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 [a] \[{{K}_{3}}[Fe\,{{(OH)}_{6}}]\] Let oxidation state of Fe in \[{{K}_{3}}[Fe\,{{(OH)}_{6}}]=x\] \[(+1\times 2)+x+(4\times -2)=0\] \[x=+6\] [b] \[{{K}_{2}}[Fe\,{{O}_{4}}]\] Let oxidation state of Fe in \[{{K}_{2}}[Fe\,{{O}_{4}}]=x\] \[(+1\times 2)+x+(4\times -2)=0\] \[x=+6\] [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 \[{{[Fe\,{{(CN)}_{6}}]}^{3-}}=x\] \[\therefore \] \[x+(6\times -1)=-3\] \[\therefore \] \[x=+3\] \[\therefore \] \[FeS{{O}_{4}}\,.\,{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\] has Fe in lowest oxidation state.
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