Question:

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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Remember the formula for spin-only magnetic moment: \(μ_{spin} =\sqrt{n(n + 2)}\) B.M., where n is the number of unpaired electrons. The strength of the ligand determines the pairing of d-electrons.

Updated On: Jan 10, 2025
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Correct Answer: 6

Solution and Explanation

The electronic configuration of Mn is \([ \text{Ar} ] \, 3d^5 \, 4s^2\).
In \([\text{Mn(H}_2\text{O})_6]^{2+}\), Mn is in \(+2\) oxidation state. Water is a weak field ligand.
Electronic configuration of \(\text{Mn}^{2+}\) is \([ \text{Ar} ] \, 3d^5\).
Since \(\text{H}_2\text{O}\) is a weak field ligand, there will be no pairing of electrons in the \(d\) orbitals.
Number of Unpaired Electrons (\(n\))
\[n = 5.\]
Spin-Only Magnetic Moment (\(\mu_\text{spin}\)) 
\[\mu_\text{spin} = \sqrt{n(n+2)} \, \text{B.M.}\]
\[\mu_\text{spin} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.}\]
Nearest Integer:
The nearest integer is:
\[6.\]

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Concepts Used:

D and F Block Elements

The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements. 

They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.

Read More: The d and f block elements