Q.Explain on the basis of valence bond theory that $[Ni(CN)_4]^{2-}$ ion with square planar structure is diamagnetic and the $[NiCl_4]^{2-}$ ion with tetrahedral geometry is paramagnetic.

The key idea is that the magnetic property of a complex depends on the number of unpaired electrons, which is determined by the hybridisation and geometry of the central metal ion under Valence Bond Theory.

Step 1: Electronic configuration of Ni²⁺

Ni (Z = 28): $[Ar] 3d^8 4s^2$

Ni²⁺: $[Ar] 3d^8$ — three 3d orbitals are doubly occupied and two are singly occupied (two unpaired electrons).

Step 2: $[Ni(CN)_4]^{2-}$ — square planar, diamagnetic

CN⁻ is a strong field ligand. It forces pairing of the two unpaired 3d electrons, leaving one 3d orbital empty. The hybridisation is $dsp^2$ (one 3d, one 4s, two 4p orbitals). All eight electrons are paired → no unpaired electrons → diamagnetic.

Step 3: $[NiCl_4]^{2-}$ — tetrahedral, paramagnetic

Cl⁻ is a weak field ligand. It does not cause pairing. The hybridisation is $sp^3$ (one 4s, three 4p orbitals), using the original 3d⁸ configuration with two unpaired electrons. Hence, the complex is paramagnetic.

Why This Happens: The Concept

Valence Bond Theory (VBT) explains bonding in terms of hybridization of atomic orbitals. But to understand magnetism and geometry, we need to see how the ligand field affects the $d$-orbital energies.

For a $Ni^{2+}$ ion, the electronic configuration is $[Ar]\,3d^8$. In a free ion, all five $d$-orbitals are degenerate (same energy). When ligands approach, they split these orbitals into different energy levels depending on the geometry.

The nature of the ligand (strong field vs weak field) decides whether electrons pair up or remain unpaired. This pairing directly determines:

• The hybridization scheme (and thus geometry)
• The magnetic property (diamagnetic = all paired, paramagnetic = unpaired electrons)

Step-by-Step Analysis

1. Identify the central metal ion and its $d$-electron count

Nickel in both complexes is in the +2 oxidation state.

$Ni$ atomic number = 28.

$Ni^{2+}$: loses two $4s$ electrons → configuration: $3d^8$.

So we have 8 electrons in the $3d$ orbitals.

2. Consider the ligand strength

• $CN^-$ is a strong field ligand (high up in the spectrochemical series). It causes a large crystal field splitting ($\Delta$).
• $Cl^-$ is a weak field ligand (low in the spectrochemical series). It causes a small crystal field splitting ($\Delta$).

This difference is the entire reason for the different outcomes.

3. Case 1: $[Ni(CN)_4]^{2-}$ — Strong field, square planar

Because $CN^-$ is a strong field ligand, the splitting between the $d$-orbitals is large. In a square planar geometry, the $d$-orbital splitting pattern (from highest to lowest energy) is approximately:

The energy gap is so large that it is energetically favourable for electrons to pair up in the lower orbitals rather than occupy the high-energy $d_{x^2-y^2}$ orbital.

So the 8 $d$-electrons fill as:

• $d_{xz}, d_{yz}$: 2 electrons each (paired)
• $d_{z^2}$: 2 electrons (paired)
• $d_{xy}$: 2 electrons (paired)
• $d_{x^2-y^2}$: empty

Now, for bonding: the empty $d_{x^2-y^2}$ orbital, along with one $4s$ and two $4p$ orbitals, undergoes $dsp^2$ hybridization (one $d$, one $s$, two $p$). This gives a square planar geometry.

