Statement-1: `[Co^(II)(NH)_(3))_(6)]^(2+)` is not readily oxidized to `[Co^(III)(NH_(3))_(6)]^(3+)` when air is bubbled through it.
Statement-2: Crystal field stabilization energy of `Co(+III)` with a `d^(6)` configuration is higher then for Co(+III) with a `d^(7)` arrangement.

Options

A. Statement-1 is true, statement-2 is true,statement-2 is a correct explanation for statement-1
B. Statement-1 is true, statement-2 is true,statement-2 is NOT a correct explanation for statement-1
C. Statement-1 is true, statement-2 is false
D. Statement-1 is false, statement-2 is true.

Explanation

To analyze the given statements, we will evaluate each one systematically.

### Step 1: Evaluate Statement-1
**Statement-1**: `[Co^(II)(NH_3)_6]^{2+}` is not readily oxidized to `[Co^(III)(NH_3)_6]^{3+}` when air is bubbled through it.

1. **Understanding the Oxidation**: Cobalt in the +2 oxidation state (Co(II)) can be oxidized to cobalt in the +3 oxidation state (Co(III)).
2. **Behavior of Ammonia**: Ammonia (NH₃) is a weak field ligand. However, when Co(II) is oxidized to Co(III), it forms a stronger field ligand complex.
3. **Crystal Field Stabilization Energy (CFSE)**: The CFSE increases when Co(II) is oxidized to Co(III) because Co(III) has a higher CFSE due to the presence of more paired electrons in the lower energy T₂g orbitals.
4. **Conclusion for Statement-1**: Therefore, the statement is **false** because `[Co^(II)(NH_3)_6]^{2+}` is readily oxidized to `[Co^(III)(NH_3)_6]^{3+}`.

### Step 2: Evaluate Statement-2
**Statement-2**: The crystal field stabilization energy of `Co(+III)` with a `d^(6)` configuration is higher than for Co(+III) with a `d^(7)` arrangement.

1. **Understanding d-Configurations**:
- For Co(III) with a d⁶ configuration, all six electrons will occupy the lower energy T₂g orbitals, leading to maximum pairing and a higher CFSE.
- For Co(III) with a d⁷ configuration, there will be one unpaired electron in the higher energy E₍g₎ orbitals, leading to a lower CFSE.
2. **Calculating CFSE**:
- For d⁶: CFSE = 6 * (-2/5 Δ₀) = -12/5 Δ₀ = -2.4 Δ₀.
- For d⁷: CFSE = 5 * (-2/5 Δ₀) + 2 * (3/5 Δ₀) = -2 Δ₀ + 6/5 Δ₀ = -0.8 Δ₀.
3. **Conclusion for Statement-2**: Since -2.4 Δ₀ (for d⁶) is more negative than -0.8 Δ₀ (for d⁷), the statement is **true**.

### Final Conclusion
- **Statement-1** is **false**.
- **Statement-2** is **true**.

Thus, the correct answer is that Statement-1 is false and Statement-2 is true.