Match the geometry (given in column A) with the complexes (given in column B) in :
`{:("Geometry : A","Complex : B"),("I Octahedral",(P)[Ni(CN)_(4)]^(2-)),("II Square planar",(Q) Ni(CO)_(4)),("III Tetrahedral",(R)[Fe(CN)_(6)]^(4-)):}`

Options

A. I - P, II - Q, III - R
B. I - R, II - P, III - Q
C. I - R, II - Q, III - P
D. I - Q, II - P, III - R

Answer

To solve the problem of matching the geometry of complexes with their respective complexes, we will analyze each complex and determine its geometry based on the hybridization of the central metal atom.

### Step 1: Analyze the first complex, [Ni(CN)₄]²⁻

1. **Determine the oxidation state of Nickel (Ni)**:
- The charge of the complex is -2.
- Each cyanide (CN) has a charge of -1, and there are 4 cyanides.
- Let the oxidation state of Ni be \( x \).
- The equation is: \( x + 4(-1) = -2 \)
- Solving gives \( x = +2 \).

2. **Determine the electron configuration of Ni²⁺**:
- Nickel has an atomic number of 28, so its electron configuration is [Ar] 3d⁸ 4s².
- For Ni²⁺, we remove 2 electrons from the 4s orbital, resulting in 3d⁸.

3. **Identify the ligand strength**:
- Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons in the d-orbitals.

4. **Determine hybridization**:
- With 4 ligands and a d⁸ configuration, the hybridization is dsp².

5. **Determine geometry**:
- dsp² hybridization corresponds to a square planar geometry.

**Conclusion for P: [Ni(CN)₄]²⁻ matches with Square planar (II)**.

### Step 2: Analyze the second complex, Ni(CO)₄

1. **Determine the oxidation state of Nickel (Ni)**:
- The complex is neutral, so the oxidation state of Ni is 0.

2. **Determine the electron configuration of Ni**:
- The electron configuration remains [Ar] 3d⁸ 4s².

3. **Identify the ligand strength**:
- Carbon monoxide (CO) is also a strong field ligand.

4. **Determine hybridization**:
- With 4 ligands and no unpaired electrons, the hybridization is sp³.

5. **Determine geometry**:
- sp³ hybridization corresponds to a tetrahedral geometry.

**Conclusion for Q: Ni(CO)₄ matches with Tetrahedral (III)**.

### Step 3: Analyze the third complex, [Fe(CN)₆]⁴⁻

1. **Determine the oxidation state of Iron (Fe)**:
- The charge of the complex is -4.
- Each cyanide (CN) has a charge of -1, and there are 6 cyanides.
- Let the oxidation state of Fe be \( x \).
- The equation is: \( x + 6(-1) = -4 \)
- Solving gives \( x = +2 \).

2. **Determine the electron configuration of Fe²⁺**:
- Iron has an atomic number of 26, so its electron configuration is [Ar] 3d⁶ 4s².
- For Fe²⁺, we remove 2 electrons from the 4s orbital, resulting in 3d⁶.

3. **Identify the ligand strength**:
- Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons in the d-orbitals.

4. **Determine hybridization**:
- With 6 ligands and a d⁶ configuration, the hybridization is d²sp³.

5. **Determine geometry**:
- d²sp³ hybridization corresponds to an octahedral geometry.

**Conclusion for R: [Fe(CN)₆]⁴⁻ matches with Octahedral (I)**.

### Final Matching:
- P: [Ni(CN)₄]²⁻ → Square planar (II)
- Q: Ni(CO)₄ → Tetrahedral (III)
- R: [Fe(CN)₆]⁴⁻ → Octahedral (I)

### Summary of Matches:
- P → II
- Q → III
- R → I