In the complex $\ce{[Ni(H2O)2(NH3)4]^2+}$ the magnetic moment ($\mu$) of $\ce{Ni}$ is –

  • Zero

  • $2.83~\mathrm{BM}$

  • $1.73~\mathrm{BM}$

  • $3.87~\mathrm{BM}$

It is easy to find the unpaired electrons and hence magnetic moment if we know the bonding strcture.

$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series. I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.

Nickel has $\ce{d^8}$ configuration here. I dont understand whether I should pair this electron or not. If I pair this electron, I get this:

enter image description here

If I pair the electron then I have to use one $\ce{3d}$ orbital and one $\ce{4d}$ orbital. I have not used both orbitals for overlap in any other question before so I don't know if that's right. Well, I think it should be because anyway these orbitals will make hybridized orbitals of equal energy for bonding. In that case, we get $\ce{d^2sp^3}$ (octahedral geometry).

If I do not pair the electron, I get $\ce{sp^3d^2}$ hybridization.

I am not sure but is this logic correct that since there are more number of strong field ligands than weak field ligands, pairing will occur?

Should I pair the electrons or not and why?

  • 1
    $\begingroup$ I don't think that the electrons would pair up. This pairing even does not happen in Ni(NH_3)_6 . Anyways, I am unsure why dsp3d is not generally observed. $\endgroup$ Dec 17, 2016 at 17:02

1 Answer 1


$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series.

No, this is not correct. I keep saying this time and time again. Ammonia is a medium-field ligand at best, and water is the defined ‘zero point’ below which weak field ligands begin.

I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.

This is simplified at best. The transition from low spin to high spin depends on both the metal and the ligands (and also the geometry). In fact, some complexes are known both in a high-spin and in a low-spin state with identical central metal and ligands. You always need to consider the big picture to determine whether a high spin or low spin complex will result.

I am also going to completely ignore the stuff you wrote about hybridisation. You should not consider hybridisation in transition metal complexes as I have posted numerous times.

To answer this question you should ask yourself how many ligands you have around the nickel centre and which geometry that implies. The complex being tetraamminediaquanickel(II) results in six ligands and thus an octahedral configuration. The next step should be to consider the molecular orbital scheme of an octahedral complex (although the d-orbitals alone carry all the necessary information so even the crystal field model can answer it).

orbital scheme octahedral complex
Figure 1: simplified MO scheme of an octahedral $\ce{[ML6]}$ complex ignoring π $\ce{L\bond{->}M}$ interactions. Originally taken from Professor Klüfers’ web scriptum of his coordination chemistry course at the LMU Munich.

The important bit to note is that the d orbitals split into a $\mathrm{t_{2g}}$ and an $\mathrm{e_g}$ part. $\mathrm{e_g}$ corresponds to the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2- y^2}$ orbitals. Thus, a perfect octahedral geometry will generate two degenerate orbitals which should be filled with parallel spins by Hund’s rule. Even if you start to assume Jahn-Teller distortions here (which are not likely, since a high-spin $\mathrm{d^8}$ octahedral configuration is not susceptible to Jahn Teller because there are no unevenly populated degenerate orbitals), that would not lead to an energy difference large enough to induce spin pairing. Remember that a typical 3d-metal octahedral complex is high-spin even in $\mathrm{d^6}$ cases; the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ is generally much larger than what can be achieved by Jahn-Teller distortion.

Now you may decide to argue using square planar complexes. However, the complex in question cannot be square planar since there are six ligands.

Therefore, a magnetic moment of zero is not an option in the case presented; $\ce{[Ni(H2O)2(NH3)4]^2+}$ cannot reasonably adopt a low-spin configuration. The high-spin complex is paramagnetic and in a triplet state ($2S+1 = 3$).

  • 3
    $\begingroup$ Can you simplify this answer considering that students are not taught jahn teller distortion, molecular orbital theory or crystal field theory in detail in high school? I first will have to learn these topics to be able to comprehend your answer. $\endgroup$
    – Arishta
    Dec 18, 2016 at 1:58
  • 3
    $\begingroup$ @Blue No, that is not possible. You cannot comprehend coordination complexes without crystal field theory. Any school teaching otherwise is teaching inconsistently and inconsequentially. $\endgroup$
    – Jan
    Dec 19, 2016 at 21:59
  • $\begingroup$ Well i said we are not taught these in detail. I know about eg and t2g orbital s but i have no idea about a1g, t1u....its not there in the school textbooks as well,...we do most of the questions using hybridization of the central atom and use crystal field theory to explain some exceptional cases, d-d transition, etc...we are only taught molecular orbital theory for homodiatomic molecules..nothing has been said about jahn teller distortion in my textbook ... $\endgroup$
    – Arishta
    Dec 20, 2016 at 6:09
  • 1
    $\begingroup$ @Arishta I presume that you had been reading NCERT Textbooks which of course don't have all this information. For class 12, you should consider reading Inorganic Chemistry by JD Lee or atleast the concise Indian version by Wiley India. All doubts regarding Jahn Teller effect and rest of the chapter will be cleared. $\endgroup$ Apr 19, 2018 at 12:35

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