Question

The green paramagnetic species formed by heating $\text{KMnO}_4$ at $513\text{ K}$ is

• A. $\text{KO}_2$
• B. $\text{K}_2\text{MnO}_4$
• C. $\text{Mn}_3\text{O}_4$
• D. MnO

Hint:

Remember the distinct colors and oxidation states of manganese anions to save time in competitive exams: - Permanganate ion ($\text{MnO}_4^{-}$): $\text{Mn}^{+7}$ ($3d^0$), Purple, Diamagnetic. - Manganate ion ($\text{MnO}_4^{2-}$): $\text{Mn}^{+6}$ ($3d^1$), Green, Paramagnetic.

Answer 1

The Correct Option is B

Concept: Potassium permanganate ($\text{KMnO}_4$) is a dark purple crystalline solid. When heated strongly to its thermal decomposition threshold around $513\text{ K}$, it undergoes a chemical breakdown reaction yielding potassium manganate ($\text{K}_2\text{MnO}_4$), manganese dioxide ($\text{MnO}_2$), and oxygen gas ($\text{O}_2$). The formal balanced chemical equation representing this thermal decomposition is written as follows: \[ 2\text{KMnO}_4 \xrightarrow{\Delta \, (513\text{ K})} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2\uparrow \]

Step 1: Identifying the specific product based on physical properties.
The question highlights two specific diagnostic criteria for the product of interest: 1) It must be a green colored species. 2) It must exhibit paramagnetic magnetic behavior. Let us evaluate the two manganese-containing compounds produced during the thermal decomposition:

• $\text{MnO}_2$ (Manganese dioxide) is a dark brown/black insoluble solid.

• $\text{K}_2\text{MnO}_4$ (Potassium manganate) forms dark green crystals and dissolves in water to produce a distinct, vibrant green solution due to the presence of the manganate anion ($\text{MnO}_4^{2-}$).
This immediately focuses our attention on $\text{K}_2\text{MnO}_4$ as the primary candidate.

Step 2: Confirming the magnetic properties via electronic configuration.
Let us rigorously verify the paramagnetic nature of the manganate ion ($\text{MnO}_4^{2-}$) by determining the oxidation state and d-electron configuration of its central manganese atom. In the ionic compound potassium manganate ($\text{K}_2\text{MnO}_4$), potassium retains its standard $+1$ oxidation state. Let $x$ represent the unknown oxidation state of Manganese (Mn), while Oxygen possesses a stable state of $-2$: \[ 2(+1) + x + 4(-2) = 0 \] \[ 2 + x - 8 = 0 \quad \Rightarrow \quad x - 6 = 0 \quad \Rightarrow \quad x = +6 \] Thus, the central manganese ion exists in a $+6$ oxidation state, denoted as $\text{Mn}^{6+}$.

Step 3: Finding the number of unpaired d-electrons.
The ground-state electronic configuration of a neutral transition metal Manganese atom ($Z = 25$) is: \[ \text{Mn} = [\text{Ar}] \, 3d^5 \, 4s^2 \] To achieve the $\text{Mn}^{6+}$ state, we remove six valence electrons sequentially (two from the $4s$ shell and four from the $3d$ subshell): \[ \text{Mn}^{6+} = [\text{Ar}] \, 3d^1 \] Since the $3d$ subshell contains exactly one solitary electron, this electron is unpaired. The presence of an unpaired electron creates a net permanent magnetic dipole moment, which dictates that the species is structurally paramagnetic.

Step 4: Concluding the analysis.
Since $\text{K}_2\text{MnO}_4$ satisfies both fundamental criteria perfectly—being deeply green-colored and containing a paramagnetic $\text{Mn}^{6+}$ center with a $3d^1$ configuration—it is the correct answer, corresponding to Option (2).