Group 2 nitrates decompose when heated. Describe how the thermal stability of Group 2 nitrates changes with increasing proton number. Explain your answer.
- (a)Group 2 nitrates decompose when heated. Describe how the thermal stability of Group 2 nitrates changes with increasing proton number. Explain your answer.[3]
- (b)Copper(II) nitrate decomposes in a similar manner to Group 2 nitrates. Write an equation for the decomposition of Cu(NO3)2.[1]
- (c)Cu(NO3)2 is added to water to form solution A. Fig. 1.1 shows some reactions of solution A. [Figure 1.1] Complete Table 1.1 to show the formula and colour of each of the copper-containing species present in A, B, C and D. [Table 1.1][4]
- (d)(i)EDTA4– is a polydentate ligand. Explain what is meant by a polydentate ligand.[2]
- (d)(ii)Group 2 metal ions can form complexes similar to those of transition elements. A solution of EDTA4– is added to water containing [Ca(H2O)6]2+ to form a new complex, [CaEDTA]2-, as shown. equilibrium 1 [Ca(H2O)6]2+ + EDTA4– ⇌ [CaEDTA]2- + 6H2O Circle on the structure of EDTA4– in Fig. 1.2 the six atoms that form bonds with the metal ion. [Figure 1.2][1]
- (d)(iii)The calcium ions in [Ca(H2O)6]2+ and [CaEDTA]2- have a coordination number of 6. Explain what is meant by coordination number.[1]
- (d)(iv)The complex [CaEDTA]2- can be used to remove toxic metals from the body. Table 1.2 shows the numerical values for the stability constants, Kstab, for some metal ions with EDTA4–. [Table 1.2] An aqueous solution containing [CaEDTA]2- is added to a solution containing equal concentrations of Cr3+(aq), Fe3+(aq) and Pb2+(aq). The resulting mixture is left to reach a state of equilibrium. State the type of reaction when [CaEDTA]²– reacts with Cr3+(aq), Fe3+(aq) and Pb2+(aq).[1]
- (d)(v)Deduce the relative concentrations of [CrEDTA]¯, [FeEDTA]¯ and [PbEDTA]2- present in the resulting mixture. Explain your answer. highest concentration lowest concentration[1]
- (e)The number of moles of water of crystallisation in a hydrated ionic salt can be determined by titration using aqueous EDTA4– ions with a suitable indicator. • 0.255g of hydrated chromium(III) sulfate, Cr2(SO4)3•nH2O, is dissolved in water and made up to 100cm³ in a volumetric flask. • 25.0cm³ of this solution requires 26.2 cm³ of 0.00800 moldm⁻³ aqueous EDTA4—– ions to reach the end-point. The reaction occurs as shown. [Cr(H2O)6]3+ + EDTA4– → [CrEDTA]– + 6H2O Use the data to calculate the value of n in the formula of Cr2(SO4)3•nH2O. Show your working.[3]
- (f)A solution of Cr3+(aq) and a solution of Fe3+(aq) have different colours. Explain why the two complexes have different colours.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [3] M1 increases (down the group)
- [3] M2 radius / size of (cat)ion / M2+ increases
- [3] M3 less polarisation / distortion of anion / nitrate ion / NO3- OR less weakening of N–O / N=O (bond)
- [1] Cu(NO3)2 → CuO + 2NO2 + 1/2O2
- [4] copper-containing species formula of copper-containing species formed colour copper-containing formed A [Cu(H2O)6]2+ (pale) blue B Cu(H2O)4(OH)2 or Cu(OH)2 (pale) blue C [Cu(NH3)4(H2O)2]2+ dark blue D CuCl42- yellow Two correct for one mark, four correct for two marks, six correct for three marks, eight correct for four marks.
- [2] M1 (a species) that donates more than two lone pairs
- [2] M2 to form dative / coordinate bonds to a metal atom or ion
- [1] six atoms circled, 2N and 4O from different CO2-
- [1] the number of co-ordinate bonds being formed by the metal ion
- [1] ligand exchange
- [1] [FeEDTA]- > [CrEDTA]- > [PbEDTA]2- highest concb lowest conch AND Kstab of [FeEDTA]- is highest
- [3] M1 moles of Cr3+ = 2.096 × 10-4 in 25.0 cm³)
- [3] M2 moles of Cr3+ = 8.384 × 10–4 (in 100.0 cm³) moles of Cr2(SO4)3•nH2O = 8.384 × 10–4/2 = 4.192 × 10–4
- [3] M3 Mr of Cr2(SO4)3•nH2O = 0.2550 / 4.192 × 10–4 = 608.3 n = (608.3 – 392.3) / 18 = 12
- [2] M1 ΔE is different
- [2] M2 different frequency (of light) is absorbed
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →