
The brown ring complex compound is formulated as \[[Fe{({H_2}O)_5}(NO)]S{O_4}\]. The oxidation state of iron is:
A. $ + 1$
B. $ + 2$
C. $ + 3$
D. Zero
Answer
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Hint: Oxidation state refers to the degree of oxidation of an atom in a chemical compound. It is the imaginary charge that an atom should have if all the bonds between the atoms of the different elements were completely ionic. In other words, the oxidation number refers to the number of electrons transferred between the atoms of a compound/ complex in a chemical reaction.
Complete step by step answer:
During the brown ring test, a coordination complex is formed which has a primary valence in which there are five molecules of water and one molecule of nitrosonium cation. In the complex, the central metal atom is iron and is surrounded by both the primary and secondary valances. There is a sulphate anion present in the secondary valence outside the coordination sphere. In order to find the oxidation state of the central metal atom, we need to determine the oxidation state of the surrounding ligands. In the case of brown ring complex of \[[Fe{({H_2}O)_5}(NO)]S{O_4}\] , the charge on the surrounding ligands are as follows:
${H_2}O = zero$
$NO = + 1$
The overall charge on the coordination sphere will be $ + 2$ as the counterpart of this coordination sphere, $SO_4^{2 - }$ is having an overall charge of $ - 2$ . The brown ring can be written in its ionic form as:
\[[Fe{({H_2}O)_5}(NO)]S{O_4} \to {[Fe{({H_2}O)_5}(NO)]^{ + 2}} + SO_4^{2 - }\]
The charge of the nitrosonium ion is $ + 1$ and the water molecule is a neutral species. Thus, in order to counter the other negative charge, the iron atom shows a $ + 1$ oxidation state to maintain the electrical neutrality of the coordination complex.
Hence, the correct option is A. $ + 1$ .
Note:
In order to carry out the brown ring test, a freshly prepared ferrous sulfate is used because if ferrous sulfate is stored or it is the old solution, then it reacts with the atmospheric oxygen and gets oxidized to form a corrosive brown-yellow coating of basic ferric sulfate, which is an adduct of ferric oxide and ferric sulfate.
Complete step by step answer:
During the brown ring test, a coordination complex is formed which has a primary valence in which there are five molecules of water and one molecule of nitrosonium cation. In the complex, the central metal atom is iron and is surrounded by both the primary and secondary valances. There is a sulphate anion present in the secondary valence outside the coordination sphere. In order to find the oxidation state of the central metal atom, we need to determine the oxidation state of the surrounding ligands. In the case of brown ring complex of \[[Fe{({H_2}O)_5}(NO)]S{O_4}\] , the charge on the surrounding ligands are as follows:
${H_2}O = zero$
$NO = + 1$
The overall charge on the coordination sphere will be $ + 2$ as the counterpart of this coordination sphere, $SO_4^{2 - }$ is having an overall charge of $ - 2$ . The brown ring can be written in its ionic form as:
\[[Fe{({H_2}O)_5}(NO)]S{O_4} \to {[Fe{({H_2}O)_5}(NO)]^{ + 2}} + SO_4^{2 - }\]
The charge of the nitrosonium ion is $ + 1$ and the water molecule is a neutral species. Thus, in order to counter the other negative charge, the iron atom shows a $ + 1$ oxidation state to maintain the electrical neutrality of the coordination complex.
Hence, the correct option is A. $ + 1$ .
Note:
In order to carry out the brown ring test, a freshly prepared ferrous sulfate is used because if ferrous sulfate is stored or it is the old solution, then it reacts with the atmospheric oxygen and gets oxidized to form a corrosive brown-yellow coating of basic ferric sulfate, which is an adduct of ferric oxide and ferric sulfate.
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