Question

Why is the first ionization of aluminum less than the first ionization of magnesium? Select all that apply.

A. The last valence electron of magnesium ends the occupation of the 3s subshell.

B. The statement is incorrect because the general trend left to right in a row is for the effective nuclear charge to increase.

C. The 3s electrons shield the 3p electrons from the nuclear charge.

D. The last valence electron of aluminum begins the occupation of the 3p subshell.

I've tried combinations A&D, and A-C-D but both were incorrect. I'm not sure what is wrong

Answer 1

Your answer choices are partially correct althought at first sight A,C and D all seems to be correct; but A and D does not provide any satisfactory explaination.

Here is my explaination - Option C should be the correct answer.

Electronic configuration of Mg is 1S2 2S2 2P6 3S2 , a filled stable configuration.

while that of Al is 1S22S22P63S2 3P1 which is an unstable configuration

Now we know that,first IE is the amount of energy needed to remove a valnce shell electron from an isolated gaseous atom so stability or electronic configuration of elements does matter.

As Aluminium's outer electron is in the 3p orbital which is of higher energy and is further from the nucleus. So it is shielded by the 3s orbital and hence requires less energy to remove the electron than the magnesium's outer electron which is closer and not shielded by the nucleus.