Question

Find the oxidation state of Fe in the brown ring complex $\left[ Fe{{({{H}_{2}}O)}_{5}}NO \right]S{{O}_{4}}$:
(a) +1
(b) -1
(c) 2
(d) 0

Answer 1

Hint : The first nitrosyl complex was discovered by J. Priestley in 1790, (${{\left[ Fe{{({{H}_{2}}O)}_{5}}(NO) \right]}^{2+}}$). Here, we will first write the oxidation state of each molecule individually leaving only Fe, which we have to find.

Complete step by step solution : This complex is formed in the qualitative test for nitrate with ferrous sulfate and sulfuric acid (“brown ring test”). This brown complex also is used in the brown ring test for detection of nitrate ions. In this test some ferrous salt is added to the solution to be tested. If nitrate ions are present in the solution, then a brown ring is formed at the interface between the ${{H}_{2}}S{{O}_{4}}$ layer and the aqueous layer. At the interface-layer, the nitrate ions oxidize ferrous ions, forming NO and ferric ions. As a result, the NO reacts with excess ferrous ions, forming the deep brown nitrosyl/iron(I) complex.

In the brown ring complex:
The five water molecules present in the equation are neutral.
NO is a Monodentate Anionic Ligand, oxidation state of NO is +1.
Sulphate ($S{{O}_{4}}$) is a divalent ion, oxidation state of $S{{O}_{4}}$is -2.
Assuming the oxidation state of Fe as x.
x+5(0) +(+1) +(-2) =0
x+1-2=0
x=+1
So, the correct answer is (a).

Note : The brown complex is not very stable. Immediately after the complex is formed, the color starts slowly fading away. As a result, after several minutes the color already is less intense as it was earlier.