What is the oxidation state of S in HSO42-? A. +9 B. +7 C. +5 D. +6
+5 right?
or is it 6.
i know its either 5 or 6
the charge of \(HSO_4\) isn't -2, it should be -1. Unless you mean just the \(SO_4\) ion, which is a -2.
there's a screenshot above
well that's just wrong, but based on that info, there's really a simple way to find the oxidation of one of the atoms in an ion
The charges of all the pieces of the ion have to add up to the total charge on the ion, and there are lots of atoms that form common "always" ions
as a cation, H is always a +1 as an anion, oxygen is "99%" a -2 (unless it's in a peroxide, which this isn't, so it's still a -2)
So now we just set up a simple algebraic statement
\[(+1) + (sulfur) + 4(-2) = -2\]
one hydrogen, (+1), plus the sulfur, whatever it is, plus 4 oxygens (4x(-2)) must equal the -2 charge on the ion
so 6
5
yea its 5
it is 5. more specifically +5
yep okay thanks
when dealing with charges, lots of times we explicitly write the positive sign, +5, just to be sure
yea i noticed. thanks again
wherever you got the question from, it's wrong though.
my chem teacher
well politely (very politely) inform them to check their charges
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