The oxidation state of Mo in its oxido-complex species ${\left[{\mathrm{Mo}}_2{\mathrm{O}}_4{\left({\mathrm{C}}_2{\mathrm{H}}_4\right)}_2{\left({\mathrm{H}}_2\mathrm{O}\right)}_2\right]}^{2-}$ is:

Let x be the molybdenum (Mo) oxidation number.

oxygenation level is,

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{O}}_2={\mathrm{C}}_2{\mathrm{H}}_4=0,\\ \mathrm{O}=-2\end{array}$

In ${\left[{\mathrm{Mo}}_2{\mathrm{O}}_4{\left({\mathrm{C}}_2{\mathrm{H}}_4\right)}_2{\left({\mathrm{H}}_2\mathrm{O}\right)}_2\right]}^{2-}$ the oxidation number is found is,

$\begin{array}{l}\Rightarrow 2\mathrm{x}+4\times (-2)+2\times 0+0=0\\ \Rightarrow 2\mathrm{x}-8=-2\\ \Rightarrow \mathrm{x}=+3\end{array}$

As a result, Molybdenum's oxidation status in its oxo complex species 

${\left[{\mathrm{Mo}}_2{\mathrm{O}}_4{\left({\mathrm{C}}_2{\mathrm{H}}_4\right)}_2\left({\mathrm{H}}_2{\mathrm{O}}_2\right)\right]}^{2-}$ is+ 3.

Hence correct option is B.