Find the oxidation state of Fe in the brown ring complex [Fe(H $_{2}$ O) $_{5}$ NO]SO $_{4}$ :

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Solution

The correct option is A 1
Let the oxidation state of $F e$ is $x$ in the complex

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

$x + 5 \times 0 + 1 = + 2 x = + 1$

Therefore, option is A.

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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}

{N O}_{3}^{-}

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

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