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Why does $\ce{[Co(NO2)6]^4-}$ ion have 3 unpaired electron as opposed to 1?

The book says that it has 3 unpaired electrons. I thought $\ce{NO2-}$ was a strong field ligand, making $\ce{Co^2+}$ a $\ce{d^7}$ low spin ion. Six electrons go in the $\mathrm{t_{2g}}$ orbital and 1 in the $\mathrm{e_g}$ orbital.

Please explain the bonding in this compound (preferably with a suitable MO diagram to go with it).

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    $\begingroup$ You can't predict which ligand is strong or weak, there are many factors which can change this. $\endgroup$ – King Tut Feb 13 '18 at 12:32
  • $\begingroup$ An easier method - instead of getting tangled in the mess of strong vs weak field ligand - would be to make deductions based on the compound's calculated paramagnetic moments. Unfortunately, a simple Google search yields too little information. Perhaps, the chemists with more efficient database searches can help out here? $\endgroup$ – Gaurang Tandon Mar 15 '18 at 15:51
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According to Elliot, Hathaway & Slade (1966) cited by this paper, Co(II) with nitrite ligands is low spin.

enter image description here

So maybe it is just a typo. For d7 low spin, you would expect one unpaired electron.

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