In this question, we're asked to find how many
valence electrons are in Si two plus.
Another example I saw said how many
valence electrons are in s minus, so I
wanted to do a positive ion and a
negative ion, and there are multiple ways
to do this. So the first and probably
arguably the easiest way is to find out
how many valence electrons are in the
original atom. So in this case Si 2 plus:
first find Si, so Si has four valence
electrons. Si 2 plus is going to have two
less electrons, so two valence electrons.
Said another way, Si 2 plus will have the
same electron configuration as magnesium: one, two less electrons, so it
will have two valence electrons. Since
it's two plus, two less electrons, so it
has two valence, which would answer
this question. For S minus, doing it the
same way, if it's negatively charged, it
would have one extra electron. So it's in
6A, so sulfur has six valence electrons.
S- would have the same electron
configuration as chlorine, and it would
have seven valence electrons. So this has seven valence
electrons. Another way of thinking about
this is that for silicon the electron
configuration is neon here, and then 3s2
3p2. 3s2 3p2 ,  but if it loses two
electrons these three p electrons go
away, so the electron configuration of Si
2 plus is neon 3s2, which has two
valence electrons. If we look at sulfur
first and then sulfur - it's again neon,
so looking at just sulfur first, the
electron configuration 3s2 and then 3
p 1 2 3 4, 3p4. However when we add an
electron to it it becomes 3p5. To find
the valence electrons add all the
electrons in N equals 3 so these
2 s electrons and these 5 p electrons for
a total of seven valence electrons. So
either way you do this you come to the
same answer.
Si two plus has two valence electrons and S
minus has seven valence electrons. I
should also note that these aren't
necessarily stable ions in all of these
examples.
