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Q5.19:
\([Cr(NH_{3})_{6}]^{3+}\) is paramagnetic while \([Ni(CN)_{4}]^{2-}\) is diamagnetic. Explain why?

Solution :

Step 1: Electronic Configuration and Ligand Field Analysis of [Cr(NH₃)₆]³⁺

First, we determine the oxidation state and electronic configuration of the central metal ion, Chromium ($Cr$). The complex is $[Cr(NH_{3})_{6}]^{3+}$.

  • The oxidation state of $Cr$ is $x + 6(0) = +3 \implies x = +3$.
  • Ground state of $Cr$ ($Z=24$): $[Ar] 3d^5 4s^1$.
  • $Cr^{3+}$ ion: $[Ar] 3d^3$.

The ligand $NH_{3}$ is a strong field ligand. However, in $d^3$ configuration, the three electrons occupy the $t_{2g}$ orbitals singly in accordance with [Hund's Rule of Maximum Multiplicity].

Orbital Set Population
$t_{2g}$ ($d_{xy}, d_{yz}, d_{xz}$) $\uparrow \uparrow \uparrow$
$e_g$ ($d_{z^2}, d_{x^2-y^2}$) (empty)

Because there are 3 unpaired electrons ($n=3$), the magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \, BM$. Since unpaired electrons are present, the complex is paramagnetic.

Step 2: Electronic Configuration and Ligand Field Analysis of [Ni(CN)₄]²⁻

Next, we analyze the complex $[Ni(CN)_{4}]^{2-}$.

  • Oxidation state of $Ni$: $x + 4(-1) = -2 \implies x = +2$.
  • Ground state of $Ni$ ($Z=28$): $[Ar] 3d^8 4s^2$.
  • $Ni^{2+}$ ion: $[Ar] 3d^8$.

$CN^{-}$ is a strong field ligand, which induces pairing of electrons in the $3d$ orbitals. In a square planar geometry [Valence Bond Theory], the $d$ orbitals split such that the $d_{x^2-y^2}$ orbital is high in energy.

The configuration $[Ar] 3d^8$ under the influence of $CN^{-}$ leads to:

  • $d_{xy}, d_{yz}, d_{xz}$ orbitals: Fully occupied (paired).
  • $d_{z^2}$ orbital: Occupied (paired).
  • $d_{x^2-y^2}$ orbital: Empty (reserved for hybridization).
3d orbitals Pairing occurs

Since all 8 electrons are paired ($n=0$), there is no net magnetic moment. Therefore, the complex is diamagnetic.

Step 3: Comparative Summary

The difference arises from the number of unpaired electrons determined by the interaction between the metal's $d$-orbital occupancy and the crystal field splitting energy ($\Delta$) imposed by the ligands.

  • $[Cr(NH_{3})_{6}]^{3+}$: Presence of 3 unpaired electrons in $d$-orbitals causes paramagnetic behavior.
  • $[Ni(CN)_{4}]^{2-}$: Strong field $CN^{-}$ causes pairing of all $d$-electrons, leaving no unpaired electrons, resulting in diamagnetic behavior.

Final Solution: [Cr(NH₃)₆]³⁺ is paramagnetic due to the presence of 3 unpaired electrons in its 3d subshell, while [Ni(CN)₄]²⁻ is diamagnetic because the strong field CN⁻ ligand forces the pairing of all 3d electrons, resulting in zero unpaired electrons.]


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