D- and f-Block Elements JEE 2024: Zinc Ion Count in Q

How many zinc ions are in precipitate Q in the JEE Advanced 2024 question?

The supplied official solution gives 2 or 3 zinc ions for the JEE Advanced 2024 zinc ferrocyanide question, depending on the precipitate’s composition. The source is JEE Advanced 2024, Paper 2, Chemistry, under d- and f-Block Elements, not JEE Main.

Zinc chloride reacts with potassium ferrocyanide to give a white precipitate called Q. The question asks: “The number of zinc ions present in the molecular formula of Q is ____.”

This is a numerical-answer question; no multiple-choice options are supplied. The question bank classifies it as medium and gives an expected solving time of 60 seconds, not an official exam limit.

Which ions should you identify before writing the precipitate formula?

Identify zinc, potassium and ferrocyanide ions before balancing the formula. Keep ferrocyanide together as one complex ion, and require the total positive and negative charges in the precipitate formula to sum to zero.

Potassium ferrocyanide contains four potassium ions for each ferrocyanide ion. Their charges balance: ${\mathrm{K}}_4[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N}{)}_6]$ $4(+1)+(-4)=0$

The six cyanide ligands belong inside the complex. They are not six separate counterions when you balance the salt formula.

Distinguish charge per zinc ion from number of zinc ions. Each zinc ion carries two positive charge units; the question asks how many such ions occur in the formula.

Why does normal zinc ferrocyanide contain two zinc ions?

One ferrocyanide ion needs four positive charge units, and each zinc ion supplies two. In this branch of the supplied solution, potassium does not remain in the precipitate.

Let the unknown zinc-ion count for one ferrocyanide ion be: $n=\text{number of zinc ions}$

Charge balance gives: $2n-4=0$ $2n=4$ $n=2$

The ionic equation is:

Including potassium and chloride gives the balanced reaction:

Check conservation before accepting the formula:

• Zinc: two atoms on each side.
• Ferrocyanide: one complete unit on each side.
• Potassium: four atoms on each side.
• Chlorine: four atoms on each side.

The zinc subscript in the precipitate formula is two. This composition contains 2 zinc ions per formula unit, not four.

How can the potassium-containing precipitate have three zinc ions?

In the presence of excess potassium ferrocyanide, the supplied solution says a double-salt-type precipitate can also form. This composition retains potassium in the precipitate, so zinc no longer supplies all the positive charge.

The supplied composition is:

Its two ferrocyanide ions have a total negative charge of: $2(-4)=-8$

Two potassium ions supply part of the positive charge: $2(+1)=+2$

The positive charge still required from zinc is: $8-2=6$

Using the unknown zinc-ion count, charge balance becomes:

$2n=6$ $n=3$ The balanced reaction is:

There are three zinc atoms and two ferrocyanide units on each side. Potassium and chlorine also balance:

Charge balance verifies each supplied composition; it does not independently establish which experimental product forms. A neutral formula alone is not evidence that a particular precipitate forms.

Compare the zinc counts in the two supplied compositions:

The supplied official answer is therefore 2 or 3. Counting only the normal salt misses the potassium-containing composition.

Why is four zinc ions the wrong charge-balance result?

Four is an illustrative wrong numerical answer, not an option from the paper. It results from seeing four negative charge units on ferrocyanide and assigning four zinc ions, without accounting for the two positive charge units carried by each zinc ion.

Test that proposed normal-salt formula: $\mathrm{Z}{\mathrm{n}}_4[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N}{)}_6]$ $4(+2)+(-4)=+4$

The result is not zero, so this cannot be the neutral formula in the normal-salt branch. Divide the required positive charge by the charge per zinc ion instead:

For the double salt, subtract potassium’s contribution before dividing. Only the remaining positive charge must come from zinc:

Which two related questions test the same foundations?

Use the questions below to check oxidation-state accounting and zinc’s electronic configuration. Both are original related practice, not additional verified JEE PYQs, with worked answers included.

What is the oxidation state of iron in ferrocyanide and ferricyanide?

Iron is Fe(II) in ferrocyanide and Fe(III) in ferricyanide. Assign each cyanide ligand a charge of minus one and make the total equal the complex-ion charge.

The two ions are:

Each cyanide ligand is: $\mathrm{C}{\mathrm{N}}^{-}$

For ferrocyanide:

For ferricyanide:

Why is zinc a d-block element but not a transition element?

Zinc belongs to the d-block, but neither its atom nor its usual zinc ion has an incomplete d subshell. Under the standard definition, a transition element must have an incomplete d subshell in its atom or a resulting cation.

The configurations are:

Zinc loses the two outer s electrons when forming its usual ion; the d subshell remains full. Write both configurations before classifying zinc, rather than treating “d-block” and “transition element” as interchangeable terms.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

For a worked example of the same idea, see Units and Measurements JEE 2022: Magnetic Field Solution.