4. Case 2: $[NiCl_4]^{2-}$ — Weak field, tetrahedral

$Cl^-$ is a weak field ligand. The splitting $\Delta$ is small. In a tetrahedral geometry, the $d$-orbital splitting is inverted compared to octahedral:

• Lower energy set: $e$ ($d_{x^2-y^2}, d_{z^2}$)
• Higher energy set: $t_2$ ($d_{xy}, d_{yz}, d_{zx}$)

The splitting $\Delta_t$ is much smaller than in octahedral complexes (roughly $\frac{4}{9}$ of $\Delta_o$). So the energy cost of pairing electrons is greater than the energy gained by occupying the lower $e$ set.

Thus, the 8 $d$-electrons fill from the bottom up:

• $e$ set (lower): 4 electrons — both orbitals doubly occupied
• $t_2$ set (higher): 4 electrons — the first three occupy the three orbitals singly (Hund's rule), and the fourth pairs up in one of them

This gives two unpaired electrons — the complex is paramagnetic.

For bonding: since the $d$-orbitals are all occupied (or partially occupied), the metal uses $sp^3$ hybridization (one $s$, three $p$ orbitals) — no $d$-orbital is empty for $dsp^2$. This yields a tetrahedral geometry.

Summary Table

Method: Valence Bond Theory (VBT) Analysis of Geometry and Magnetism

Concept: Valence bond theory links ligand field strength → electron pairing → hybridisation → geometry and magnetic behaviour.

Step 1: Determine the oxidation state and electron configuration of Ni

• For $[Ni(CN)_4]^{2-}$:

  Let Ni oxidation state be $x$.

  $x + 4(-1) = -2 \Rightarrow x = +2$

  Ni in +2 state: Atomic number 28, configuration $[Ar]\,3d^8 4s^2$

  Remove 2 electrons → $3d^8$

• For $[NiCl_4]^{2-}$:

  Same calculation → Ni is also in $+2$ state → $3d^8$

Step 2: Identify ligand field strength and decide hybridization

• CN⁻ is a strong field ligand → causes pairing of electrons

  • In $3d^8$, one $3d$ orbital is emptied by pairing
  • Hybridization: $dsp^2$ (one $d$, one $s$, two $p$ orbitals)
  • Geometry: Square planar

• Cl⁻ is a weak field ligand → no pairing

  • All $3d$ orbitals remain singly occupied as far as possible
  • Hybridization: $sp^3$ (one $s$, three $p$ orbitals)
  • Geometry: Tetrahedral

Step 3: Count unpaired electrons and determine magnetism

Step 4: Final explanation

• In $[Ni(CN)_4]^{2-}$, strong CN⁻ forces pairing of $3d$ electrons → all electrons paired → diamagnetic (repelled by magnetic field)
• In $[NiCl_4]^{2-}$, weak Cl⁻ cannot cause pairing → two unpaired electrons remain → paramagnetic (attracted by magnetic field)

Key takeaway: The same metal ion ($Ni^{2+}$) with $3d^8$ configuration gives different geometries and magnetic properties depending on ligand strength — a direct consequence of valence bond theory.

Here are the common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Forgetting to check the oxidation state of the central metal ion first.

• The Mistake: Students jump straight to the geometry or the ligand without first calculating the oxidation state of Nickel (Ni). This leads to the wrong $d$-electron count ($d^n$ configuration).
• Why it’s wrong: The number of $d$-electrons determines how the electrons will fill the orbitals, which directly controls the magnetic property (diamagnetic vs. paramagnetic).
• How to Avoid:
  1. Always start with the charge balance.
  2. For $[Ni(CN)_4]^{2-}$:
     • Let the oxidation state of Ni be $x$.
     • Charge of CN⁻ is $-1$ each. Total ligand charge = $4 \times (-1) = -4$.
     • Overall complex charge = $-2$.
     • Equation: $x + (-4) = -2 \implies x = +2$.
     • Result: Ni is in the $+2$ oxidation state.
  3. For $[NiCl_4]^{2-}$:
     • Charge of Cl⁻ is $-1$ each. Total ligand charge = $4 \times (-1) = -4$.
     • Overall complex charge = $-2$.
     • Equation: $x + (-4) = -2 \implies x = +2$.
     • Result: Ni is also in the $+2$ oxidation state.
  4. Key fact: Ni in the ground state is $[Ar] 3d^8 4s^2$. In the $+2$ state, it loses the two $4s$ electrons, giving a $3d^8$ configuration.

