MCQ
$KMn{O_4}$ in basic medium is reduced to
- A${K_2}Mn{O_4}$
- ✓$Mn{O_2}$
- C$Mn{(OH)_2}$
- D$M{n^{2 + }}$
$KMn{O_4}$ is first reduced to manganate and then to insoluble manganese dioxide. Colour changes first from purple to green and finally becomes colourless.
$2KMn{O_7} + 2KOH \to 2{K_2}Mn{O_4} + {H_2}O + O$
$2{K_2}Mn{O_4} + 2{H_2}O \to 2Mn{O_2} + 4KOH + 2O$
$\overline {2KMn{O_2} + {H_2}O\xrightarrow{{{\text{alkaline}}}}2Mn{O_2} + 2KOH + 3[O]} $
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$(a)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$
$(b)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{-}$
$(c)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right]$
$(d)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right]^{2+}$
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$1\,L$ Solution $( X )+ BaCl _2$ solution (excess) $\rightarrow Z$
The number of moles of $Y$ and $Z$ respectively are
Rate of the reaction is the highest for