Velence bond theroy describes the bonding in complexs in terms of coordinate -covalent bond resulting from overlap filled ligand orbitals with vacant metal hybrid orbitals This theory explains magnetic behaviour and geometrical shape of coordination compounds Magnetic moment of a complex compound can be determined experimentally and theoretically by using spin only formula
Magnetic moment `sqrtn (n+2)BM` (where n = No. unpaired electrons) .
`Ni^(2+)` cation combines with a uninegative monodentate ligand `X^(Θ)` to form paramagnetic complex `[NiCI_(4)]^(2-)` The number of unpaired electrons(s) in central metal cation and geometry of this complex respectively are
(a) One,tetrahedral
(b) Two,tetrahedral
(c ) One,square planar
(d) Two, square planar .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the oxidation state of Nickel in the complex
- The complex is given as \([NiCl_4]^{2-}\).
- Each chloride ion (Cl) has a charge of -1, and there are 4 chloride ions, contributing a total of -4.
- Let the oxidation state of Nickel (Ni) be \(x\).
- The overall charge of the complex is -2, so we can set up the equation:
\[ x + 4(-1) = -2 \]
\[ x - 4 = -2 \implies x = +2 \]
- Therefore, the oxidation state of Nickel is +2.

**Hint:** Remember that the oxidation state can be calculated by balancing the total charge of the complex with the charges of the ligands.

### Step 2: Write the electronic configuration of Ni²⁺
- The atomic number of Nickel (Ni) is 28. Its ground state electronic configuration is:
\[ [Ar] 3d^8 4s^2 \]
- For Ni²⁺, we remove 2 electrons from the 4s orbital:
\[ Ni^{2+}: [Ar] 3d^8 \]

**Hint:** When forming cations, electrons are removed from the outermost shell first (4s before 3d).

### Step 3: Fill the d-orbitals for Ni²⁺
- The 3d subshell can hold a maximum of 10 electrons. For Ni²⁺, we have 8 electrons to place in the 3d orbitals:
- The filling of the 3d orbitals will be as follows:
\[ \uparrow \downarrow \quad \uparrow \downarrow \quad \uparrow \downarrow \quad \uparrow \quad \uparrow \]
- This results in 2 paired electrons and 6 unpaired electrons.

**Hint:** Remember to follow Hund's rule while filling the orbitals to maximize unpaired electrons.

### Step 4: Identify the type of ligand and its effect on hybridization
- Chlorine (Cl) is a weak field ligand. Weak field ligands do not cause pairing of electrons in the d-orbitals.
- Therefore, the hybridization will involve the 3d, 4s, and 4p orbitals to accommodate the 4 ligands.
- Since we have 4 ligands, the hybridization will be \(sp^3\).

**Hint:** The strength of the ligand determines whether pairing occurs and the type of hybridization.

### Step 5: Determine the geometry of the complex
- The \(sp^3\) hybridization corresponds to a tetrahedral geometry.
- Since we have 2 unpaired electrons, the complex is paramagnetic.

**Hint:** The number of ligands and the type of hybridization can help you deduce the geometry of the complex.

### Conclusion
- The number of unpaired electrons in the central metal cation (Ni²⁺) is 2.
- The geometry of the complex \([NiCl_4]^{2-}\) is tetrahedral.

**Final Answer:** The correct option is (b) Two, tetrahedral.