Complex ions `[NiCl_(6)]^(4-),[Ni(CN)_(6)]^(4-)` similar in their given properties:

Options

A. oxidation state, geometryq
B. co-ordination number, EAN
C. magnetic moment, geometry
D. stability, colour

Answer

To analyze the complex ions \([NiCl_6]^{4-}\) and \([Ni(CN)_6]^{4-}\), we will follow a systematic approach to determine their properties, including oxidation state, hybridization, geometry, magnetic nature, and stability.

### Step-by-Step Solution:

**Step 1: Determine the oxidation state of Nickel in both complexes.**

For \([NiCl_6]^{4-}\):
- Let the oxidation state of Ni be \(x\).
- The charge of Cl is \(-1\) and there are 6 Cl ligands.
- The overall charge of the complex is \(-4\).

The equation becomes:
\[ x + 6(-1) = -4 \]
\[ x - 6 = -4 \]
\[ x = +2 \]

For \([Ni(CN)_6]^{4-}\):
- Similarly, let the oxidation state of Ni be \(x\).
- The charge of CN is \(-1\) and there are 6 CN ligands.

The equation becomes:
\[ x + 6(-1) = -4 \]
\[ x - 6 = -4 \]
\[ x = +2 \]

**Step 2: Write the electronic configuration of Nickel.**

Nickel (Ni) has an atomic number of 28. The electronic configuration is:
\[ [Ar] 4s^2 3d^8 \]

**Step 3: Analyze the hybridization and geometry of both complexes.**

For \([NiCl_6]^{4-}\):
- Chloride (Cl\(^-\)) is a weak field ligand, which does not cause pairing of electrons.
- The hybridization will be \(sp^3d^2\) (6 orbitals for 6 ligands).
- Geometry: Octahedral.

For \([Ni(CN)_6]^{4-}\):
- Cyanide (CN\(^-\)) is a strong field ligand, which causes pairing of electrons.
- The hybridization will also be \(sp^3d^2\) (6 orbitals for 6 ligands).
- Geometry: Octahedral.

**Step 4: Determine the magnetic nature of both complexes.**

For \([NiCl_6]^{4-}\):
- Since there are unpaired electrons (due to weak field ligand), it is paramagnetic.

For \([Ni(CN)_6]^{4-}\):
- All electrons are paired (due to strong field ligand), so it is diamagnetic.

**Step 5: Calculate the coordination number and effective atomic number.**

Both complexes have a coordination number of 6 (as there are 6 ligands).

Effective Atomic Number (EAN) can be calculated as:
\[ \text{EAN} = Z - \text{oxidation state} + \text{number of donating electrons from ligands} \]
Where \(Z\) is the atomic number of Ni (28).

For both complexes:
\[ \text{EAN} = 28 - 2 + (6 \times 2) = 28 - 2 + 12 = 38 \]

**Step 6: Discuss the stability and color of both complexes.**

- The stability of \([Ni(CN)_6]^{4-}\) is greater than that of \([NiCl_6]^{4-}\) due to the stronger field strength of cyanide compared to chloride.
- The color of the complexes will also differ due to the different ligand field strengths affecting the d-d transitions.

### Summary of Properties:
- **\([NiCl_6]^{4-}\)**:
- Oxidation state: +2
- Hybridization: \(sp^3d^2\)
- Geometry: Octahedral
- Magnetic nature: Paramagnetic
- Coordination number: 6
- EAN: 38

- **\([Ni(CN)_6]^{4-}\)**:
- Oxidation state: +2
- Hybridization: \(sp^3d^2\)
- Geometry: Octahedral
- Magnetic nature: Diamagnetic
- Coordination number: 6
- EAN: 38