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Hg 2+ ions and iodide ions, I – , form a complex with the formula [HgI 4 ] 2– . Write the expression for K stab of [HgI 4 ] 2– . State its units. K stab = units =
See it on the paper →Define bidentate ligand.
See it on the paper →Some Cd 2+ (aq) is added to a solution containing equal concentrations of CH 3 NH 2 and en. Predict which of the two complexes in Table 6.1 forms at the higher concentration. Explain your answer. complex that forms at the higher concentration explanation (ii)
See it on the paper →Complete the expression for the K stab of [Cd(CH 3 NH 2 ) 4 ] 2+ . K stab =
See it on the paper →Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl (aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, K stab .
See it on the paper →Complete the expression for the K stab of [Ni(en) 3 ] 2+ . K stab =
See it on the paper →Predict which complex ion, [Ni(NH 3 ) 6 ] 2+ or [Ni(en) 3 ] 2+ , is present in the resulting mixture in the highest concentration. Explain your answer. complex ion present in largest concentration = (i)
See it on the paper →Complete the equation for the ligand exchange reaction occurring in (i). 2+
See it on the paper →Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl (aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, K stab .
See it on the paper →Use the stability constant data in Table 3.1 to calculate the value of the equilibrium constant, K c , for the following equilibrium. [Fe(H 2 O) 5 (H 2 PO 4 )] 2+ + SCN – [Fe(H 2 O) 5 SCN] 2+ + H 2 PO 4– value of K c =
See it on the paper →Complete Table 4.1 to suggest formulae for X and Y. Table 4.1 X Y formula of complex
See it on the paper →Suggest how pyrazine is able to bond to two separate ruthenium ions.
See it on the paper →The overall charge of the ruthenium complex in Fig. 4.1 is 5+. Deduce the possible oxidation states of the two ruthenium ions in the complex.
See it on the paper →Write an expression for the K stab of [Cu(CN) 4 ] 3− . K stab =
See it on the paper →In a solution the concentrations of CN − and [Cu(CN) 4 ] 3− are both 0.0010 mol dm −3 . Use your expression from (c)(i) and the value of K stab to calculate the concentration of Cu + (aq) in this solution. concentration of Cu + (aq) = mol dm –3
See it on the paper →Define stability constant.
See it on the paper →Identify the most stable complex in Table 4.1. Explain your choice.
See it on the paper →In a solution at equilibrium at 298 K, [[Cu(H 2 O) 6 ] 2+ ] = 3.00 × 10 –10 mol dm –3 and [EDTA 4– ] = 5.00 × 10 –12 mol dm –3 . Use the expression for K stab to calculate the concentration of [Cu(EDTA)] 2– in this solution. Show your working. [[Cu(EDTA)] 2– ] = mol dm –3
See it on the paper →Write an expression for the K stab of [Cu(CN) 4 ] 3− . K stab =
See it on the paper →In a solution the concentrations of CN − and [Cu(CN) 4 ] 3− are both 0.0010 mol dm −3 . Use your expression from (c)(i) and the value of K stab to calculate the concentration of Cu + (aq) in this solution. concentration of Cu + (aq) = mol dm –3
See it on the paper →Comprehensive responses • Most candidates were able to define a bidentate ligand. Limited responses • Some common errors were to omit any reference to lone pairs or the term ‘metal’ with atom/ion in their answer.
— Examiner report, February/March 2026, 9701/42 Q6(c)(i) (Cambridge)
This question was answered correctly. Some candidates put charges outside the concentration brackets or had the inverse of the expression.
— Examiner report, October/November 2023, 9701/41 Q5(c)(i) (Cambridge)
This was often answered well. Some candidates were unable to use a correct expression for Kstab.
— Examiner report, October/November 2023, 9701/42 Q4(e) (Cambridge)
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