A-LevelChemistryAtomic structure9701/43 · Oct/Nov 2018 · Q319 Marks

(a) Complete the table to show the total number of unpaired electrons in the 3d and 4s orbitals of each isolated gaseous atom. number of unpaired electrons ---|--- | 3d | 4s Cr | | Mn | | Fe | |

(a)Complete the table to show the total number of unpaired electrons in the 3d and 4s orbitals of each isolated gaseous atom. number of unpaired electrons ---|--- | 3d | 4s Cr | | Mn | | Fe | | [2][2]
(b)Solid potassium manganate(VII), KMnO4, decomposes on heating to form manganese(IV) oxide, potassium manganate(VI) and a colourless gas. Construct an equation for this reaction. [2][2]
(c)Explain the origin of colour in transition element complexes. [3][3]
(d)The reaction scheme shows some reactions of [Cu(H2O)6]2+. [Figure: Reaction scheme starting from [Cu(H2O)6]2+. Reaction 1 with NaOH(aq) forms precipitate A. Reaction 2 from precipitate A with excess NH3(aq) forms solution of B. Reaction 3 from [Cu(H2O)6]2+ forms CuCO3(s). Reaction 4 from CuCO3(s) with CH3CO2H(aq) forms solution of C.]
(d)(i)Write the formulae of precipitate A, ........................................................................................................................... complex ion B, ......................................................................................................................... compound C. ............................................................................................................................. [3][3]
(d)(ii)Identify a suitable reagent for reaction 3. [1][1]
(d)(iii)Write an equation for reaction 4. [1][1]
(d)(iv)Describe two visual observations that would be made during reaction 4. [1][1]
(e)Platin, Pt(NH3)2Cl2, is a neutral complex of platinum(II). Explain why Pt(NH3)2Cl2 has no charge. [1][1]
(f)(i)Pt(NH3)2Cl2, displays cis-trans isomerism. Draw the structure of trans-platin. State its shape and the Cl–Pt–Cl bond angle. shape ..................................................... Cl–Pt–Cl bond angle ..................................................... [2][2]
(f)(ii)Cis-platin is an effective anti-cancer drug. Describe the action of cis-platin in this role. [2][2]
(g)The use of cis-platin can cause side effects so nedaplatin has been developed. Nedaplatin can be synthesised from cis-platin, Pt(NH3)2Cl2, by replacing the two chloride ion ligands with a single bidentate ligand as shown. Suggest the structure for nedaplatin. Pt(NH3)2Cl2 + [Figure: Bidentate ligand with two -O- groups] [Figure: Box for nedaplatin structure] nedaplatin [1][1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[1] Cr: 3d = 5, 4s = 1
[1] Mn: 3d = 5, 4s = 0
[1] Fe: 3d = 4, 4s = 0
[1] formulae of K2MnO4 and O2
[1] rest of the equation (2KMnO4 → K2MnO4 + O2 + MnO2)
[1] M1 d orbitals split into two levels / lower and upper orbitals
[1] M2 visible light is absorbed and the complementary colour observed
[1] M3 electron(s) promoted / excited
[1] precipitate A [Cu(H2O)4(OH)2] OR Cu(OH)2
[1] solution B [Cu(NH3)4(H2O)2]2+
[1] solution C Cu(CH3CO2)2
[1] Na2CO3 or CO32-
[1] CuCO3 + 2CH3CO2H → Cu(CH3CO2)2 + CO2 + H2O
[1] any two for one mark: fizzing / bubbles / effervescence OR solid disappears OR green / blue solution (formed)
[1] sum of the charges of the (four) ligands equals the oxidation number / charge of Pt OR a calculation Pt +2, NH3 neutral / no charge, both Cl¯'s −1 (so no overall charge)
[1] structure correct
[1] square planar and 180°
[1] M1 this can bond / bind with DNA
[1] M2 which prevents replication of the DNA / strand OR prevents cell division
[1] structure of nedaplatin