Question:
Which of the following transition metal complexes is expected to be diamagnetic?
Which of the following transition metal complexes is expected to be diamagnetic?
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CN$^-$ is a strong field ligand that often produces low-spin, diamagnetic complexes especially for d$^8$ metals like Ni$^{2+}$.
Updated On: Jun 10, 2026
- [Ni(CN)$_4$]$^{2-}$
- [NiCl$_4$]$^{2-}$
- [Fe(H$_2$O)$_6$]$^{3+}$
- [CoF$_6$]$^{3-}$
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The Correct Option is A
Solution and Explanation
Concept:
Magnetic behavior of coordination complexes depends on the presence of unpaired electrons. Strong field ligands cause pairing (low spin), while weak field ligands produce unpaired electrons (high spin).
Step 1: Oxidation state and d-electron count
• [Ni(CN)$_4$]$^{2-}$: Ni$^{2+}$ = 3d$^8$
• [NiCl$_4$]$^{2-}$: Ni$^{2+}$ = 3d$^8$
• [Fe(H$_2$O)$_6$]$^{3+}$: Fe$^{3+}$ = 3d$^5$
• [CoF$_6$]$^{3-}$: Co$^{3+}$ = 3d$^6$
Step 2: Ligand field strength analysis
• CN$^-$ is a strong field ligand → causes electron pairing.
• Cl$^-$, F$^-$, H$_2$O are weak field ligands → no pairing.
Step 3: Magnetic behavior
• [Ni(CN)$_4$]$^{2-}$: strong field → all electrons paired → diamagnetic
• [NiCl$_4$]$^{2-}$: paramagnetic (unpaired electrons)
• [Fe(H$_2$O)$_6$]$^{3+}$: paramagnetic (high spin d$^5$)
• [CoF$_6$]$^{3-}$: paramagnetic (high spin)
Thus, only [Ni(CN)$_4$]$^{2-}$ is diamagnetic.
Step 1: Oxidation state and d-electron count
• [Ni(CN)$_4$]$^{2-}$: Ni$^{2+}$ = 3d$^8$
• [NiCl$_4$]$^{2-}$: Ni$^{2+}$ = 3d$^8$
• [Fe(H$_2$O)$_6$]$^{3+}$: Fe$^{3+}$ = 3d$^5$
• [CoF$_6$]$^{3-}$: Co$^{3+}$ = 3d$^6$
Step 2: Ligand field strength analysis
• CN$^-$ is a strong field ligand → causes electron pairing.
• Cl$^-$, F$^-$, H$_2$O are weak field ligands → no pairing.
Step 3: Magnetic behavior
• [Ni(CN)$_4$]$^{2-}$: strong field → all electrons paired → diamagnetic
• [NiCl$_4$]$^{2-}$: paramagnetic (unpaired electrons)
• [Fe(H$_2$O)$_6$]$^{3+}$: paramagnetic (high spin d$^5$)
• [CoF$_6$]$^{3-}$: paramagnetic (high spin)
Thus, only [Ni(CN)$_4$]$^{2-}$ is diamagnetic.
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