
The order of increasing oxidation number of \[{\mathbf{S}}\] in \[{{\mathbf{S}}_{\mathbf{8}}},{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}},{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}},{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}}\;\] is given below :
A.\[{{\mathbf{S}}_{\mathbf{8}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}}\]
B.\[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{8}}}\]
C.\[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{8}}} < {{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}}\]
D.\[{{\mathbf{S}}_{\mathbf{8}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}}\]
Answer
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Hint: To answer this question, recall the concept of oxidation numbers. The oxidation state of an atom is defined as the number of electrons lost and therefore, describes the extent of oxidation of the atom. For example, the oxidation state of carbon in \[{\text{C}}{{\text{O}}_{\text{2}}}\] would be $ + 4$ since the hypothetical charge held by the carbon atom if both of the carbon-oxygen double bonds were completely ionic would be equal to $ + 4$ .
Complete step by step answer:
The most frequent terms Oxidation state and oxidation number are terms frequently used interchangeably. They are defined and described as the number of electrons lost in an atom. The values can be zero, positive, or negative. Analysing each of the options systematically:
The oxidation number of sulphur \[{{\mathbf{S}}_{\mathbf{8}}}\] will be O. The oxidation number of sulphur \[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}}\] will be
\[2x - 16 = - 2\]
\[ \Rightarrow x = 7\]
This value of oxidation number is not possible as the maximum oxidation state of S can be 6. Hence, this compound has a peroxy-linkage which has not been considered in the above equation.
Hence its oxidation state is $ + 6$.
The oxidation number of sulphur in \[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}}\]
\[2x - 6 = - 2\]
\[ \Rightarrow x = + 2\]
The oxidation number of sulphur in \[{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}}\]
\[4x - 12 = - 2\]
$ \Rightarrow x = 2.5$
Hence, the order of increasing oxidation state is \[{{\mathbf{S}}_{\mathbf{8}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}}\].
Thus, the correct option is option D.
Note:
In most of the compounds, the oxidation number of oxygen is $ - 2$ . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $ - 1$ . Example, \[N{a_2}{O_2}\]
Superoxide- Every oxygen atom is allocated an oxidation number of \[ - \dfrac{{{\text{ }}1}}{2}\] . Example, \[K{O_2}\]
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $ + 1$ .
Complete step by step answer:
The most frequent terms Oxidation state and oxidation number are terms frequently used interchangeably. They are defined and described as the number of electrons lost in an atom. The values can be zero, positive, or negative. Analysing each of the options systematically:
The oxidation number of sulphur \[{{\mathbf{S}}_{\mathbf{8}}}\] will be O. The oxidation number of sulphur \[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}}\] will be
\[2x - 16 = - 2\]
\[ \Rightarrow x = 7\]
This value of oxidation number is not possible as the maximum oxidation state of S can be 6. Hence, this compound has a peroxy-linkage which has not been considered in the above equation.
Hence its oxidation state is $ + 6$.
The oxidation number of sulphur in \[{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}}\]
\[2x - 6 = - 2\]
\[ \Rightarrow x = + 2\]
The oxidation number of sulphur in \[{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}}\]
\[4x - 12 = - 2\]
$ \Rightarrow x = 2.5$
Hence, the order of increasing oxidation state is \[{{\mathbf{S}}_{\mathbf{8}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{ - {\mathbf{2}}} < {{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{8}}}^{ - {\mathbf{2}}}\].
Thus, the correct option is option D.
Note:
In most of the compounds, the oxidation number of oxygen is $ - 2$ . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $ - 1$ . Example, \[N{a_2}{O_2}\]
Superoxide- Every oxygen atom is allocated an oxidation number of \[ - \dfrac{{{\text{ }}1}}{2}\] . Example, \[K{O_2}\]
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $ + 1$ .
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