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Port Hedland and Mt Lawley Senior High Schools Chem 12 Revision Week 1.

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Presentation on theme: "Port Hedland and Mt Lawley Senior High Schools Chem 12 Revision Week 1."— Presentation transcript:

1 Port Hedland and Mt Lawley Senior High Schools Chem 12 Revision Week 1

2 Welcome meaghersclasses@hotmail.com http://meaghersclasses.freeforums.org/ www.lawley.wa.edu.au username: chemistry password: student http://www.meaghersclasses.podomatic.com/

3 Purpose: to revise the course to provide revision resources to enable you to maximise your TEE results to help you with assessment strategies.

4 The course You must have a copy of the syllabus statement. This is available at www.curriculum.wa.edu.au

5 Another document that you should read before the TEE is the examiner’s report. This is also available from curriculum council website. http://www.curriculum.wa.edu.au/internet/Communications/Reports_Statistics/TEE_Examiners_Reports/

6 Electron configuration. This concept, like ALL of chemistry, is explained with electrostatics. ie opposite charges attract and like charges repel

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21 ionisation energy explanation and trends

22 The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms (to produce 1 mole of gaseous ions each with a charge of 1+).

23 This is more easily seen in symbol terms. It is the energy needed to carry out this change per mole of X.

24 Ionisation energies are measured in kJmol -1. 1 st ionisation energies range from 381 kJmol -1 to 2370 kJmol -1. Q. Which element do you think has the highest 1 st ionisation energy?

25 First ionisation energies display periodicity.

26 A number of factors can be used to explain an elements ionisation energy. The charge on the nucleus. (ie the bigger the nuclear charge the more strongly the electrons are held.)

27 The distance of the electron from the nucleus. (ie Attraction falls off very rapidly with distance. An electron close to the nucleus will be much more strongly attracted than one further away.)

28 The number of electron between the outer electrons and the nucleus. (ie The pull of the protons is screened or shielded by each level of electrons between the nucleus and the outer electron.)

29 Whether the electron is on its own in an orbital or paired with another electron. ( ie Two electrons in the same orbital experience a bit of repulsion from each other. This offsets the attraction of the nucleus, so that paired electrons are removed rather more easily than you might expect. This will be explained later.)

30 eg He 2370 kJ mol -1. Li 519 kJ mol -1

31 The general trend is for ionisation energies to increase across a period.

32 Be 1s 2 2s 2 1st I.E. = 900 kJ mol -1 B 1s 2 2s 2 2p x 1 1st I.E. = 799 kJ mol -1 Offsetting the fact that boron has one extra proton is the fact that its outer electron is in a 2p orbital rather than a 2s. 2p orbitals have a slightly higher energy than the 2s orbital, and the electron is, on average, further from the nucleus. ie more shielding ….at a greater dist.

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34 N 1s 2 2s 2 2p x 1 2p y 1 2p z 1 1st I.E. = 1400 kJ mol -1 O 1s 2 2s 2 2p x 2 2p y 1 2p z 1 1st I.E. = 1310 kJ mol -1 Screening due to inner electrons is identical. Oxygen’s electron comes from the 2p x pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be.

35 The general trend is for ionisation energies to decrease down a group.

36 Every one of the transition metals loses its outer electron from the 4s orbital. The increase across the period is due to …increasing nuclear charge.

37 Before next session: What do you want/need from these classes? Send me an email. Register on the forum site.


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