The brown ring complex compound [Fe(H2O)5NO]SO4 has the following components:

1. Iron (Fe)
2. Water molecules (H2O)
3. Nitric oxide (NO)
4. Sulfate ion (SO4)

To determine the oxidation state of iron in this compound, we need to first understand the concept of oxidation state.

Oxidation state is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. It is also known as oxidation number.

To determine the oxidation state of iron in this compound, we need to assign oxidation states to all the other atoms in the compound first.

Oxidation state of oxygen (O) in water (H2O) is -2.
Oxidation state of nitrogen (N) in nitric oxide (NO) is +2.
Oxidation state of sulfur (S) in sulfate ion (SO4) is +6.

Using this information, we can calculate the oxidation state of iron as follows:

Let x be the oxidation state of iron (Fe).

The total charge of the compound must be zero, so we can write:

(+5) x [(oxidation state of Fe)] + (-10) x [(oxidation state of O in H2O)] + (2) x [(oxidation state of N in NO)] + (-2) x [(oxidation state of S in SO4)] = 0

Simplifying the above equation, we get:

(+5)x + (-20) + (2) x (+2) + (-2) x (+6) = 0

(+5)x - 20 + 4 - 12 = 0

(+5)x - 28 = 0

(+5)x = 28

x = +5.6

Since the oxidation state of iron cannot be a fractional number, we need to round it to the nearest whole number. In this case, the oxidation state of iron is +6.

However, we need to take into account the fact that Fe is coordinated to five water molecules and one NO molecule. The sum of the oxidation states of all ligands in a complex is equal to the charge of the complex. In this case, the complex has a charge of +1 (from the sulfate ion), so the sum of the oxidation states of the ligands must be -1.

Water molecules are neutral, so their oxidation state is 0.

The oxidation state of NO is +2, so the sum of the oxidation states of the five water molecules must be -2.

Therefore, the sum of the oxidation states of the water molecules and the NO molecule is:

(5 x 0) + (+2) = +2

Since the sum of the oxidation states of the ligands is -1, the oxidation state of iron must be:

-1 - (+2) = -3

However, we need to remember that the oxidation state of iron in a complex is not always the same as its formal oxidation state. In this case, the iron atom is in a low-spin state, which means that the d electrons are paired up in the t2g orbitals. This indicates that the iron atom has lost two electrons, which corresponds to an oxidation state of +2.

Therefore, the oxidation state of iron in the brown ring complex compound [Fe(H2O)5NO]