Question

The brown ring complex compound is formulated as $\left[Fe\right(H_{2}O{)}_5\left(NO\right)]S{O}_4$]. The oxidation state of iron in this complex is:

• A. $0$
• B. $+1$
• C. $+2$
• D. $+3$

Answer 1

Answer: (B)

$+1$

The oxidation number of iron in the above compound can be calculate as follow:

[Fe(NO)(H2O)${}_5$]SO${}_4$ dissociate in water to liberate[Fe(NO)(H2O)${}_5$]${}^{2+}$ ions and ${S{O}_4}^{2-}$ ions.

The charge on NO is +1 i.e. NO${}^{+}$, whereas H${}_2$O ligand is neutral.

Hence the oxidation state of iron in above complex is

$X+(+1)+5\left(0\right)=+2$

$\Rightarrow X=+1$

Therefore, iron is present in +1 oxidation state in the above complex.

Hence, the correct option is $B$