Amongst the following ions, which is considered as most stable in `M^(2+)` state ?

Options

A. `Ti^(2+)//Ti(-1.63V)`
B. `V^(2+)//V(-1.11V)`
C. `Cr^(2+)//Cr(-0.90V) `
D. `Mn^(2+)//Mn(1.18V)`

Answer

To determine which ion is the most stable in the \( M^{2+} \) state among the given transition metal ions, we can follow these steps:

### Step 1: Understand the concept of oxidation states and stability
- Transition metals can exhibit multiple oxidation states, and the stability of these states can vary. The stability of the \( M^{2+} \) state in transition metals is influenced by their standard electrode potentials (\( E^\circ \)).

### Step 2: Review the standard electrode potentials
- The standard electrode potential (\( E^\circ \)) indicates the tendency of a species to be reduced. A lower (more negative) \( E^\circ \) value suggests that the ion is more stable in its oxidized state because it has a lower tendency to be reduced.

### Step 3: Analyze the given ions
- Identify the \( E^\circ \) values for the ions in question. The ion with the lowest \( E^\circ \) value will be the most stable in the \( M^{2+} \) state.

### Step 4: Compare the \( E^\circ \) values
- For example, if we have ions such as \( Ti^{2+} \), \( Fe^{2+} \), \( Cu^{2+} \), and \( Mn^{2+} \):
- Let’s say their \( E^\circ \) values are as follows:
- \( Ti^{2+} \): -1.63 V
- \( Fe^{2+} \): -0.44 V
- \( Cu^{2+} \): +0.34 V
- \( Mn^{2+} \): +1.51 V
- The ion with the lowest (most negative) \( E^\circ \) value is \( Ti^{2+} \).

### Step 5: Conclusion
- Based on the analysis, the ion that is most stable in the \( M^{2+} \) state is \( Ti^{2+} \) because it has the lowest standard electrode potential.

### Final Answer
- The most stable ion in the \( M^{2+} \) state is **Titanium (\( Ti^{2+} \))**.

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