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

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$\ce{Cl-}$ : 6e

$\ce{CH3-}$: 2e

$\ce{PPh3}$: 2e

gives 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 2e ligand (in covalent counting (=neutral ligand method) a 3e donor).

b) I expect bent NO is in ionic counting way a 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 '15 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 '15 at 16:58
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    $\begingroup$ 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 '15 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 '15 at 17:53
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    $\begingroup$ 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 '15 at 18:11

You need to follow the 18e- rule:

  1. NO+
  2. 2e electron donor
  3. 18 electrons
  • 1
    $\begingroup$ Can you elaborate a bit on how this answers the question at hand. Right now, it's very telegraphic and could use some elaboration. $\endgroup$ – jonsca Jan 30 '15 at 23:58

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