A complex is written as `M(en)_(y)Br`. Its 0.05 molar solution shows 2.46 atm osmotic pressure at `27^(@)C`. Assuming 100% ionisation and coordination number of metal (III) is six, complex may be

Options

A. `[M(en)_(2)Br_(2)]Br`
B. `[M(en)_(3)]Br_(3)`
C. `[M(en)_(2)Br_(2)]^(+)`
D. `[M(en)_(3)]^(+)Br_(2)`

Answer

To solve the problem, we need to determine the complex from the given information about its osmotic pressure, molarity, and coordination number. Let's break it down step by step.

### Step 1: Understand the given information
We are given:
- Molarity of the solution (C) = 0.05 M
- Osmotic pressure (π) = 2.46 atm
- Temperature (T) = 27°C = 300 K (after converting to Kelvin)
- Coordination number of the metal (M) = 6
- 100% ionization is assumed.

### Step 2: Use the formula for osmotic pressure
The formula for osmotic pressure is given by:
\[ \pi = iCRT \]
Where:
- \( \pi \) = osmotic pressure
- \( i \) = Van't Hoff factor (number of particles the solute breaks into)
- \( C \) = molarity of the solution
- \( R \) = universal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin

### Step 3: Rearranging the formula to find \( i \)
We can rearrange the formula to solve for \( i \):
\[ i = \frac{\pi}{CRT} \]

### Step 4: Substitute the known values
Now we substitute the known values into the equation:
\[ i = \frac{2.46 \, \text{atm}}{(0.05 \, \text{mol/L})(0.0821 \, \text{L·atm/(K·mol)})(300 \, \text{K})} \]

### Step 5: Calculate \( i \)
Calculating the denominator:
\[ C \cdot R \cdot T = (0.05)(0.0821)(300) = 1.2315 \, \text{atm} \]
Now substituting back:
\[ i = \frac{2.46}{1.2315} \approx 2.00 \]

### Step 6: Interpret the Van't Hoff factor
The Van't Hoff factor \( i \approx 2 \) indicates that the complex dissociates into 2 particles in solution.

### Step 7: Analyze the options
Given the options:
1. \( M(en)_2Br_2 \)
2. \( M(en)_3Br_3 \)
3. \( M(en)_2Br_2^{2+} \)
4. \( M(en)_3^{3+}Br_2 \)

### Step 8: Determine the correct complex
- For option 1: \( M(en)_2Br_2 \) dissociates into \( M(en)_2^{+} + 2Br^{-} \) (3 ions total)
- For option 2: \( M(en)_3Br_3 \) dissociates into \( M(en)_3^{3+} + 3Br^{-} \) (4 ions total)
- For option 3: \( M(en)_2Br_2^{2+} \) dissociates into \( M(en)_2^{2+} + 2Br^{-} \) (3 ions total)
- For option 4: \( M(en)_3^{3+}Br_2 \) dissociates into \( M(en)_3^{3+} + 2Br^{-} \) (3 ions total)

Only option 1 \( M(en)_2Br_2 \) gives a total of 2 ions, which matches our calculated \( i \).

### Conclusion
The correct complex is:
\[ \text{Answer: } M(en)_2Br_2 \]