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When I tried to figure out why manganate ion is paramagnetic, given that its oxidation number is +6, I am not able to understand why there are unpaired electrons. Is it the case that six of the seven valence electrons of Mn are taken by oxygen by ionic means, and a unpaired electron remains?

Secondly, for manganese to make 7 bonds, I believe an electron should be promoted from the 4s orbital. Where does this electron move to, 5s or 4p?

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    $\begingroup$ Could you please edit your question to make it more intelligible, using punctuation where appropriate? $\endgroup$ – Buck Thorn Mar 30 '19 at 8:31
  • $\begingroup$ @NightWriter Hi, I just am asking about its electron sharing while formation of manganate ion. I find difficulty to imagine how manganese has managed to make 6 bonds! Where and to which orbital does the electron jump to. Kindly answer it by providing diagrammatic explaination of orbitals $\endgroup$ – Atharv Mar 30 '19 at 15:01
  • $\begingroup$ You might want to consider using ligand Field Theory. $\endgroup$ – ANZGC FlyingFalcon Mar 31 '19 at 12:59
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We've already had an answer from the point of view of crystal field theory. This is actually more or less equivalent to your solution that six electrons are taken by ionic means (which is an oversimplification of the meaning of oxidation number +6, but not a bad place to start thinking about this) leaving one unpaired electron to explain the paramagnetism.

At the other extreme of oversimplification, you could model this using a simple valence-bond model based on the Lewis structure (see Wikipedia), which I think is what you may have in mind in your second question. In this model, since there are formally two single and two double bonds, you would say that there are four sigma bonds formed by Mn $4s3d^3$ hybrid orbitals, and two pi bonds formed by the remaining two Mn $3d$ orbitals. These six orbitals together contain seven electrons from the Mn and four from the four O atoms, for a total of eleven electrons, again leaving one unpaired.

In practice, for what it's worth most research chemists would use neither of these pictures, but instead a molecular orbital model.

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I’ll try and rationalise this using only crystal field theory, for the sake of simplicity.

In the presence of a tetrahedral ligand field, the metal d ions split into a E set and a T2 set. (The E set involves the x^2 - y^2 and z^2 orbitals, and the T2 set involves the other three).

The T2 set is raised in energy, and the E set is lowered in energy. The splitting is particularly low (O2- is a pi donor ligand, and it’s a tetrahedral complex) so this would be high-spin.

In other words, there would be one electron populating the E set, so it’s paramagnetic.

Hope this helped.

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    $\begingroup$ Even if you are restricting yourself to crystal field theory, I am not sure talking about high and low spin helps for a nominally $d^1$ complex. $\endgroup$ – Aant Mar 31 '19 at 17:00
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    $\begingroup$ Fair enough. I Guess the HS/LS just came out naturally since I was too used to invoking it. $\endgroup$ – ANZGC FlyingFalcon Apr 3 '19 at 6:31

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