Magnetic moment, ionic conductance and colligative properties are useful in deciding structure/constitution of a given unknown complex compound.
Q. A metal M having elecftronic configuration `(n-1)d^(8)ns^(2)` forms complexes with co-ordination No.=4 and 6, if it forms diamagnetic complexes then permissible oxidation states of metal cation and geometry is:

Options

A. `+2`, octahedral
B. `+4`, octahedral
C. `+2`, square planar
D. (b) and (c) both

Answer

To solve the problem, we need to analyze the given metal M with the electronic configuration (n-1)d^8 ns^2 and determine the permissible oxidation states and geometries for its diamagnetic complexes with coordination numbers of 4 and 6.

### Step 1: Determine the oxidation states of the metal cation
The electronic configuration of the metal is (n-1)d^8 ns^2. The possible oxidation states can be derived from the loss of electrons from the outermost shell (ns) and the (n-1)d subshell.

- **Oxidation State +2**:
- When the metal loses 2 electrons from the ns^2, the configuration becomes (n-1)d^8.
- **Oxidation State +4**:
- When the metal loses 4 electrons (2 from ns and 2 from (n-1)d), the configuration becomes (n-1)d^6.

### Step 2: Analyze the coordination number 4
For a coordination number of 4, the geometry can either be tetrahedral or square planar. However, since the complex is stated to be diamagnetic, we need to consider strong field ligands that cause pairing of electrons.

- **Diamagnetic Complex**:
- In the +2 oxidation state with a d^8 configuration, the strong field ligands will cause pairing in the d orbitals. The square planar geometry is favored in this case, which involves dsp^2 hybridization.

### Step 3: Analyze the coordination number 6
For a coordination number of 6, the most common geometry is octahedral.

- **Diamagnetic Complex**:
- In the +4 oxidation state with a d^6 configuration, strong field ligands again will cause pairing of electrons. In this case, the t2g orbitals will be filled first, leading to a configuration where the eg orbitals are vacant. The hybridization involved here is d^2sp^3, resulting in octahedral geometry.

### Conclusion
- **Permissible Oxidation States**: +2 and +4
- **Geometries**:
- For +2: Square planar (coordination number 4)
- For +4: Octahedral (coordination number 6)

### Final Answer
The permissible oxidation states of the metal cation are +2 and +4, with geometries being square planar for +2 and octahedral for +4.

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