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Intext Questions · 4.8

Q.Calculate the 'spin only' magnetic moment of M2+(aq)M^{2+}(aq) ion (Z = 27).

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For a M2+M^{2+} ion with atomic number 27 (cobalt), the spin-only magnetic moment is calculated from the number of unpaired electrons in the 3d73d^7 configuration. The result is μ=15 BM≈3.87 BM\mu = \sqrt{15} \, \text{BM} \approx 3.87 \, \text{BM}.

The magnetic moment of a transition metal ion in aqueous solution is often dominated by the spin of unpaired electrons, because the orbital angular momentum is largely "quenched" by the surrounding water molecules. This is why we use the spin-only formula:

μ=n(n+2) BM\mu = \sqrt{n(n+2)} \, \text{BM}

where nn is the number of unpaired electrons and BM stands for Bohr magneton.

The key is to first determine the electronic configuration of the ion, then find how many electrons remain unpaired in the dd-subshell.

  1. Identify the element and its ground-state configuration Atomic number Z=27Z = 27 corresponds to cobalt (Co). The neutral atom has the configuration:

1s2 2s2 2p6 3s2 3p6 4s2 3d71s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^7

or in condensed form: [Ar] 4s2 3d7[\text{Ar}] \, 4s^2 \, 3d^7.

  1. Form the M2+M^{2+} ion When forming a +2+2 cation, electrons are removed first from the 4s4s orbital (since it is higher in energy than 3d3d for neutral atoms, though the order flips for ions). So:

M2+:[Ar] 3d7M^{2+}: [\text{Ar}] \, 3d^7

The 4s4s electrons are gone; we are left with seven electrons in the 3d3d subshell.

  1. Determine the number of unpaired electrons in 3d73d^7 For a free ion (or in a weak-field environment like aqueous solution), the dd-orbitals are degenerate to a first approximation, and Hund's rule applies: electrons occupy each orbital singly before pairing begins. The dd-subshell has five orbitals. Filling seven electrons:
    • First five electrons: one in each orbital, all parallel spins (↑ ↑ ↑ ↑ ↑).
    • Next two electrons: each pairs up in an orbital, giving two paired electrons and three unpaired. So the arrangement is:

↑↓↑↓↑↑↑\uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow

That gives three unpaired electrons (n=3n = 3).

Watch out

A common mistake is to forget that the 4s4s electrons are removed first when forming M2+M^{2+}. Some students incorrectly keep the 4s24s^2 and remove from 3d3d, leading to a wrong dd-count. Always remove from the outermost shell (4s4s before 3d3d for cations). …

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