The d- and f-Block Elements JEE 2019: Manganese Oxoanions

What is the answer to the JEE Advanced 2019 manganese oxoanion question?

A, B and D are correct in the d- and f-block elements JEE 2019 manganese oxoanion question. This is a JEE Advanced 2019, Paper 1, Chemistry, multiple-correct MCQ, not a JEE Main question.

Manganese dioxide is fused with potassium hydroxide under oxygen to produce salt W. Electrolytic oxidation of an alkaline solution of W produces salt X. Their manganese-containing anions are Y and Z, respectively.

Select every correct statement:

How do you identify W, X, Y and Z and count their d electrons?

W is potassium manganate and X is potassium permanganate. Y is the manganate ion, with one formal d electron; Z is the permanganate ion, with none. Keep the salt formulas distinct from the anion formulas throughout the calculation.

Oxygen oxidises manganese dioxide during fusion in alkaline medium:

The salt and its manganese-containing anion are:

Taking each oxygen’s oxidation state as minus two gives:

Electrolytic oxidation removes one electron from manganate:

Electron loss raises manganese’s oxidation state from plus six to plus seven. The resulting salt and anion are:

Neutral manganese has the configuration:

For formal electron counting, remove electrons in this order: $4s\text{before}3d$

Removing the two outer-shell electrons and then four d electrons leaves one d electron. Removing seven electrons leaves none:

These are formal metal d counts within the oxoanions. They do not imply that isolated manganese cations are present.

Why does manganate disproportionate in acid, making A correct?

A is correct because manganese in manganate is both oxidised and reduced in acidic solution. Some manganese centres form permanganate, while others form manganese dioxide.

The oxidation-state changes are:

Two manganese centres each lose one electron, supplying the two electrons accepted by one centre. This fixes the permanganate-to-manganese-dioxide product ratio at two to one:

Check the atoms:

• Manganese: three atoms on each side.
• Oxygen: twelve atoms on each side.
• Hydrogen: four atoms on each side.

The total charge is also balanced:

This is acidic disproportionation, not the earlier alkaline electrolytic oxidation. In the electrolytic step, manganate loses electrons to form permanganate without also forming manganese dioxide.

Why are B and D correct but C incorrect?

B correctly describes pi bonding, C reverses the magnetic properties, and D correctly assigns colour and tetrahedral geometry. Check these properties independently: visible colour does not prove that an ion has unpaired electrons.

For B, the official solution uses: $p\pi \text{–}d\pi$

Oxygen p orbitals and manganese d orbitals participate in pi bonding in both oxoanions. A formal count of zero metal d electrons does not prevent empty metal orbitals from participating in this bonding description. B is correct.

For C, manganate’s one d electron is unpaired, making it paramagnetic. Permanganate has no unpaired d electrons and is diamagnetic. C is incorrect because it reverses these assignments.

For D, manganate is green and permanganate is purple. Both ions are tetrahedral, so D is correct.

Compare the ions:

• Y, manganate: paramagnetic, green and tetrahedral.

• Z, permanganate: diamagnetic, purple and tetrahedral.

Permanganate’s colour arises from charge-transfer absorption, involving electron transfer from mainly oxygen-based to manganese-based orbitals. It does not require a transition between occupied and empty metal d orbitals, so zero formal d electrons does not mean colourless.

The independent checks give correct options A, B and D. Colour supports D, not the magnetic assignments in C.

What electron-counting mistake produces option C?

Assigning zero d electrons to manganese(VI) predicts diamagnetism; assigning one to manganese(VII) predicts paramagnetism. Swapping the formal d counts therefore produces exactly the reversed magnetic properties in C.

That reasoning fails because increasing the positive oxidation state removes another electron. It cannot increase the d-electron count.

For these manganese ions, subtract the oxidation state from group number seven: $\text{Formal d count}=7-\text{oxidation state}$

Calculate oxidation state before assigning magnetism. Overall ion charge, colour and the similar names “manganate” and “permanganate” are not substitutes for that calculation.

How can you apply the method to two related oxoanion questions?

Use oxidation-state changes to count transferred electrons, then use group number to obtain the formal d count. These are original related practice questions from The d- and f-Block Elements, not additional verified PYQs.

Question 1: Write the balanced half-reaction for reduction of permanganate to manganese(II) in acidic solution. How many electrons are accepted per manganese atom?

Manganese changes from plus seven to plus two: $+7-(+2)=5$

Each manganese atom accepts five electrons. Balance oxygen with four water molecules, then hydrogen with eight hydrogen ions:

Verify the charge on the left: $-1+8-5=+2$

Both sides have total charge plus two. For more practice with electrode reactions, use How to Study Electrochemistry JEE: A Step-by-Step Plan.

Question 2: Determine chromium’s oxidation state, formal d count and magnetic behaviour in chromate: $\mathrm{C}\mathrm{r}{\mathrm{O}}_4^{2-}$

Calculate the oxidation state:

Chromium belongs to group six:

Chromate is therefore diamagnetic. Cover both answers and reproduce the electron balance for Question 1 and the oxidation-state calculation for Question 2 before checking your work.

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