin to form oxyhaemoglobin (water is deposited and expelled from body when breathing out)
oxyhaemoglobin is transported around the body and gives up an oxygen to where it is needed, another water molecule is picked up (e.g. product of respiration in muscles) and haemoglobin returns to the lungs to repeat the process
if carbon monoxide is inhaled, a water ligand is replaced with a carbon monoxide ligand, instead of an oxygen ligand
carbon monoxide bonds strongly to the central iron ion, so it is not readily replaced by oxygen or water
since carbon monoxide bonds irreversibly, less oxygen can be transported around the body, leading to oxygen starvation in organs, which causes headaches, unconsciousness, and death
isomerism in complex ions
octahedral complexes with 3 bidentate ligands display optical isomerism
octahedral complexes with 4 ligands of the same type and 2 ligands of a different type display cis-trans isomerism (2 different ligands opposite each other - trans isomer, 2 different ligands adjacent to each other - cis isomer)
square planar complexes with 2 ligands of the same type and 2 ligands of a different type display cis-trans isomerism (2 different ligands opposite each other a trans isomer, 2 different ligands adjacent to each other - cis isomer)
we can see the significance of isomerism in complex ions since cis-[Pt(NH₃)₂(Cl)₂] is used to treat cancer and trans-[Pt(NH₃)₂(Cl)₂] is not
d-subshell splitting
when ligands are attached to a central metal ion, some d-orbitals gain energy, this creates an energy gap, ΔE
when electrons absorb light energy, some move from the ground state (lower energy level orbital) to the excited state (higher energy level orbital)
in order for this to happen, the energy from light must equal ΔE
ΔE equation
ΔE = hv = hc / λ
ΔE - change in energy (J)
h - Planck’s constant
v - frequency of light absorbed (Hz)
hc - speed of light (3.00x10⁸ ms⁻¹)
λ - wavelength of light absorbed (m)
what is the size of ΔE dependent on?
the central metal ion and its oxidation state
e.g. [Fe(H₂O)₆]²⁺ (aq) → [Fe(H₂O)₆]³⁺ (aq)
the colour of an iron 2+ complex ion is pale green, with an octahedral shape
when the central iron ion has a 3+ oxidation state, the colour is yellow, with an octahedral shape
[V(H₂O)₆]²⁺ (aq) → [V(H₂O)₆]³⁺ (aq)
when a vanadium complex ion has an overall charge of 2+, the colour observed is violet, and the complex forms an octahedral shape
when the vanadium ion has a 3+ charge, the complex is a green colour with an octahedral shape
we can see that a change in the oxidation state of the central metal ion alone affects ΔE, since the ligand type and coordination number remains the same
type of ligand
e.g. [Co(H₂O)₆]²⁺ (aq) + 6NH₃ (aq) → [Co(NH₃)₆]²⁺ (aq) + 6H₂O (l)
the colour of the [Co(H₂O)₆]²⁺ complex is pink, with an octahedral shape
the colour of the [Co(NH₃)₆]²⁺ complex is straw, with an octahedral shape
we can see that ligand substitution has changed the size of ΔE, and therefore the frequency of light absorbed, but coordination number and complex shape have remained the same
coordination number
e.g. [Cu(H₂O)₆]²⁺ (aq) + 4Cl⁻ (aq) → [CuCl₄]²⁻ (aq) + 6H₂O (l)
[Cu(H₂O)₆]²⁺ is a pale blue colour and an octahedral shape
[CuCl₄]²⁻ is a yellow complex with a tetrahedral shape
we can see that ligands substitution has caused a change in coordination number and shape, which has had an effect on ΔE
explain how transition metals display coloured complexes, and why other complexes cannot
the larger the ΔE the higher the frequency of light absorbed (red is lowest frequency, purple is highest frequency)
when white light (all frequencies combined) hits a transition metal, only one frequency is absorbed
any frequencies which are not absorbed are reflected or transmitted, and the combination of all these frequencies produces the colour we observe, which is complementary to the colour absorbed
e.g. [Cu(H₂O)₆]²⁺ absorbs frequencies that produce red light, so we observe the complementary colour of light blue/cyan
in complexes with a full or empty d-subshell, electrons cannot migrate to the higher energy level, therefore we see these complexes as colourless solutions or white compounds
