The species having tetrahedral shape is

Options

A. `[PdCl_(4)]^(2-)`
B. `[Ni(CN)_(4)]^(2)`
C. `[Pd(CN)_(4)]^(2-)`
D. `[NiCl_(4)]^(2-)`

Answer

To determine which species has a tetrahedral shape, we will analyze the coordination complexes step by step.

### Step 1: Identify the Complexes
We will consider the complexes provided in the video transcript:
1. **PdCl4**
2. **NiCl4^2-**

### Step 2: Analyze PdCl4
- **Oxidation State Calculation**:
- Let the oxidation state of Pd be \( x \).
- The charge from 4 chloride ions (each -1) is \( -4 \).
- Therefore, the equation is:
\[ x + (-4) = 0 \implies x = +4 \]
- **Electron Configuration**:
- The electron configuration of Pd (Palladium) is \( [Kr] 4d^{10} \).
- In the +4 oxidation state, it loses 4 electrons, resulting in \( 4d^8 \).
- **Hybridization**:
- Since chloride is a weak field ligand, it does not cause pairing of electrons.
- The hybridization is \( dsp^2 \), leading to a square planar geometry, not tetrahedral.

### Step 3: Analyze NiCl4^2-
- **Oxidation State Calculation**:
- Let the oxidation state of Ni be \( x \).
- The charge from 4 chloride ions is \( -4 \).
- The overall charge of the complex is -2, so:
\[ x + (-4) = -2 \implies x = +2 \]
- **Electron Configuration**:
- The electron configuration of Ni (Nickel) is \( [Ar] 4s^2 3d^8 \).
- In the +2 oxidation state, it loses 2 electrons from the 4s orbital, resulting in \( 3d^8 \).
- **Hybridization**:
- The presence of weak field ligands (chloride) means that there will be no pairing of electrons.
- The hybridization is \( sp^3 \), leading to a tetrahedral geometry.

### Conclusion
Based on the analysis:
- **PdCl4** has a square planar geometry.
- **NiCl4^2-** has a tetrahedral geometry.

Thus, the species having a tetrahedral shape is **NiCl4^2-**.

### Final Answer
**The species having tetrahedral shape is NiCl4^2-.**

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