I was met with a question that required me to find which complex is an outer orbital complex, and two of the options included $\ce{ [Ni(NH3)6]^2+}$ and $\ce{[Co(NH3)6]^3+}$,

I want to know why does $\ce {NH3}$ acts as a strong field ligand and cause pairing of electrons in d orbital in the case of $\ce{[Co(NH3)6]^3+}$ but acts as a weak field ligand and doesnt cause pairing in the case of $\ce{ [Ni(NH3)6]^2+}$ (hence making it a outer orbital complex)?

I know that $\ce{NH3}$ lies in the middle of spectrochemical series, so how do I identify if it will cause pairing in a complex or not?

  • 3
    $\begingroup$ The concept of "outer orbital complex" is way outdated and not taught in most places, though India seems to be an exception (maybe that's where you are?). It is now known that both high spin and low spin complexes use "inner orbitals" for bonding (that is (n-1)d orbitals, so 3d for cobalt rather than ever using 4d). $\endgroup$ – Andrew May 21 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.