Question

The oxidation number of Cr is +6 in:
This question has multiple correct options
A. \[{\text{FeC}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
B. \[{\text{KCr}}{{\text{O}}_{\text{3}}}{\text{C1}}\]
C. \[{\text{Cr}}{{\text{O}}_5}\]
D. \[{\left[ {{\text{Cr}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }\]

Answer 1

Hint:In a neutral atom, the sum of the oxidation numbers of all the atoms is equal to zero. However, in the ion, the sum of the oxidation number of all the atoms is equal to the charge on the ion.

Complete answer:
Let x be the oxidation number of chromium in \[{\text{FeC}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
The oxidation numbers of iron and oxygen are +2 and -2 respectively.
In a neutral atom, the sum of the oxidation numbers of all the atoms is equal to zero. Hence, the sum of the oxidation numbers of one iron atom, two chromium atoms and four oxygen atoms will be equal to zero.
Calculate the oxidation number of chromium in \[{\text{FeC}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
\[
\Rightarrow + 2 + 2x + 4\left( { - 2} \right) = 0 \\
\Rightarrow 2 + 2x - 8 = 0 \\
\Rightarrow 2x - 6 = 0 \\
 \]
\[
\Rightarrow 2x = 6 \\
\Rightarrow x = + 3 \\
 \]
Hence, the oxidation number of chromium in \[{\text{FeC}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is \[ + 3\] .

Let x be the oxidation number of chromium in \[{\text{KCr}}{{\text{O}}_{\text{3}}}{\text{C1}}\]
The oxidation numbers of potassium, chlorine and oxygen are +1,-1 and -2 respectively.
In a neutral atom, the sum of the oxidation numbers of all the atoms is equal to zero. Hence, the sum of the oxidation numbers of one potassium atom, one chromium atom, one chlorine atom and three oxygen atoms will be equal to zero.
Calculate the oxidation number of chromium in \[{\text{KCr}}{{\text{O}}_{\text{3}}}{\text{C1}}\]
\[
\Rightarrow + 1 + x + 3\left( { - 2} \right) - 1 = 0 \\
\Rightarrow 1 + x - 6 - 1 = 0 \\
\Rightarrow x - 6 = 0 \\
\Rightarrow x = + 6 \\
 \]
Hence, the oxidation number of chromium in \[{\text{KCr}}{{\text{O}}_{\text{3}}}{\text{C1}}\] is \[ + 6\] .

Let x be the oxidation number of chromium in \[{\text{Cr}}{{\text{O}}_5}\]
The oxidation numbers of oxygen of peroxide linkage and normal oxygen are -1 and -2 respectively. In \[{\text{Cr}}{{\text{O}}_5}\]molecule, two peroxide linkages are present. Each peroxide linkage contains two oxygen atoms. Thus, out of five oxygen atoms, four oxygen atoms are present as two peroxide linkages. These oxygen atoms have oxidation numbers of -1 each. The remaining one oxygen atom has an oxidation number of -2.
In a neutral atom, the sum of the oxidation numbers of all the atoms is equal to zero. Hence, the sum of the oxidation numbers of one chromium atom and five oxygen atoms will be equal to zero.
Calculate the oxidation number of chromium in \[{\text{Cr}}{{\text{O}}_5}\]
\[
\Rightarrow x + 2\left( { - 1 - 1} \right) - 2 = 0 \\
\Rightarrow x - 4 - 2 = 0 \\
\Rightarrow x - 6 = 0 \\
\Rightarrow x = + 6 \\
 \]
Hence, the oxidation number of chromium in \[{\text{Cr}}{{\text{O}}_5}\] is \[ + 6\] .

Let x be the oxidation number of chromium in \[{\left[ {{\text{Cr}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }\]
The oxidation numbers of iron and oxygen are +2 and -2 respectively.
In the ion, the sum of the oxidation number of all the atoms is equal to the charge on the ion. Hence, the sum of the oxidation numbers of one chromium atom, four hydrogen atoms and four oxygen atoms will be equal to -1.
Calculate the oxidation number of chromium in \[{\left[ {{\text{Cr}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }\]
\[
\Rightarrow x + 4\left( {1 - 2} \right) = - 1 \\
\Rightarrow x + 4\left( { - 1} \right) = - 1 \\
\Rightarrow x - 4 = - 1 \\
\Rightarrow x = + 3 \\
 \]
Hence, the oxidation number of chromium in \[{\left[ {{\text{Cr}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }\] is \[ + 3\] .
The oxidation number of Cr is +6 in \[{\text{KCr}}{{\text{O}}_{\text{3}}}{\text{C1}}\] and \[{\text{Cr}}{{\text{O}}_5}\]

Thus, the correct answers are options (B) and (C).

Note:Oxidation number represents the number of electrons gained or lost to form an ion. A cation is obtained when a neutral atom loses either one or more electrons. An anion is formed when a neutral atom gains either one or more electrons.