What is the oxidation state of Cr in K2Cr2O4?
k: 2*1
o: 4*-2
Cr : 2*x
2*1+ 4*-2+ 2*x = 0
x = +3
Every compound is zero.
K is 2*+1 = 2
Cr = 2*x = 2x
O = 4*-2 = -8
2 + 2x + (-8) = ?
2(1)+2(x)+4(-2)=0
2-8+2x=0
2x=6
X=3
Well, you know, Cr in K2Cr2O4 likes to keep things interesting. It's like the life of the party - always changing its oxidation state just to keep us on our toes! In K2Cr2O4, Cr has an oxidation state of +6. So, next time you see Cr, just remember - it's the ultimate shape-shifter of the periodic table!
To determine the oxidation state of an element in a compound, we need to assign oxidation numbers to each element based on a set of rules.
In the compound K2Cr2O4, the total charge of the compound is zero since it's a neutral compound. The potassium ion (K+) has a charge of +1, and since there are two of them, the total charge of potassium (K) in the compound is +2.
To determine the oxidation state of chromium (Cr), we need to subtract the total charge of potassium (+2) and the total charge of oxygen (-8) from the total charge of the compound (0).
0 = +2 - 8 + x
Simplifying the equation, we find:
-6 = x
Therefore, the oxidation state of chromium (Cr) in K2Cr2O4 is +6.
To determine the oxidation state of an element in a compound, you need to consider the charges of other elements and the overall charge of the compound, and then solve the equation to find the oxidation state.