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OpenStudy (anonymous):

What is the oxidation state of S in HSO42-? A. +9 B. +7 C. +5 D. +6

OpenStudy (anonymous):

+5 right?

OpenStudy (anonymous):

or is it 6.

OpenStudy (anonymous):

i know its either 5 or 6

OpenStudy (jfraser):

the charge of \(HSO_4\) isn't -2, it should be -1. Unless you mean just the \(SO_4\) ion, which is a -2.

OpenStudy (anonymous):

OpenStudy (anonymous):

there's a screenshot above

OpenStudy (jfraser):

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

OpenStudy (jfraser):

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

OpenStudy (jfraser):

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)

OpenStudy (jfraser):

So now we just set up a simple algebraic statement

OpenStudy (jfraser):

\[(+1) + (sulfur) + 4(-2) = -2\]

OpenStudy (jfraser):

one hydrogen, (+1), plus the sulfur, whatever it is, plus 4 oxygens (4x(-2)) must equal the -2 charge on the ion

OpenStudy (anonymous):

so 6

OpenStudy (anonymous):

5

OpenStudy (anonymous):

yea its 5

OpenStudy (jfraser):

it is 5. more specifically +5

OpenStudy (anonymous):

yep okay thanks

OpenStudy (jfraser):

when dealing with charges, lots of times we explicitly write the positive sign, +5, just to be sure

OpenStudy (anonymous):

yea i noticed. thanks again

OpenStudy (jfraser):

wherever you got the question from, it's wrong though.

OpenStudy (anonymous):

my chem teacher

OpenStudy (jfraser):

well politely (very politely) inform them to check their charges

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