What is the oxidation number of chromium in HCrO4?
The correct answer is +7, but I don't understand how to come up with this. I tried adding the charges together, but it wasn't working out for me.
2007-06-22 08:16:54 · 5 answers · asked by JMEB 2 in Science & Mathematics ➔ Chemistry
The correct answer was +7, but that wasn't the only answer
The other answers I had to choose from were +1, -1, +2,+7, +6
2007-06-22 08:41:36 · update #1
Start with what you know.
H is a +1 charge in this case. O is a -2 and there are four of them. Net charge is 0 (the charge on the entire molecule)
(+1) +x -2(4) = 0
-7 +x = 0
x = 7
2007-06-22 08:21:02 · answer #1 · answered by Wheels 3 · 0⤊ 0⤋
OK...In nearly all compounds, oxygen atoms are assigned an oxidation number of -2. Hydrogen in most compounds is +1. Since this is a compound, the sum of the oxidation numbers has to be 0.
So, 1 + Cr + 4(-2) = 0
Cr = +7
2007-06-22 15:26:42 · answer #2 · answered by hcbiochem 7 · 0⤊ 0⤋
H is almost always +1 (just assume it is)
O is always -2, except when its with a Fluorine, then it becomes +2
The other atoms left have to add up to zero if it's a neutral molecule, or the charge on the molecule if it's an ion.
So there is one H. that = +1
There are four oxygens. That = -8
Then there is one chromium. +1 + -8 = -7
so to make the charge = 0, chromium must have a +7 charge.
2007-06-22 15:25:47 · answer #3 · answered by Joeseph 2 · 0⤊ 0⤋
H (+1)
O (-2)
+1-2-2-2-2 = -7 so Cr = +7
2007-06-22 15:36:04 · answer #4 · answered by skipper 7 · 1⤊ 0⤋
there are some standard chosen in computation of oxidation number.
o.n. of H = +1 ( except in compounds like, NaH, KH etc)
o.n. of O = -2 ( except in OF2)
let o.n. of Cr = x
=> 1 + x + 4 x (-2) = 0
=> x = +7
there u r.
2007-06-22 15:44:33 · answer #5 · answered by s0u1 reaver 5 · 0⤊ 0⤋