colorimetry
we can use a colorimeter to measure the concentration of transition metal ions in solution by measuring their absorbance of light
we must first plot a calibration curve by measuring the absorption of a range of known concentrations of the metal ion solution (use a pipette to accurately measure volumes), and we can use this calibration graph to work out the concentration of the unknown sample
- set the colorimeter to zero by measuring the absorbance of a blank sample, usually the solvent the transition metal ion is being dissolved in (normally water)
- white light is filtered into a narrow range of frequencies to produce monochromatic light (choice of filter is importance as the colour produced from the filter must green absorbed by the metal ion solution)
- the sample is held in a cuvette (2 frosted sides, 2 clear sides), and monochromatic light passes through the sample where some light is absorbed
- light not absorbed by the sample travels to a detector which measures the level of absorbance of the sample by comparing it to the absorbance of the blank sample
substitution reactions
H₂O and NH₃ are similar sized ligands and both uncharged, ligand substitution can occur without a change in shape or coordination number
e.g. [Co(H₂O)₆]²⁺ (aq) + 6NH₃ (aq) → [Co(NH₃)₆]²⁺ (aq) + 6H₂O (l)
colour change from pink to straw, shape stays octahedral
ligands substitution between ligands of similar size may be incomplete/partial
e.g. [Cu(H₂O)₆]²⁺ (aq) + 4NH₃ (aq) → [Cu(NH₃)₄(H₂O)₂]²⁺ (aq) + 4H₂O (l)
this reaction occurs when [Cu(H₂O)₆]²⁺ reacts with excess ammonia
a colour change from blue to dark blue is observed, shape remains octahedral
we can also have ligand substitution with larger monodentate ligands, these also cause a change in coordination number and shape of complex ion
e.g. [Co(H₂O)₆]²⁺ (aq) + 4Cl⁻ (aq) → [CoCl₄]²⁻ (aq) + 6H₂O (l)
colour change from pink to blue, shape change from octahedral to tetrahedral
[Cu(H₂O)₆]²⁺ (aq) + 4Cl⁻ (aq) → [CuCl₄]²⁻ (aq) + 6H₂O (l)
colour change from blue to yellow, shape change from octahedral to tetrahedral
[Fe(H₂O)₆]³⁺ (aq) + 4Cl⁻ (aq) → [FeCl₄]⁻ (aq) + 6H₂O (l)
no colour change (remains yellow), but shape changes from octahedral to tetrahedral
different ligands can form stronger bonds to the central metal ion, these substitution reactions are not easily reversed
e.g. [Fe(H₂O)₆]³⁺ (aq) + 6CN⁻ (aq) → [Fe(CN)₆]³⁻ (aq) + 6H₂O (l)
cyanide ligands form stronger bonds than water ligands, the new complex formed is more stable
bidentate and multidentate ligands form more stable complexes than monodentate ligands, so these ligand exchange reactions are hard to reverse
e.g. [Cu(H₂O)₆]²⁺ (aq) + 3NH₂CH₂CH₂NH₂ (aq) → [Cu(NH₂CH₂CH₂NH₂)₃]²⁺ (aq) + 6H₂O (l)
explain the chelate effect
in ligand exchange reactions, the enthalpy change is usually very small as the energy needed to break the bonds is similar to the energy released when forming new bonds
despite this, ligand exchange reactions between monodentate ligands and bidentate/multidentate ligands are not considered reversible since the products are much more stable, this is due to entropy
having more moles on the product side than the reactant side is entropically a favoured process, reactions with an increase in entropy are more likely to happen
when we substitute monodentate ligands with bidentate or multidentate ligands, the number of particles increases as we have up to six moles on the product side just from the molecules that used to be ligands
this increase in stability is known as the chelate effect
give the oxidation states and the colours of the different ions of vanadium
V²⁺ - violet, +2
V³⁺ - green, +3
VO²⁺ - blue, +4
VO₂⁺ (vanadate(V) ion) - yellow, +5
reducing vanadium
VO₂⁺ ions can be reduced to V²⁺ ions using zinc in an acidic solution
2VO₂⁺ (aq) + Zn (s) + 4H⁺ (aq) → 2VO²⁺ (aq) + Zn²⁺ (aq) + 2H₂O (l)
colour change from yellow to blue
2VO²⁺ (aq) + Zn (s) + 4H⁺ (aq) → 2V³⁺ (aq) + Zn²⁺ (aq) + 2H₂O (l)
colour change from blue to green
2V³⁺ (aq) + Zn (s) → 2V²⁺ (aq) + Zn²⁺ (aq)
colour change from green to violet
what factors influence redox potentials of transition metal ion complexes?
