Number of unpaired e- in [Cr(NH3)6] 2+

Question

As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of College Chemistry, 10th Edition.

1. What the question is asking

The student wants to know how many unpaired electrons are present in the complex ion ([\text{Cr(NH}_3)_6]^{2+}) and, consequently, what its spin‑only magnetic moment should be.
The possible answers they have seen are “2 unpaired” (low‑spin) or “4 unpaired” (high‑spin). We have to decide which situation actually occurs for this ion and show the reasoning step‑by‑step.

2. Step‑by‑step solution

Step 1 – Determine the oxidation state of chromium

[ [\text{Cr(NH}_3)_6]^{2+}\;:\;\; \text{NH}_3 \text{ is a neutral ligand} ]

[ \text{overall charge}=+2 = \text{oxidation state of Cr} + 0 ]

[ \boxed{\text{Cr is in the +2 oxidation state}} ]

Step 2 – Count the d‑electrons on Cr(II)

Chromium ground‑state configuration: ([\text{Ar}]\,3d^5 4s^1).
Remove two electrons (the +2 oxidation state) → both are taken from the 4s and one 3d:

[ \text{Cr}^{2+}: 3d^{4} ]

So the metal ion is a (d^{4}) system.

Step 3 – Write the octahedral crystal‑field diagram

For an octahedral field the five d orbitals split into:

lower‑energy (t_{2g}) set (3 orbitals)
higher‑energy (e_g) set (2 orbitals)

          e_g   (higher)   ––  Δ_oct
          |
          |
          |
          t_2g (lower)   ––  0

The magnitude of the splitting is Δ_oct. Whether the electrons pair in the lower set or occupy the higher set depends on the competition between:

Δ_oct (crystal‑field splitting energy)
P (pairing energy, the extra energy required to pair two electrons in the same orbital)

If Δ_oct  <  P → high‑spin (maximum unpaired electrons)
If Δ_oct  >  P → low‑spin (electrons pair as early as possible)

Step 4 – Assess the ligand field strength of NH₃

NH₃ is a borderline (medium‑strong) field ligand. In the spectrochemical series it lies between H₂O and CN⁻. For most 3d metal ions in low oxidation states (e.g., Mn²⁺, Fe²⁺, Cr²⁺) the ligand field generated by NH₃ is not strong enough to make Δ_oct larger than the pairing energy.

Empirically:

Metal ion	Ligand (NH₃)	Observed spin state
Mn²⁺ (d⁵)	NH₃	high‑spin (5 unpaired)
Fe²⁺ (d⁶)	NH₃	high‑spin (4 unpaired)
Cr²⁺ (d⁴)	NH₃	high‑spin (4 unpaired)

Thus for ([\text{Cr(NH}_3)_6]^{2+}) we expect Δ_oct < P, giving the high‑spin arrangement.

Step 5 – Fill the d‑orbitals for a high‑spin (d^{4}) ion

High‑spin filling (Hund’s rule) places one electron in each of the five d orbitals before any pairing:

Orbital set	Occupancy (high‑spin)
(t_{2g})	3 electrons (each singly)
(e_g)	1 electron (singly)

Result: 4 unpaired electrons.

Step 6 – Calculate the spin‑only magnetic moment

The spin‑only magnetic moment (in Bohr magnetons, BM) is given by:

[ \mu_{\text{so}} = \sqrt{n(n+2)}\;\text{BM} ]

where (n) = number of unpaired electrons.

[ n = 4 \;\Longrightarrow\; \mu_{\text{so}} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM} ]

Step 7 – Compare with experimental data (optional check)

Magnetic susceptibility measurements for ([\text{Cr(NH}3)_6]^{2+}) give (\mu{\text{eff}}) ≈ 4.9 BM, confirming the high‑spin, 4‑unpaired‑electron description.

3. Final answer

The complex ion ([\text{Cr(NH}_3)_6]^{2+}) is high‑spin. It contains 4 unpaired electrons, and its spin‑only magnetic moment is ≈ 4.9 BM (μ = √24 BM).

4. Common Mistakes

Mistake	Why it’s wrong	How to avoid it
Assuming NH₃ is always a strong‑field ligand and automatically giving a low‑spin configuration.	NH₃ is only medium‑strong; for 3d metals in low oxidation states (e.g., Cr²⁺) Δ_oct is usually smaller than the pairing energy.	Check the oxidation state and compare typical Δ_oct values (or use experimental magnetic data) before deciding spin state.
Counting the d‑electrons incorrectly (e.g., using Cr⁰ or Cr³⁺).	The overall charge of the complex must be accounted for; ligands like NH₃ are neutral.	Write the oxidation‑state equation: overall charge = oxidation state + Σ(ligand charges).
Using the wrong formula for magnetic moment (e.g., μ = n BM).	The correct spin‑only formula is μ = √[n(n+2)] BM.	Memorise the spin‑only equation; plug in the correct n.
Neglecting possible orbital contribution for ions where it is significant (e.g., Co²⁺, Fe³⁺).	For most high‑spin d⁴–d⁷ octahedral 3d ions the orbital contribution is quenched, but not always.	Recognise that for Cr²⁺ the orbital contribution is negligible; therefore the spin‑only value matches experiment.
Mixing up high‑spin vs low‑spin electron counts (thinking d⁴ high‑spin = 2 unpaired).	High‑spin d⁴ has 4 unpaired; low‑spin d⁴ has 2 unpaired.	Draw the crystal‑field diagram for both possibilities and count electrons carefully.

Original question: Number of unpaired e- in [Cr(NH3)6] 2+ on Chemistry Stack Exchange, licensed CC BY-SA.