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 (1.18 V)`

Answer

To determine which ion is the most stable in the \( M^{2+} \) state among titanium, vanadium, chromium, and manganese, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Elements and Their Positions:**
- The elements in question are titanium (Ti), vanadium (V), chromium (Cr), and manganese (Mn). These elements are located in the same period of the periodic table.

2. **Understand the Trend Across the Period:**
- As we move across a period in the periodic table, the electronegativity increases, and the non-metallic character also increases. This means that the tendency to lose electrons (oxidation) decreases.

3. **Analyze the Stability of \( M^{2+} \) State:**
- The stability of a metal in a particular oxidation state is influenced by its ability to lose electrons. As the non-metallic character increases across the period, the stability of the \( +2 \) oxidation state decreases.

4. **Consider the Oxidation Potential:**
- The oxidation potential refers to the tendency of an element to lose electrons. Elements with a higher negative oxidation potential are more stable in their positive oxidation states.

5. **Evaluate Each Element:**
- **Titanium (Ti):** Has a high negative oxidation potential, making it relatively stable in the \( +2 \) state.
- **Vanadium (V):** Less stable than titanium in the \( +2 \) state.
- **Chromium (Cr):** Has a more complex electron configuration and is less stable in the \( +2 \) state compared to titanium.
- **Manganese (Mn):** Generally less stable in the \( +2 \) state due to its higher non-metallic character.

6. **Conclusion:**
- Based on the analysis, titanium \( (Ti^{2+}) \) is considered the most stable in the \( M^{2+} \) state among the given options.

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

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