Which element in the following set would you expect to have the lowest IE3? Why?

Question

1. Na, Mg, Al
2. Al, Li, B

Answer 1

The IE is the energy needed to remove the outermost electron. In this question, IE3 is the energy needed to remove the 3rd electron (after removing the first 2 outermost electrons, IE1 and IE2). So, let's look at the electron configurations for the elements in question and see if we can determine which has the lowest IE3.

1). Na, Mg, Al

Na: [Ne]3s¹ Removing the outer electrons (IE1) results in Na⁺ which is equivalent to Ne and is very stable. Removing any more electrons would require a lot of energy.

Mg: [Ne]3s² Removing 2 electrons (IE1 and IE2) results in Mg²⁺ which is equivalent to Ne and is very stable. Removing a third electron would require a lot of energy.

Al: [Ne]3s²3p¹ Removing 2 electrons (IE1 and IE2) results in Al²⁺ which is equivalent to Ne with the 3s¹ electron (like Na) and so it is easier to remove this electron because it is still outside of the noble gas core.

Conclusion: Al has the lowest IE3

2). Al, Li, B

Al: [Ne]3s²3p¹ Refer to explanation above in part (1)

Li: 1s²2s¹ Removing the outer electron results in 1s² which is equivalent to He and is very stable. Removing the next 2 electrons would require a lot of energy

B: 2s²2s²2p¹ Removing 2 electrons (IE1 and IE2) results in B²⁺ which is equivalent to [He]2s¹. Compare removing the 2s¹ electron (IE3) to removing the 3s¹ electron from Al (IE3). The electron in Al is further from the nucleus, thus easier to remove.

Conclusion: Al has the lowest IE3