Q60 / 180 · KCET 2009 · Chemistry

The oxidation state of $\mathrm{Fe}$ in the brown ring complex: $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right] \mathrm{SO}_{4}$ is

  • A
    $+3$
  • B
    0
  • C
    $+2$
  • Correct answer
    D
    $+1$

Solution

Let the oxidation state of $\mathrm{Fe}$ in $$ \begin{array}{cc} & {\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right]^{2+}} \\ \Rightarrow & \mathrm{x}+0+1=2 \\ \therefore & \mathrm{x}=+1 \end{array} $$ Here $\mathrm{NO}$ exists as nitrosyl ion $\left(\mathrm{NO}^{+}\right)$.

Hence, the correct answer is option: 'D'.

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