Magnetic moment of `(NH_4)_2 [MnBr_4]` is ……. BM

Options

A. 5.91
B. 4.91
C. 3.91
D. 2.46

Explanation

To determine the magnetic moment of the compound \((NH_4)_2[MnBr_4]\), we will follow these steps:

### Step 1: Determine the oxidation state of manganese in \((NH_4)_2[MnBr_4]\)
The ammonium ion \((NH_4)^+\) has a charge of +1. Since there are two ammonium ions, the total positive charge contributed by ammonium is +2. The bromide ion \((Br^-)\) has a charge of -1, and there are four bromide ions, contributing a total charge of -4.

Setting up the equation for the overall charge:
\[ \text{Charge from Mn} + 2 - 4 = 0 \]
Let the oxidation state of Mn be \(x\):
\[ x + 2 - 4 = 0 \implies x - 2 = 0 \implies x = +2 \]
Thus, manganese is in the +2 oxidation state.

### Step 2: Write the electronic configuration of \(Mn^{2+}\)
The atomic number of manganese (Mn) is 25. The electronic configuration of neutral manganese is:
\[ [Ar] 4s^2 3d^5 \]
When manganese loses two electrons to form \(Mn^{2+}\), it loses the two 4s electrons:
\[ Mn^{2+}: [Ar] 3d^5 \]

### Step 3: Determine the number of unpaired electrons
In the \(3d^5\) configuration, all five d electrons are unpaired because manganese is in a high-spin state due to the presence of bromide ions, which are weak field ligands. Therefore, the number of unpaired electrons \(n\) is 5.

### Step 4: Calculate the magnetic moment
The magnetic moment (\(\mu\)) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
Substituting \(n = 5\):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \]

### Step 5: Calculate the numerical value of the magnetic moment
Calculating \(\sqrt{35}\):
\[ \mu \approx 5.91 \, \text{BM} \]

### Final Answer
The magnetic moment of \((NH_4)_2[MnBr_4]\) is approximately **5.91 BM**.

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