Transition metals exam past paper qs Flashcards
2023 7C
Chromium(III) sulfate, Cr2
(SO4)3, dissolves in water to form the complex ion [Cr(H2O)6]3+(aq).
(i) State the colour of this complex ion.
(ii) Explain why the aqueous solution of this complex ion has an acidic pH by
considering the interaction between the metal ion and the ligands.
(i) Green
(ii) the (highly charged, small) Cr3+ ion weakens / polarises /
distorts the O-H bonds (in the water ligands) (1)
- allowing some water ligands to release hydrogen ions / to
form the oxonium ion
or
water molecule (from the solution) deprotonates (one of) the
water ligands
2023 7D
A student researching the role of dichromate(VI) ions, Cr2
O7
2– , as an oxidising
agent made the statement shown.
‘Standard electrode potential data shows that it is never feasible
for a 1.00 mol dm–3 solution of potassium dichromate(VI) to oxidise
the chloride ions in hydrochloric acid.’
Comment on this statement using the data and equilibria shown.
Equilibrium 1
Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2
O(l) Ed = +1.33 V
Equilibrium 2
Cl2 (aq) + 2e– 2Cl–(aq) Ed = +1.36 V
Eocell = −0.03V and so reaction / oxidation of chloride
ions is not feasible (under standard conditions) (1)
- (this is small so) changing the conditions may make
Eocell positive so the oxidation reaction becomes feasible
(1) - (it is possible under non-standard conditions if you)
increase the concentration of HCl / H+ / Cl‒ (1) - so equilibrium 1 moves to RHS
or
equilibrium 2 moves to LHS
(so oxidation of Cl− ions is more likely / feasible)
Transition metals and their compounds can act as catalysts in many reactions such as
the ones shown:
* platinum, Pt, in the catalytic converters of vehicles
* manganese(II) ions, Mn2+(aq), in the oxidation of ethanedioate ions, C2O4 2–(aq),
by manganate(VII) ions, MnO4–(aq).
Compare and contrast the role of the catalysts in these reactions (6)
IP1 Similarity
both (increase the rate of a reaction) by providing an alternative mechanism / route
with lower activation energy (and are chemically unchanged at the end of the
reaction)
IP2 Heterogeneous / homogeneous
platinum is heterogenous / different physical state to reactants
and
Mn2+ is homogenous / same physical state to reactants
Platinum
IP3 reactants adsorb onto platinum (surface)
IP4 the reactant bonds are weakened and the products are desorbed
Mn2+
IP5 the negative ions repel so the reaction starts slowly
IP6 Mn2+ acts as an autocatalyst
or
the reaction then speeds up as Mn2+ catalyst forms
or
the reaction is faster between positive ions / Mn2+ and negative ions
A MnO2(s) + 4H+(aq) + e– Mn3+(aq) + 2H2O(l)
E-cell= +0.95
B Mn3+(aq) + e– Mn2+(aq)
E-cell= +1.51
Explain why Mn3+ ions are unstable in aqueous solution.
Include an equation and the type of reaction that occurs. (4)
(some) Mn3+ (ions from half-cell B) will oxidise (other) Mn3+ (ions
from half-cell A)
and
(some) Mn3+ (ions) will reduce (other) Mn3+ ions (1)
- as Eo for half-cell B is more positive / higher than Eo for half-cell A
or
as Eocell = (+)0.56 V (1) - 2Mn3+(aq) + 2H2O(l) → Mn2+(aq) + MnO2(s) + 4H+(aq) (1)
- disproportionation (reaction)
(i) The standard electrode potentials for two half-equations involving bromine are given.
Br2(aq) + 2e– 2Br–(aq) Ed = +1.09 V
2HOBr(aq) + 2H+(aq) + 2e– Br2
(aq) + 2H2O(l)
Ed = +1.57 V
(i) Explain why the disproportionation of bromine in water is not
thermodynamically feasible under standard conditions.
Include the overall equation for the disproportionation and its Ed cell value (3)
(ii) Bromine disproportionates in water to a small extent at 298 K.
