Select correct statement(s) regarding given complexes:

Options

A. `[Fe(CO)_(5)]`, the orbitals used for hybridisation in Fe atom are s, `p_(x),p_(y),p_(z),d_(z^(2))` and it is high spin complex
B. `[Pt(NH_(3))_(2)Cl_(2)]`, the orbtials used for hybridization on Pt atom are s,`p_(x),p_(y),d_(x^(2)-y^(2)) ` and it is low spin complex
C. `[Cr(H_(2)O)_(6)]^(3+)`, the orbitals used for hybridization in Cr atom are s,`p_(x),p_(y),p_(z),p_(x^(2)),d_(x^(2)-y^(2))` and it is high spin complex
D. `Ni(CO)_(4)`, the orbitals used of rhybridization in Ni atom are s,`p_(x),p_(y),p_(z)`, and it is low spi complex

Answer

To solve the question regarding the correct statements for the given complexes, we will analyze each complex step by step, considering the oxidation state of the metal, the nature of the ligands, and the hybridization involved.

### Step 1: Analyze the first complex (FeCO5)

1. **Identify the oxidation state of Iron (Fe)**:
- CO is a neutral ligand, so the oxidation state of Fe in FeCO5 is 0.

2. **Determine the electronic configuration of Fe**:
- The electronic configuration of Fe (atomic number 26) is [Ar] 3d6 4s2. Since Fe is in the 0 oxidation state, it retains all its electrons.

3. **Hybridization**:
- CO is a strong field ligand, which causes pairing of electrons in the d orbitals. Thus, the hybridization involves:
- 1 s orbital
- 3 p orbitals
- 2 d orbitals (dxy, dyz, dxz, dx²-y², dz²)
- Therefore, the hybridization is dsp³.

4. **Spin state**:
- Since CO is a strong field ligand, it results in a low-spin complex.

**Conclusion for FeCO5**: The statement regarding it being a high-spin complex is incorrect.

### Step 2: Analyze the second complex (Pt(NH3)2Cl2)

1. **Identify the oxidation state of Platinum (Pt)**:
- NH3 is a neutral ligand and Cl is a -1 ligand, so the oxidation state of Pt is +2.

2. **Determine the electronic configuration of Pt**:
- The electronic configuration of Pt (atomic number 78) is [Xe] 4f14 5d9 6s1. For Pt²⁺, it is [Xe] 4f14 5d8.

3. **Hybridization**:
- NH3 is a weak field ligand and does not cause pairing of electrons. Thus, the hybridization involves:
- 1 s orbital
- 3 p orbitals
- 2 d orbitals
- Therefore, the hybridization is dsp².

4. **Spin state**:
- Since NH3 is a weak field ligand, it results in a high-spin complex.

**Conclusion for Pt(NH3)2Cl2**: The statement regarding it being a low-spin complex is incorrect.

### Step 3: Analyze the third complex ([Cr(H2O)6]³⁺)

1. **Identify the oxidation state of Chromium (Cr)**:
- Water (H2O) is a neutral ligand, so the oxidation state of Cr is +3.

2. **Determine the electronic configuration of Cr**:
- The electronic configuration of Cr (atomic number 24) is [Ar] 3d5 4s1. For Cr³⁺, it is 3d3.

3. **Hybridization**:
- Water is a weak field ligand, so the electrons will not pair. Thus, the hybridization involves:
- 1 s orbital
- 3 p orbitals
- 2 d orbitals
- Therefore, the hybridization is d²sp³.

4. **Spin state**:
- Since water is a weak field ligand, it results in a high-spin complex.

**Conclusion for [Cr(H2O)6]³⁺**: The statement regarding it being a high-spin complex is correct.

### Step 4: Analyze the fourth complex (Ni(CO)4)

1. **Identify the oxidation state of Nickel (Ni)**:
- CO is a neutral ligand, so the oxidation state of Ni is 0.

2. **Determine the electronic configuration of Ni**:
- The electronic configuration of Ni (atomic number 28) is [Ar] 3d8 4s2. Since Ni is in the 0 oxidation state, it retains all its electrons.

3. **Hybridization**:
- CO is a strong field ligand, which causes pairing of electrons. Thus, the hybridization involves:
- 1 s orbital
- 3 p orbitals
- 2 d orbitals
- Therefore, the hybridization is sp³d².

4. **Spin state**:
- Since CO is a strong field ligand, it results in a low-spin complex.

**Conclusion for Ni(CO)4**: The statement regarding it being a low-spin complex is correct.

### Final Summary of Correct Statements:
- The correct statements are:
- [Cr(H2O)6]³⁺ is a high-spin complex.
- Ni(CO)4 is a low-spin complex.