Compounds of Manganese NEET 2026: KMnO₄ Decomposition

Which green paramagnetic compound forms when KMnO₄ is heated?

Option A, K₂MnO₄ (potassium manganate), is correct. In this compounds of manganese NEET 2026 question, the solution runs from thermal decomposition to green product, manganese oxidation state and unpaired-electron count. Checking magnetism alone cannot establish the answer.

When potassium permanganate decomposes on heating at 513 K, which listed compound matches both the green colour and paramagnetic behaviour?

This Chemistry MCQ belongs to d- and f-Block Elements and is rated medium on this question bank’s scale. Its expected solving time is 45 seconds, a question-bank benchmark, not an official examination time limit.

What is the balanced thermal decomposition reaction at 513 K?

Heating potassium permanganate produces potassium manganate, manganese dioxide and oxygen. Potassium manganate, K₂MnO₄, is the green product. Start with this reaction, as the official solution does: the question asks about a product formed under a stated condition, not just any compound with unpaired electrons.

2 K M n O_{4} \overset{Δ, 513 K}{\to} K_{2} M n O_{4} + M n O_{2} + O_{2}

Check every element before using the equation:

Potassium: $2 = 2$
Manganese: $2 = 1 + 1$
Oxygen: $8 = 4 + 2 + 2$

The atom check confirms that the stated reaction is balanced; it does not, by itself, prove which products form. The known decomposition reaction identifies the products, and the green colour picks out K₂MnO₄.

Next, check whether its electronic structure explains the magnetic clue. Starting with magnetism instead can lead you to a compound that this reaction does not produce.

How do you calculate manganese’s oxidation state in K₂MnO₄?

Manganese has oxidation state +6 in potassium manganate, but +7 in potassium permanganate. The extra potassium in the manganate formula changes the charge balance. Calculate the oxidation state for the green product, not for the starting compound.

For neutral K₂MnO₄, assign potassium and oxygen their usual oxidation states:

K : + 1, O : - 2

Let the manganese oxidation state be represented by: $x$

Charge balance then gives:

2 (+ 1) + x + 4 (- 2) = 0

$2 + x - 8 = 0$ $x = + 6$ Cross-check using the manganate ion alone:

x + 4 (- 2) = - 2 \Rightarrow x = + 6

The ion names differ because their charges differ:

\underset{manganate}{\underset{⏟}{M n O_{4}^{2 -}}} \underset{permanganate}{\underset{⏟}{M n O_{4}^{-}}}

For KMnO₄, the calculation instead gives:

(+ 1) + x + 4 (- 2) = 0 \Rightarrow x = + 7

Both ions contain four oxygen atoms, but their overall charges are different. That difference changes manganese’s formal electron count and therefore the magnetic behaviour.

Why is potassium manganate paramagnetic, and why is A correct?

Neutral manganese has the starting configuration shown below. Formally removing six electrons leaves one d electron for Mn(VI), so manganate has one unpaired electron and is paramagnetic.

M n : [A r] 3 d^{5} 4 s^{2}

For oxidation state +6, remove the two 4s electrons first, followed by four 3d electrons. This leaves the formal configuration:

[A r] 3 d^{5} 4 s^{2} \overset{remove two 4s electrons}{\to} [A r] 3 d^{5} \overset{remove four 3d electrons}{\to} [A r] 3 d^{1}

This is formal electron counting for Mn(VI) in manganate. It does not mean K₂MnO₄ contains isolated manganese ions carrying six positive charges.

A single d electron has no partner, so it remains unpaired. The magnetic clue therefore agrees with the reaction identity and green colour.

Final answer: A) K₂MnO₄, potassium manganate. All three checks support this option:

thermal decomposition product \to green manganate \to M n (+ 6) \to d^{1} \to one unpaired electron

How can checking only magnetism lead to option C?

Option C, MnO, illustrates an incomplete method: its manganese electron count can appear to satisfy the unpaired-electron clue, but that does not identify it as the required decomposition product. One matching property cannot replace the reaction evidence.

For manganese monoxide, oxidation-state arithmetic gives:

x + (- 2) = 0 \Rightarrow x = + 2

The corresponding formal electron count is:

M n (I I) : [A r] 3 d^{5}

Stopping here would mistake an unpaired-electron argument for proof of the answer. The official thermal decomposition equation contains K₂MnO₄ and MnO₂ as the manganese products, not MnO.

The mistake is treating one matching property as sufficient when reaction identity, colour and magnetic behaviour must all agree. Reject C on the reaction evidence: its formal electron count cannot make it a product of the stated decomposition.

Can you solve three related manganese-compound questions?

Use charge balance and formal electron counting for the first two questions, and reaction coefficients for the third. These are original related practice questions, not additional verified NEET PYQs.

What are the oxidation state and formal d-electron count of Mn in KMnO₄?

Answer: +7 and zero d electrons, consistent with diamagnetism. One potassium contributes one positive charge, while four oxygens contribute eight negative charges:

(+ 1) + x - 8 = 0 \Rightarrow x = + 7

Formally removing seven electrons from neutral manganese removes both 4s electrons and all five 3d electrons:

[A r] 3 d^{5} 4 s^{2} \overset{formal removal of seven electrons}{\to} [A r] 3 d^{0}

There are no unpaired d electrons. This differs from manganate’s one-electron count.

How many electrons are lost when manganate becomes permanganate?

One electron is lost per manganese centre, because manganese changes from oxidation state +6 to +7. An increase in oxidation state means oxidation:

M n O_{4}^{2 -} \to M n O_{4}^{-} + e^{-}

Check the charge balance: $- 2 = (- 1) + (- 1)$

Manganese and oxygen atoms are already balanced. The electron belongs on the product side because it is lost.

How many moles of K₂MnO₄ and O₂ form from 2 mol of KMnO₄?

One mole of potassium manganate and one mole of oxygen form, assuming complete decomposition according to the stated equation. Read the coefficient ratio in reaction order:

K M n O_{4} : K_{2} M n O_{4} : M n O_{2} : O_{2} = 2 : 1 : 1 : 1

For each requested product:

n (K_{2} M n O_{4}) = 2 m o l \times \frac{1}{2} = 1 m o l

n (O_{2}) = 2 m o l \times \frac{1}{2} = 1 m o l

Use the balanced coefficients for mole ratios, not the subscripts within a compound’s formula.

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Related on NEET JEEnius AI: First Law of Thermodynamics NEET 2026: 600 J Solution.

Frequently asked questions

What happens when KMnO₄ is heated at 513 K?

Potassium permanganate decomposes into potassium manganate, manganese dioxide and oxygen: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂. The green, paramagnetic product is K₂MnO₄.

Why is K₂MnO₄ paramagnetic?

Manganese has oxidation state +6 in K₂MnO₄, giving a formal d¹ electron count. This single d electron is unpaired, so potassium manganate is paramagnetic.

What is the difference between manganate and permanganate?

Manganate is MnO₄²⁻, with manganese in oxidation state +6 and a formal d¹ count. Permanganate is MnO₄⁻, with manganese in oxidation state +7 and a formal d⁰ count. Potassium manganate is paramagnetic, whereas potassium permanganate is diamagnetic.

Why is MnO not the correct answer for the KMnO₄ heating question?

The stated thermal decomposition produces K₂MnO₄ and MnO₂ as the manganese-containing products, not MnO. Although Mn(II) in MnO has a formal d⁵ count, an unpaired-electron argument alone cannot establish the answer. The reaction product, green colour and magnetic behaviour must all agree.