Give a possible reason why this reaction occurs (1)
M1
Br2(aq) + H2O(l) → HOBr(aq) + HBr(aq)
M2
Eocell = 1.09 – 1.57 = −0.48 (V)
M3
Eocell / answer is negative / <0 and
the reaction is not
(thermodynamically)
feasible
(ii) disproportionation is an equilibrium system
(and although K is very small, there is still a
small concentration of disproportionation
products)
or
excess water is used
or
concentration is not 1 mol dm−3
or
HOBr undergoes further disproportionation
The common oxidation numbers of chromium are +2, +3 and +6.
Give a reason, in terms of ionisation energies, why chromium can show variable
oxidation numbers (1)
there is only a gradual / steady increase in (successive
ionisation energies)
The bonding in chromate(VI) ions, CrO4
2–, is similar to that in sulfate(VI) ions, SO42–.
Draw a possible dot-and-cross diagram for a chromate(VI) ion (2)
2 double bonds and 2 single bonds shown as dots
and crosses (1)
- Another 4 electrons around each oxygen involved in
the double bond and another 6 electrons around
each oxygen involved in the single bond with one
different symbol on each of two oxygens to indicate
the extra electrons in the ion
A student added some pieces of zinc to an acidified solution of
potassium dichromate(VI).
Some standard electrode potentials are given in the table.
Zn2+(aq) + 2e– Zn(s) –0.76
Cr3+(aq) + e– Cr2+(aq) –0.41
Cr2O7 2–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l) +1.33
(i) Write the overall equation for the reduction of dichromate(VI) ions to
chromium(III) ions by zinc in acid conditions.
State symbols are not required. (2)
(ii) Calculate Ed cell for the reaction in (c)(i). (1)
(iii) Predict whether or not a further reduction of chromium(III) ions to
chromium(II) ions will occur. Justify your answer (1)
(i) Cr2O72− + 14H+ + 3Zn → 2Cr3+ + 7H2O + 3Zn2+
(ii) (Eocell = 1.33 – (−0.76) )
= (+) 2.09 (V)
(iii) yes/zinc and acid will reduce chromium(III) ions
to chromium(II) ions
and because
Eocell for the reaction between Zn and Cr3+ is (+)
0.35 (V)
or
Zn2+ / Zn electrode potential / SEP / Eo value is
more negative / less positive / lower than the
Cr3+ / Cr2+ value
or
Zn/ Zn2+ electrode potential / SEP / Eo value is
less negative / more positive / higher than the
Cr3+ / Cr2+
Aqueous solutions containing chromium(III) ions and chromium(II) ions have different colours.
Explain why these solutions differ in colour.
An explanation of the origin of the colours is not required (2)
the energy difference between the two sets of d
orbitals is different in the two ions / Cr3+ and Cr2+
or
there is different splitting of the d orbitals / d
subshell (1)
- electrons undergo different d-d transitions/ are
promoted to a higher d-orbital absorbing/requiring a
different amount of energy
or
a different amount of energy is absorbed the
frequency / wavelength/colour of (visible) light
absorbed is different
2022
A fuel cell produces a voltage from the reaction between a fuel and oxygen.
The reaction occurring at one electrode in a methanol fuel cell is
CH3OH(g) + H2O(l) → HCOOH(aq) + 4H+(aq) + 4e
Which reaction occurs at the other electrode?
A 4H+(aq) + O2(g) + 4e– → 2H2O(l)
B 2H2(g) + 2O2(g) + 4e– → 4OH–(aq)
C 4OH–(aq) → 2H2(g) + 2O2(g) + 4e
D 2H2O(l) → 4H+(aq) + O2(g) + 4e
A
2022
Lead-acid batteries are used as storage cells in some cars.
The electrolyte is sulfuric acid, one electrode is lead and the other is
lead(IV) oxide, PbO2.
As the cell discharges, the lead and the lead(IV) oxide are both converted to solid lead(II) sulfate, PbSO4 , and the concentration of the sulfuric acid decreases.
Deduce, using the information given, the two half-equations occurring in the
lead-acid battery.
