JEE Chemistry 2025 Question with Solution

Given: The cations are $Cu^{2+}, Fe^{3+}, Ba^{2+}, Ca^{2+}, NH_4^+, Mg^{2+}, Zn^{2+}$ and the reagent used is $K_4[\mathrm{Fe}(\mathrm{CN})_6]$.

Find: The number of cations that give a characteristic precipitate in the identification test.

Cations that give characteristic precipitates with $K_4[\mathrm{Fe}(\mathrm{CN})_6]$ are:

• $Cu^{2+}$: forms $Cu_2[\mathrm{Fe}(\mathrm{CN})_6]$
• $Fe^{3+}$: forms $Fe_4[\mathrm{Fe}(\mathrm{CN})_6]_3$
• $Zn^{2+}$: forms $Zn_2[\mathrm{Fe}(\mathrm{CN})_6]$

The following do not give characteristic precipitates in this test:

• $Ba^{2+}$
• $Ca^{2+}$
• $NH_4^+$
• $Mg^{2+}$

Therefore, the total number of cations giving characteristic precipitate is $3$.

Given: Test reagent is $K_4[\mathrm{Fe}(\mathrm{CN})_6]$.

Find: How many among the listed cations respond with a characteristic precipitate.

Check each cation one by one:

1. $Cu^{2+}$ gives a characteristic precipitate.
2. $Fe^{3+}$ gives a characteristic precipitate.
3. $Ba^{2+}$ does not give a characteristic precipitate.
4. $Ca^{2+}$ does not give a characteristic precipitate.
5. $NH_4^+$ does not give a characteristic precipitate.
6. $Mg^{2+}$ does not give a characteristic precipitate.
7. $Zn^{2+}$ gives a characteristic precipitate.

So the reacting cations are $Cu^{2+}, Fe^{3+}, Zn^{2+}$.

Hence, the required number is $3$.