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

Step 1: Write the decomposition of KMnO4.
On strong heating near \(513\ K\), potassium permanganate breaks down: \[ 2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2\uparrow \] The two manganese products are \(K_2MnO_4\) and \(MnO_2\).
Step 2: Use the colour clue.
The species we want is green. \(MnO_2\) is brown-black, while \(K_2MnO_4\) (potassium manganate) is green. So our candidate is \(K_2MnO_4\).
Step 3: Find the oxidation state of Mn in K2MnO4.
With \(K = +1\) and \(O = -2\): \[ 2(+1)+x+4(-2)=0 \Rightarrow x = +6 \] So manganese is \(Mn^{6+}\).
Step 4: Get the d-electron count.
Neutral \(Mn\) is \([Ar]3d^5 4s^2\). Removing six electrons gives \(Mn^{6+}=[Ar]3d^1\).
Step 5: Check the magnetic nature.
A single \(3d\) electron is unpaired, so \(K_2MnO_4\) is paramagnetic. Both the green colour and the paramagnetism match.
Step 6: Rule out the others and conclude.
\(KO_2\), \(Mn_3O_4\) and \(MnO\) are not the green manganate formed here, so the answer is \(K_2MnO_4\).
\[ \boxed{K_2MnO_4} \]