The P-Block Elements JEE 2024: PCl5 Fluorination

What is the answer to the phosphorus halide question in JEE Advanced 2024?

Option B is correct for the p-block elements JEE 2024 fluorination question: both products retain the tetrachlorophosphonium cation. The anions differ, one contains chlorine and fluorine, while the other contains only fluorine.

The question appeared in JEE Advanced 2024, Paper 2, Chemistry, as a single-correct MCQ. It is rated medium on the question bank’s scale, not on an official JEE difficulty scale.

The prompt asks you to identify the pair of ionic products obtained when phosphorus pentachloride undergoes fluorination in a polar organic solvent. The supplied choices are:

• A)

• B)

• C)

• D)

Why does the polar organic solvent matter?

In the stated polar organic solvent, phosphorus pentachloride forms ionic species rather than existing only as neutral covalent molecules. The solvent phrase tells you which form of the products to consider. It is chemical information, not incidental wording.

Follow the supplied official solution by looking for cation–anion products, rather than starting with neutral phosphorus pentafluoride. Options C and D list neutral molecules, so they do not represent the ionic products identified by that solution.

This is a reaction-specific statement. It does not mean every covalent compound becomes ionic in every polar solvent. Here, the stated behaviour of phosphorus pentachloride directs the first elimination: retain A and B for closer examination.

Which cation remains unchanged, and what are the two anions?

The reaction-specific fact to recall is that the cation remains tetrachlorophosphonium. The official solution retains this ion in both products: $[{\mathrm{P}\mathrm{C}\mathrm{l}}_4{]}^{+}$

The first anion contains both chlorine and fluorine. Pairing it with the retained cation gives the mixed-halide salt:

The supplied solution describes this salt as a colourless crystal. That description is a secondary identification fact, not the basis for choosing the answer.

The second anion is fully fluorinated. Pairing it with the same cation gives:

This product is described as a white crystal. “Fully fluorinated” refers to the second anion, not to replacement of every chlorine atom in the entire salt.

These ion identities are recalled chemical facts from the supplied solution. Charge arithmetic can check them, but cannot establish which products form.

How do I check the formulas in option B?

Both salts are electrically neutral, and phosphorus has oxidation state five in each constituent ion. These checks verify that the identified formulas are consistent. They do not replace the reaction-specific fact about which cation remains unchanged.

For each salt, the charges add to zero:

Assign each halogen an oxidation state of minus one. Let the unknown represent the oxidation state of phosphorus:

The two verified ion pairs match option B exactly. Option A also contains charge-balanced ion pairs, so charge balance alone cannot distinguish the options. Electrical neutrality and oxidation-state checks establish consistency; they do not predict which products form under the stated conditions.

Why is option A wrong even though its charges balance?

Option A fails because its first salt uses a fluorinated cation instead of the chlorine-containing cation retained in the official solution. The incorrect assumption is that fluorination must replace chlorine in the cation as well as in the anion.

That assumption introduces: $[{\mathrm{P}\mathrm{F}}_4{]}^{+}$

Pairing it with the fully fluorinated anion produces the first salt in A:

But the supplied solution retains: $[{\mathrm{P}\mathrm{C}\mathrm{l}}_4{]}^{+}$

A and B share the other product:

Recognising that shared salt cannot settle the question. Both members of a proposed pair must match the reaction. One correct salt does not rescue a pair whose other salt changes the cation without justification.

Can you apply the method to three related phosphorus-halide questions?

Use formula arithmetic to check charge, but use the stated chemistry to identify products. These are original practice questions, not additional verified JEE PYQs. They test charge calculation, the difference between neutral and ionic formulas, and whether a complete product pair matches the reaction.

1. With phosphorus in oxidation state five, what charge does the following species have? ${\mathrm{P}\mathrm{C}\mathrm{l}}_4{F}_2$

Calculate the total charge:

The ion is therefore:

Four chlorine atoms and two fluorine atoms contribute six negative units. Phosphorus contributes five positive units, leaving one negative unit overall.

2. Why are these formulas not interchangeable when phosphorus has oxidation state five in both?

Their net charges differ:

The first has five fluorine atoms and is neutral. The second has six fluorine atoms and a negative charge, so it requires a counterion in a neutral salt.

3. Is recognising the first salt enough to accept this proposed product pair for the reaction?

No. The second salt uses a cation not retained in the supplied solution. Before selecting an option, check the cation and anion in each listed salt, not just the product you recognise.

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