Ionization energy question

Question

the first ionization energy(IE1) for sulfur is less than the IE1 for phosphorus because;

a. hund's rule is violated

b. P has a p3 configuration

c. P is below S on the periodic table

d. 1st ionization energies decrease across a period

e. P is to the right of S on the periodic table

i do not think b, c, d, e is correct

so i am going towards a.

but im not sure why..so please explain

Answer 1

I actually think the correct answer is (B). If you look at the electron configurations of sulfur and phosphorus, they are as follows:

S 1s^2 2s^2 2p^6 3s^2 3p^4

P 1s^2 2s^2 2p^6 3s^2 3p^3

The p orbital can hold a maximum of 6 electrons (3 with 'up sign' and 3 with 'down spin'). The way you can think of it is like this:

___   ___   ____

p x    p y    p z

In each of those, you can have a maximum of 2 electrons, thus 6 in total. For phosphorus, there is one electron in each while for sulfur, in one case, two electrons must be grouped together. Since electrons have the same charge (-1), they prefer not to be near one another. Thus, with sulfur, it is easier to get rid of one the electrons that is paired because it would rather be alone so that is by itself, thereby becoming like phosphorus.

Phosphorus likes the way it is--3 electrons, by themselves--so you would need more energy to get rid of one of its electron because its normal state is the stable state.

Make sense?