The spin only magnetic moment of$\left[ Mn \left( H _2 O \right)_6\right]^{2+}$ complexes is ____ BM (Nearest integer) (Given: Atomic no of $Mn$ is 25 )

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

The spin-only magnetic moment of a complex is calculated using the formula:
$\mu = \sqrt{n(n+2)}$ Bohr Magneton (BM), where n is the number of unpaired electrons.

Step 1: Determine the electron configuration of Mn.
Manganese (Mn) has the atomic number 25, so its electron configuration is [Ar] 3d^5 4s^2.

Step 2: Analyze the complex $[Mn(H_2O)_6]^{2+}$.
In this complex, Mn is in the +2 oxidation state. This means it loses two electrons, typically from the 4s sublevel, resulting in the configuration:
[Ar] 3d^5.

Step 3: Calculate the number of unpaired electrons.
The 3d^5 configuration implies five unpaired electrons because each 3d orbital contains one electron.
Thus, n = 5.

Step 4: Calculate the magnetic moment.
Using the formula:
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.
The nearest integer is 6 BM.

Verification: Check if this value fits the given range (6,6). It does, confirming our solution.

The spin-only magnetic moment of $[Mn(H_2O)_6]^{2+}$ is 6 BM.

The spin only magnetic moment of\(\left[ Mn \left( H _2 O \right)_6\right]^{2+}\) complexes is ____ BM (Nearest integer) (Given: Atomic no of $Mn$ is 25 )

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Correct Answer: 6

Solution and Explanation

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