State symbols are required. (3)
Pb(s) + SO42−(aq) ⇌ PbSO4(s) + 2e−
PbO2(s) + 4H+(aq) + SO42−(aq) + 2e− ⇌ PbSO4(s) + 2H2O(l)
2021
Transition metals form complex ions.
(a) Complex ions have a central metal ion surrounded by ligands.
(i) Give a reason why the ammonium ion cannot act as a ligand (1)
ammonium ions do not have a lone pair (of electrons for bonding)
2021
(ii) Explain why the complex ions [Co(NH3)6]2+ and [Co(H2O)6]2+ are coloured and
have different colours (2)
d orbitals/d sub-shell split (into two different energies) (1)
- difference in energy depends on the ligands (1)
- difference in energy leads in different
frequencies/wavelengths/photons of light absorbed (1) - (so) the unabsorbed frequencies/wavelengths/photons are
reflected/transmitted
2021
Compare and contrast the complex ions formed by cobalt(III) ions with the ligand ethane-1,2-diamine and with the ligand EDTA4–.
Ignore any difference in colour (6)
(Similarities)
At least one from
* both ligands form dative covalent bonds with the cobalt(III) ions (1)
- both have coordination number 6 (1)
- both complex ions will be octahedral (1)
(Differences)
At least one from
* EDTA is hexadentate, ethane-1,2-diamine is bidentate
OR
ratio of cobalt(III) to EDTA is 1:1,
with ethane-1,2-diamine it is 1:3 (1)
- complex with EDTA will be anionic / negatively charged, with
ethane-1,2-diamine will be cationic / positively charged (1) - complex of EDTA is more stable than the complex with
ethane-1,2-diamine because there is an increase in entropy (1)
2021
Hydrated chromium(III) chloride, CrCl3.6H2O, dissolves in water to form a number of different complex ions containing both chloride and water ligands.
The general formula of these complex ions is Cr(H2O)x(Cl)y+
In an experiment, 0.10 mol of a complex reacted with excess silver nitrate solution to produce 0.20 mol of silver chloride.
Chloride ions which are ligands within the complex do not react with silver nitrate.
Deduce the formula of this chromium(III) complex ion. Justify your answer (2)
two moles of chloride ions in aqueous solution so one mole of chloride
ion is in the complex
- complex ion formula
[Cr(H2O)5(Cl)]2+
2021
A catalytic converter decreases the emissions of gases, such as carbon monoxide
and nitrogen monoxide, from an internal combustion engine.
Describe the stages in a catalytic converter that result in this decrease.
No equations are required (3)
adsorption of CO and/or NO molecules on the catalytic surface (1)
- weakening of bonds
(and chemical reaction between CO and NO) (1) - desorption of CO2 and/or N2 /product (molecules) from the
catalytic surface
2020
A student stated that ‘the elements scandium and zinc are d-block elements but are not transition metals’.
Discuss this statement, using appropriate electronic configurations to support
your answer (4)
both elements / atoms have the last added electron in the
d-subshell / d orbital (so are d-block elements) (1)
but neither forms a (stable) ion with an incomplete d-subshell / d orbital
(so are not transition metals) (1)
Zn2+ is 1s22s22p63s23p63d10 (so d subshell is full) (1)
Sc3+ is 1s22s22p63s23p6 (so d subshell is empty)
2020
When chromium(III) sulfate dissolves in water, a green solution containing the
[Cr(H2O)6]3+ ion forms.
(i) Give the shape of this complex ion
(ii) Explain why the chromium complex ion is coloured.(3)
(i) Octahedral
(ii) (ligand / water molecule causes) d orbitals to split (into 2 energy levels) (1)
light/energy (in the visible region) absorbed to promote electrons (to
higher d orbitals) (1)
the remaining light / unabsorbed light / complementary colour / green light
is transmitted (1)
2020
When a solution of EDTA4− is added to a solution of [Cr(H2O)6]3+ ions, a new complex ion is formed.
[Cr(H2O)6]3+ + EDTA4− ⇌ [Cr(EDTA)]− + 6H2O
The equilibrium constant for this equilibrium is 2.51 × 1023 dm3 mol−1.
