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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

All of these

Vedclass · Answer

(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 $D$.

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

Similar Questions

When potassium permanganate $(KMnO_4)$ is heated at $513 \ K$,it decomposes to form two manganese-based products. The physical properties of the product in which manganese has a higher oxidation state than the other are:

The formula of corrosive sublimate is

Which one of the following is lunar caustic?

The transition metal ions responsible for colour in ruby and emerald are,respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module