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

The key idea is that for $Z = 27$, the element is cobalt (Co). The $M^{2+}$ ion means we remove two electrons from the neutral atom.

Step 1: Write the electronic configuration of neutral Co ($Z = 27$):

$[Ar]\, 3d^7 4s^2$.

Step 2: For $Co^{2+}$, remove the two $4s$ electrons first (standard for transition metals):

$[Ar]\, 3d^7$. …

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:

where $n$ 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 $d$-subshell.

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

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

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

The $4s$ electrons are gone; we are left with seven electrons in the $3d$ subshell.

3. Determine the number of unpaired electrons in $3d^7$ For a free ion (or in a weak-field environment like aqueous solution), the $d$-orbitals are degenerate to a first approximation, and Hund's rule applies: electrons occupy each orbital singly before pairing begins. The $d$-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:

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

Method: Spin-Only Magnetic Moment Formula

This uses the spin-only formula (valid for first-row transition metal ions where orbital angular momentum is quenched):

where:

• $\mu$ = magnetic moment in Bohr Magnetons (BM)
• $n$ = number of unpaired electrons

Step-by-Step Solution

Step 1: Identify the element and its electronic configuration

• Atomic number $Z = 27$ → Cobalt (Co)
• Ground state configuration: $\text{Co} : [\text{Ar}]\, 3d^7 4s^2$

Step 2: Write configuration for $M^{2+}$ ion

• Remove two electrons: first from $4s$, then from $3d$ $\text{Co}^{2+} : [\text{Ar}]\, 3d^7$

Step 3: Find number of unpaired electrons using Hund's rule

For $3d^7$:

• Fill each of the five $d$ orbitals singly first (Hund's rule)
• Then pair up

Orbital diagram: …

Here are the common mistakes students make when calculating the spin-only magnetic moment for $M^{2+}(aq)$ (Z = 27), along with how to avoid each.

1. ✗ Mistake: Wrong electronic configuration of the atom

What students do wrong:

They write the configuration for Z = 27 as $1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2$ — but then forget to remove electrons from the correct orbital for the $+2$ ion.

How to avoid:

• For transition metals, electrons are removed first from the 4s orbital, not the 3d.
• For Z = 27 (Cobalt, Co):
  • Co atom: $[Ar] 3d^7 4s^2$
  • Co$^{2+}$: remove two 4s electrons → $[Ar] 3d^7$

Key rule: In ions, the 4s orbital empties before 3d.

2. ✗ Mistake: Counting unpaired electrons incorrectly

What students do wrong:

They fill the 3d orbitals without applying Hund’s rule (maximum multiplicity), leading to wrong number of unpaired electrons.

How to avoid:

• For $3d^7$:
  • Fill each of the five d orbitals singly first (Hund’s rule).
  • Then pair up only after all orbitals have one electron.
  • Correct filling:
    • 5 orbitals get 1 electron each → 5 unpaired
    • Remaining 2 electrons pair in two orbitals → 3 unpaired electrons left.

Result: $n = 3$ unpaired electrons.

3. ✗ Mistake: Using the wrong formula

What students do wrong:

They use $\mu = \sqrt{n(n+2)}$ but plug in the total number of d electrons (7) instead of the number of unpaired electrons (3).

How to avoid:

• The spin-only formula is:

where $n$ = number of unpaired electrons, not total electrons.

• For $n = 3$:

Always double-check: $n$ is the count of unpaired electrons.

4. ✗ Mistake: Forgetting the units

What students do wrong:

They write the answer as just a number (e.g., 3.87) without specifying $\mu_B$.

How to avoid:

• Magnetic moment is expressed in Bohr magnetons ($\mu_B$).
• Always write:

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