P-Block Elements NEET 2026: Manganese Reduction Explained

What is the correct answer to the two manganese statements in NEET 2026?

Option D is correct: manganese is reduced in both mixtures. This P-Block Elements NEET 2026 question is a medium-difficulty, two-statement single-correct MCQ. The question bank gives an expected solving time of 75 seconds, which is a practice benchmark, not an official exam time limit.

Sodium chloride and sodium iodide are separately heated with concentrated sulphuric acid and manganese dioxide. Judge these claims:

Statement I: Manganese’s oxidation state increases in the sodium chloride mixture.
Statement II: Manganese’s oxidation state decreases in the sodium iodide mixture.

The options map to these truth values:

The task concerns manganese, not merely whether the halide undergoes oxidation. Track the named element before selecting an option.

Why is manganese reduced when sodium chloride forms HCl?

Manganese gains two electrons in the chloride reaction, so Statement I is false. Follow the official solution’s sequence: concentrated sulphuric acid first releases hydrogen chloride from sodium chloride. Manganese dioxide then oxidises chloride to chlorine while its own manganese undergoes reduction.

The initial acid-displacement step is:

N a C l + H_{2} S O_{4} \to N a H S O_{4} + H C l

The subsequent reaction is:

M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}

Calculate manganese’s oxidation number in manganese dioxide. Each oxygen contributes minus two, and the compound is neutral:

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

In manganese chloride, each chloride contributes minus one:

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

The change is:

Δ (oxidation number) = (+ 2) - (+ 4) = - 2

That decrease of two units means a gain of two electrons per manganese atom. The balanced reduction half-reaction shows this explicitly:

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

Chloride supplies those electrons: $2 C l^{-} \to C l_{2} + 2 e^{-}$

C l : - 1 \to 0

Manganese dioxide is the oxidising agent because it causes chloride to lose electrons. The agent itself is reduced, so Statement I is false.

Why does iodide also reduce manganese dioxide?

Iodide acts as a strong reducing agent and reduces manganese dioxide in acidic medium, so Statement II is true. The official method again starts with hydrogen-halide formation. This represents initial hydrogen iodide formation, not isolation of pure hydrogen iodide.

The initial acid-displacement step is:

N a I + H_{2} S O_{4} \to N a H S O_{4} + H I

Concentrated sulphuric acid can also oxidise iodide. That competing chemistry makes this unsuitable as a clean preparation of hydrogen iodide, but it does not reverse the stated manganese reduction.

A balanced representation using the official manganese sulphate product is:

M n O_{2} + 2 H I + H_{2} S O_{4} \to M n S O_{4} + I_{2} + 2 H_{2} O

Calculate manganese’s final oxidation number using the sulphate ion’s overall charge of minus two: $S O_{4}^{2 -}$

x - 2 = 0 \Rightarrow x = + 2

Sulphate is the counter-ion, not evidence that manganese remains in its original oxidation state. Comparing manganese in the starting oxide and the sulphate product gives:

M n : + 4 \to + 2 reduction

The complementary iodine change is: $2 I^{-} \to I_{2} + 2 e^{-}$

I : - 1 \to 0 oxidation

Two iodide ions together lose two electrons, matching the two electrons gained per manganese atom. Manganese’s oxidation state decreases, making Statement II true.

How do the two conclusions give option D?

D: Statement I is incorrect and Statement II is correct. Both mixtures reduce manganese, but Statement I claims oxidation while Statement II claims reduction.

NaCl mixture: Statement I is false.

M n : + 4 \to + 2 reduction

NaI mixture: Statement II is true.

M n : + 4 \to + 2 reduction

Use this check: identify the named element, calculate its initial and final oxidation states, classify the change, then map the truth values to the options. Judge the two statements separately; this is not an assertion–reason causal test.

How does confusing an oxidising agent with oxidation lead to option A?

Option A follows if you wrongly mark Statement I true while correctly marking Statement II true. The error starts with recognising manganese dioxide as an oxidising agent, then assuming that manganese must undergo oxidation. That confuses what an agent does to another species with what happens to the agent itself.

An oxidising agent causes another species to lose electrons. It gains those electrons and is reduced.

The worked numbers settle the issue in both mixtures:

M n : + 4 \to + 2

The halide being oxidised does not change that manganese result. Use oxidation-number arithmetic, not the word “oxidising”, to judge Statement I.

How can you test the same method on three related halogen-redox questions?

Identify the species that gains electrons, the species that loses them, and the electron count in each reaction below. These are original practice questions, not additional verified NEET PYQs. Each answer follows its question immediately.

Which species are the oxidising and reducing agents in the chloride reaction?

Answer: Manganese dioxide is the oxidising agent; chloride supplied by hydrogen chloride is the reducing agent. In the reaction below, manganese gains electrons while chloride loses them:

M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}

M n : + 4 \to + 2 gains electrons

C l : - 1 \to 0 loses electrons

How do you balance the HBr reaction and classify manganese’s change?

Answer: Manganese is reduced, and balancing requires four hydrogen bromide units and two water molecules. The supplied reaction is:

M n O_{2} + H B r \to M n B r_{2} + B r_{2} + H_{2} O

Its balanced form is:

M n O_{2} + 4 H B r \to M n B r_{2} + B r_{2} + 2 H_{2} O

M n : + 4 \to + 2 reduction

B r : - 1 \to 0 oxidation

Two bromide ions form bromine; the other two remain as counter-ions in manganese bromide. The two oxidised bromide ions supply the two electrons gained by manganese.

Which ion reduces chlorine, and how many electrons are transferred?

Answer: Iodide is the reducing agent, and two electrons are transferred in the equation as written. Chlorine gains the electrons that iodide loses:

C l_{2} + 2 I^{-} \to 2 C l^{-} + I_{2}

2 I^{-} \to I_{2} + 2 e^{-}

I : - 1 \to 0 C l : 0 \to - 1

Each iodide loses one electron. Multiply by its coefficient before reporting the total electron transfer.

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Frequently asked questions

What is the correct option for the NEET 2026 manganese statements question?

Option D is correct: Statement I is false and Statement II is true. Manganese changes from +4 to +2 in both mixtures, so it is reduced in both. Statement I incorrectly claims an increase in oxidation state, while Statement II correctly claims a decrease.

Why is MnO2 reduced if it is an oxidising agent?

An oxidising agent causes another species to lose electrons and gains those electrons itself. In the chloride reaction, MnO2 oxidises chloride to chlorine while manganese changes from +4 to +2. This two-electron gain is reduction.

What is the oxidation state of manganese in MnSO4?

Manganese has an oxidation state of +2 in MnSO4 because the sulphate ion has an overall charge of -2 and the compound is neutral. In the iodide mixture, manganese therefore changes from +4 in MnO2 to +2 in MnSO4. Sulphate is the counter-ion; its presence does not mean manganese avoids reduction.

Why can't concentrated sulphuric acid be used to prepare pure HI from NaI?

Although the initial acid-displacement step forms HI, concentrated sulphuric acid can also oxidise iodide. This competing reaction makes the method unsuitable for clean preparation of HI. In the stated mixture containing MnO2, manganese is still reduced from +4 to +2.