Question

$\underset{–}{X}M{n}_2^{7+}+\underset{–}{Y}{O}_7^{2-}\to \underset{–}{Z}M{n}^{4+}+\underset{–}{A}{O}_2$
X,Y,Z and A are stoichiometric coefficients.
Here, X,Y,Z and A are respectively

• A. 20, 12, 56, 42
• B. 12, 28, 42, 56
• C. 44, 28, 12, 56
• D. None of these

Answer 1

The correct option is A 20, 12, 56, 42

Let’s calculate how many electrons to be added to either side.
$\stackrel{+7X2+4}{6e^{-}+M{n}_2\to 2Mn}$

i. $\stackrel{+7X20}{{20}_7\to {70}_2+28e^{-}}$

ii. Multiply Eq. (i) by 28 and Eq. (ii) by and add them.
$28\text{/}x6e^{-}+28M{n}_2^{7+}\to 56M{n}^{4+}{120}_7^{2-}\to {420}_2+28\text{/}x6e^{-}\underset{–}{}28M{n}_2^{7+}+{120}_7^{2-}\to 56M{n}^{4+}+{420}_2$