Which of the following complex is diamagnetic?

  1. $\ce{[Cr(Cl3)(NMe3)]}$
  2. $\ce{[Co(EDTA)]-}$
  3. $\ce{[Cu(CN)(NO2)(NH3)(Py)]}$
  4. $\ce{[Fe(NH3)6]^2+}$

For (4) I found that since $\ce{NH3}$ is a weak field ligand it cannot pair the electrons of $\ce{Fe^2+}$. Thus, (4) will remain paramagnetic.

But, for the rest of the options, I'm unable to clearly understand if they are paramagnetic or diamagnetic. I need help here.


1) Cr3+ has 3 electrons in 3d orbital. Obviously, it is a tetrahedral splitting, so configuration will be eg2 t2g1 which shows it is paramagnetic.

2) Here you need to know that for Co3+ all complexes are low spin except [CoF6]3- and [Co(H2O)3F3]. So, the complex given will be low spin with all 6 electrons paired & thus it will be diamagnetic.

3) Cu2+ has 3d9 configuration. So, it can never be diamagnetic as one electron will always be unpaired.

4) What you thought is right. But I would like to add that NH3 doesn't always behave like a weak field or strong field ligand. It depends on many factors which can't be explained using CFT.

Hope it helps!

  • $\begingroup$ If complexes [Co(F6)3- and [Co(H20)3F3] are diamagentic exceptionally why is [Co(EDTA)]- diamagnetic ?It should also be paramagnetic $\endgroup$ – kishan sharma Mar 17 '18 at 16:51
  • $\begingroup$ Why do you think the first one has an octahedral splitting? There are only four ligands. $\endgroup$ – Justanotherchemist Mar 17 '18 at 21:58
  • $\begingroup$ @kishansharma the two complexes I mentioned are the only high spin complexes of Co3+, so other complexes including the one you asked will be low spin (with all 6 electrons paired and therefore diamagnetic) $\endgroup$ – FreakyLearner Mar 18 '18 at 8:07
  • $\begingroup$ @Justanotherchemist Thanks...i have corrected it $\endgroup$ – FreakyLearner Mar 18 '18 at 8:11

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