# How can the outermost electrons in silicon be 2s and 2p electrons?

My book says the number of outermost electrons in silicon is $4$. Yes, right. But then it says those are $\ce{2s}$ and $\ce{2p}$ electrons. The electronic configuration of silicon is $\ce{1s^2 2s^2 2p^6 3s^2 3p^2}$.

How are the outermost electrons $\ce{2s}$ and $\ce{2p}$ electrons. Those belong to third shell.

Then it says the maximum number of electrons in the outermost shell is $8$. Correct. But then it says those are $\ce{2s + 6p}$ electrons. How?

• The electron configuration you give is correct, 2 electrons each in 3s and 3p orbitals. Sometimes this is written in shorthand as $\ce{(Ne) 3s^2 3p^2}$ where the (Ne) represents Neon's configuration. I suspect that there is /are typos in your text. (btw, don't hate Chemistry: you can be interested in all sorts of subjects while still having favourites) – porphyrin Jan 11 '18 at 11:51
• Please add citations to your post, it is not helpful to anyone further if the source of this is unclear. – Martin - マーチン Jan 11 '18 at 12:10

The book means to say that there are 2 electrons in the $\mathrm{3s}$ subshell and and 2 electrons in the $\mathrm{3p}$ subshell. You can see this in the electron configuration as

$\mathrm{...3s^2 3p^2}$

and that the maximum configuration has 2 electrons in the $\mathrm{3s}$ subshell and 6 electrons in the $\mathrm{3p}$ subshell which would be

$\mathrm{...3s^2 3p^6}$

Concur with this point:

"$\ce{2s}$ and $\ce{2p}$ valence electrons" is not the same thing as "two $\ce{s}$ and two $\ce{p}$ valence electrons".

Your book probably used the latter, but you interpreted it as the former. Though, it is entirely possible that the book wrote it the way you say it had, but unless you quote that particular line and cite the book (as Mod. Martin pointed out), it will be impossible to ascertain that.

So for the record, silicon does have a valence configuration of $$\ce{3s^{2} 3p^{2}}$$

You must be attentive while reading such things ;-)

Now for the second part of your question,

Then it says the maximum number of electrons in the outermost shell is $\ce{8}$

Well, technically, the maximum possible number of electrons in the outermost shell (which is $\ce{n = 3}$ in this case) is $\ce{18}$ and not $\ce{8}$.

That's because the third shell is constituted by $\ce{s, p, d}$ subshells. (Granted, ground-state silicon doesn't use those $\ce{d}$ orbitals, but your post only mentions "outermost shell", so I thought mentioning this should be of help)

The third shell would therefore have one $\ce{s}$ orbital (holds 2 electrons), three $\ce{p}$ orbitals (holds 6 electrons) and five $\ce{d}$ orbitals (holds 10 electrons). So,

$$\mathrm{ 2 + 6 + 10 = 18 }$$