Question

${[Fe{({H_2}O)_5}NO]^{2 + }}$ is a complex formed during the brown ring test for $N{O_3}^ - $ ion. In this complex :
a.) There are three unpaired electrons so that its magnetic moment is 3.87 BM.
b.) NO transfer its electron to $F{e^{2 + }}$ so that iron as $F{e^{( - )}}$ and NO as $N{O^ + }$.
c.) The colour is because of charge transfer.
d.) All of the above statements are correct.

Answer 1

Hint: The brown ring test is used for identification of $N{O_3}^ - $ in the compound. During this brown ring test, ${[Fe{({H_2}O)_5}NO]^{2 + }}$ complex is formed which has been isolated and studied. The complex ${[Fe{({H_2}O)_5}NO]^{2 + }}$ is unstable in nature. In this complex, Fe is in +1 oxidation state while NO is also in +1 oxidation state. There is transfer of charges in the complex which results in change in physical appearance.

Complete answer:
> We know that brown ring test is used for identification of $N{O_3}^ - $ in the compound given. The results of this test show the presence of a ring around the test tube which is brown in colour.
> Initially a test tube is taken. We put freshly prepared ferrous sulphate and then nitric acid. After this, concentrated sulphuric acid is added on sides of the test tube. Then what we see is a brown coloured ring which confirms the presence of $N{O_3}^ - $ ion.
> During this brown ring test, ${[Fe{({H_2}O)_5}NO]^{2 + }}$ complex is formed which has been isolated and studied. This complex has three unpaired electrons and thus has a magnetic moment of 3.87 BM. So, our option a.) is correct.
> Now, for option b.) it is true that NO transfers its one electron to $F{e^{2 + }}$ but after transfer of electron Fe has an oxidation state of +1 and not (-1). So, the option b.) is not completely true and will account for incorrect options.
> The option c.) says that the colour is due to charge transfer which is true. Thus, option c.) is also the correct answer.

So, the correct options are a.) and c.).

Note: Fe (+1) has electronic configuration as - $[Ar]3{d^6}4{s^1}$
NO is a strong ligand. So, one pairing of 3d and 4s orbital will take place and the complex will have three unpaired electrons.
Almost all the nitrates dissolve in water. So, their test in liquid state is not easy. The Brown ring test is used to detect the presence of nitrate in soil, water and food samples.