Mistake 2: Assuming the same geometry leads to the same magnetic property.

• The Mistake: Students think that because both complexes have the same metal ion (Ni²⁺) and the same coordination number (4), they will have the same magnetic behavior.
• Why it’s wrong: The geometry (square planar vs. tetrahedral) and the strength of the ligand (strong field CN⁻ vs. weak field Cl⁻) completely change how the $d$-orbitals split and how electrons fill them.
• How to Avoid:
  • Remember the rule: Strong field ligands (like CN⁻, CO, NH₃) cause pairing of electrons. Weak field ligands (like Cl⁻, Br⁻, H₂O) cause no pairing (Hund's rule is followed).
  • Visualize the splitting:
    • Tetrahedral ($T_d$): Splitting is small. Electrons fill all orbitals singly first (Hund's rule).
    • Square planar ($D_{4h}$): Splitting is large. Electrons pair up in the lower energy orbitals.

Mistake 3: Drawing the wrong orbital filling diagram for square planar geometry.

• The Mistake: Students use the tetrahedral or octahedral splitting diagram for the square planar complex.
• Why it’s wrong: Square planar geometry has a very specific $d$-orbital splitting pattern. The energy order is: $d_{x^2-y^2} \gg d_{xy} > d_{z^2} > d_{xz} = d_{yz}$.
• How to Avoid:
  • Draw the correct diagram:
    • For $[Ni(CN)_4]^{2-}$ (Square planar, strong field):
      • The $d_{x^2-y^2}$ orbital is very high in energy (empty).
      • The $d_{xy}$ orbital is next highest.
      • The $d_{z^2}$, $d_{xz}$, and $d_{yz}$ are lower.
    • Filling for $d^8$: All 8 electrons pair up in the four lower orbitals ($d_{xy}$, $d_{z^2}$, $d_{xz}$, $d_{yz}$). The $d_{x^2-y^2}$ is empty.
    • Result: No unpaired electrons → Diamagnetic.

Mistake 4: Drawing the wrong orbital filling diagram for tetrahedral geometry.

• The Mistake: Students forget that in tetrahedral geometry, the $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals are higher in energy than the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals.
• Why it’s wrong: The splitting is inverted compared to octahedral.
• How to Avoid:
  • Draw the correct diagram:
    • For $[NiCl_4]^{2-}$ (Tetrahedral, weak field):
      • The $d_{xy}$, $d_{xz}$, $d_{yz}$ (the $t_2$ set) are higher.
      • The $d_{x^2-y^2}$, $d_{z^2}$ (the $e$ set) are lower.
    • Filling for $d^8$: First, fill the lower $e$ set with 4 electrons (paired). Then, place the remaining 4 electrons in the higher $t_2$ set. According to Hund's rule, they will occupy all three orbitals singly before pairing.
    • Result: Two unpaired electrons (in the $t_2$ set) → Paramagnetic.

Mistake 5: Confusing the terms "diamagnetic" and "paramagnetic".

• The Mistake: Students write the correct electron configuration but then state the wrong magnetic property.
• Why it’s wrong: It's a direct loss of marks for a simple definition.
• How to Avoid:
  • Memorize:
    • Diamagnetic: All electrons are paired. The substance is repelled by a magnetic field.
    • Paramagnetic: Has one or more unpaired electrons. The substance is attracted by a magnetic field.
  • Check your final diagram: Count the unpaired electrons. If the count is zero, it's diamagnetic. If non-zero, it's paramagnetic.

Summary Table to Avoid Mistakes