How many electrons with l = 1 does Si in its ground state have?

Question

I was solving practice problems for electron configuration and periodic table, and I got stuck through a question:

An atom of silicon in its ground state has how many electrons with quantum number $l = 1$?
a) 14
b) 2
c) 8
d) 6
c) 28

According to the solution the answer is c: 8 electrons, but that wasn't my answer.

My analysis:

1. Silicon is a chemical element with symbol $\ce{Si}$ and atomic number 14, this is its electron configuration:

$$\mathrm{1s^22s^22p^63s^23p^2}$$

2. The quantum number $l = 1$ corresponds to $\mathrm{p}$ levels only. So we have:

$$\mathrm{2p^63p^2}$$

3. The total number of orbitals is: $6 + 2 = 8$ orbitals.

4. Since each orbital has 2 electrons, the total number of electrons is $8 \times 2 = 16$ electrons.

What's wrong with my answer? Maybe the question has a wrong answer?

Answer 1

You're on the right track, but those superscripts in the electronic configuration don't refer to orbitals. They refer to the number of electrons in that type of orbitals.

For more information, you might want to look at this question: What are the maximum number of electrons in each shell?

Answer 2

I was also stuck with questions like these in our exams and I learned through the internet that the superscripts are actually the number of electrons and that $l=1$ refers to p-orbital. In the electronic configuration of Silicon, consider only the p-orbital, since it is aking for electrons that have p-orbitals (l=1)

$2\mathrm{p}^6$ and $3\mathrm{p}^2$ are sub-orbitals of p-orbital. Since 2p have 6 electrons in it and 3p have 2 electrons in it, there are total of 8 electrons that have $l=1$ as one of their quantum numbers.