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# The oxidation state of $Fe$ in ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ is:A.$+ 2$B.$+ 3$C.$+ 4$D.$+ 6$

Last updated date: 09th Sep 2024
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:Hint:This question gives knowledge about oxidation number. Oxidation number is the charge present on an atom when it forms ionic bonds with different heteroatoms. Oxidation number is also termed as oxidation state.

Let the oxidation state of iron be $x$.
Consider ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ to determine the oxidation state of iron. In this molecule, the oxidation state of cyanide ion is $- 1$ and potassium is $+ 1$ . The overall charge on this compound is $0$. On summing up all the charges along with their molecularity we observe that the oxidation number of iron is $+ 2$.
Therefore, the oxidation state of iron in the complex ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ is $+ 2$.
Hence, option $A$ is the correct option.