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 $\ce {NH3}$ 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 doesn't cause pairing in the case of $\ce{ [Ni(NH3)6]^2+}$ (hence making it an 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?

  • 4
    $\begingroup$ The concept of "outer orbital complex" is way outdated and not taught in most places, though India seems to be an exception (maybe that's where you are?).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
    Commented May 21, 2021 at 18:43
  • $\begingroup$ @Andrew For cobalt, it is easy to see that it will use 3d, 4s and 4p orbitals to form 6, d2sp3 hybridised orbitals (I know that it has been stressed many times in ChemSE that we should not use hybridisation in case for coordination compounds but use of MOT is beyond my knowledge). The problem is with nickel complex, if $\ce{NH3}$ causes pairing of electrons then we left with only one vacant 3d orbital and thus, cannot form d2sp3 orbitals. $\endgroup$
    – Apurvium
    Commented Dec 17, 2021 at 6:18
  • $\begingroup$ So, can we say that "because $\ce{NH3}$ cannot form d2sp3 orbitals, it will not cause pairing of $\ce{e-}$ and use 4s, 4p and 4d orbitals to form 6, sp3d2 orbitals". As chemistry theory somehow tries to explain the experimental facts. $\endgroup$
    – Apurvium
    Commented Dec 17, 2021 at 6:21
  • $\begingroup$ @Apurvium There’s no reason you can’t use 3d orbitals just because they’re full. It just means that you end up with mostly filled anti bonding orbitals. These anti bonding orbitals are the ones we look at when we talk about splitting in the spectrochemical series. The 6 bonding orbitals are filled by the six electron pairs from the NH3 ligands $\endgroup$
    – Andrew
    Commented Dec 17, 2021 at 12:48
  • 1
    $\begingroup$ Also, the actual answer here is trivial. As $\ce{Ni^2+}$ is $d^8$, there is only one possible d electron configuration for an octahedral complex. It will have completely filled $t_{2g}$ orbitals and unpaired electrons in the two $e_g$ orbitals regardless of the strength of the ligand. $\endgroup$
    – Andrew
    Commented Dec 17, 2021 at 14:12

1 Answer 1


In case of $\ce{[Ni(NH3)6]^2+}$, the configuration is $3\text d^8$. After splitting from spherical to octahedral field, the lower energy orbitals, i.e., $\text t_{2g}$ are already filled. Therefore, pairing of $\ce{e-}$ will not occur whether the ligand is strong or weak. So, the compound is a high-spin complex.

$\uparrow$ $\downarrow$ $\uparrow$ $\downarrow$ $\uparrow$ $\downarrow$ $\uparrow$ $\uparrow$

${|<------------\text t_{2g}---------->}|{<----\text e_g----->|}$

In case of $\ce{[Co(NH3)6]^3+}$, the configuration is $3\text d^6$. After splitting from spherical to octahedral field, the lower energy orbitals, i.e., $\text t_{2g}$ are not fully occupied. Therefore, pairing of $\ce{e-}$ can occur depending on the ligand. In this case, the compound is a low-spin complex.

$\text {CFSE}_\text {high spin}=-\frac{2}{5}\times 6 +\frac{3}{5}\times 2 + 0\cdot P$

Before pairing:

$\uparrow$ $\downarrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$

${|<-----------\text t_{2g}--------->}|{<-----\text e_g------->|}$

After pairing:

$\uparrow$ $\downarrow$ $\uparrow$ $\downarrow$ $\uparrow$ $\downarrow$ $~~$ $~~$

${|<-----------\text t_{2g}----------->}|{<----\text e_g----->|}$

$\text {CFSE}_\text {low spin}=-\frac{2}{5}\times 6 +\frac{3}{5}\times 0 + 2\cdot P$

  • $\begingroup$ I'm quite confident that the layout is completely broken on my device for these attempts at making an image. The site design is responsive, so it won't look the same for everyone that reads it. Unfortunately, if you need images, you should prepare them on your device and upload them. $\endgroup$ Commented Jun 16, 2022 at 8:11

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