` Mn^(2+)` forms a complex with Br– ion. The magnetic moment of the complex is 5.92 B. M. What could not be the probable formula and geometry of the complex?

Options

A. `[MnBr_6]^(4-)`, octahedral
B. `[MnBr_4]^(2-)` square planner
C. `[MnBr_4]^(2-)` tetrahedral
D. `[MnBr_5]^(3-)` trigonal bipyramidal

Answer

To solve the problem, we need to analyze the magnetic moment of the complex formed by \( \text{Mn}^{2+} \) and \( \text{Br}^- \) ions, and determine the probable formula and geometry of the complex.

### Step-by-Step Solution:

1. **Understanding Magnetic Moment**:
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Given Magnetic Moment**:
We are given that the magnetic moment of the complex is \( 5.92 \, \text{B.M.} \).

3. **Calculating \( n \)**:
We will square the magnetic moment to find \( n \):
\[ (5.92)^2 = n(n + 2) \]
\[ 35 = n^2 + 2n \]
Rearranging gives us:
\[ n^2 + 2n - 35 = 0 \]

4. **Solving the Quadratic Equation**:
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 2, c = -35 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-35)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 140}}{2} \]
\[ n = \frac{-2 \pm \sqrt{144}}{2} \]
\[ n = \frac{-2 \pm 12}{2} \]
This gives us two possible values for \( n \):
\[ n = 5 \quad \text{(taking the positive root)} \]

5. **Determining the Electron Configuration**:
The electron configuration for \( \text{Mn}^{2+} \) (which has lost two electrons) is:
\[ \text{Mn}^{2+}: [\text{Ar}] 3d^5 \]
Since \( n = 5 \), all five electrons in the \( 3d \) subshell are unpaired.

6. **Analyzing Possible Complexes**:
We need to determine the possible formulas and geometries for the complexes formed with \( \text{Br}^- \):
- **For \( \text{MnBr}_6^{4-} \)**:
- \( \text{Mn}^{2+} \) has an oxidation state of +2, and \( \text{Br}^- \) is a weak field ligand. The geometry would be octahedral.
- **For \( \text{MnBr}_4^{2-} \)**:
- Similar analysis shows that it would also likely be octahedral, but with fewer ligands.
- **For \( \text{MnBr}_5^{3-} \)**:
- This would lead to a trigonal bipyramidal geometry, which is consistent with the unpaired electrons.

7. **Identifying the Incorrect Option**:
The geometry and electronic configuration must match the number of unpaired electrons and the nature of the ligands. The geometry of \( \text{MnBr}_4^{2-} \) cannot be square planar because \( \text{Br}^- \) is a weak field ligand and does not cause pairing of electrons. Thus, the geometry cannot be square planar.

### Conclusion:
The probable formula and geometry that could NOT be correct for the complex formed by \( \text{Mn}^{2+} \) and \( \text{Br}^- \) is:
- **Formula**: \( \text{MnBr}_4^{2-} \)
- **Geometry**: Square planar