HIGHER CHEMISTRY REVISION. Unit 1:- Patterns in the Periodic Table 1. The three statements below are taken from a note made by a student studying trends.

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HIGHER CHEMISTRY REVISION. Unit 1:- Patterns in the Periodic Table 1. The three statements below are taken from a note made by a student studying trends in the Periodic Table. 1.First Ionisation Energy The energy required to remove one mole of electrons from one mole of atoms in the gaseous state. 2.Second Ionisation Energy The energy required to remove a second mole of electrons The measure of the attraction an atom has for the shared electrons in a bond. (a) Complete the note above to give the heading for the third statement. (b) What is the trend in the first ionisation energy across a period from left to right. (c) Why is the second ionisation energy of sodium so much greater than its first ionisation energy? (a) Electronegativity. (b) Increases from left to right. (c) After 1 st electron removed the next electron has to be taken from energy level below it, closer to nucleus. As the 2 nd electron comes from energy level closer to the nucleus it is more firmly attracted to the nucleus.

2. On crossing the Periodic Table, there are trends in the sizes of atoms and ions. (a)Why is the atomic size of chlorine less than that of sodium? (b) Ion Ionic radius / pm Si P Why is there a large increase in ionic radius on going from Si 4+ + to P 3- ? (a) Chlorine has a larger nuclear charge (17 protons) than sodium (11 protons). Both atoms have 3 occupied electron energy levels. The electrons are attracted more by the larger charge in the chlorine nucleus and so are brought closer to the nucleus. (b) Si 4+ has an electron arrangement of 2,8 – two occupied electron energy levels. P 3- has electron arrangement of 2,8,8 – three occupied electron energy levels. The third energy level is further from the nucleus.

3. Ionisation energies provide information about the structure of atoms. Write the equation, showing state symbols, for the first ionisation energy of sodium. Na(g)  Na + (g) + e -