2KMnO_4 xrightarrow{200^{circ}C} K_2MnO_4 + M | vedclass

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(D) The thermal decomposition of $KMnO_4$ at $200^{\circ}C$ is given by: $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$. Thus,'$X$' is $K_2MnO_4$ (potas

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All of these

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(D) The thermal decomposition of $KMnO_4$ at $200^{\circ}C$ is given by: $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$. Thus,'$X$' is $K_2MnO_4$ (potassium manganate).$1$. In $K_2MnO_4$,the oxidation state of $Mn$ is $+6$ and the $MnO_4^{2-}$ ion has a tetrahedral geometry. So,statement $C$ is correct.$2$. $K_2MnO_4$ undergoes disproportionation in acidic or neutral medium to form $KMnO_4$ (purple colour). $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$. Thus,reaction with $H_2O$ (neutral/acidic) gives a purple colour. So,statement $B$ is correct.$3$. $K_2MnO_4$ reacts with $Cl_2$ to form $KMnO_4$ (purple colour): $2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl$. So,statement $A$ is correct.Since all statements are correct,the answer is $D$.

List-$I$ (Aqueous solution of salt)	List-$II$ (Color)
$A$. $FeSO_4 \cdot 7H_2O$	$i$. Green
$B$. $NiCl_2 \cdot 4H_2O$	$ii$. Pale pink
$C$. $MnCl_2 \cdot 4H_2O$	$iii$. Blue
$D$. $CoCl_2 \cdot 6H_2O$	$iv$. Light green
$E$. $Cu_2Cl_2$	$v$. Pink
	$vi$. Colorless

$2KMnO_4 \xrightarrow{200^{\circ}C} K_2MnO_4 + MnO_2 + O_2$. Here,'$X$' is $K_2MnO_4$. Which of the following statements is correct for '$X$'?

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