morsagh
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[Re2Cl8] electron configuration
If Re(III) has 4f14 5d4 than 4 free electrons from d orbital will make 4 M-M bond. According to Sidgwick´s rule there is free
orbital s than px,y,z and one in d. To radon there are 10e- missing. So why it is [Re2Cl8] and not [Re2Cl10] when 4 ligands can
donate only 8 electrons to each Re3+?
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blogfast25
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Quote: Originally posted by morsagh  | | If Re(III) has 4f14 5d4 than 4 free electrons from d orbital will make 4 M-M bond. According to Sidgwick´s rule there is free
orbital s than px,y,z and one in d. To radon there are 10e- missing. So why it is [Re2Cl8] and not [Re2Cl10] when 4 ligands can
donate only 8 electrons to each Re3+? |
Did you mean [Re<sub>2</sub>Cl<sub>8</sub>]<sup>2-</sup> ?
According to 'Inorganic Chemistry' (Shriver and Atkins, third edition, p.302), there is indeed a quadruple
σ<sup>2</sup>π<sup>4</sup>δ<sup>2</sup> bond between the Re atoms.
Bond angle between the Cl-Re and Re-Re axis is 104 degrees.
I think the empty 6s and 6p orbitals hydridise to sp<sup>2</sup> and then each Cl<sup>-</sup> donates one of its lone p
electron pairs. Non-existence of such an anion with 10 Cl could be explained by excessive steric hindrance?
Note that Re(0) = [Xe] 4f<sup>14</sup> 5d<sup>5</sup> 6s<sup>2</sup>
[Edited on 23-9-2015 by blogfast25]
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Ozone
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Yes, it's typically prepared as the tetrabutylammonium salt. We made this in a senior/grad level lab. After the Chemist survives the HCl gas, the
synthesis yields beautiful emerald-green crystals. A quadruple bond is evident in the UV-VIS spectrum.
See:
http://www.sigmaaldrich.com/catalog/product/aldrich/250201?l...
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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