…………is a low spin complex.

Options

A. `[Ni(CN)_4]^(2-)`
B. `[CoF_6]^(3-)`
C. `[NiCl_4]^(3-)`
D. `[FeF_6]^(3-)`

Answer

To determine which complex is a low spin complex, we need to analyze the given options based on the nature of the ligands and the oxidation state of the central metal ion. Here’s a step-by-step solution:

### Step 1: Identify the Metal and its Oxidation State
- For the first complex, Ni(CN)₄²⁻, we need to calculate the oxidation state of nickel (Ni).
- Let the oxidation state of Ni be \( x \).
- The charge of the complex is -2, and CN is a -1 ligand. Therefore, we have:
\[ x + 4(-1) = -2 \]
\[ x - 4 = -2 \implies x = +2 \]
- The oxidation state of Ni in this complex is +2.

### Step 2: Determine the Electron Configuration
- Nickel (Ni) has an atomic number of 28, which means its electron configuration is [Ar] 3d⁸ 4s².
- For Ni²⁺, we remove the two 4s electrons:
\[ \text{Ni}^{2+} : [Ar] 3d^8 \]

### Step 3: Analyze the Ligand
- The ligand CN⁻ (cyanide) is a strong field ligand. Strong field ligands tend to cause pairing of electrons in the d-orbitals.
- Since CN⁻ is a strong field ligand, it will pair the 3d electrons.

### Step 4: Electron Pairing
- In the case of Ni²⁺ with 3d⁸ configuration, the electrons will be arranged as follows:
- The 3d orbitals will fill up as: ↑↓ ↑↓ ↑↓ ↑ ↑
- All electrons will be paired due to the strong field nature of CN⁻.

### Step 5: Determine Hybridization
- The hybridization of Ni(CN)₄²⁻ can be determined as follows:
- Since all electrons are paired, the hybridization will be \( sp^3d^2 \) (considering 4 CN ligands).

### Conclusion
- Since there are no unpaired electrons in Ni(CN)₄²⁻, it is classified as a **low spin complex**.

### Final Answer
The complex that is a low spin complex is **Ni(CN)₄²⁻**.

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