Statement-1: `[Co(NO_(2))_(6)]^(4-)" and "[Co(NO_(2))_(6)]^(3-)` both complex involve `d^(2)sp^(3)` hybridization but former is paramagnetic and later one diamagnetic.
Statement-2: In `[Co(NO_(2))_(6)]^(4-)` one of the 3d electrons jumps to 4d orbital making the two d-orbitals empty for `d^(2)sp^(3)` hybridization. `NO_(2)` is a stronger ligand which favours the pairing of electrons and therefore, `d^(2)sp^(3)` hybridization.

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.

Answer

To analyze the statements given in the question, we will break down the information regarding the complexes `[Co(NO₂)₆]⁴⁻` and `[Co(NO₂)₆]³⁻`, focusing on their electronic configurations, hybridization, and magnetic properties.

### Step 1: Determine the oxidation states of cobalt in both complexes.
- For `[Co(NO₂)₆]⁴⁻`, the overall charge is -4. Each `NO₂` ligand has a charge of -1, giving a total of -6 from the ligands. Let the oxidation state of cobalt be \( x \).
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
- For `[Co(NO₂)₆]³⁻`, the overall charge is -3. Using the same reasoning:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

### Step 2: Write the electronic configurations of cobalt in both oxidation states.
- Cobalt in the +2 oxidation state (3d⁷):
- Electronic configuration: \( [Ar] 3d^7 4s^2 \)
- Cobalt in the +3 oxidation state (3d⁶):
- Electronic configuration: \( [Ar] 3d^6 4s^2 \)

### Step 3: Analyze the hybridization of both complexes.
- **For `[Co(NO₂)₆]⁴⁻` (Co²⁺)**:
- Cobalt has 7 electrons in the 3d subshell. Since `NO₂` is a strong field ligand, it causes pairing of electrons. One of the 3d electrons can jump to the 4s orbital, allowing for the formation of `d²sp³` hybridization.

- **For `[Co(NO₂)₆]³⁻` (Co³⁺)**:
- Cobalt has 6 electrons in the 3d subshell. The strong field ligand `NO₂` will also cause pairing of electrons here, resulting in `d²sp³` hybridization as well.

### Step 4: Determine the magnetic properties of both complexes.
- **For `[Co(NO₂)₆]⁴⁻`**:
- After the electron jump, there will be one unpaired electron remaining in the 3d orbitals, making it paramagnetic.

- **For `[Co(NO₂)₆]³⁻`**:
- All electrons are paired in the 3d orbitals, making it diamagnetic.

### Conclusion:
- **Statement 1**: Both complexes involve `d²sp³` hybridization; however, `[Co(NO₂)₆]⁴⁻` is paramagnetic due to one unpaired electron, while `[Co(NO₂)₆]³⁻` is diamagnetic due to all paired electrons. This statement is true.

- **Statement 2**: The explanation provided is correct. The strong field ligand `NO₂` causes pairing and allows for `d²sp³` hybridization in both cases. The electron jump in `[Co(NO₂)₆]⁴⁻` is necessary for this hybridization. This statement is also true.

### Final Answer:
Both statements are true.