Question

4. Consider an element of Period 2 which has the following successive ionization energies, in kJ/mol. What is the charge of the common ion? IE1, 1314 IE2, 3389 IE3, 5298 IE4, 7471 IEs, 10992 IE6, 13329 IE7, 71345 IE8, 84087 (1) -1 (2) –2 (3) –3 (4) +6 (5) +7

Answer 1

The successive IE are (dividing each by 1000 to easily analyze the results) : we have to check where a big jump is coming in these successive IE's. That would mean a noble gas configuration is reached at that point. At that point, we would know the element.

For easy comparison, just take first two digits and an approximate value:

Let the element be X, than X+z is the charge on ion on successive removal of electrons.

IE	1.3	3.4	5.3	7.5	11	13	71	84
X+z	X+	X+2	X+3	X+4	X+5	X+6	X+7	X+8

The big jump is coming between 13 and 71, that is IE6 and IE7. It means there is a nobel gas configuration at IE6. For second period element, that nobel gas is He (atomic number =2). So the element is with atomic number 2 + 6 = 8, that is Oxygen.

Charge of the common ion of oxygen is -2, for example in H2O.

hence, option 2) is correct.

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