Potassium permanganate on heating at 513 K gives a product which is :
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To determine if a transition metal compound is paramagnetic or diamagnetic, always find the oxidation state of the metal, write down its d-electron configuration, and check for unpaired electrons. Any unpaired electrons will result in paramagnetism.
Step 1: Understanding the Question:
We need to identify the properties (color and magnetic nature) of the product formed when potassium permanganate (KMnO\(_4\)) is heated. Step 2: Detailed Explanation: 1. The Reaction:
When potassium permanganate is heated to about 513 K (240 \(^\circ\)C), it undergoes thermal decomposition. The balanced chemical equation for the reaction is:
\[ 2\text{KMnO}_4(s) \xrightarrow{\Delta} \text{K}_2\text{MnO}_4(s) + \text{MnO}_2(s) + \text{O}_2(g) \]
The main solid products are potassium manganate (K\(_2\)MnO\(_4\)) and manganese dioxide (MnO\(_2\)). The question most likely refers to the major ionic product, potassium manganate. 2. Color of the Product:
Potassium manganate (K\(_2\)MnO\(_4\)) is a well-known compound that is green in color. Potassium permanganate itself is purple, so a distinct color change occurs. 3. Magnetic Nature of the Product:
The magnetic nature depends on the presence of unpaired electrons in the manganese ion.
- In potassium manganate, K\(_2\)MnO\(_4\), the oxidation state of manganese needs to be determined. Let it be x.
2(+1) + x + 4(-2) = 0 \(\implies\) 2 + x - 8 = 0 \(\implies\) x = +6.
- The atomic number of Mn is 25, and its ground-state electron configuration is [Ar] 3d\(^5\) 4s\(^2\).
- For Mn\(^{6+}\), the electron configuration is [Ar] 3d\(^1\).
- Since the Mn\(^{6+}\) ion has one unpaired electron in its d-orbital, the manganate ion (MnO\(_4\)\(^{2-}\)) and thus the compound K\(_2\)MnO\(_4\) is paramagnetic. Step 3: Final Answer:
Combining the properties, the product (potassium manganate) is paramagnetic and green.
