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The oxidation number of iron in brown ri...

The oxidation number of iron in brown ring complexes +x. What is value of x.

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To determine the oxidation number of iron in the brown ring complex, we can follow these steps: ### Step 1: Identify the complex The brown ring complex is represented as [Fe(H2O)5(NO)]SO4. Here, we have iron (Fe) coordinated with five water molecules (H2O) and one nitric oxide (NO) molecule, along with a sulfate ion (SO4^2-). ### Step 2: Assign oxidation states to the ligands - Water (H2O) is a neutral ligand, so it has an oxidation state of 0. - The sulfate ion (SO4^2-) has a charge of -2. - The nitric oxide (NO) acts as a donor and, in this complex, it is considered to have a charge of +1 (as it donates one electron to iron). ### Step 3: Set up the equation Let the oxidation number of iron be +x. We can now write the equation based on the overall charge of the complex. The total charge of the complex must equal 0, since it is neutral. The equation can be set up as follows: \[ \text{Total charge} = \text{Oxidation state of Fe} + \text{Oxidation states of ligands} \] This gives us: \[ x + (5 \times 0) + (+1) + (-2) = 0 \] ### Step 4: Simplify the equation Now, simplify the equation: \[ x + 0 + 1 - 2 = 0 \] This simplifies to: \[ x - 1 = 0 \] ### Step 5: Solve for x Now, we can solve for x: \[ x = 1 \] ### Conclusion The oxidation number of iron in the brown ring complex is +1.
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