Q.Write down the number of 3d electrons in each of the following ions: $Ti^{2+}$, $V^{2+}$, $Cr^{3+}$, $Mn^{2+}$, $Fe^{2+}$, $Fe^{3+}$, $Co^{2+}$, $Ni^{2+}$ and $Cu^{2+}$. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Concept: Magnetic Moment Calculation — the number of unpaired electrons in 3d orbitals determines the magnetic moment, and in an octahedral field the five d‑orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets.

Step 1: Determine the 3d electron count for each ion.

Remove electrons from the 4s orbital first, then from 3d.

• $Ti^{2+}$: $[Ar]3d^2$
• $V^{2+}$: $[Ar]3d^3$
• $Cr^{3+}$: $[Ar]3d^3$
• $Mn^{2+}$: $[Ar]3d^5$
• $Fe^{2+}$: $[Ar]3d^6$
• $Fe^{3+}$: $[Ar]3d^5$
• $Co^{2+}$: $[Ar]3d^7$
• $Ni^{2+}$: $[Ar]3d^8$
• $Cu^{2+}$: $[Ar]3d^9$

Step 2: For hydrated ions (weak field, octahedral), fill $t_{2g}$ first with one electron each, then pair, then fill $e_g$.

Hund’s rule applies — each orbital gets one electron before pairing.

Concept and Intuition

To find the 3d electron count for a transition metal ion, you must remember the Aufbau principle for neutral atoms: for elements in the 3d series, the 4s orbital fills before 3d (e.g., $[Ar]4s^2 3d^x$). When forming a positive ion, electrons are removed first from the 4s orbital, not the 3d — this is a common mistake. So for $Ti^{2+}$, the neutral Ti has $[Ar]4s^2 3d^2$; removing two electrons takes both 4s electrons, leaving $3d^2$.

Once we know the 3d count, we consider the hydrated ion in an octahedral crystal field. Water is a weak-field ligand, so the splitting energy $\Delta_o$ is small. This means electrons fill all five orbitals singly before pairing (Hund’s rule) — the high-spin configuration. The five d orbitals split into a lower-energy triplet ($d_{xy}, d_{xz}, d_{yz}$ — called $t_{2g}$) and a higher-energy doublet ($d_{z^2}, d_{x^2-y^2}$ — called $e_g$). For weak fields, electrons occupy $t_{2g}$ first, then $e_g$, all with parallel spins as far as possible.

Let’s work through each ion step by step.

1. $Ti^{2+}$ (Titanium, Z = 22)

Neutral Ti: $[Ar]4s^2 3d^2$. Remove 2 electrons → both from 4s.

3d electrons = 2.

In octahedral field: two electrons go into $t_{2g}$ (lower energy), both unpaired.

Configuration: $t_{2g}^2 e_g^0$ (2 unpaired electrons).

2. $V^{2+}$ (Vanadium, Z = 23)

Neutral V: $[Ar]4s^2 3d^3$. Remove 2 electrons → both from 4s.

3d electrons = 3.

Three electrons: all occupy $t_{2g}$ singly (Hund’s rule).

Configuration: $t_{2g}^3 e_g^0$ (3 unpaired).

3. $Cr^{3+}$ (Chromium, Z = 24)

Neutral Cr: $[Ar]4s^1 3d^5$ (exception: half-filled d gives stability). Remove 3 electrons → first the 4s electron, then two from 3d.

3d electrons = 3.

Same as V²⁺: $t_{2g}^3 e_g^0$ (3 unpaired).

4. $Mn^{2+}$ (Manganese, Z = 25)

Neutral Mn: $[Ar]4s^2 3d^5$. Remove 2 electrons → both from 4s.

3d electrons = 5.

Five electrons: fill all five orbitals singly — $t_{2g}^3 e_g^2$ (5 unpaired). This is a half-filled d shell, extra stable.

5. $Fe^{2+}$ (Iron, Z = 26)

Neutral Fe: $[Ar]4s^2 3d^6$. Remove 2 electrons → both from 4s.

3d electrons = 6.

Six electrons: first five singly occupy all orbitals, the sixth pairs in a $t_{2g}$ orbital.

Configuration: $t_{2g}^4 e_g^2$ (4 unpaired electrons).

6. $Fe^{3+}$ (Iron, Z = 26)

Neutral Fe: $[Ar]4s^2 3d^6$. Remove 3 electrons → both 4s and one 3d.

3d electrons = 5.

Same as Mn²⁺: $t_{2g}^3 e_g^2$ (5 unpaired).

7. $Co^{2+}$ (Cobalt, Z = 27)

Neutral Co: $[Ar]4s^2 3d^7$. Remove 2 electrons → both from 4s.

3d electrons = 7.

Seven electrons: fill $t_{2g}$ with three, then $e_g$ with two (all singly), then the remaining two pair in $t_{2g}$.

Configuration: $t_{2g}^5 e_g^2$ (3 unpaired electrons).

