lonisation Energies grnpn below represents the successive lonization energies of
ID: 543166 • Letter: L
Question
lonisation Energies grnpn below represents the successive lonization energies of an element X plotted against the number of the electron removed. X is not the symbol for the element 4.5 Logarithm ionization energy 3.5 2.5 02461o 12 Electron removed (a) From this graph it is possible to deduce the group in the Periodic Table to which belongs. X is in (b) From the graph it is possible to deduce that the most stable ion of X will be (b) The ionization energies of sodium, Na, are shown in the table below Show with a tick (v). in the third row of the table below, all the ionization numbers that involve the removal of an electron from an s-orbital lonization kJ mol- lonization 1st 2nd 3rd 4th Sth 6th 7th 8th 9th 10th 11th energy 496 4563 6913 9544 13352 16611 20115 25491 8934 141367 159079 numberExplanation / Answer
Q1.a) From the graph, it is seen that change in inflation energy is highest between the removal of 3rd electron and 4th electron. This indicates that the element X had a stable ion when 3rd electron was removed. This means that element X belongs to group 13 or group 3 (from old periodic table)
Q1.b) Since the change in inflation energy is highest between removal of third and 4th electron this means that element X had a stable ion when 3rd electron was removed i.e. X+3
b. The electronic configuration of Na is 1s2 2s22p63s1. From the ionization enenergiesenergies mentioned in the table, the electron removal happens as per the configuration with 1st electron being removed is the outermost, i.e. 3s1 then electrons in p shell I.e. 2p6 2p5 2p4 2p3 2p2 2p11s2 and 1s1. Thus, with removal of each electron the ionization energy increases. The place where change in ionization energy increases drastically indicates the change in shell or subshell. Thus, from table the ionization number indicating removal of s electrons are:
1st,8th,9th,10th and 11th.
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