Read the passage given below and answer the following questions :
According to Valence Bond Theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals ar allowed to overlap with ligand rbitals that can donate electron pairs for bonding.
Assertion (A) : In the square planar complexes, the hybridisation involved is `dsp^(2)`.
Reason (R) : In `[NiCl)_(4)]^(2-)`. Here nickel is in +2 oxidation state and has the electronic configuration `3d^(8)`.

Options

A. Assertion and reason both are correct statements and reason is correct explanation for assertion.
B. Assertion and reason both are correct statements and reason is not correct explanation for assertion.
C. Assertion is correct statement but reason is wrong statement
D. Assertion is wrong statements but reason is correct statement.

Answer

To solve the question, we need to analyze the assertion and reason provided in the context of the passage related to Valence Bond Theory and hybridization in coordination compounds.

### Step-by-Step Solution:

1. **Understanding the Assertion (A)**:
- The assertion states that in square planar complexes, the hybridization involved is `dsp^2`.
- Square planar complexes typically arise from the hybridization of one d-orbital, one s-orbital, and two p-orbitals, which corresponds to `dsp^2` hybridization.

2. **Understanding the Reason (R)**:
- The reason states that in `[NiCl4]^(2-)`, nickel is in the +2 oxidation state and has the electronic configuration `3d^8`.
- To verify this, we need to calculate the oxidation state of nickel in `[NiCl4]^(2-)`.
- Chlorine (Cl) has a charge of -1. Therefore, for four Cl atoms, the total charge contributed is -4.
- The overall charge of the complex is -2, so we can set up the equation:
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]
- Thus, the oxidation state of nickel is indeed +2.

3. **Electronic Configuration of Ni**:
- Nickel (Ni) has an atomic number of 28. Its ground state electronic configuration is:
\[ [Ar] 4s^2 3d^8 \]
- When Ni is in the +2 oxidation state, it loses the two 4s electrons, resulting in:
\[ 3d^8 \]

4. **Conclusion**:
- Both the assertion (A) and the reason (R) are correct.
- However, the reason does not explain why the hybridization is `dsp^2` for square planar complexes. It simply states the oxidation state and electronic configuration of nickel.

### Final Answer:
- Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion.