3. Chemical bonding
A section of Chemistry, 9701
Listing 10 of 239 questions
Sketches of the shapes of some atomic orbitals are shown. Identify the type of orbital, s, p, or d. x y z x y z x y z shape of orbital type of orbital  Cadmium forms the two ions, Cd2 2+ and Cd2+. The electronic configuration of cadmium in these ions is shown. ● ● 4d105s1 ● ● 4d10 Use this information to explain why cadmium is not a transition element. Methylamine, CH3NH2, is a monodentate ligand. State what is meant by the term monodentate in this context. 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 = –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. Cd Cd  State what is meant by the term stability constant. 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 equilibrium1. Explain your answer. decreases no change increases Kstab [[Cd(CH3NH2)4(H2O)2]2+] explanation  EDTA4– is a polydentate ligand. When a solution of EDTA4– is added to [Cd(H2O)6]2+ a new complex 2– is formed. The values for the stability constants for two Cd2+ complexes are shown. Kstab [Cd(CH3NH2)4(H2O)2]2+ 4.0 × 106 2– 4.0 × 1016 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. Methylamine is a Brønsted-Lowry base. Write an equation showing how methylamine dissolves in water to give an alkaline solution. Methylamine is a useful reagent in organic chemistry. Write an equation for the reaction of ethanoyl chloride with methylamine. Methylamine also reacts with propanone to form compound P as shown. H3C H3C C O H2N CH3 CH3 + H2O + H3C H3C C N P Deduce the type of reaction shown here. 
9701_s19_qp_43
THEORY
2019
Paper 4, Variant 3
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Carbonmonoxide gas, CO, and nitrogen gas, N2, are both diatomic molecules. The diagram shows the arrangement of outer electrons in a molecule of CO. C O State one similarity and one difference in the way the atoms in a carbon monoxide molecule are bonded together compared to the atoms in a nitrogen molecule. The table states the electronegativity values of carbon, nitrogen and oxygen atoms. C N O electronegativity 2.5 3.0 3.5 Use the electronegativity values and relevant details from the Data Booklet to complete the table below. N2 CO number of electrons per molecule typeof intermolecular (van der Waals’) force  N2is less reactive than COeven though N2has a lower bond energy than CO. Suggest why COis more reactive than N2. Both carbonmonoxide and nitrogen are gases at room temperature and pressure. They both behave like ideal gases under certain conditions. State the two conditions necessary for these two gases to approach ideal gas behaviour. Explain why N2behaves more like an ideal gas than COdoes at 20.0 °C and 101 kPa. Calculate the amount, in mol, of pure nitrogen gas which occupies 100 cm3 at 101 kPa and 20.0 °C. Use relevant information from the Data Booklet. Show your working. Assume nitrogen behaves as an ideal gas.  mol  
9701_s21_qp_21
THEORY
2021
Paper 2, Variant 1
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The 3d orbitals in an isolated Fe2+ ion are degenerate. Complete the diagram to show the splitting of the 3d orbital energy levels in an isolated Fe2+ ion and when Fe2+ forms an octahedral complex. energy Fe2+ in an octahedral complex isolated Fe2+ ion  Bipyridine, bipy, is a bidentate ligand. N N bipy Explain what is meant by bidentate ligand. The complex 2+ exists as two stereoisomers. Complete the three-dimensional diagrams to show the two stereoisomers of 2+. State the type of stereoisomerism shown. Use N N to represent bipy in your diagrams. Fe Fe type of stereoisomerism  Standard electrode potentials can be used to compare the stability of different complex ions for a given transition element. Table 4.1 lists electrode potentials for some electrode reactions for Fe3+ / Fe2+ systems. Table 4.1 electrode reaction E o / V [Fe(H2O)6]3+ + e– [Fe(H2O)6]2+ +0.77 [Fe(CN)6]3– + e– [Fe(CN)6]4– +0.36 3+ + e– 2+ +0.96 Use relevant data from Table 4.1 to state which iron(complex is hardest to reduce. Explain your choice. iron(complex explanation  The ligand bipyridine consists of two pyridine rings. Pyridine, C5H5N, and benzene, C6H6, have similar planar, cyclic structures. N pyridine By reference to the hybridisation of the carbon atoms and the nitrogen atom, and orbital overlap, suggest how the σ and π bonds are formed in a pyridine molecule. Pyridine reacts with Cl 2 in the presence of Al Cl 3 as shown in . N pyridine N 3-chloropyridine + Cl 2 + HCl Al Cl 3 Cl The mechanism of this reaction is similar to that of the chlorination of benzene. Al Cl 3 reacts with chlorine to generate an electrophile, Cl +. Complete the diagram to show the mechanism for the reaction of pyridine with Cl +. Include all relevant charges, dipoles, lone pairs of electrons and curly arrows as appropriate. N Cl +  
9701_s22_qp_41
THEORY
2022
Paper 4, Variant 1
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The 3d orbitals in an isolated Fe2+ ion are degenerate. Complete the diagram to show the splitting of the 3d orbital energy levels in an isolated Fe2+ ion and when Fe2+ forms an octahedral complex. energy Fe2+ in an octahedral complex isolated Fe2+ ion  Bipyridine, bipy, is a bidentate ligand. N N bipy Explain what is meant by bidentate ligand. The complex 2+ exists as two stereoisomers. Complete the three-dimensional diagrams to show the two stereoisomers of 2+. State the type of stereoisomerism shown. Use N N to represent bipy in your diagrams. Fe Fe type of stereoisomerism  Standard electrode potentials can be used to compare the stability of different complex ions for a given transition element. Table 4.1 lists electrode potentials for some electrode reactions for Fe3+ / Fe2+ systems. Table 4.1 electrode reaction E o / V [Fe(H2O)6]3+ + e– [Fe(H2O)6]2+ +0.77 [Fe(CN)6]3– + e– [Fe(CN)6]4– +0.36 3+ + e– 2+ +0.96 Use relevant data from Table 4.1 to state which iron(complex is hardest to reduce. Explain your choice. iron(complex explanation  The ligand bipyridine consists of two pyridine rings. Pyridine, C5H5N, and benzene, C6H6, have similar planar, cyclic structures. N pyridine By reference to the hybridisation of the carbon atoms and the nitrogen atom, and orbital overlap, suggest how the σ and π bonds are formed in a pyridine molecule. Pyridine reacts with Cl 2 in the presence of Al Cl 3 as shown in . N pyridine N 3-chloropyridine + Cl 2 + HCl Al Cl 3 Cl The mechanism of this reaction is similar to that of the chlorination of benzene. Al Cl 3 reacts with chlorine to generate an electrophile, Cl +. Complete the diagram to show the mechanism for the reaction of pyridine with Cl +. Include all relevant charges, dipoles, lone pairs of electrons and curly arrows as appropriate. N Cl +  
9701_s22_qp_43
THEORY
2022
Paper 4, Variant 3
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Questions Discovered
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