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
615.3k+ views
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.
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.
Recently Updated Pages
The given figure shows two endocrine glands marked class 11 biology NEET_UG

Match columnI with columnII and select the correct class 11 biology NEET

Match column I with column II and select the correct class 11 biology NEET_UG

Which floral family has left 9 right + 1 arrangement class 11 biology NEET_UG

Which is not a variety of sheep A Lohi B Beetal C Nellore class 11 biology NEET_UG

Match column I with column II and select the correct class 11 biology NEET_UG

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

