Question:
Consider the following complex ions:
$[\text{Ni(CN)}_4]^{2-}$, $[\text{Fe(CN)}_6]^{3-}$, $[\text{Cu(H}_2\text{O)}_6]^{2+}$, and $[\text{Co(CN)}_6]^{3-}$.
The complex ions that are expected to show diamagnetic behavior at room temperature are:
Consider the following complex ions:
$[\text{Ni(CN)}_4]^{2-}$, $[\text{Fe(CN)}_6]^{3-}$, $[\text{Cu(H}_2\text{O)}_6]^{2+}$, and $[\text{Co(CN)}_6]^{3-}$.
The complex ions that are expected to show diamagnetic behavior at room temperature are:
$[\text{Ni(CN)}_4]^{2-}$, $[\text{Fe(CN)}_6]^{3-}$, $[\text{Cu(H}_2\text{O)}_6]^{2+}$, and $[\text{Co(CN)}_6]^{3-}$.
The complex ions that are expected to show diamagnetic behavior at room temperature are:
Show Hint
$\text{d}^8$ systems with strong-field ligands almost always form square-planar, diamagnetic complexes, while $\text{d}^6$ systems with strong-field ligands form low-spin, diamagnetic octahedral complexes.
Updated On: Jun 16, 2026
- $[\text{Ni(CN)}_4]^{2-}$ and $[\text{Co(CN)}_6]^{3-}$
- $[\text{Ni(CN)}_4]^{2-}$ and $[\text{Cu(H}_2\text{O)}_6]^{2+}$
- $[\text{Co(CN)}_6]^{3-}$ and $[\text{Fe(CN)}_6]^{3-}$
- $[\text{Cu(H}_2\text{O)}_6]^{2+}$ and $[\text{Co(CN)}_6]^{3-}$
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to determine which of the given transition metal coordination complex ions exhibit diamagnetic behavior (no unpaired electrons) at room temperature.
Step 2: Key Formula or Approach:
We must determine the oxidation state, coordination geometry, and d-electron configuration of the central metal in each complex, and then apply Crystal Field Theory (CFT) to evaluate electron pairing.
Step 3: Detailed Explanation:
• Let us evaluate each complex systematically:
• 1. $[\text{Ni(CN)}_4]^{2-}$:
itemize
• Nickel is in the $+2$ oxidation state ($\text{Ni}^{2+}$, a $\text{d}^8$ system).
• $\text{CN}^-$ is a very strong-field ligand, causing square planar coordination geometry.
• In a square planar crystal field, the $\text{d}_x^2\text{-y}^2$ orbital is pushed very high in energy, while the other four d-orbitals are lower.
• The $8$ electrons completely fill the four lower energy d-orbitals ($\text{d}_{xz}$, $\text{d}_{yz}$, $\text{d}_{z^2}$, and $\text{d}_{xy}$), with all spins paired. Thus, it is diamagnetic.
2. $[\text{Co(CN)}_6]^{3-}$:
• Cobalt is in the $+3$ oxidation state ($\text{Co}^{3+}$, a $\text{d}^6$ system).
• $\text{CN}^-$ is a strong-field ligand, forming an octahedral geometry.
• This strong crystal field splitting causes the electrons to adopt a low-spin configuration: all $6$ electrons go into the lower-energy $\text{t}_{2\text{g}}$ orbitals, completely filling them:
\[ \text{t}_{2\text{g}}^6\ \text{e}_{\text{g}}^0 \]
• Since all electrons are paired, this complex is diamagnetic.
3. $[\text{Fe(CN)}_6]^{3-}$:
• Iron is in the $+3$ oxidation state ($\text{Fe}^{3+}$, a $\text{d}^5$ system).
• With the strong-field ligand $\text{CN}^-$, it adopts a low-spin $\text{t}_{2\text{g}}^5\ \text{e}_{\text{g}}^0$ configuration, which leaves $1$ unpaired electron. Thus, it is paramagnetic.
4. $[\text{Cu(H}_2\text{O)}_6]^{2+}$:
• Copper is in the $+2$ oxidation state ($\text{Cu}^{2+}$, a $\text{d}^9$ system).
• A $\text{d}^9$ configuration always has exactly $1$ unpaired electron, regardless of the ligand field strength. Thus, it is paramagnetic.
itemize
Step 4: Final Answer:
Therefore, only $[\text{Ni(CN)}_4]^{2-}$ and $[\text{Co(CN)}_6]^{3-}$ are diamagnetic (Option A).
The question asks us to determine which of the given transition metal coordination complex ions exhibit diamagnetic behavior (no unpaired electrons) at room temperature.
Step 2: Key Formula or Approach:
We must determine the oxidation state, coordination geometry, and d-electron configuration of the central metal in each complex, and then apply Crystal Field Theory (CFT) to evaluate electron pairing.
Step 3: Detailed Explanation:
• Let us evaluate each complex systematically:
• 1. $[\text{Ni(CN)}_4]^{2-}$:
itemize
• Nickel is in the $+2$ oxidation state ($\text{Ni}^{2+}$, a $\text{d}^8$ system).
• $\text{CN}^-$ is a very strong-field ligand, causing square planar coordination geometry.
• In a square planar crystal field, the $\text{d}_x^2\text{-y}^2$ orbital is pushed very high in energy, while the other four d-orbitals are lower.
• The $8$ electrons completely fill the four lower energy d-orbitals ($\text{d}_{xz}$, $\text{d}_{yz}$, $\text{d}_{z^2}$, and $\text{d}_{xy}$), with all spins paired. Thus, it is diamagnetic.
2. $[\text{Co(CN)}_6]^{3-}$:
• Cobalt is in the $+3$ oxidation state ($\text{Co}^{3+}$, a $\text{d}^6$ system).
• $\text{CN}^-$ is a strong-field ligand, forming an octahedral geometry.
• This strong crystal field splitting causes the electrons to adopt a low-spin configuration: all $6$ electrons go into the lower-energy $\text{t}_{2\text{g}}$ orbitals, completely filling them:
\[ \text{t}_{2\text{g}}^6\ \text{e}_{\text{g}}^0 \]
• Since all electrons are paired, this complex is diamagnetic.
3. $[\text{Fe(CN)}_6]^{3-}$:
• Iron is in the $+3$ oxidation state ($\text{Fe}^{3+}$, a $\text{d}^5$ system).
• With the strong-field ligand $\text{CN}^-$, it adopts a low-spin $\text{t}_{2\text{g}}^5\ \text{e}_{\text{g}}^0$ configuration, which leaves $1$ unpaired electron. Thus, it is paramagnetic.
4. $[\text{Cu(H}_2\text{O)}_6]^{2+}$:
• Copper is in the $+2$ oxidation state ($\text{Cu}^{2+}$, a $\text{d}^9$ system).
• A $\text{d}^9$ configuration always has exactly $1$ unpaired electron, regardless of the ligand field strength. Thus, it is paramagnetic.
itemize
Step 4: Final Answer:
Therefore, only $[\text{Ni(CN)}_4]^{2-}$ and $[\text{Co(CN)}_6]^{3-}$ are diamagnetic (Option A).
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