CAPE Chemistry June 2009 U1 P1 - Answers And Explanations

 Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

 

Download Paper Here

1. B

2. D

3. A

4. D

5. C

6. D

7. B

8. C

9. C

10. B

11. C

12. D

13. C

14. B

15. D

16. A

17. B

18. C

19. B

20. A

21. D

22. B

23. A

24. C

25. B

26. A

27. D

28. C

29. B

30. B - Detailed solution below.

31. A

32. B

33. D

34. B

35. C

36. A

37. D

38. C

39. D

40. A

41. D

42. A

43. D

44. C

45. B

Question # 30 - Detailed Solution

Hydrogen Emission Spectrum

Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

 

Theory for this syllabus section, U1 M1 SS 1.7 is broken up into 6 small, and easily manageable parts. After going through them, you will be shown how to use each part to answer the only three CAPE past paper questions on the topic. They are:

1. 2019 U1 P2 Q1(a) - Link to paper
2. 2009 U1 P2 Q1(a) - Link to paper
3. 1999 U1 P2 Q4(a)

____________________________________________________________________________

Part 1

Each element has a characteristic 'fingerprint' line emission spectrum due to its unique arrangement of electrons/electronic configuration. The simplest of these is produced by hydrogen.

____________________________________________________________________________

Part 2

• When sufficient energy is supplied to the atoms of hydrogen, an electron is promoted (excited) from a lower energy level to a higher energy level where it becomes unstable.
• The unstable electron will emit the excess energy as radiation and drop back to the lower energy level. Each line in the emission spectrum occurs due to the transitions of electrons moving from a higher energy level to a lower energy level.
• The energy difference between the higher energy level and lower energy level is fixed. The fixed frequencies of each line provide evidence for discrete energy levels.
• Each line represents emitted radiation of a specific wavelength and frequency. Electronic transitions from higher levels to the level, n = 2 give rise to a series of lines known as the Balmer series.

____________________________________________________________________________

Part 3

In summary,

• Electrical or thermal energy is passed through and absorbed by a sample of the element.
• Radiation is emitted at certain wavelengths and frequencies.
• The emission spectrum consists of separate lines at the particular frequencies.

____________________________________________________________________________

Part 4

Electron transitions occurring between energy levels associated with the hydrogen emission spectrum are shown in the diagram below.

These transitions produce the hydrogen spectrum which follows.

____________________________________________________________________________

Part 5

The three main regions of the hydrogen emission spectrum, as related to series of lines are:

Lyman series < 400 nm - ultraviolet region.

Balmer series 400 nm to 700 nm - visible region.

Paschen series > 700 nm - infrared region.

____________________________________________________________________________

Part 6

To calculate the energy, E, of a quantum of radiation with a frequency (ν) of 4.57 × 10¹⁴ Hz where h = 4 ×10^-13 kJ s mol^-1, we use the formula E = hν as illustrated next.

E = h × ν

   = 4 ×10^-13 kJ s mol^-1 × 4.57 × 10¹⁴ Hz

   = 182.8 kJ mol^-1

____________________________________________________________________________

How To Answer The Past Paper Questions

2019 U1 P2 Q1(a)(i) - Each point in Part 3 is one mark each.

2019 U1 P2 Q1(a)(ii) - Part 5 says it is the Balmer series.

2019 U1 P2 Q1(a)(iii) - First three points and first sentence in 4th point of Part 2 are one mark each.

2019 U1 P2 Q1(a)(iv) - Lines in Balmer series, only, of Part 4 for full marks.

2009 U1 P2 Q1(a)(i) - Answer is found in Part 1.

2009 U1 P2 Q1(a)(ii) - Draw just Balmer part of diagram in Part 5 and label as required.

2009 U1 P2 Q1(a)(iii) - First, second and fourth points in Part 2 will get you full marks.

2009 U1 P2 Q1(a)(iv) - Part 5 says it is in the visible region.

2009 U1 P2 Q1(a)(v) - Part 6 is the solution.

1999 U1 P2 Q4(a)(i) - Same as 2009 U1 P2 Q1(a)(ii)

1999 U1 P2 Q4(a)(ii) - Same as 2019 U1 P2 Q1(a)(ii)

1999 U1 P2 Q4(a)(iii) - All four points in Part 2 for full marks.

Transition Elements

Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

Exam Tip For 2021

Vanadium is most likely to come in 2021 (last seen 11 years ago) and copper is second most likely (last seen 12 years ago).

