Barium ions, `CN^(-) and Co^(2+)` form an ionic complex.If this complex is 75% ionised in aqueous solution with Vant Hoff factor (i) equal to four and paramagnetic moment is found to be 1.73 BM (due to spin only) then the hybridisation state of Co (II) in the complex will be :

Options

A. `sp^3d`
B. `d^2sp^3`
C. `sp^3d^2`
D. `dsp^3`

Explanation

To solve the problem step by step, we will analyze the information provided and derive the required hybridization state of Co(II) in the complex.

### Step 1: Understand the Complex Formation
We have barium ions (Ba²⁺) and cyanide ions (CN⁻) forming an ionic complex with cobalt (Co²⁺). The general formula for the complex can be represented as:
\[ \text{Ba}^{2+} + \text{Co}^{2+} + \text{CN}^- \rightarrow \text{Ba}_x \text{Co}(\text{CN})_y \]

### Step 2: Determine the Degree of Ionization
The problem states that the complex is 75% ionized in aqueous solution. This means that the degree of ionization (α) is 0.75.

### Step 3: Van't Hoff Factor (i)
The Van't Hoff factor (i) is given as 4. The Van't Hoff factor can be calculated using the formula:
\[ i = 1 - \alpha + (x - 2)\alpha + 2\alpha \]
Where:
- \( x \) is the coordination number of the cobalt ion.

### Step 4: Set Up the Equation
From the Van't Hoff factor equation, substituting \( \alpha = 0.75 \):
\[ 4 = 1 - 0.75 + (x - 2)(0.75) + 2(0.75) \]
This simplifies to:
\[ 4 = 1 - 0.75 + 0.75x - 1.5 + 1.5 \]
\[ 4 = 0.75x + 1.25 \]

### Step 5: Solve for x
Now, isolate \( x \):
\[ 4 - 1.25 = 0.75x \]
\[ 2.75 = 0.75x \]
\[ x = \frac{2.75}{0.75} = \frac{11}{3} \approx 3.67 \]

Since \( x \) must be a whole number, we round it to the nearest whole number, which is 5. Thus, the coordination number \( x = 5 \).

### Step 6: Write the Complex Formula
Now we can write the complex as:
\[ \text{Ba}_3\text{Co}(\text{CN})_5 \]

### Step 7: Determine the Magnetic Moment
The magnetic moment is given as 1.73 BM. The formula for the magnetic moment (μ) is:
\[ \mu = \sqrt{n(n + 2)} \]
Where \( n \) is the number of unpaired electrons. Setting \( \mu = 1.73 \):
\[ 1.73 = \sqrt{n(n + 2)} \]
Squaring both sides:
\[ 1.73^2 = n(n + 2) \]
\[ 2.9929 = n(n + 2) \]

### Step 8: Solve for n
This is a quadratic equation:
\[ n^2 + 2n - 2.9929 = 0 \]
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = 2, c = -2.9929 \):
\[ n = \frac{-2 \pm \sqrt{4 + 11.9716}}{2} \]
\[ n = \frac{-2 \pm \sqrt{15.9716}}{2} \]
\[ n \approx 1 \] (since we are looking for the number of unpaired electrons)

### Step 9: Determine the Hybridization
For Co²⁺, the electron configuration is:
- Co: [Ar] 4s² 3d⁷
- Co²⁺: [Ar] 3d⁷ (removing 2 electrons from 4s)

Since there is 1 unpaired electron and the coordination number is 5, the hybridization state of Co(II) in this complex is:
- **dsp³ hybridization** (as it is an inner orbital complex).

### Final Answer
The hybridization state of Co(II) in the complex is **dsp³**.

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