a large redox potential E⦵ means a less stable ion that is more likely to be reduced
written in already reduced form
e.g. Zn²⁺ (aq) + 2e⁻ ⇌ Zn (s) E⦵ = -0.76
Cu²⁺ (aq) + 2e⁻ ⇌ Cu (s) E⦵ = +0.34
(Cu²⁺ is less stable)
E⦵ values follow standard conditions (298K, 100kPa, concentration of ions at 1moldm⁻³), there may be a difference between theoretical redox potentials recorded and actual redox potential calculated
ligands - standard redox potentials are measured in aqueous solutions, therefore the central metal ion is surrounded by water molecules, however ligands other than water may form stronger bonds with the central metal ion, so the redox potential could be higher or lower than the standard redox potential
pH - generally, if the solution is more acidic, redox potential will be larger and therefore the ion more easily reduced
e.g. MnO₂ (aq) + 4H⁺ (aq) + 2e⁻ ⇌ Mn²⁺ (aq) + 2H₂O (l)
this reaction is occurring in acidic conditions as H⁺ ions are needed for reduction, the low pH increases the redox potential, making MnO₂ more easily reduced
PbO₂ (aq) + H₂O (l) + 2e⁻ ⇌ PbO (s) + 2OH⁻ (aq)
as OH⁻ ions are produced as a result of reduction in this reaction, it is occurring in alkaline conditions, and the high pH decreases the redox potential, meaning PbO₂ is less easily reduced and the backward reaction is more likely
explain how Tollen’s reagent is used to distinguish between aldehydes and ketones
when we react enough aqueous ammonia to aqueous silver nitrate, we form colourless [Ag(NH₃)₂]⁺
the half equation for [Ag(NH₃)₂]⁺ has a large redox potential, therefore Ag⁺ is easily reduced to Ag
Ag⁺ (aq) + e⁻ ⇌ Ag (s) E⦵ = +0.80V
so when we add [Ag(NH₃)₂]⁺ to an aldehyde, the Ag⁺ ions are readily reduced to produce a positive test result of a silver mirror, and consequently the aldehyde is oxidised to a carboxylic acid
ketones are not oxidised by [Ag(NH₃)₂]⁺
RCHO (aq) + 2[Ag(NH₃)₂]⁺ (aq) + 3OH⁻ (aq) → RCOO⁻ (aq) + 2Ag (s) + 4NH₃ (aq) + 2H₂O (l)
redox titrations and examples
- add reducing/oxidising agent with unknown concentration but known volume to conical flask, and add excess sulfuric acid (H₂SO₄) to ensure there are sufficient H⁺ ions for the reduction of the oxidising agent
- add an oxidising/reducing agent with known concentration in burette and add until the faint colour of the ion appears/disappears (add drop-wise towards end point)
- read volume of oxidising/reducing agent used from bottom of meniscus at eye level and record results to 2 decimal places, repeat until 2 concurrent results are achieved
the redox titration between the reducing agents C₂O₄²⁻ (oxalate ions) or Fe²⁺ and the oxidising agent KMnO₄ (potassium permanganate) results in a faint purple colour in the reducing agents, which represents that all of the reducing agent has been oxidised and there is a slight excess of KMnO₄
redox titration calculations e.g. 18.3cm³ of 0.0250moldm⁻³ MnO₄⁻ reacted with 25cm³ of acidified iron(II) sulfate solution, calculate the concentration of Fe²⁺ ions
- write out balanced equation
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e⁻
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O - find moles of known reagent
moles of MnO₄⁻ = conc * vol = 0.0250 * 18.3x10⁻³ = 0.000458mol - use the molar ratio to find the miles of unknown reagent
moles of Fe²⁺ = 0.000458 * 5 = 0.00229mol - find the concentration of unknown reagent
conc = mol / vol = 0.00229 / 25x10⁻³ = 0.0916moldm⁻³
heterogenous catalysts
catalyst is in a different phase from the reactants
increasing surface area of a heterogenous catalyst will increase the rate of reaction as there are more active sites for more particles to react with the catalyst at the same time
this can be done by finely coating an inert support medium such as SiO₂ with the catalyst, which can minimise costs as less of the catalyst is required to cover more surface area
