00:01
So we're asked to find the oxidation state of si in each of these.
00:07
So for part a, we have si -o -2.
00:09
So if we look at si, we don't know the oxidation number of it, but we do know the number of o2, or specifically o.
00:20
We know that o has the charge of 2 minus, so that's 2 minus, and it's two of them.
00:26
And then we add that to our unknown, x, which is the oxidation state or number of and this should be equal to zero which means this net overall charge is usual.
00:40
So it's all for this.
00:41
We have x plus two times two or negative two, so that i give is negative four is equal to zero.
00:46
So we have that x is equal to four.
00:49
So the oxidation number of si in this case is positive four.
00:53
Now for part b we're going to do something similar.
00:57
So we have x which is the oxidation number of f i plus that of there's four oxygens, which has a 2 minus charge for each oxygen, which is equal to the net overall charge...