Question

The oxidation state of $Fe$ in the brown ring complex : $[Fe(H_2O)_5 NO]SO_4$ is

• A. +2
• B. +1
• C. +3
• D. 0

Answer 1

Answer: (B)

Let the oxidation state of $Fe$ in

$\left[ Fe \left( H _{2} O \right)_{5} NO \right] SO _{4}$ is $ x$ .

$\left[ Fe \left( H _{2} O \right)_{5} NO \right]^{2+} $

$\Rightarrow \,\,\,\,\,\,x+0+1=2$

$\therefore \,\,\,\,\,\,x=+1$

Here $NO$ exists as nitrosyl ion $(NO^+)$