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The **oxidation state of Fe** in $K_4[Fe(CN)_6]$ is:
A+1
B+2
C+3
D+4
Answer & Solution
Correct answer: B. +2
Charge balance: 4(+1) + [Fe + 6(-1)] = 0 ⇒ Fe + (-6) = -4 ⇒ Fe = **+2**. K₄[Fe(CN)₆] is potassium ferrocyanide.
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