an example is the use of a solid iron catalyst in the Haber process (N₂ (g) + 3H₂ (g) → 2NH₃ (g))
homogenous catalysts
catalyst is in the same phase as the reactants, generally aqueous catalysts in aqueous reactants
form intermediate species, reactants combine with catalyst, which reacts to form products
catalyst is always reformed
an example is using sulfuric acid to make an ester
the contact process
the manufacture of sulfuric acid uses V₂O₅ as a heterogenous catalyst to catalyse SO₂ to SO₃
V₂O₅ + SO₂ → V₂O₄ + SO₃
V₂O₄ + 1/2O₂ → V₂O₅
SO₃ then bubbled through H₂O to form H₂SO₄
first, vanadium (V) oxidises sulfur dioxide to form our product, and is itself reduced, but secondly V₂O₄ is oxidised by oxygen to reform the catalyst
catalytic poisoning and example
heterogenous catalysts can be poisoned by impurities, which can bind to the surface of a catalyst and block the active sites for reactants to adsorb, reducing the surface area and slowing down the reaction
a poisoned catalyst means that less product is made, the catalyst needs to be replaced or cleaned more often, increasing the cost and time of the chemical process
e.g. the hydrogen required for the Haber process is made from methane, which contains sulfur impurities, and any sulfur that is not removed will adsorb to the surface of the iron catalyst and form iron sulfide, making the catalyst less efficient
energy profile diagrams of homogenous catalysts
homogenous catalysts form an energy profile diagram with a lower peak, but they also contain a dip and then another, even lower, peak
the first part can be explained as the catalyst lowering the activation energy so more particles have enough energy to react
at the dip, an intermediate species is formed (reactants combine with catalyst)
at the second peak, the intermediate species react to form products and the catalyst is reformed
oxidation of iodide ions
S₂O₈²⁻ (aq) + 2I⁻ (aq) → I₂ (aq) + 2SO₄²⁻ (aq)
this reaction is very slow and has a very high activation energy as we are trying to react negatively charged ions together, which repel
we can use Fe²⁺ as a catalyst to lower the activation energy and speed up the reaction
S₂O₈²⁻ (aq) + 2Fe²⁺ → 2Fe³⁺ (aq) + 2SO₄²⁻ (aq)
peroxodisulfate is reduced and catalyst is oxidised to produce our first product
2I⁻ (aq) + 2Fe³⁺ (aq) → 2Fe²⁺ (aq) + I₂ (aq)
the Fe³⁺ intermediate produced reacts with I⁻ ions to produce or second product and reform the catalyst
we can test the success of this reaction by adding a few drops of starch solution, a colour change to black-blue indicates a positive result
example of autocatalysis
2MnO₄⁻ (aq) + 16H⁺ (aq) + 5C₂O₄²⁻ (aq) → 2Mn²⁺ (aq) + 8H₂O (l) + 10CO₂ (g)
this reaction is very slow with a high activation energy as two negatively charged ions together will repel
Mn²⁺ is a product of the reaction, but also acts as a catalyst, meaning that as the reaction proceeds, the amount of product and rate of reaction increases
MnO₄⁻ (aq) + 8H⁺ (aq) + 4Mn²⁺ (aq) → 5Mn³⁺ (aq) + 4H₂O (l)
Mn²⁺ autocatalyses the reaction, converting MnO₄⁻ to Mn³⁺, producing the intermediate species
2Mn³⁺ (aq) + C₂O₄²⁻ (aq) → 2Mn²⁺ (aq) + 2CO₂ (g)
the intermediate formed reacts with the C₂O₄²⁻ to reform the catalyst and make product
why do transition metals make good catalysts
transition metals have variable oxidation states, which makes them good catalysts since their electrons sit in the 3d and 4s sub-shells, which are very close in energy, meaning they can easily give and receive electrons to form temporary bonds without being used up themselves
in general, catalysts are used to make products faster and can be used to lower the temperature required for a reaction to happen, which saves time, energy, and money, and is also usually better for the environment
