| (i)\[M{{n}^{2+}}\] |
| (ii) \[Mn{{O}_{2}}\] |
| (iii) \[KMn{{O}_{4}}\] |
| (iv) \[{{K}_{2}}Mn{{O}_{4}}\] |
A) \[(i)>(ii)>(iii)>(iv)\]
B) \[(i)<(ii)<(iv)>(iii)\]
C) \[(ii)<(iii)<(i)<(iv)\]
D) \[(iii)>(i)>(iv)>(ii)\]
Correct Answer: B
Solution :
Key Idea: the sum of oxidation states of all elements in a compound is always zero. (i) Oxidation state of Mn in \[M{{n}^{2+}}=+2\] (ii) Let oxidation state of Mn \[Mn{{O}_{2}}=x\] \[\therefore \] \[x+(2x+2)=0\] \[\therefore \] \[x=+\,4\] (iii) Let oxidation state of Mn in \[KMn{{O}_{4}}=x\] \[\therefore \] \[+1+x+(-2\times 4)=0\] \[\therefore \] \[x=+\,7\] (iv) Let oxidation state of Mn in \[{{K}_{2}}Mn{{O}_{4}}=x\] \[\therefore \] \[(+1\times 2)+x+(-2\times 4)=0\] \[\therefore \] \[x=+\,6\] \[\therefore \] Increasing order of oxidation states is (i) < (ii) < (iv) < (iii)
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