Question

Arrange the following in correct order of oxidation number of Mn
(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)

Answer 1

Answer: (B)

(i) < (ii) < (iv)< (iii)

i) Oxidation number of Mn in $M{n}^{2+}=+2$
(ii) Let oxidation number of Mn in $Mn{O}_2=x$
$\therefore x+(-2\times 2)=0,x=+4$
(iii) Let oxidation number of Mn in KMno_4 = x
$\therefore +1+x+(-1\times 4)=0,\therefore x+=+7$
(iv) Let oxidation number of Mn in $K_{2}Mn{O}_4$ = x
$\therefore (+1\times 2)+x+(-2\times \times 4_{=}0$
or x + 2 - 8 = 0
$\therefore$ x = +6
$\therefore$ the order of oxidation state is (i) < (ii) < (iv) < (iii)