Question
The oxidation states of metal in the compounds $$F{e_{0.94}}O$$ and $$\left[ {Cr{{\left( {PP{h_3}} \right)}_3}{{\left( {CO} \right)}_3}} \right]$$ respectively are
A.
$$\frac{{200}}{{94}},0$$
B.
$$0,\frac{{94}}{{200}}$$
C.
$$2,1$$
D.
$$1,\frac{{200}}{{94}}$$
Answer :
$$\frac{{200}}{{94}},0$$
Solution :
$$\eqalign{
& F{e_{0.94}}O:x \times 0.94 - 2 = 0 \cr
& \therefore x = \frac{2}{{0.94}} = \frac{{200}}{{94}} \cr
& \therefore {\text{Oxidation state of}}\,\,Fe = \frac{{200}}{{94}} \cr} $$
$$\left[ {Cr{{\left( {PP{h_3}} \right)}_3}{{\left( {CO} \right)}_3}} \right]:$$ $$x + \left( {0 \times 3} \right) + \left( {0 \times 3} \right) = 0,x = 0$$
$$\therefore {\text{Oxidation state of}}\,\,Cr = 0.$$