| (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 in \[Mn{{O}_{2}}=x\] \[\therefore \] \[x+(2x-2)=0\] \[\therefore \] \[x=+4\] (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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