question_answer

The oxidation states of sulphur in the anions \[SO_{3}^{2-},\,{{S}_{2}}O_{4}^{2-}\] and \[{{S}_{2}}O_{6}^{2-}\] follow the order:                                                                                                              [AIPMT 2003]

A)  \[{{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}<SO_{3}^{2-}\]

B)  \[{{S}_{2}}O_{6}^{2-}<{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}\]

C)  \[{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}<{{S}_{2}}O_{6}^{2-}\]

D)  \[SO_{3}^{2-}<{{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}\]

Correct Answer: C

Solution :

                       

Oxidation state of \['S'\] in

\[SO_{3}^{2-},\,\,x+\,(-2\times 3)=-2,\,\,x=+6-2=+4\]

Oxidation state of \['S'\] in

\[{{S}_{2}}O_{4}^{2-}\,2x\,+(-2\times 4)=-2\]

\[2x=+8-2=+\,6\]

\[x=\frac{+6}{2}=+3\]

Oxidation state of \['S'\] in

\[{{S}_{2}}O_{6}^{2-}\,2x+(-2\times 6)=-2\]

\[2x=12-2=10\]

\[x=\frac{10}{2}=+5\]

On the basis of structures

Hence, increasing order of oxidation state of is

\[{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}<{{S}_{2}}O_{6}^{2-}\]