Sketches of the shapes of some atomic orbitals are shown. Identify the type of orbital, s, p, or d. [Figure 4.a]

1. (a)Identify the type of orbital, s, p, or d. shape of orbital [Figure 4.a (orbital 1)] [Figure 4.a (orbital 2)] [Figure 4.a (orbital 3)] type of orbital[1]
2. (b)Cadmium forms the two ions, Cd22+ and Cd2+. The electronic configuration of cadmium in these ions is shown. • [Kr] 4d105s1 • [Kr] 4d10 Use this information to explain why cadmium is not a transition element.[1]
3. (c)(i)Methylamine, CH3NH2, is a monodentate ligand. State what is meant by the term monodentate in this context.[1]
4. (c)(ii)In the presence of aqueous methylamine, [Cd(H2O)6]²+ reacts to form a mixture of two isomeric octahedral complexes. equilibrium 1 [Cd(H2O)6]2+ + 4CH3NH2 ⇌ [Cd(CH3NH2)4(H2O)2]2+ + 4H2O ∆H⦵ = −57 kJ mol⁻¹ Complete the three-dimensional diagrams to show the isomers of [Cd(CH3NH2)4(H2O)2]2+. Use L to represent CH3NH2 in your diagrams. [Figure 4.c.ii][2]
5. (d)(i)State what is meant by the term stability constant.[1]
6. (d)(ii)Complete the table by placing one tick (✔) in each row to suggest how increasing temperature will affect Kstab and the equilibrium concentration of the cadmium complex, [[Cd(CH3NH2)4(H2O)2]²+], for equilibrium 1. Explain your answer. [Table 4.d.ii][2]
7. (d)(iii)EDTA⁴⁻ is a polydentate ligand. When a solution of EDTA⁴⁻ is added to [Cd(H2O)6]²+ a new complex [CdEDTA]²⁻ is formed. The values for the stability constants for two Cd²+ complexes are shown. [Table 4.d.iii: Cd(CH3NH2)4(H2O)2]2+ Kstab 4.0 x 10^6, [CdEDTA]2- Kstab 4.0 x 10^16] A solution containing equal numbers of moles of CH3NH2 and EDTA is added to [Cd(H2O)6]2+. Predict which complex is formed in the larger amount. Explain your answer.[1]
8. (e)Methylamine is a Brønsted-Lowry base. Write an equation showing how methylamine dissolves in water to give an alkaline solution.[1]
9. (f)(i)Methylamine is a useful reagent in organic chemistry. Write an equation for the reaction of ethanoyl chloride with methylamine.[2]
10. (f)(ii)Methylamine also reacts with propanone to form compound P as shown. [Figure 4.f.ii] Deduce the type of reaction shown here.[1]

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.