Transition Metals Test PRi Flashcards
Q1a) Identify a suitable reagent, give the formula for copper containing species and write equation for reaction of [Cu(H2O)6]2+ ⟶ Pale Blue ppt.
Reagent – Ammonia
Species – [Cu(H2O)4(OH)2]
Equation – [Cu(H2O)6]2+ + 2NH3 ⟶ [Cu(H2O)3(OH)3] + 2NH4+
Q1b) Reagent, species and equation for Reaction 2
Reagent – Ammonia in excess
Species – [Cu(H2O)2(NH3)4]2+
Equation – [Cu(H2O)4(OH)2] + 4NH3 ⟶ [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-
Q1c) Reagent, species and equation for Reaction 3
Reagent – Na2CO3
Species – [Cu(H2O)6]2+
Equation – [Cu(H2O)6]2+ + CO3(2-) ⟶ CuCO3 + 6H2O
Q1d) Reagent, species and equation for Reaction 4
Reagent – HCl (Conc/xs)
Species – [Cu(H2O)6]2+
Equation – [Cu(H2O)6]2+ + 4Cl- ⟶ [CuCl4]2- + 6H2O
Q2a) Explain why the reaction between I- ions and S2O82- ions is slow
They are both negatively charged
-ve ions repel one another
Q2b) Other than having variable oxidation states, explain why Fe2+ ions are good catalysts for this reaction
Positive ions attract -ve ions in catalysed process
Q2c) Write a half-equation for the reduction of S2O8(2-) ions to SO4(2-) ions
S2O8(2-) + 2e- ⟶ 2SO4(2-)
Q2d) Construct an overall equation for the reaction between S2O8(2-) ions and I- ions
S2O8(2-) + 2I- ⟶ 2SO4(2-) + I2
Q3a) Calculate the entropy change for the reaction between ammonia and oxygen
∆S = S(s products) - Σ(s reactants)
((4x211) + (6x189)) - ((4x193) + (5x205)) = 181 J/K/mol
Q3b) Calculate a value for the Gibbs free-energy change (∆G), in kJ/mol, for the reaction between ammonia and oxygen at 600ºC
∆G = ∆H - T∆S 600ºC = 873K ∆G = -905 - 873(181x10-3) = -1063 kJ/mol
Q3c) The reaction between ammonia and oxygen was carried out at a higher temperature.
Explain how this change affects the value of ∆G for this reaction
1) ∆G becomes more negative
2) Entropy change/∆S is +ve/T∆S gets bigger
Q3d) Platinum acts as a heterogeneous catalyst in the reaction between ammonia and oxygen. It provides an alternative reaction route with a lower activation energy.
Describe the stages of this alternative route
1) Reactants absorbed onto Platinum surface/Pt provides surface/active site
2) Reaction on surface. Bond breaking/forming occurs on surface
3) Desorption of the product or words to that effect (wtte)
Q3e) Deduce the change in oxidation state of Nitrogen, when NH3 is oxidised to NO
Oxidation state changes from -3 to +2 (or +5 overall)
Q3f) When ammonia reacts with oxygen, Nitrous Oxide (N2O) can be produced instead of NO
Give an equation for this reaction
2NH3 + 2O2 ⟶ N2O + 3H2O
Q4a) Explain, in terms of electrons, why the complexes are different colours. (You are NOT required to explain why the observed colours are red-violet and green)
1) In each of P & Q, oxidation state of Cr = +3
2) In each of P & Q, electron config is same/3d3
3) Ligands are different
4) Different energies of electrons/different orbital energies
5) Different wavelengths of light absorbed
6) Different wavelengths of light transmitted/reflected
Q4b) Write an equation to show how the [Co(NH3)6]2+ ion reacts with 1,2-diaminoethane
Explain the thermodynamic reasons why this reaction occurs
1) 4 particles form 7 particles
2) Entropy increases
3) ∆H approx. 0/no net change in bond enthalpies
4) ∆G -ve
5) [Co(NH3)6]2+ + 3(NH2CH2CH2NH2) ⟶ [Co(NH2CH2CH2NH2)3]2+ + 6NH3
Q4c) i) Draw displayed structure of Cisplatin. Show value of one of the bond angles at Platinum. State the charge if any
90º. Charge = 0
Q4c) ii) When Cisplatin is ingested, an initial reaction involves one of the Chloride ligands being replaced by water.
Write an equation for this reaction
(NH3)2PtCl2 + H2O ⟶ [(NH3)2PtCl(H2O)]+ + Cl
Q4c) iii) Suggest how the risk associated with the use of this drug can be minimised
Use in small amounts/short bursts/target application/monitor the patients
Q4d) Explain, with the aid of equations, how and why Vanadium (V) oxide is used in the Contact Process
1) V2O5 + SO2 ⟶ V2O4 + SO3
2) V2O4 + 1/2 O2 ⟶ V2O5
3) Act as a catalyst
4) Speeds up overall reaction b/w SO2 & O2
Q5a) Use information to calculate a value for the mass of FeSO4.7H2O in the sample of X
X dissolved in 250cm3 of water
After xs acid added, 25cm3 reacted with 21.3cm3 of 0.015M K2Cr2O7
1) Mol of Cr2O7(2-) = (21.3÷1000) x 0.0150 = 3.195x10-4
2) Cr2O7(2-) : Fe2+ = 1:6
3) Mol of Fe2+ = 6 x (3.195x10-4) = 1.917x10-3
4) Original is 250cm3 = (1.917x10-3) x 10 = 1.917x10-2
5) Mass of FeSO4.7H2O = (1.917x10-2) x 277.9 = 5.33g
Q5b) Suggest ONE property of an impurity that would cause the calculated mass of FeSO4.7H2O in X to be greater than the actual mass of X.
Explain your answer
Impurity is a reducing agent
Q6) Which one of the following reactions in aqueous solution has the most +ve change in entropy?
A - hexaaquacopper (II) + Ammonia ⟶ tetraaminodiaquacopper + Water
B - hexaaquacopper (II) + Chloride ⟶ tetrachlorocuprate + water
C - hexaaquacopper (II) + EDTA ⟶ EDTAcuprate + water
D - hexaaquacopper (II) + 1,2-diaminoethane ⟶ [Cu(EDTA)2(H2O2]2+ + Water
C – [Cu(H2O)6]2+ + EDTA4- ⟶ [Cu(EDTA)]2- + 6H2O
Q7) The Vanadium does NOT have an oxidation state of +3 in: A - [V(H2O)6]3+ B - [V(C2O4)3]3- C - [V(OH)3(H2O)3] D - [VCl4]3-
D – [VCl4]3-
Q8) Which one of the following would lead to an inaccurate result?
A - Transferring weighed sample of Iron (II) Sulphate into a wet conical flask
B - Failing to measure accurately the volume of water used to dissolve each weighed sample of Iron (II) Sulphate
C - Transferring standard solution of K2Cr2O7 from its original container to burette using a wet beaker
D - Failing to measure accurately the volume of dilute H2SO4 added to the mixture before titration
C – Transferring standard solution of K2Cr2O7(2-) from its original container to burette using a wet beaker
Q9) Which one of the following would lead to the greatest error in the calculation of the percentage of Iron (II) in the sample?
A - An error of 0.005g made when weighing out a sample of mass 0.987g
B - An end point error of 0.1cm3 in 25cm3
C - Error of 5cm3 when measuring out 25cm3 of dilute H2SO4
D - Using the average of the Titration values 25.4, 25.7 and 25.9 when the correct value is 25.5cm3
D – Using the average of the titration values 25.4, 25.7 and 25.9 when the correct value is 25.5cm3
Q10)
D