What is the XeOF4 Lewis structure

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konster19



Registration date: 01/25/2012
Posts: 7
Place of residence: Hamburg
Posted: Jul 19, 2012 3:38 PM Title: Valence line formula from XeF2

hello just looked at the repetition of valence formulas and so far everything is clear, use:

(2 * (number of H atoms) + 8 * (number of remaining atoms) - (total number of valence electrons) = x

x / 2 = number of bonds

but do not understand why I e.g. with XeF2, XeF4 or I3 there are not enough connections

thank you in advance for your efforts
magician4
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Registration date: 05.10.2009
Posts: 11677
Place of residence: Hamburg
Posted: Jul 19, 2012 7:05 pm Subject:

Quote:
but do not understand why I e.g. with XeF2, XeF4 or I3 there are not enough connections

because the rule you quoted is something like "baby's very first attempts at walking", at best. therefore: forget them (as you can see from the counterexamples you have given)


greeting

ingo
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konster19



Registration date: 01/25/2012
Posts: 7
Place of residence: Hamburg
Posted: Jul 20, 2012 2:40 PM Subject:

Thanks in advance...

But how do I come up with the valence formula from the exceptions?
magician4
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Registration date: 05.10.2009
Posts: 11677
Place of residence: Hamburg
Posted: Jul 21, 2012 6:35 am Subject:

for "fluorine with some structure" it's simple: fluorine as the most electronegative element can only make a single bond, and will therefore always make a single bond

Consequently:

- XeF2 is to be drawn as F-Xe-F
since both Xe and each of the fluorine atoms contribute an electron for the respective joint bond, 8 - (2 * 1) = 6 electrons = 3 electron pairs remain on Xe
Xe then has 5 substituents: 2 fluorine atoms and 3 residual electron pairs
It therefore belongs to the structure class AX according to VSEPR2E.3 -> pseudo structure: trigonal-bipyramidal, real structure: linear

- XeF4 is like XeF2 to draw, just with two more Xe-F on it.
Regarding the electronic situation, the same argument applies, only this time the Xe contributes 4 of its own electrons (one for each Xe-F bond). This leaves 8 - (4 * 1) = 4 electrons = 2 electron pairs in the Xe
Xe then has 6 substituents: 4 fluorine atoms and 2 residual electron pairs
It therefore belongs to the structure class AX according to VSEPR4E.2 -> pseudo-structure: octahedral, real structure: square-planar
(link to 2 drawings)

iodine as an element of the 7th main group is extremely reluctant to appear in multiple bonds, and that is where the I also does3- -ion ​​(you forgot the negative charge in your posting) no exception.
First of all, it is important to recognize that this molecule consists of iodine atoms in different oxidation states: the "central" iodine is in the + I oxidation state (possible for higher halogens), whereas the two outer iodine atoms are each in the "halogen-typical" one "Hang around the oxidation level (-I)

geometric charge distribution: (-1) - (+1) - (-1)
... with which the entire unit then responds to the external charge -1 + 1-1 = -1.
iodine has 7 external electrons, which is why I.(+ I) of which logically only has 6 remaining, i.e. 3 electron pairs
every "outer" iodide I.- contributes one of its 4 electron pairs to the central iodine - (+) - ion, which is why it then has 5 substitutes in the result: 2 iodine atoms, 3 free electron pairs
According to VSEPR, this is now the structure class AX2E.3, and that goes for the XeF2 Said: pseudo-structure trigonal-bipyramidal, real structure linear
I.3- is therefore the XeF2 completely isoelectronic, and therefore structurally isomorphic


greeting

ingo
_________________
a month in the laboratory saves you a quarter of an hour in the library!