Colours and formulae of all species required in CAPE past papers for at least the last 15 years are shown below:

Learn these and you'll be able to handle the following questions:

1. 2019 U1 P2 Q3(d)(iii)
2. 2014 U1 P2 Q3(d)(i)
3. 2010 U1 P2 Q3(a)
4. 2009 U1 P2 Q3(d)
5. 2008 U1 P2 - Resit Q3(b)(i)
6. 2004 U2 P2 - Q2(b)(i)

Past papers from 2005 to 2016 can be found here: Past Paper Booklet

Definitions To Learn

Transition element - A d-block element whose atom has an incomplete d sub-shell or forms at least one stable ion with a partially filled d sub-shell.

Ligand - A molecule, atom or ion that bonds with a central metal atom/ion by donating an electron pair.

These are worth a maximum of 2 marks each and are the answers to:

1. 2019 U1 P2 Q3(d)(i)
2. 2014 U1 P2 Q3(a)
3. 2007 U1 P2 Q6(a)(i)

Difference Between Reactivity Series And Electrochemical Series

Reactivity And Electrochemical Series - Metals

The reactivity series can be used to compare the relative reactivities of different metals. It lists metals in order of general chemical reactivity. Metals generally react by losing electrons to form positive ions. The more readily a metal loses electrons, the more reactive it is – and the greater its strength as a reductant. Metals higher up in the series can reduce the ions of those lower down.

The standard electrode potential of a metal also indicates its strength as a reductant. The more negative the value of the standard electrode potential of a metal, the greater is its strength as a reductant. Hence, you might expect the metal reactivity series and standard electrode potentials to list metals in the same order. However, you must remember that the metal reactivity series is based on observing a range of reactions, such as displacement reactions between solid metals and solid metal oxides. Standard electrode potentials refer specifically to reactions taking place in aqueous solution.

Below is a comparison of the reactivity series with the electrochemical series, which ranks metals according to their standard electrode potentials.

The obvious discrepancy is the relative positions of sodium and calcium. Calcium is a stronger reductant than sodium according to Eo (standard electrode potential) values, but the metal reactivity series suggests that calcium is less reactive than sodium. This discrepancy arises because calcium reacts at a much slower rate, in displacement reactions for example, which in turn happens because two electrons must be removed, not one as for sodium.

Note also that aluminium reacts readily with oxygen in the air, forming a layer of stable aluminium oxide on its surface. This impervious oxide coat often causes aluminium to exhibit lower reactivity than its position in the metal reactivity series indicates.

Group IV Elements - CAPE Chemistry Unit 1

Elements - Structure And Bonding

Main Points To Be Used In Answering A Past Paper Question:

• Down the group there is a change in structure from giant molecular to giant metallic and a change in bonding from covalent to metallic.
• From C to Ge elements exhibit a giant molecular structure.
• Sn and Pb exhibit a giant metallic lattice structure.
• From C to Ge bond length between group IV atoms increases and bond energy/strength decreases.

Elements - Electrical Conductivity

Trend: Going down the group there is a general increase in electrical conductivity.

Reason: Down the group there is a gradual increase in metallic character due to an increase in delocalisation of electrons throughout the structure.

Note: There is an increase in electrical conductivity from C(diamond) to Sn. There is a decrease from Sn to Pb which has never been addressed in the past papers. Hence we use the phrase "general increase" when talking about the trend in electrical conductivity down the group. 

CAPE Chemistry Past Paper Answer - 2016 Unit 2 Paper 2 Question 5

Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

5(a)(i)

As applied to completely miscible binary systems, this states that the partial pressure of a constituent of a binary mixture at any given temperature is equal to the normal vapor pressure of that constituent at the stated temperature multiplied by the mole fraction of the constituent in the mixture.

