A) \[{{K}_{3}}[Fe{{(OH)}_{6}}]\]
B) \[{{K}_{2}}[Fe{{O}_{4}}]\]
C) \[FeS{{O}_{4}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O\]
D) \[{{[Fe{{(CN)}_{6}}]}^{3-}}\]
Correct Answer: C
Solution :
Key Idea: Find oxidation state of Fe in all compound to find which has Fe in least oxidation state \[{{\text{K}}_{\text{3}}}\text{ }\!\![\!\!\text{ Fe(OH}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }\] Let oxidation state of Fe in \[{{K}_{3}}[Fe{{(OH)}_{6}}]=x\] \[(+\,1\,\times \,3)+x+(-1\times 6)=0\] \[\therefore \] \[x=+\,3\] (b) \[{{K}_{2}}[Fe{{O}_{4}}]\] Let oxidation state of Fe in \[{{K}_{2}}[Fe{{O}_{4}}]=x\] \[(+1\times 2)+x+(4x-2)=0\] \[x=+\,6\] (c) \[FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O\] Let oxidation state of Fe in \[FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O=x\] \[\therefore \] \[x+(-2)+2+(-2)=0\] \[x=+\,2\] (d) \[{{[Fe{{(CN)}_{6}}]}^{3-}}\] Let oxidation state of Fe in \[{{[Fe{{(CN)}_{6}}]}^{3-}}=x\] \[\therefore \] \[x+(6x-1)=-3\] \[\therefore \] \[x=+\,3\] \[\therefore \]\[FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\]has Fe in lowest oxidation state.
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