Oxidation Number of Iron in [Fe(CN)2(H2O)4]NO3

The complex [Fe(CN)2(H2O)4]NO3 consists of a central iron atom bonded to two cyanide ligands, four water molecules, and a nitrate ion. To determine the oxidation number of iron in this complex, we need to consider the oxidation states of the other atoms involved.

1. Oxidation Number of Cyanide (CN) Ligands
- The cyanide ligand is a strong field ligand, which means it forms a coordinate bond with the central metal atom by donating its lone pair of electrons.
- The oxidation state of carbon (C) in CN is -2, and the oxidation state of nitrogen (N) is -3.
- Since there are two cyanide ligands, the total oxidation number contributed by CN is -2 x 2 = -4.

2. Oxidation Number of Water (H2O) Ligands
- The oxygen (O) atom in water has an oxidation state of -2.
- Since there are four water molecules, the total oxidation number contributed by H2O is -2 x 4 = -8.

3. Oxidation Number of Nitrate (NO3) Ion
- The nitrate ion (NO3-) has a net charge of -1.
- The oxygen (O) atoms in NO3 have an oxidation state of -2.
- We can assign the oxidation state of nitrogen (N) in NO3 as +5 to balance the overall charge of the ion.
- Therefore, the total oxidation number contributed by NO3 is (+5) + (-2 x 3) = +5 - 6 = -1.

4. Calculation of Oxidation Number of Iron (Fe)
- Let the oxidation number of iron (Fe) be x.
- The sum of the oxidation numbers of all the atoms in the complex must equal the overall charge of the complex, which is 0.
- We can write the equation as: x + (-4) + (-8) + (-1) = 0
- Simplifying the equation: x - 4 - 8 - 1 = 0
- Combining like terms: x - 13 = 0
- Adding 13 to both sides: x = 13

5. Conclusion
The oxidation number of iron (Fe) in the complex [Fe(CN)2(H2O)4]NO3 is +3.