Q.Predict the number of unpaired electrons in the square planar $[Pt(CN)_4]^{2-}$ ion.

The key idea is that the magnetic moment depends on the number of unpaired electrons, which in turn depends on the metal's oxidation state, geometry, and ligand field splitting.

Step 1: Determine the oxidation state and electron configuration of Pt.

Pt is in group 10. In $[Pt(CN)_4]^{2-}$, each CN⁻ is −1, so Pt has oxidation state $+2$.

Pt(0): $[Xe]\,4f^{14}5d^9 6s^1$

Pt(II): remove two electrons (from 6s first), giving $[Xe]\,4f^{14}5d^8$.

Step 2: Consider the geometry and ligand field. …

Why This Approach Works

Magnetic moment tells us about unpaired electrons. To predict them, we need two things: the number of d-electrons on the central metal ion, and how those electrons arrange themselves under the influence of the ligands.

Square planar geometry is a special case. It arises most commonly for $d^8$ metal ions with strong-field ligands — think Ni²⁺, Pd²⁺, Pt²⁺. The crystal field splitting in a square planar complex is essentially an extreme version of octahedral splitting where two trans ligands are removed, causing one set of d-orbitals to rise dramatically in energy. The result is a large energy gap between the lower and upper d-orbitals, forcing electrons to pair up.

CN⁻ is a strong-field ligand (high up in the spectrochemical series). So we expect maximum pairing.

Let’s walk through it step by step.

1. Find the oxidation state of platinum. The complex ion is $[Pt(CN)_4]^{2-}$. Each CN⁻ ligand carries a –1 charge. Let the oxidation state of Pt be $x$.

So platinum is in the +2 oxidation state.

2. Determine the d-electron count for Pt(II).

   Platinum (Pt) has atomic number 78. Its ground-state electron configuration is $[Xe]\,4f^{14}\,5d^9\,6s^1$.

   When Pt loses two electrons to become Pt²⁺, it loses the 6s electron first, then one 5d electron.

   So Pt²⁺ has the configuration $[Xe]\,4f^{14}\,5d^8$.

   That’s a $d^8$ system.

3. Recall the crystal field splitting pattern for square planar geometry.

   In a square planar complex, the d-orbital energies (from lowest to highest) are:

   • $d_{xz}, d_{yz}$ (degenerate, lowest)
   • $d_{z^2}$ (slightly higher)
   • $d_{xy}$ (higher still)
   • $d_{x^2-y^2}$ (highest, by a large margin)

   The energy gap between the $d_{xy}$ and $d_{x^2-y^2}$ orbitals is very large — comparable to or larger than the pairing energy for a $d^8$ ion with strong-field ligands.

   For square planar $d^8$ with strong-field ligands, the splitting is so large that all eight electrons occupy the four lower orbitals, leaving the $d_{x^2-y^2}$ orbital empty.

4. Fill the electrons according to Hund’s rule and the Aufbau principle. …

Method: Crystal Field Theory (CFT) for Square Planar Geometry

Why this method?

Square planar complexes arise from $d^8$ configurations (like $Pt^{2+}$) when the ligand field is very strong. CFT explains how $d$-orbital energies split, which directly tells us the electron arrangement and number of unpaired electrons.

Step 1: Determine the oxidation state and $d$-electron count

• Platinum (Pt) is in Group 10, atomic number 78.
• Each cyanide ligand ($CN^-$) has a charge of $-1$. Four ligands give $-4$.
• Complex charge is $2-$.
• Let oxidation state of Pt be $x$:

• $Pt^{2+}$: neutral Pt has $[Xe]\,4f^{14}5d^9 6s^1$. Removing two electrons (from 6s first) gives $5d^8$.
• $d^8$ configuration.

Step 2: Recall the square planar $d$-orbital splitting

For a strong-field square planar complex, the energy order (lowest to highest) is:

• The gap between $d_{xy}$ and $d_{x^2 - y^2}$ is very large (strong field).
• This forces pairing of electrons in the lower orbitals.

Step 3: Fill the $d$-orbitals (Aufbau + Hund's rule)

For $d^8$ in strong-field square planar: …

Here are the most common mistakes students make when calculating magnetic moments and unpaired electrons for square planar complexes like $[Pt(CN)_4]^{2-}$, along with how to avoid each.

1. Forgetting to Check the Oxidation State of the Metal

The Mistake:

Students jump straight to the geometry without first finding the oxidation number of platinum. This leads to using the wrong $d^n$ configuration.

How to Avoid:

Always start with the charge balance.

• Ligand charge: Each $CN^-$ is $-1$, so total from 4 ligands = $-4$.
• Complex charge: $-2$.
• Let oxidation state of Pt be $x$:

So Pt is in the +2 oxidation state.

2. Using the Wrong Electronic Configuration for the Metal Ion

The Mistake:

Assuming Pt has the same configuration as Ni or Pd, or forgetting that Pt is a 5d series element.

How to Avoid:

• Pt atomic number = 78.
• Neutral Pt: $[Xe]\,4f^{14}\,5d^9\,6s^1$ (or $5d^8\,6s^2$ — but the $5d^9\,6s^1$ is more stable).
• For $Pt^{2+}$, remove two electrons — remove the $6s$ electrons first:

So the $d$-electron count is $d^8$.

3. Assuming Square Planar Always Means Low-Spin

The Mistake:

Students think all square planar complexes are diamagnetic (no unpaired electrons). This is not always true — it depends on the metal and ligand field strength.

How to Avoid:

• For $d^8$ ions, square planar geometry is strongly favoured by strong-field ligands like $CN^-$.
• Strong field → large splitting → electrons pair up in the lower orbitals.
• For $Pt^{2+}$ (a 5d metal), the splitting is very large due to relativistic effects.
• Result: All 8 electrons are paired → zero unpaired electrons.

4. Confusing Square Planar with Tetrahedral Geometry

The Mistake:

Some students treat $[Pt(CN)_4]^{2-}$ as tetrahedral (common for $d^8$ with weak-field ligands like $Cl^-$), leading to a different $d$-orbital splitting and wrong unpaired electron count.

How to Avoid:

• $Pt^{2+}$ with strong-field ligands like $CN^-$ always forms square planar complexes.
• Tetrahedral $d^8$ would have 2 unpaired electrons; square planar $d^8$ has 0 unpaired electrons.
• Memorise: $Ni^{2+}$ can be tetrahedral or square planar; $Pt^{2+}$ and $Pd^{2+}$ are always square planar.

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