Problem 1 - A Complete Resource Book in Chemistry for JEE Main

Question

Among which of the following pairs of ions, the lower oxidation state in aqueous solution is more stable than the other? (a) \(\mathrm{Ti}^{+}, \mathrm{Ti}^{3+}\) (b) \(\mathrm{V}^{2+}, \mathrm{VO}^{2+}\) (c) \(\mathrm{Cr}^{2+}, \mathrm{Cr}^{3+}\) (d) \(\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\)

Step-by-Step Solution

Answer

The lower oxidation state is more stable for the pair \(V^{2+}, VO^{2+}\) (option b).

Step 1: Understanding Oxidation States

Oxidation states indicate the degree of oxidation (loss of electrons) of an ion. A lower oxidation state means the ion has more electrons compared to a higher oxidation state.

Step 2: Stability in Aqueous Solutions

In general, ions in lower oxidation states are more stable in the presence of reducing agents, while higher oxidation states are stabilized by oxidizing environments.

Step 3: Comparison of Each Pair

Examine each pair of ions to determine which one has a more stable lower oxidation state. This involves considering the electron configuration and the common oxidation states based on the chemistry of each element: - **(a) Ti^{+} vs. Ti^{3+}:** Ti typically shows stable states of Ti^{3+} and Ti^{4+} rather than Ti^{+}. - **(b) V^{2+} vs. VO^{2+}:** V^{2+} is a common, stable oxidation state for vanadium compared to VO^{2+}. - **(c) Cr^{2+} vs. Cr^{3+}:** Cr^{3+} is more stable than Cr^{2+}. Cr^{3+} is commonly seen in aqueous solution. - **(d) Cu^{+} vs. Cu^{2+}:** Cu^{2+} is more stable than Cu^{+} in aqueous solutions.

Step 4: Concluding Observation

From the analysis of all pairs, we observe that in option (b), V^{2+} is more stable in aqueous solution than VO^{2+}. The stabilization is due to the relative ease with which V^{2+} can form complexes compared to VO^{2+}.

Key Concepts

Aqueous Solution Stability

When discussing aqueous solution stability, we're referring to how well ions remain in solution without participating in additional chemical reactions. Stability is influenced by several factors, including the ion's oxidation state and the nature of its environment.
In an aqueous solution, ions with lower oxidation states often find stability in the presence of reducing agents, which help maintain their electron-rich status. By contrast, higher oxidation states require an oxidizing environment to remain stable due to their electron-deficient nature.
For instance, in the exercise, we see that V\(^{2+}\) is more stable than VO\(^{2+}\) in aqueous solution. This is because V\(^{2+}\) can readily form complexes, contributing to its stability. In summary, the stability of ions in aqueous solutions hinges upon the matching of their oxidation states with the environmental conditions they are in, as well as their ability to participate in complex formation.

Transition Metal Ions

Transition metal ions are fascinating due to their variable oxidation states and complex-forming abilities. These metals, found in the 'd-block' of the periodic table, possess d orbitals that enable them to lose or gain electrons with relative ease.
Consider vanadium (V) featured in the exercise. It can exist as V\(^{2+}\), V\(^{3+}\), and other states. Transition metals like V have multiple accessible oxidation states because of their electron configurations, where d orbitals can be involved in bonding.
This versatility allows transition metal ions to bind with various ligands forming complexes that contribute to their stability in different oxidation states. Transition metals exhibit unique characteristics such as colored solutions and catalytic properties due to these varying oxidation states.

Oxidation State Comparison

Oxidation state comparison involves examining possible electron configurations of ions to determine stability and behavior in solutions. The oxidation state, essentially the charge on an ion that results from the loss or gain of electrons, influences properties such as color, magnetic behavior, and reactivity.
For example, the exercise illustrates that comparing the oxidation states of ion pairs helps us understand which states are more stable. Cu\(^{2+}\) is more stable than Cu\(^{+}\) in aqueous solutions due to its favorable ionic interactions and ability to form stable complexes easier than Cu\(^{+}\).
Such comparisons are crucial in predicting the chemical behavior of ions, especially when they interact with other substances. By understanding oxidation states, chemists can manipulate conditions for desired reactions, optimize yields, or isolate specific elements from mixtures. The thoughtfully balanced nature of oxidation states thus plays a pivotal role in chemistry, helping predict stability and guiding reactions.