By considering the equilibrium for this reaction and changes in entropy, comment
on the value of the equilibrium constant. No calculations are required. (3)
(the equilibrium constant is large) so the equilibrium lies well / far to
the right hand side / [Cr(EDTA)]– is more stable (1)
as ∆Ssystem is positive / increases entropy (of the system) (1)
as 2 mol of reactants form 7 mol of products
2020
State the initial and final colours you would see as the chlorine bubbles
through the aqueous vanadium(II) chloride, VCl2 (aq). (2)
purple / lilac / violet to yellow (solution)
2019
This is a question about catalysis.
(a) The reaction between iodide ions and peroxodisulfate ions is catalysed by iron(II) ions.
2I– + S2O8 2– (with catalyst Fe2+) ->
I2+ 2SO42
(i) Give a reason why the reaction between iodide ions and peroxodisulfate ions has a high activation energy and is therefore very slow without a catalyst.(1)
(ii) Explain, with the aid of two equations, how the iron(II) ions catalyse this
reaction. State symbols are not required. (3)
(i) the two negative ions repel each other
(ii) 2Fe2+ + S2O82−→ 2Fe3+ + 2SO42−
2Fe3+ + 2I− → 2Fe2+ + I2
(catalysis is possible because) variable oxidation state/iron
has more than one oxidation state/number
or
both steps now involve oppositely charged ions
2019
The oxidation of sulfate(IV) ions to sulfate(VI) ions is catalysed by cobalt(II) ions in
acidic solution. The role of cobalt(II) ions is similar to that of iron(II) ions in (a).
SO3 2– + ½O2
Co2+ SO4 2-
Deduce two ionic equations to show how cobalt(II) ions catalyse the reaction in acidic solution. State symbols are not required. (2)
Oxidation
2Co2+ + ½O2 + 2H+ → 2Co3+ + H2O
Reduction
2Co3+ + SO32− + H2O → 2Co2+ + SO42− + 2H+
The rate of oxidation of ethanedioate ions by manganate(VII) ions starts slowly
and then rapidly increases.
2MnO4– + 16H+ + 5C2O4
2– ->2Mn2+ + 8H2
O + 10CO2
What is the catalyst in this reaction?(1)
A CO2
B H+
C Mn2+
D MnO4
C
2019
The trend in the strength of gaseous adsorption by three transition elements is
tungsten > platinum > silver
Silver is not suitable as a replacement for platinum in a catalytic converter
because the adsorption of gases is too weak to allow significant chemical reaction.
Give a possible reason why tungsten would also not be a suitable replacement
for platinum in a catalytic converter. Refer to the mechanism of heterogenous
catalysis in your answer. (1)
Tungsten
(because) adsorption is too strong and so
desorption would be too slow
2019
Colour is often used in chemistry to identify substances.
Compare and contrast the origin of the colour of a copper(II) complex with the origin of the colour of the copper(II) ion in a flame test.
You do not need to state any specific colours. (6)
Similarities
(IP1) the differences in energy levels determines the
colour of the flame test and complex ion
Differences
Flame test
(IP2) heat (energy) results in electron promotion
(IP3) return of an (excited) electron to a lower (energy)
state
Complex ion
(IP4) d orbitals are split (in energy by the ligands)
(IP5) light (energy) is needed for electron promotion
(IP6) the colour not absorbed is the colour seen
2018
Which of these ions has the electronic configuration [Ar]3d5?
(1)
A Cr3+
B Fe2+
C Mn2+
D Mn3+
C
In which of these complex ions does the transition metal have the
oxidation number +3?
(1)
A [Ag(CN)2]−
B [CuCl4]2−
C [Fe(CN)6 ]3−
D [Ni(EDTA)]2−
C
Which type or types of bonding exist within the complex ion [Cr(H2O)6]3+?
(1)
A dative covalent only
B dative covalent and covalent only
C dative covalent and ionic only
D dative covalent, covalent and ionic
B
Which best explains why [Cu(NH3)2]+ ions are colourless?
(1)
A all complex ions having a metal ion with a +1 charge are colourless
B no electronic transitions can take place between d-orbitals
C the d-orbitals cannot split in energy
D there are no electrons in the d-subshell
B
