Question

$F_2$ can be prepared by reacting hexafluoromanganate $\left(IV\right)$ with antimony pentafluoride as follows:
$K_{2}Mn{F}_6+Sb{F}_5\stackrel{{150}^{o}C}{\to }KSb{F}_6+Mn{F}_3+F_2$
The number of equivalents of $K_{2}Mn{F}_6$ required to react completely with one mole of $Sb{F}_5$ in the given reaction is:

• A. $1.52$
• B. $5.0$
• C. $0.5$
• D. $4.0$

Answer 1

The correct option is C $0.5$

$K_2\stackrel{+4}{Mn}{F}_6\rightleftharpoons \stackrel{+4}{Mn}{F}_4+2KF$ ....(i)
$2KF+2Sb{F}_5\to 2KSb{F}_6$ .....(ii)
$\stackrel{+4}{Mn}{F}_4\rightleftharpoons \stackrel{+3}{Mn}{F}_3+\frac{1}{2}{F}_2$ ....(iii)
Adding equations (i), (ii) and (iii), we get
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$K_{2}Mn{F}_6+2Sb{F}_5\to 2KSb{F}_6+Mn{F}_3+\frac{1}{2}{F}_2$ .......(iv)
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In equation (iii),
$e^{-}+\stackrel{+4}{Mn}\to \stackrel{+3}{Mn}$ ($n$- factor $=1$)
$F^{⊝}\to \frac{1}{2}{F}_2+e^{-}$($n$-factor $=1$)
Therefore, according to equation (iv),
$2$ mol of $Sb{F}_5\equiv 1$ mol of $K_{2}Mn{F}_6$ $\equiv 1$ Eq of $K_{2}Mn{F}_6$
$\therefore 1$ mol of $Sb{F}_5=\frac{1}{2}$ eq. $=0.5$ eq. of $K_{2}Mn{F}_6$