Nickel `(Z=28)` combines with a uninegative monodentate ligand `X^(-)` to form a paramagnetic complex `[NiX_4]^(2-)`. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively:

Options

A. (a) One, tetrahedral
B. (b) Two, tetrahedral
C. (c) One, square planar
D. (d) Two, square planar

Explanation

To solve the question, we need to determine the number of unpaired electrons in the nickel ion and the geometry of the complex ion `[NiX_4]^{2-}`.

### Step-by-Step Solution:

1. **Identify the oxidation state of nickel in the complex:**
The complex is `[NiX_4]^{2-}`. Since each ligand `X^(-)` has a charge of -1 and there are four ligands, the total negative charge contributed by the ligands is -4. To balance this with the overall charge of -2, nickel must have a charge of +2.
\[ \text{Oxidation state of Ni} = +2 \]

2. **Determine the electronic configuration of Ni:**
Nickel (Ni) has an atomic number of 28. The ground state electronic configuration of Ni is:
\[ \text{Ni: } [Ar] 3d^8 4s^2 \]
When nickel is in the +2 oxidation state, it loses two electrons, which are removed from the 4s orbital first. Therefore, the electronic configuration of Ni in the +2 state is:
\[ \text{Ni}^{2+}: [Ar] 3d^8 4s^0 \]

3. **Analyze the d-orbital configuration:**
In the +2 oxidation state, nickel has the configuration of `3d^8`. The distribution of electrons in the d-orbitals (considering that halides are weak field ligands) will not cause pairing of electrons. The 3d orbitals will be filled as follows:
\[ \text{3d: } \uparrow \uparrow \uparrow \uparrow \uparrow \downarrow \downarrow \downarrow \downarrow \]
This means there are 2 unpaired electrons in the 3d orbitals.

4. **Determine the hybridization and geometry:**
Since the complex has four ligands, we can deduce the hybridization. The coordination number is 4, which typically leads to either sp³ hybridization (tetrahedral) or dsp² hybridization (square planar). However, since we have a weak field ligand (X^(-)), it will not cause pairing, leading to sp³ hybridization and a tetrahedral geometry.
\[ \text{Hybridization: } sp^3 \quad \text{(Tetrahedral)} \]

5. **Final answer:**
The number of unpaired electrons in the nickel ion is 2, and the geometry of the complex ion `[NiX_4]^{2-}` is tetrahedral.

### Summary:
- **Number of unpaired electrons:** 2
- **Geometry of the complex:** Tetrahedral