Which is/are correct statement(s)?

Options

A. `[Ag(NH_(3))_(2)]^(+)` is linear with sp hybridisation `Ag^(+)` ions
B. `NiCl_(4)^(2-),CrO_(4)^(2-)` annd `MnO_(4)^(-)` have tetrahedral geometry
C. `[Cu(NH_(3))_(4)]^(2+),[[Pt(NH_(3))_(4)]^(2+) and [Ni(CN)_(4)]^(2-)` have `dsp^(2)` hybridisation of the metal ion
D. `Fe(CO)_(5)` has trigonal bipyramidal structure with `d_(z^(2))sp^(3)` hybridised ion

Answer

To determine which statements are correct regarding coordination compounds, let's analyze each statement step by step.

### Step 1: Analyze the first statement - Ag(NH3)2^+
- **Central Metal**: Silver (Ag)
- **Oxidation State Calculation**: Let the oxidation state of Ag be \( x \). Since NH3 is a neutral ligand, we have:
\[ x + 2(0) = +1 \implies x = +1 \]
- **Electronic Configuration**: Ag has atomic number 47, and its electronic configuration is [Kr] 4d10 5s1. In the +1 oxidation state, it has 4d10.
- **Hybridization**: With two NH3 ligands, the hybridization is sp (linear geometry).
- **Conclusion**: This statement is correct.

### Step 2: Analyze the second statement - NiCl4^2-
- **Central Metal**: Nickel (Ni)
- **Oxidation State Calculation**: Let the oxidation state of Ni be \( x \). With 4 Cl^- ligands:
\[ x - 4 = -2 \implies x = +2 \]
- **Electronic Configuration**: Ni has atomic number 28, with electronic configuration [Ar] 3d8 4s2. In the +2 oxidation state, it has 3d8.
- **Hybridization**: Since Cl is a weak field ligand, the hybridization is sp3 (tetrahedral geometry).
- **Conclusion**: This statement is correct.

### Step 3: Analyze the third statement - CrO4^2-
- **Central Metal**: Chromium (Cr)
- **Oxidation State Calculation**: Let the oxidation state of Cr be \( x \):
\[ x - 8 = -2 \implies x = +6 \]
- **Electronic Configuration**: Cr has atomic number 24, with electronic configuration [Ar] 3d5 4s1. In the +6 oxidation state, it has no d-electrons.
- **Hybridization**: The hybridization is sp3 (tetrahedral geometry).
- **Conclusion**: This statement is correct.

### Step 4: Analyze the fourth statement - MnO4^-
- **Central Metal**: Manganese (Mn)
- **Oxidation State Calculation**: Let the oxidation state of Mn be \( x \):
\[ x - 8 = -1 \implies x = +7 \]
- **Electronic Configuration**: Mn has atomic number 25, with electronic configuration [Ar] 3d5 4s2. In the +7 oxidation state, it has no d-electrons.
- **Hybridization**: The hybridization is sp3 (tetrahedral geometry).
- **Conclusion**: This statement is correct.

### Step 5: Analyze the fifth statement - Copper, Platinum, and Nickel have dsp2 hybridization.
- **Copper (Cu)**: In the +2 oxidation state, Cu has the configuration 3d9. The hybridization is dsp2 due to the presence of strong field ligands.
- **Platinum (Pt)**: In the +2 oxidation state, Pt can utilize d-orbitals for hybridization, leading to dsp2.
- **Nickel (Ni)**: In the +2 oxidation state, Ni can also exhibit dsp2 hybridization.
- **Conclusion**: This statement is correct.

### Step 6: Analyze the sixth statement - Fe(CO)5 has a trigonal bipyramidal structure with d2sp3 hybridization.
- **Central Metal**: Iron (Fe)
- **Oxidation State**: In Fe(CO)5, Fe is in the zero oxidation state.
- **Electronic Configuration**: Fe has atomic number 26, with electronic configuration [Ar] 3d6 4s2. In the zero oxidation state, it retains all its electrons.
- **Hybridization**: The presence of CO, a strong field ligand, leads to pairing of electrons and d2sp3 hybridization.
- **Conclusion**: This statement is correct.

### Final Conclusion:
All statements A, B, C, D, E, and F are correct.