Sketch the shape of a d orbital.
- (a)Sketch the shape of a d orbital.[1]
- (b)(i)Explain what is meant by the term transition element.[1]
- (b)(ii)Name the type of bonding that occurs between a ligand and a transition element.[1]
- (c)Give the formulae of two oxides of iron. State the oxidation number of iron in each compound.[1]
- (d)CO and CN⁻ are monodentate ligands. Complete the table for the following two complexes. [Table with metal ion, ligand, co-ordination number, formula of complex ion, charge of complex ion. Rows for Ni²⁺/CO/4 and Fe³⁺/CN⁻/ /3⁻][2]
- (e)(i)Transition element complexes can exhibit stereoisomerism. [Cu(H₂O)₄(NH₃)₂]²⁺ and Pt(NH₃)₂Cl₂ show the same type of isomerism. Name this type of isomerism.[1]
- (e)(ii)Complete the three-dimensional diagrams of the two isomers for [Cu(H₂O)₄(NH₃)₂]²⁺ and Pt(NH₃)₂Cl₂. [Figure X.X][2]
- (f)(i)Copper can form complexes with the ligands ammonia and en, H₂NCH₂CH₂NH₂, as shown. [Cu(H₂O)₆]²⁺(aq) + en(aq) ⇌ [Cu(H₂O)₄(en)]²⁺(aq) + 2H₂O(l) Kstab = 3.98 x 10¹⁰ equilibrium 4 [Cu(H₂O)₆]²⁺(aq) + 2NH₃(aq) ⇌ [Cu(H₂O)₄(NH₃)₂]²⁺(aq) + 2H₂O(l) Kstab = 5.01 × 10⁷ equilibrium 5 Write an expression for the stability constant, Kstab, for equilibrium 5. State its units. Kstab = units =[2]
- (f)(ii)The standard entropy change, ΔSᵒ, for equilibrium 4 is +23 J K⁻¹ mol⁻¹ and for equilibrium 5 is –8.4 J K⁻¹ mol⁻¹. Suggest an explanation for this difference by reference to both equilibria.[1]
- (f)(iii)Of the three copper complexes in equilibria 4 and 5, state the formula of the copper complex that is the most stable and explain your choice. copper complex explanation[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [1] any diagram
- [1] (elements) forming one or more (stable) ions with incomplete / partially filled d orbital(s) / sub-shell
- [1] dative covalent / coordinate
- [1] FeO and +2; Fe₂O₃ and +3 all ALLOW Fe₃O₄ and +3 and +2
- [2] mark as ✓✓ (Ni²⁺, CO, 4, Ni(CO)₄, 2+; Fe³⁺, CN⁻, 6, Fe(CN)₆, 3-)
- [1] cis-trans isomerism ALLOW geometric(al)
- [2] one correct pair [1] two correct pairs [2]
- [1] Kstab = [Cu(H₂O)₄(NH₃)₂²⁺] / ([Cu(H₂O)₆²⁺][NH₃]²)
- [1] units = dm⁶ mol⁻² ECF from M1
- [1] equilibrium 4 has a (net) increase in moles of product / 2 moles goes to 3 moles whereas equilibrium 5 has same number of moles of reactants and products / 3 moles vs 3 moles
- [1] [Cu(H₂O)₄(en)]²⁺ and (equilibrium) constant / Kstab is the largest / highest ALLOW [Cu(H₂O)₄(en)]²⁺ and constant / Kstab of eqm 4 is greater/higher
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