# Is my analysis of quantum numbers describing the ground state of At correct?

Which series of quantum numbers describes the highest (energy) occupied orbital in a ground state of At atom?

a) n = 6, l = 0
b) n = 6, l = 2
c) n = 5, l = 2
d) n = 4, l = 3
e) n = 6, l = 1

According to my teacher, the answer is e.

My analysis:

• The electron configuration of At atom:

\begin{aligned}1s^22s^22p^63s^23p^6\ce{3d^10}4s^24p^64\ce{d^10}5s^25p^64\ce{f^14}\ce{5d^10}6s^26p^5 \end{aligned}

Knowing for levels 0,1,2,3 correspond to s,p,d,f respectively we can say that:

a) \begin{aligned}6s^2\end{aligned} b) \begin{aligned}Nothing \end{aligned} c) \begin{aligned}\ce{5d^10}\end{aligned} d) \begin{aligned}\ce{4f^14}\end{aligned} e) \begin{aligned}\ce{6p^5}\end{aligned}

What I thought here first is that d is the answer since 4f14 has the largest number of electrons, and thus will occupy the highest energy around.

Then I saw on another question that the orbitals are filled so that the ones of lowest energy are filled first.

$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s ...$$

In this case, e is the answer since 6p5 has the highest energy regardless of the number of electrons it is holding?

• Your second way of thinking about it is correct. The number of electrons in the orbital is not the determining factor here. Well written question. – Jason Patterson Oct 8 '14 at 18:15