What will be the hybridisation of `Ni(CN)_(5)^(-3)` ?

Options

A. `sp^(3)d^(2)`
B. `sp^(3)d`
C. `dsp^(3)`
D. `d^(2)sp^(3)`

Answer

To determine the hybridization of the complex ion \( \text{Ni(CN)}_5^{-3} \), we will follow these steps:

### Step 1: Determine the Oxidation State of Nickel
The overall charge of the complex is -3. The cyanide ion (\( \text{CN} \)) has a charge of -1. Since there are 5 cyanide ions, the total contribution from the cyanide ions is:
\[ 5 \times (-1) = -5 \]
Let the oxidation state of nickel be \( x \). The equation for the oxidation state is:
\[ x + (-5) = -3 \]
Solving for \( x \):
\[ x - 5 = -3 \implies x = -3 + 5 = +2 \]
Thus, the oxidation state of nickel in this complex is +2.

### Step 2: Determine the Electron Configuration of Nickel
The electron configuration of nickel (Ni) is:
\[ \text{Ni: } [\text{Ar}] 3d^8 4s^2 \]
When nickel is in the +2 oxidation state, it loses two electrons, typically from the 4s orbital first:
\[ \text{Ni}^{2+}: [\text{Ar}] 3d^8 4s^0 \]

### Step 3: Analyze the Ligands
The ligand \( \text{CN}^- \) is a strong field ligand. Strong field ligands cause pairing of electrons in the d-orbitals. Since we have 5 cyanide ligands, they will cause the 3d electrons to pair up.

### Step 4: Electron Pairing in the d-Orbitals
In the case of \( \text{Ni}^{2+} \) with the configuration \( 3d^8 \):
- The 3d orbitals will be filled as follows:
\[ \text{3d: } \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \]
This means all 8 electrons are paired in the 3d orbitals.

### Step 5: Determine the Hybridization
Since there are 5 ligands and the strong field ligand causes pairing, the hybridization can be determined as follows:
- The 3d orbitals will hybridize with one s and three p orbitals to accommodate the 5 electron pairs from the 5 ligands.

Thus, the hybridization of \( \text{Ni(CN)}_5^{-3} \) is:
\[ \text{Hybridization: } \text{DSP}^3 \]

### Conclusion
The hybridization of \( \text{Ni(CN)}_5^{-3} \) is \( \text{DSP}^3 \).

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