8. $Ni^{2+}$ (Nickel, Z = 28)

Neutral Ni: $[Ar]4s^2 3d^8$. Remove 2 electrons → both from 4s. …

Method: Crystal Field Theory + Spin-Only Magnetic Moment Calculation

Method Name: Spin-Only Magnetic Moment using Crystal Field Splitting (for octahedral $d^n$ ions)

Step 1: Determine the number of 3d electrons for each ion

For transition metal ions, remove electrons from the 4s orbital first, then from 3d.

Key exam point: For Cr and Cu, the neutral atom has a half-filled or fully-filled d-subshell (exception to Aufbau), but ions follow normal removal order.

Step 2: Occupancy of five 3d orbitals in octahedral field

In an octahedral crystal field, the five d-orbitals split into:

• $t_{2g}$ (lower energy): $d_{xy}$, $d_{yz}$, $d_{zx}$
• $e_g$ (higher energy): $d_{x^2-y^2}$, $d_{z^2}$

Hund's rule applies: electrons fill degenerate orbitals singly before pairing.

Note: $H_2O$ is a weak field ligand, so every hydrated ion here is high-spin — each keeps the maximum number of unpaired electrons Hund's rule allows for its $d$-electron count. (For $d^8$ ($Ni^{2+}$) and $d^9$ ($Cu^{2+}$) the octahedral occupancy is the same whatever the field strength.)

Step 3: Calculate spin-only magnetic moment …

Here are the most common mistakes students make when calculating magnetic moments and determining 3d orbital occupancy for hydrated transition metal ions, along with how to avoid each.

1. Mistake: Forgetting to Account for the Charge When Writing Electronic Configuration

The Error: Students often write the configuration for the neutral atom (e.g., Fe: $[Ar] 3d^6 4s^2$) and then directly use the same $d^n$ count for the ion without removing electrons from the correct subshell.

Example of Mistake: For $Fe^{2+}$, writing $3d^6$ but thinking it comes from simply removing two electrons from the 4s orbital after the 3d is filled — which is wrong because in ions, the 4s empties first.

How to Avoid:

• Rule: For transition metal ions, remove electrons from the 4s orbital before the 3d orbital.
• Correct method:
  • Write neutral atom: Fe = $[Ar] 3d^6 4s^2$
  • Remove 2 electrons: first from 4s → $[Ar] 3d^6$
  • So $Fe^{2+}$ has 6 3d electrons.

Quick check for all ions asked:

2. Mistake: Ignoring Hund’s Rule When Filling the Five 3d Orbitals

The Error: Students fill orbitals in pairs before all five are singly occupied, especially for $d^4$, $d^5$, $d^6$, and $d^7$ configurations.

Example of Mistake: For $Mn^{2+}$ ($d^5$), writing $\uparrow\downarrow$ in one orbital and three singles — instead of all five orbitals singly occupied.

How to Avoid:

• Hund’s Rule: Electrons occupy degenerate orbitals singly first, with parallel spins, before pairing.
• For free ions (no ligand field): Fill all five orbitals with one electron each before pairing.
• For hydrated ions (octahedral field): The same rule applies for high-spin complexes (which is the case for all these hydrated ions because water is a weak field ligand).

Correct occupancy for each (high-spin octahedral):

3. Mistake: Confusing High-Spin vs Low-Spin for Hydrated Ions

The Error: Students assume all octahedral complexes are low-spin, or they apply strong-field rules (like for $CN^-$) to water complexes.

Example of Mistake: For $Fe^{2+}$ ($d^6$) in water, writing $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$ _ _ (low-spin, 0 unpaired electrons) instead of the correct high-spin arrangement with 4 unpaired electrons.

How to Avoid:

• Remember: Water ($H_2O$) is a weak field ligand — it causes a small crystal field splitting ($\Delta_o$).
• Weak field → high-spin (electrons prefer to occupy all orbitals singly before pairing).
• Strong field ligands (like $CN^-$, $CO$) cause low-spin.
• For all ions listed, hydrated = high-spin — including $Co^{2+}$, which is high-spin with water like the rest.

Unpaired electron count for each (high-spin):

4. Mistake: Using the Wrong Formula for Magnetic Moment

The Error: Students use $\mu = \sqrt{n(n+2)}$ but forget that $n$ = number of unpaired electrons, not total d electrons.

Example of Mistake: For $Fe^{2+}$ ($d^6$), using $n=6$ → $\mu = \sqrt{6 \times 8} = \sqrt{48} \approx 6.93$ BM (wrong). Correct: $n=4$ → $\mu = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM.

How to Avoid:

• Formula: $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons.
• Always count unpaired electrons from the orbital diagram first.
• Memorize common values:
  • $n=1$ → $\mu \approx 1.73$ BM
  • $n=2$ → $\mu \approx 2.83$ BM
  • $n=3$ → $\mu \approx 3.87$ BM
  • $n=4$ → $\mu \approx 4.90$ BM
  • $n=5$ → $\mu \approx 5.92$ BM

5. Mistake: Thinking $Cr^{3+}$ and $V^{2+}$ (Both $d^3$) Have Different Magnetic Moments — They Are the Same …