Among `[Ni(CN)_(4)]^(4-), [Ni(PPh_(3))_(3)Br] and [Ni(dmg)_(2)]` species, the hybridisation state of the Ni-atoms are respectively:

Options

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

Explanation

To determine the hybridization state of the Ni atoms in the given coordination compounds, we will analyze each compound step by step.

### Step 1: Analyze `[Ni(CN)_(4)]^(4-)`
1. **Identify the oxidation state of Ni**: In `[Ni(CN)_(4)]^(4-)`, CN is a neutral ligand. Therefore, the oxidation state of Ni is +2.
2. **Determine the electron configuration of Ni**: The atomic number of Ni is 28. The electron configuration is `[Ar] 3d^8 4s^2`.
3. **Count the valence electrons**: Ni in the +2 oxidation state has lost 2 electrons (from 4s), resulting in `[Ar] 3d^8`.
4. **Determine the hybridization**: Since CN is a strong field ligand, it causes pairing of electrons in the 3d orbital. The hybridization involves the 3d, 4s, and 4p orbitals. The hybridization is `dsp^2` (which corresponds to square planar geometry).

### Step 2: Analyze `[Ni(PPh_(3))_(3)Br]`
1. **Identify the oxidation state of Ni**: In `[Ni(PPh_(3))_(3)Br]`, PPh3 is a neutral ligand and Br is a -1 ligand. Therefore, the oxidation state of Ni is +1.
2. **Determine the electron configuration of Ni**: As before, Ni has an atomic number of 28, and in the +1 oxidation state, it has the configuration `[Ar] 3d^8 4s^1`.
3. **Count the valence electrons**: Ni in the +1 oxidation state has 9 valence electrons (8 from 3d and 1 from 4s).
4. **Determine the hybridization**: The hybridization involves the 3d, 4s, and 4p orbitals. The hybridization is `dsp^3` (which corresponds to trigonal bipyramidal geometry).

### Step 3: Analyze `[Ni(dmg)_(2)]`
1. **Identify the oxidation state of Ni**: In `[Ni(dmg)_(2)]`, dmg (dimethylglyoxime) is a neutral ligand. Therefore, the oxidation state of Ni is +2.
2. **Determine the electron configuration of Ni**: Again, Ni has an atomic number of 28, and in the +2 oxidation state, it has the configuration `[Ar] 3d^8`.
3. **Count the valence electrons**: Ni in the +2 oxidation state has 8 valence electrons (from 3d).
4. **Determine the hybridization**: The hybridization involves the 3d and 4s orbitals. The hybridization is `sp^2` (which corresponds to planar geometry).

### Summary of Hybridization States
- For `[Ni(CN)_(4)]^(4-)`: Hybridization is `dsp^2`.
- For `[Ni(PPh_(3))_(3)Br]`: Hybridization is `dsp^3`.
- For `[Ni(dmg)_(2)]`: Hybridization is `sp^2`.

### Final Answer
The hybridization states of the Ni atoms in the respective species are:
1. `[Ni(CN)_(4)]^(4-)`: `dsp^2`
2. `[Ni(PPh_(3))_(3)Br]`: `dsp^3`
3. `[Ni(dmg)_(2)]`: `sp^2`