The hybrisation states of the central atom ion in the complex ions `[FeF_(6)]^(3-) ,[Fe(H_(2)O))_(6)]^(3+)` and `[Ni(NH_(3))_(6)]^(2+)` are
(a) `sp^(3)d^(2),dsp^(2)` and `d^(4)s^(2)` respectively
(b) all `3d^(2)4s4p^(3)`
(c ) all `4s 4p^(3) 4d^(2)`
(d) `sp^(3)d^(2), dsp^(3)` and `p^(4)d^(2)` respectively .

Explanation

To determine the hybridization states of the central atom in the complex ions \([FeF_6]^{3-}\), \([Fe(H_2O)_6]^{3+}\), and \([Ni(NH_3)_6]^{2+}\), we will follow these steps:

### Step 1: Identify the oxidation states of the central metal ions.
- For \([FeF_6]^{3-}\):
- Fluoride (F) has a charge of -1. Therefore, for 6 fluorides, the total negative charge is -6.
- Let the oxidation state of Fe be \(x\). The equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- For \([Fe(H_2O)_6]^{3+}\):
- Water (H2O) is neutral, so the oxidation state of Fe is +3.
- For \([Ni(NH_3)_6]^{2+}\):
- Ammonia (NH3) is also neutral, so the oxidation state of Ni is +2.

### Step 2: Determine the electron configuration of the metal ions.
- For \(Fe^{3+}\):
- Atomic number of Fe = 26. The electron configuration is \( [Ar] 3d^6 4s^2 \).
- For \(Fe^{3+}\), it loses 3 electrons (2 from 4s and 1 from 3d), resulting in \(3d^5\).

- For \(Ni^{2+}\):
- Atomic number of Ni = 28. The electron configuration is \( [Ar] 3d^8 4s^2 \).
- For \(Ni^{2+}\), it loses 2 electrons (both from 4s), resulting in \(3d^8\).

### Step 3: Determine the hybridization based on the ligands and electron configuration.
- For \([FeF_6]^{3-}\):
- Fluoride is a weak field ligand, so it does not cause pairing of electrons. The \(3d^5\) configuration will use \(3d\), \(4s\), and \(4p\) orbitals for hybridization.
- The hybridization is \(sp^3d^2\) (1 s, 3 p, and 2 d orbitals).

- For \([Fe(H_2O)_6]^{3+}\):
- Water is a weak field ligand, similar to fluoride, so pairing does not occur.
- The hybridization is also \(sp^3d^2\).

- For \([Ni(NH_3)_6]^{2+}\):
- Ammonia is a strong field ligand, which can cause pairing. The \(3d^8\) configuration will use \(4s\) and \(4p\) orbitals for hybridization.
- The hybridization is \(d^2sp^3\) (1 d, 2 s, and 3 p orbitals).

### Step 4: Summarize the hybridization states.
- \([FeF_6]^{3-}\): \(sp^3d^2\)
- \([Fe(H_2O)_6]^{3+}\): \(sp^3d^2\)
- \([Ni(NH_3)_6]^{2+}\): \(d^2sp^3\)

### Conclusion:
The correct answer is (a) \(sp^3d^2\), \(sp^3d^2\), and \(d^2sp^3\) respectively.