K = +1*4 = +4
CN = -1*6 = -6
To make K4[Fe(CN)6] neutral, Fe must be ??
what is the oxidation state of iron in K4[Fe(CN)6]?
3 answers
2
To find the oxidation state of iron (Fe) in K4[Fe(CN)6], we can use the following equation:
Charge of the overall compound = Sum of charges of all the atoms in the compound
The charge of the potassium ion (K+) is +1 and we have four of them, so their total charge is +4.
The charge of the cyanide ion (CN-) is -1 and we have six of them, so their total charge is -6.
Let X be the oxidation state of iron (Fe).
The charge of the entire compound K4[Fe(CN)6] is 0, so we can write:
+4 + X + (-6) = 0
Simplifying and solving for X, we get:
X = +2
Therefore, the oxidation state of iron (Fe) in K4[Fe(CN)6] is +2.
Charge of the overall compound = Sum of charges of all the atoms in the compound
The charge of the potassium ion (K+) is +1 and we have four of them, so their total charge is +4.
The charge of the cyanide ion (CN-) is -1 and we have six of them, so their total charge is -6.
Let X be the oxidation state of iron (Fe).
The charge of the entire compound K4[Fe(CN)6] is 0, so we can write:
+4 + X + (-6) = 0
Simplifying and solving for X, we get:
X = +2
Therefore, the oxidation state of iron (Fe) in K4[Fe(CN)6] is +2.