Assertion A solution of `[Ni(H_(2)O)_(6)]^(2+)` is green but a solution of `[Ni(CN)_(4)]^(2+)` is colourless
Reason `[Ni(CN)_(4)]^(2+)` is square planar complex .

Options

A. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .
B. If both `(A)` and ` (R )` are correct and `(R )` is the correct explanation of `(A ) ` .
C. If`(A)` is correct, but `( R )` is incorrect .
D. Both `(A)` and `(R ) ` are incorrect .

Answer

### Step-by-Step Solution

1. **Understanding the Assertion**: The assertion states that a solution of \([Ni(H_2O)_6]^{2+}\) is green while a solution of \([Ni(CN)_4]^{2+}\) is colorless. We need to analyze why this is the case.

**Hint**: Consider the role of ligands and their strength in determining the color of coordination compounds.

2. **Analyzing the Coordination Compounds**:
- For \([Ni(H_2O)_6]^{2+}\):
- Nickel is in the +2 oxidation state.
- Nickel has an atomic number of 28, leading to an electronic configuration of \([Ar] 3d^8 4s^2\).
- In the +2 state, the configuration becomes \(3d^8\).
- Water is a weak field ligand, which does not cause pairing of electrons in the \(3d\) orbitals.
- Therefore, there are 2 unpaired electrons in \([Ni(H_2O)_6]^{2+}\), which leads to the absorption of certain wavelengths of light, resulting in the green color.

**Hint**: Remember that the presence of unpaired electrons is crucial for the color in transition metal complexes.

3. **Analyzing the Second Coordination Compound**:
- For \([Ni(CN)_4]^{2+}\):
- Nickel is still in the +2 oxidation state.
- The electronic configuration remains \(3d^8\).
- Cyanide (\(CN^-\)) is a strong field ligand, which causes pairing of electrons in the \(3d\) orbitals.
- As a result, all \(3d\) electrons are paired, leading to no unpaired electrons in \([Ni(CN)_4]^{2+}\).
- Consequently, this complex is colorless because there are no unpaired electrons to absorb visible light.

**Hint**: Identify the type of ligand (weak vs strong) and its effect on electron pairing.

4. **Understanding the Geometry**:
- The geometry of \([Ni(CN)_4]^{2+}\) is square planar due to the presence of four ligands and the pairing of electrons.
- The hybridization involved is \(d^2sp^2\), which is characteristic of square planar complexes.

**Hint**: Relate the hybridization and geometry to the arrangement of ligands around the central metal ion.

5. **Conclusion**:
- The assertion is correct: \([Ni(H_2O)_6]^{2+}\) is green due to unpaired electrons, while \([Ni(CN)_4]^{2+}\) is colorless due to paired electrons.
- The reason is also correct: \([Ni(CN)_4]^{2+}\) is a square planar complex.
- However, the reason does not explain the assertion correctly, as the color difference is primarily due to the presence or absence of unpaired electrons, not solely due to geometry.

**Final Answer**: The assertion is correct, the reason is correct, but the reason is not the correct explanation for the assertion.

### Final Answer Option: B

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