5(a)(ii)

1. The interactions between components are similar to those in pure components.

2. There is no volume change on mixing the components.

5(b)(i)

An azeotropic mixture is one which boils or distils without change in composition, and in general it has a boiling point higher or lower than that of any of its pure constituents.

5(b)(ii)

An azeotrope is not a compound because its composition varies with pressure.

5(b)(iii)

Mixture X boils at a temperature T1 and at equilibrium gives off a vapor of composition Y richer in the more volatile component A. Continued distillation produces the azeotrope as the distillate while pure B is the residue.

5(c)

At equilibrium, mass of compound in water = (5-x) g and mass of compound in solvent = x g. 

Partition coefficient = concentration in water/concentration in ether, therefore  ((5-x)/100)/(x/25) = 0.2 i.e. x = 2.8 g.

Unit 1 Module 1 SS 2 Notes

Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

2.1

State the various forces of attraction between particles.

Ionic bonds, covalent bonds, metallic bonds, van der Waals' forces.

Both bonding (intramolecular) forces and intermolecular forces arise from electrostatic attractions between opposite charges. Bonding forces are due to the attraction between cations and anions (ionic bonding), nuclei and electron pairs (covalent bonding), or metal cations and delocalized valence electrons (metallic bonding). Intermolecular forces, on the other hand, are due to the attraction between molecules as a result of partial charges, or the attraction between ions and molecules. The two types of forces differ in magnitude, and Coulomb's law explains why:

• Bonding forces are relatively strong because they involve larger charges that are closer together.
• Intermolecular forces are relatively weak because they typically involve smaller charges that are farther apart.

Ion-Dipole Forces

When an ion and a nearby polar molecule (dipole) attract each other, an ion-dipole force results. The most important example takes place when an ionic compound dissolves in water. The ions become separated because the attractions between the ions and the oppositely charged poles of the H2O molecules overcome the attractions between the ions themselves.

Dipole-Dipole Forces

When polar molecules lie near one another, as in liquids and solids, their partial charges act as tiny electric fields that orient them and give rise to dipole-dipole forces: the positive pole of one molecule attracts the negative pole of another (diagram below).

Polar molecules and dipole-dipole forces. In a solid or a liquid, the polar molecules are close enough for the partially positive pole of one molecule to attract the partially negative pole of a nearby molecule. The orientation is more orderly in the solid (left) than in the liquid (right) because, at the lower temperatures required for freezing, the average kinetic energy of the particles is lower. (Interparticle spaces are increased for clarity.)

For molecular compounds of approximately the same size and molar mass, the greater the dipole moment, the greater the dipole-dipole forces between the molecules are, and so the more energy it takes to separate them. Consider the boiling points of the compounds in the next diagram. Methyl chloride, for instance, has a smaller dipole moment than acetaldehyde, so less energy is needed to overcome the dipole-dipole forces between its molecules and it boils at a lower temperature.

Dipole moment and boiling point. For compounds of similar molar mass, the boiling point increases with increasing dipole moment. (Note the increasing color intensities in the electron density models.) The greater dipole moment creates stronger dipole-dipole forces, which require higher temperatures to overcome.

The Hydrogen Bond

A special type of dipole-dipole force arises between molecules that have an H atom bonded to a small, highly electronegative atom with lone electron pairs. The most important atoms that fit this description are N,0, and F. The H-N, H-O,
and H- F bonds are very polar, so electron density is withdrawn from H. As a result, the partially positive H of one molecule is attracted to the partially negative lone pair on the N, 0, or F of another molecule, and a hydrogen bond (H bond) forms. Thus, the atom sequence that allows an H bond (dotted line) to form is -B:····H-A-, where both A and B are N, O, or F. Three examples are

The small sizes of N, O, and F are essential to H bonding for two reasons:

1. It makes these atoms so electronegative that their covalently bonded H is
    highly positive.


2. It allows the lone pair on the other N, O, or F to come close to the H.

The Significance of Hydrogen Bonding

Hydrogen bonding has a profound impact in many systems. Here we'll examine one major effect on physical properties and preview its enormous importance in biological systems.