Question

The oxidation states of metal in the compounds $F{e}_{0.94}O$ and $\left[Cr\right(PP{h}_3{)}_3\left(CO{)}_3\right]$respectively are:

• A. $\frac{200}{94},0$
• B. $0,\frac{94}{200}$
• C. $2,1$
• D. $1,\frac{200}{94}$

Answer 1

The correct option is A $\frac{200}{94},0$

The oxidation state of $Fe$ in $F{e}_{0.94}O$:

Let the oxidation number of $Fe$ be $x$.

Oxidation number of $O=-2$.

$x\times 0.94+(-2)=0$

$x=+\frac{2}{0.94}$ or $x=+\frac{200}{94}$

$\therefore$ Oxidation state of $Fe$ $=\frac{200}{94}$

Oxidation state of $Cr$ in $\left[Cr\right(PP{h}_3{)}_3\left(CO{)}_3\right]$

Let the oxidation number of $Cr$ be $x$.

Oxidation number of $CO=0,PP{h}_3=0$.

$x+(0\times 3)+(0\times 3)=0$ or $x=0$

$\therefore$ Oxidation state of Cr $=0$