Consider the reaction $Cu^{2+} + X^- → Cu_2X_2 + X_2$.
X² will be predominantly?

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

To determine which halogen ($X_2$) will predominantly form in the reaction $Cu^{2+} + X^- → Cu_2X_2 + X_2$, we need to consider the redox potential of the halogens and their relative ability to oxidize copper ions.

The order of oxidizing power among the halogens is F₂ > Cl₂ > Br₂ > I₂. Thus, iodine has the least oxidizing power among common halogens like Cl₂, Br₂, and I₂. This means iodine would be more easily reduced and tend to form in the reaction over time when compared to other halogens.

According to the reaction, copper (II) ions are reduced, and halide ions are oxidized. Iodine is less reactive compared to chlorine and bromine, thus making it the more stable form in such reactions, meaning it will easily form from iodide ions ($I^-$) being oxidized.

This explain why I₂ will predominantly form in the given chemical reaction.

Correct Answer: I₂

Answer 2

To determine which halogen ($X_2$) will predominantly form in the reaction $Cu^{2+} + X^- → Cu_2X_2 + X_2$, we need to consider the redox potential of the halogens and their relative ability to oxidize copper ions.

The order of oxidizing power among the halogens is F₂ > Cl₂ > Br₂ > I₂. Thus, iodine has the least oxidizing power among common halogens like Cl₂, Br₂, and I₂. This means iodine would be more easily reduced and tend to form in the reaction over time when compared to other halogens.

According to the reaction, copper (II) ions are reduced, and halide ions are oxidized. Iodine is less reactive compared to chlorine and bromine, thus making it the more stable form in such reactions, meaning it will easily form from iodide ions ($I^-$) being oxidized.

This explain why I₂ will predominantly form in the given chemical reaction.

Correct Answer: I₂

Consider the reaction \(Cu^{2+} + X^- → Cu_2X_2 + X_2\).
X² will be predominantly?

The Correct Option is C

Solution and Explanation

Top Questions on d -and f -Block Elements

Questions Asked in JEE Main exam

Consider the reaction \(Cu^{2+} + X^- → Cu_2X_2 + X_2\).X2 will be predominantly?

The Correct Option is C

Solution and Explanation

Top Questions on d -and f -Block Elements

Questions Asked in JEE Main exam

Consider the reaction \(Cu^{2+} + X^- → Cu_2X_2 + X_2\).
X² will be predominantly?