Question

Arrange the following as increase in oxidation number : (i) $Mn ^{2+}$ (ii) $MnO _{2}$ (iii) $KMnO _{4}$ (iv) $K _{2} MnO _{4}$

• A. $(i)>(ii)>(iii)>(iv)$
• B. $(i)(iii)$
• C. $ (ii)
• D. $(iii)>(i)>(iv)>(ii)$

Answer 1

Answer: (B)

$(i)(iii)$

Answer 2

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 $Mn MnO _{2}=x$
$\therefore x+(2 x+2)=0$
$\therefore x=+4$
(iii) Let oxidation state of $Mn$ in $KMnO _{4}=x$
$\therefore+1+x+(-2 \times 4)=0$
$\therefore x=+7$
(iv) Let oxidation state of $Mn$ in $K _{2} MnO _{4}=x$
$\therefore(+1 \times 2)+x+(-2 \times 4)=0$
$\therefore x=+6$
$\therefore$ Increasing order of oxidation states is (i) $