What is the oxidation number of Fe in $[F e (H_{2} O) 5 N O] S O_{4}$ ?

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Solution

The oxidation number of Fe in this complex, is +1. This complex is formed in the 'brown ring' test for the nitrate ion in which Fe is in +1 state, and NO exists as $N O^{+}$ .

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[F e (H_{2} O)_{5} N O^{+}] S O_{4}

The oxidation state of Fe in browns ring complex $[F e (H_{2} O)_{5} N O^{+}] S O_{4}$ is:

[F e (H_{2} O)_{5} N O] S O_{4}

[F e (H_{2} O)_{5} N O] S O_{4}

F e

[F e (H_{2} O)_{5} (N O)]^{2 +}

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