Q60 / 180 · NEET 2026 (Re-NEET) · Chemistry
The green paramagnetic species formed by heating KMnO$_4$ at $513$ K is
- A$\text{KO}_2$
- Correct answerB$\text{K}_2\text{MnO}_4$
- C$\text{Mn}_3\text{O}_4$
- D$\text{MnO}$
Solution
When potassium permanganate ($\text{KMnO}_4$) is heated at $513\text{ K}$, it decomposes to form potassium manganate, manganese dioxide, and oxygen gas.
$2\text{KMnO}_4 \xrightarrow{513\text{ K}} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$
The species $\text{K}_2\text{MnO}_4$ (potassium manganate) is green in color.
In $\text{K}_2\text{MnO}_4$, the oxidation state of $\text{Mn}$ is $+6$. The electronic configuration of $\text{Mn}^{6+}$ is $[\text{Ar}] 3d^1$.
Since it contains one unpaired electron in the $d$-orbital, it is paramagnetic in nature.
Thus, the green paramagnetic species formed is $\text{K}_2\text{MnO}_4$.
Answer: $\text{K}_2\text{MnO}_4$
$2\text{KMnO}_4 \xrightarrow{513\text{ K}} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$
The species $\text{K}_2\text{MnO}_4$ (potassium manganate) is green in color.
In $\text{K}_2\text{MnO}_4$, the oxidation state of $\text{Mn}$ is $+6$. The electronic configuration of $\text{Mn}^{6+}$ is $[\text{Ar}] 3d^1$.
Since it contains one unpaired electron in the $d$-orbital, it is paramagnetic in nature.
Thus, the green paramagnetic species formed is $\text{K}_2\text{MnO}_4$.
Answer: $\text{K}_2\text{MnO}_4$
Hence, the correct answer is option: 'B'.