Which of the following ions is the most stable in aqueous solution ?
`(At . No . Ti = 22, V = 23, Cr = 24, Mn = 25)`

Options

A. `Ti^(3+)`
B. `Mn^(3+)`
C. `Cr^(3+)`
D. `V^(3+)`

Explanation

To determine which of the given ions (Ti³⁺, V³⁺, Cr³⁺, Mn³⁺) is the most stable in aqueous solution, we can follow these steps:

### Step 1: Write the electronic configurations of the ions
We first need to write the electronic configurations for each of the ions after losing three electrons.

- **Ti³⁺ (Atomic number 22)**:
- Neutral Ti: [Ar] 3d² 4s²
- Ti³⁺: [Ar] 3d¹ (losing 2 electrons from 4s and 1 from 3d)

- **V³⁺ (Atomic number 23)**:
- Neutral V: [Ar] 3d³ 4s²
- V³⁺: [Ar] 3d² (losing 2 electrons from 4s and 1 from 3d)

- **Cr³⁺ (Atomic number 24)**:
- Neutral Cr: [Ar] 3d⁵ 4s¹
- Cr³⁺: [Ar] 3d³ (losing 1 electron from 4s and 2 from 3d)

- **Mn³⁺ (Atomic number 25)**:
- Neutral Mn: [Ar] 3d⁵ 4s²
- Mn³⁺: [Ar] 3d⁴ (losing 2 electrons from 4s and 1 from 3d)

### Step 2: Analyze the stability based on electron configuration
Now, we analyze the stability of these ions based on their electronic configurations. The stability of transition metal ions in aqueous solution is often related to the electron configuration of their d-orbitals.

- **Ti³⁺**: [Ar] 3d¹
- **V³⁺**: [Ar] 3d²
- **Cr³⁺**: [Ar] 3d³
- **Mn³⁺**: [Ar] 3d⁴

### Step 3: Determine the most stable configuration
The most stable configurations in transition metals are those that are either fully filled or half-filled.

- **Cr³⁺ (3d³)** has a half-filled d-subshell, which provides extra stability due to exchange energy and symmetry.
- **Mn³⁺ (3d⁴)** is not half-filled but is close to it, while Ti³⁺ and V³⁺ have fewer electrons in the d-orbitals.

### Conclusion
Among the ions listed, **Cr³⁺** is the most stable in aqueous solution due to its half-filled d-orbital configuration.

### Answer:
**Cr³⁺ is the most stable ion in aqueous solution.**

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