Select the correct statements among the following.

Options

A. Co (III) is stabilised in presence of weak field ligands,
while Co (II) is stabilised in presence of
strong field ligand
B. Four coordinated complexes
of Pd(II) and Pt(II) are paramagnetic and square planar.
C. `[Ni(CN)_(4)]^(4-)` ion and `[Ni(CO)_(4)]` are
tetrahedral and square planar respectively.
D. `Ni^(2+)` ion does not form inner orbital
octahedral complexes in presence of weak ligand fields.

Answer

To solve the question, we need to analyze the statements provided and determine their correctness based on our understanding of coordination chemistry, specifically regarding the behavior of transition metal complexes with different ligands.

### Step-by-Step Solution:

1. **Analyze Statement 1:**
- **Statement:** Copper (Co) is stabilized in the presence of weak field ligands while Co²⁺ is stabilized in the presence of strong field ligands.
- **Analysis:** Cobalt (Co) has an atomic number of 27, and its oxidation states Co²⁺ and Co³⁺ have different electronic configurations. Co²⁺ (3d⁷) does not undergo pairing in the presence of weak field ligands, leading to an outer orbital complex (sp³d²). Co³⁺ (3d⁶) is stabilized by strong field ligands due to pairing, forming an inner orbital complex (d²sp³). Thus, this statement is **incorrect**.

2. **Analyze Statement 2:**
- **Statement:** Four coordinated complexes of Pd²⁺ and Pt²⁺ are paramagnetic and square planar.
- **Analysis:** Palladium (Pd) has an electronic configuration of 4d⁸, and when it loses two electrons to become Pd²⁺, it has 4d⁸. Platinum (Pt) has an electronic configuration of 5d⁸. Both Pd²⁺ and Pt²⁺ in a square planar geometry (dsp² hybridization) will have paired electrons and thus are **diamagnetic**, not paramagnetic. Therefore, this statement is **incorrect**.

3. **Analyze Statement 3:**
- **Statement:** Ni(CN)₄²⁻ is tetrahedral and Ni(CO)₄ is square planar.
- **Analysis:** Ni(CN)₄²⁻ is actually square planar due to the strong field nature of cyanide ligands, which causes pairing of electrons (dsp² hybridization). Ni(CO)₄, on the other hand, is tetrahedral (sp³ hybridization). Thus, this statement is **incorrect**.

4. **Analyze Statement 4:**
- **Statement:** Ni²⁺ does not form inner orbital octahedral complexes in the presence of weak field ligands.
- **Analysis:** Ni²⁺ has an electronic configuration of 3d⁸. In the presence of weak field ligands, it does not undergo pairing and forms outer orbital complexes (sp³d²). Therefore, this statement is **correct**.

### Conclusion:
The correct statements among the options provided are:
- **Option D:** Ni²⁺ does not form inner orbital octahedral complexes in the presence of weak field ligands.