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The pair of compounds having metals in their highest oxidation state is:
A. ${\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }}\;{\rm{and}}\;{\left[ {{\rm{Cu}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{2 - }}$
B. ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}\;{\rm{and}}\;{\rm{Cr}}{{\rm{O}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}$
C. ${\left[ {{\rm{FeC}}{{\rm{l}}_4}} \right]^ - }\;{\rm{and}}\;{\rm{C}}{{\rm{o}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}$
D. \[{\left[ {{\rm{NiC}}{{\rm{l}}_{\rm{4}}}} \right]^{2 - }}\;{\rm{and}}\;{\left[ {{\rm{CoC}}{{\rm{l}}_{\rm{4}}}} \right]^{2 - }}\]

Answer
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Hint: We know that the oxidation state shows the degree of oxidation of any compound. It is expressed in integers; it may be positive, negative and zero. It usually gives the knowledge about the loss of electrons in any compound or element.

Complete step-by-step answer: In option A, the oxidation state of metal Fe in ${\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }}$ is calculated as shown below:
Let us consider the oxidation state of metal Fe as $x$. Therefore,
$\begin{array}{c}
x + 6\left( { - 1} \right) = - 3\\
x = - 3 + 6\\
x = + 3
\end{array}$
Therefore, the oxidation state of metal Fe in ${\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }}$ is $ + 3$ and similarly, the oxidation state of metal Cu in ${\left[ {{\rm{Cu}}{{\left( {{\rm{CN}}} \right)}_4}} \right]^{2 - }}$ is $ + 2$.
Similarly, in option B the oxidation state of metal Mn in ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$ is $ + 4$ and the oxidation state of metal Cr in ${\rm{Cr}}{{\rm{O}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}$ is $ + 6$.
In option C the oxidation state of metal Fe in ${\left[ {{\rm{FeC}}{{\rm{l}}_4}} \right]^ - }$ is $ + 3$ and the oxidation state of metal Co in ${\rm{C}}{{\rm{o}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}$ is also $ + 3$.
In option D the oxidation state of metal Ni is ${\left[ {{\rm{NiC}}{{\rm{l}}_{\rm{4}}}} \right]^{2 - }}$ is $ + 2$ and the oxidation state of metal Co in \[{\left[ {{\rm{CoC}}{{\rm{l}}_{\rm{4}}}} \right]^{2 - }}\] is $ + 2$.
So, according to the individual oxidation state of each compound, the highest oxidation state of Mn and Cr is highest. Thus, the compound of option B has the highest oxidation state.

Therefore, the correct option for this question is B that is ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}\;{\rm{and}}\;{\rm{Cr}}{{\rm{O}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}$.

Note: The oxidation state depends on the color property of any element. The oxidation statement also describes the compound is paramagnetic, ferromagnetic and diamagnetic.