The magnetic moment of a complex `(A)` of Co was found to be `4.89BM` and the `EAN as 36` co alos forms complex `(B)` with magnetic moment `3.47BM` and `EAN` as37 and complex `(C )` with `EAN` as `36but` diamagnetic Which of the following statements is true regarding the above observation?

Options

A. The oxidation states of Co in (A),(B) and `(C )` are `+3,+2` and `+3` respectively .
B. Complexes `(A)` and `(B)` have `sp^(3)d^(2)` hybridisation state while `(C )` has `dsp^(3)` hybrisation state .
C. The spin multiplicities of Co in `(A),(B)` and `(C )` are 3,2 and 1,respectively .
D. The oxidation states of `Co` in `(A),(B)` and `(C )` are `+6,+8` and `+1` respectively .

Answer

To solve the question regarding the complexes of cobalt and their magnetic moments and effective atomic numbers (EAN), we will analyze each complex step by step.

### Step 1: Analyze Complex A
- **Given Data**:
- Magnetic moment = 4.89 BM
- EAN = 36

- **Interpretation**:
The magnetic moment (μ) can be used to determine the number of unpaired electrons (n) in the complex using the formula:
\[ \mu = \sqrt{n(n+2)} \text{ BM} \]
Setting μ = 4.89 BM, we can solve for n:
\[ 4.89 = \sqrt{n(n+2)} \]
Squaring both sides:
\[ 24.00 = n(n+2) \]
Rearranging gives:
\[ n^2 + 2n - 24 = 0 \]
Solving this quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm 10}{2} \]
This gives us \( n = 4 \) (since n cannot be negative).

- **Conclusion for Complex A**:
Complex A has 4 unpaired electrons, indicating that cobalt is in the +3 oxidation state (3d^6 configuration). The hybridization is sp³d², indicating it is an outer orbital complex.

### Step 2: Analyze Complex B
- **Given Data**:
- Magnetic moment = 3.47 BM
- EAN = 37

- **Interpretation**:
Using the same formula for magnetic moment:
\[ 3.47 = \sqrt{n(n+2)} \]
Squaring both sides:
\[ 12.04 = n(n+2) \]
Rearranging gives:
\[ n^2 + 2n - 12.04 = 0 \]
Solving this quadratic equation:
\[ n = \frac{-2 \pm \sqrt{4 + 48.16}}{2} = \frac{-2 \pm 8.0}{2} \]
This gives us \( n = 3 \).

- **Conclusion for Complex B**:
Complex B has 3 unpaired electrons, indicating that cobalt is in the +2 oxidation state (3d^7 configuration). The hybridization is also sp³d².

### Step 3: Analyze Complex C
- **Given Data**:
- EAN = 36
- Diamagnetic (magnetic moment = 0)

- **Interpretation**:
Since the complex is diamagnetic, it has no unpaired electrons. The EAN of 36 suggests that cobalt is likely in the +3 oxidation state (3d^6 configuration).

- **Conclusion for Complex C**:
The hybridization must be d²sp³, indicating that the inner d orbitals are being used.

### Final Conclusion
- **Summary of Complexes**:
- Complex A: 4 unpaired electrons, sp³d² hybridization
- Complex B: 3 unpaired electrons, sp³d² hybridization
- Complex C: 0 unpaired electrons, d²sp³ hybridization

### True Statement
The correct statement regarding the observations is that Complex A and B have sp³d² hybridization, while Complex C has d²sp³ hybridization.