I'd like to count the total valence electron of following neutral complex in the ionic counting (=donor-pair):

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Electron count:

$$\begin{align} \ce{Cp-} : \ce{6e-} \\ \ce{CH3-}: \ce{2e-} \\ \ce{PPh3} : \ce{2e-} \end{align}$$

which gives $\ce{10e-}$ from these three ligands.

1. Question: How to decide the charge of NO

a) if I set $\ce{NO+}$ the charge of Re must be +, therefore Re(I): d6

b) if I set $\ce{NO-}$ the charge of Re must be 2+, therefore Re(III): d4

Probably only a) gives tetragonal geometry whereas b) gives square-planar geometry, correct?

2. Question: How many electrons for NO (linear/bent)

a) I expect linear NO is in ionic counting way a $\ce{2e-}$ ligand (in covalent counting (=neutral ligand method) a $\ce{3e-}$ donor).

b) I expect bent NO is in ionic counting way a $\ce{2e-}$ ligand (in covalent counting (=neutral ligand method) a 1e donor).

3. Question: Total valence electron number

I remember: When we go from linear to bent the VE number decreases with -2, therefore I expect:

a) linear goes along with $\ce{NO+}$, therefore: 10 + 6 + 2 = 18 VE

b) bent goes along with $\ce{NO-}$, therefore: 10 + 4 + 2 = 16 VE

Especially with 1. Question I feel pretty uncomfortable.

  • $\begingroup$ 1) square planar is for d8 only (this is mostly Co-group +1 and Ni-group +2, with virtually no examples outside these two columns) or ligands like porphyrins 2)this is somewhat simplified, but linear NO is NO+ (with extra electron given to central atom) and 2 electrons, but stable only with strong electron-donating systems, while bent NO is NO- and also 2 electrons. 3) without NO we have 7(Re(0)) + 5 (Cp) + 1 (CH3) + 2 (PPh3) = 15, 13 from NO. This is consistent with linear NO. $\endgroup$
    – permeakra
    Jan 30, 2015 at 13:35
  • $\begingroup$ what do you mean with "13 from NO." I expect the difference in VE number between bent and linear is only 1 (17 or 18 VE). $\endgroup$
    – laminin
    Feb 2, 2015 at 16:58
  • 1
    $\begingroup$ it's a typo, I meant 3 (three) And no, the difference is 2 (if you assume NO to be neutral) - either 3 (linear) or 1 (bent). Another common view is that NO is either NO+ (isoelectronic to linear CO or CN-) or bent NO- (and behaving like amine nitrogen). $\endgroup$
    – permeakra
    Feb 2, 2015 at 17:01
  • $\begingroup$ square planar is in normal cases only for d8. There are examples for Cu(+II) d9 in [Cu(NH3)4]2+ (reason: weak field) is also square planare. a) Would the case case for a very weak field be true for d10? I expect not. b) So d7 never gives square planar? $\endgroup$
    – laminin
    Feb 2, 2015 at 17:53
  • 1
    $\begingroup$ Cu2+ is usually octahedral in water solutions. The two remaining ligands are water molecules and pretty easily exchange, but still. $d7$ usually are, to my knowledge, mostly unstable. Actually, I recalled 1 $d8$ outside Co/Ni group: Au3+ in $ce{[AuCl4]^{-}}$ - like complexes, it is square planar. $\endgroup$
    – permeakra
    Feb 2, 2015 at 18:11


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