Select correct statement(s) regarding octahedron complex having CFSE`=-1.2Delta_(0)`.

Options

A. Compound is neither low spin nor high spin complex
B. Type of hybridisation complex does not depend upon nature of ligands
C. Magnetic moment of complex compounds is either `sqrt(15)B.M or sqrt(8)B.M.`
D. All are incorrect statements

Answer

To solve the question regarding the octahedral complex with a crystal field stabilization energy (CFSE) of -1.2Δ₀, we will analyze the implications of this value step by step.

### Step 1: Understanding CFSE
CFSE is a measure of the energy difference between the electrons in the lower energy t₂g orbitals and the higher energy eₕ orbitals in an octahedral complex. The formula for CFSE in an octahedral field is given by:
\[ \text{CFSE} = (n_{t2g} \times -0.4Δ₀) + (n_{eg} \times 0.6Δ₀) \]
where \( n_{t2g} \) is the number of electrons in the t₂g orbitals and \( n_{eg} \) is the number of electrons in the eₕ orbitals.

### Step 2: Setting Up the Equation
Given that CFSE = -1.2Δ₀, we can set up the equation:
\[ -1.2Δ₀ = (n_{t2g} \times -0.4Δ₀) + (n_{eg} \times 0.6Δ₀) \]

### Step 3: Analyzing Possible Configurations
1. **Case 1: 3 Electrons in t₂g**
- If \( n_{t2g} = 3 \) and \( n_{eg} = 0 \):
\[ \text{CFSE} = 3 \times -0.4Δ₀ + 0 \times 0.6Δ₀ = -1.2Δ₀ \]
This configuration is valid.

2. **Case 2: 6 Electrons in t₂g and 2 in eₕ (Low Spin)**
- If \( n_{t2g} = 6 \) and \( n_{eg} = 0 \):
\[ \text{CFSE} = 6 \times -0.4Δ₀ + 0 \times 0.6Δ₀ = -2.4Δ₀ \] (not valid)
- If \( n_{t2g} = 4 \) and \( n_{eg} = 2 \):
\[ \text{CFSE} = 4 \times -0.4Δ₀ + 2 \times 0.6Δ₀ = -1.6Δ₀ \] (not valid)
- If \( n_{t2g} = 2 \) and \( n_{eg} = 4 \):
\[ \text{CFSE} = 2 \times -0.4Δ₀ + 4 \times 0.6Δ₀ = 2.0Δ₀ \] (not valid)
- If \( n_{t2g} = 4 \) and \( n_{eg} = 2 \):
\[ \text{CFSE} = 4 \times -0.4Δ₀ + 2 \times 0.6Δ₀ = -1.2Δ₀ \] (valid)

### Step 4: Determining Spin State
From the configurations, we can conclude:
- The complex can be either a low-spin or high-spin complex based on the arrangement of electrons.

### Step 5: Magnetic Moment Calculation
The magnetic moment (μ) can be calculated using the formula:
\[ μ = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.
- For 2 unpaired electrons:
\[ μ = \sqrt{2(2 + 2)} = \sqrt{8} \]
- For 3 unpaired electrons:
\[ μ = \sqrt{3(3 + 2)} = \sqrt{15} \]

### Conclusion
Based on the analysis, we can conclude:
1. The complex can exhibit both low-spin and high-spin configurations.
2. The type of hybridization depends on the nature of ligands and the number of ligands.
3. The magnetic moment can either be √15 or √8 Bohr magneton.

### Final Statements
All the statements regarding the octahedral complex having CFSE = -1.2Δ₀ are correct.