The magnetic moment of Mn²⁺ ion is a) 5.92 BM b) 2.80 BM c) 8.95 BM d) 3.90 BM

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Q. The magnetic moment of Mn²⁺ ion is

a) 5.92 BM

b) 2.80 BM

c) 8.95 BM

d) 3.90 BM

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Concept understanding — Magnetic Properties of Transition Metals

Magnetic Properties of Transition Metals

The Intuition: Why Do Some Metals Get Pulled Into a Magnetic Field?

Imagine a tiny bar magnet. If you bring it near a strong magnet, it either gets pulled in or pushed away. Transition metals behave like collections of these tiny magnets — but the "magnet" inside each atom is the electron itself.

An electron is not just a charged particle; it also spins. That spin makes it behave like a microscopic current loop, which generates a magnetic field. In most atoms, electrons pair up with opposite spins, and their magnetic effects cancel out. But in transition metals, the d-orbitals are being filled one electron at a time. When an orbital has an unpaired electron, that electron's magnetic field is not cancelled. The atom as a whole becomes a tiny permanent magnet.

Now bring a bunch of these atomic magnets near an external magnetic field. They will try to align with it, and the material gets pulled into the field. That is paramagnetism. The more unpaired electrons an ion has, the stronger the pull.

Note

This is completely different from ferromagnetism (like iron nails). Ferromagnetism requires the atomic magnets to cooperate and align permanently with each other, even after the external field is removed. Paramagnetism is weaker and only exists while the external field is present.

The Precise Statement

The magnetic moment of a transition metal ion arises almost entirely from the spin of its unpaired electrons. The orbital contribution (the electron's motion around the nucleus) is usually "quenched" — suppressed by the electric field of surrounding atoms in a crystal or solution.

The magnitude of this spin-only magnetic moment is given by:

\mu = \sqrt{n(n+2)} \ \text{Bohr magnetons}

where $n$ is the number of unpaired electrons.

\mu_s = \sqrt{n(n+2)} \ \mu_B

This formula comes from quantum mechanics. Each unpaired electron has a spin quantum number $s = 1/2$ . For $n$ unpaired electrons, the total spin quantum number $S = n/2$ . The magnetic moment in Bohr magnetons is:

\mu = g \sqrt{S(S+1)}

where $g \approx 2$ for a free electron (the gyromagnetic ratio). Substituting $S = n/2$ gives:

\mu = 2 \sqrt{\frac{n}{2}\left(\frac{n}{2}+1\right)} = \sqrt{n(n+2)}

How to Use It: A Quick Table

Number of unpaired electrons ( $n$ )	$\mu$ (Bohr magnetons)
1	$\sqrt{3} \approx 1.73$
2	$\sqrt{8} \approx 2.83$
3	$\sqrt{15} \approx 3.87$
4	$\sqrt{24} \approx 4.90$
5	$\sqrt{35} \approx 5.92$

Tip

Memorise the pattern: the values are roughly $n+1$ for small $n$ , but the exact formula is what you need in exams. For $n=5$ , the moment is nearly 6 — the maximum possible for a first-row transition metal ion.

A Common Mistake to Avoid …

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Step 1. Mn (Z=25) is [Ar]3d⁵4s²; forming Mn²⁺ removes both 4s electrons, giving [Ar]3d⁵ -- an exactly half-filled d sub-shell with n=5 unpaired electrons (one in each of the five d orbitals, by Hund's rule).

Step 2. Apply the spin-only magnetic-moment formula, μ = √[n(n+2)] μ_B, with n=5: μ = √[5×7] = √35 ≈ 5.916 μ_B, rounded to 5.92 BM in the standard reference table. …

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Q.The magnetic moment of Mn²⁺ ion is

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