The Lewis dot structure for the ion IF4O+ shows an expanded octet on iodine. Some elements in the third period or below form compounds in which they have 8 or 10 valence electrons as the central atom. Oxygen and fluorine aren't able to have an expanded octet so iodine must the central atom. This is also consistent with the least electronegative atom being in the center.
After determining the central atom we need to determine the total number of valence electrons by adding the valence electrons of the individual atoms:
I: 7
F: 4(7) = 28
O: 6
Total = 7+28+6=41
We must now subtract one electron, because the overall charge of +1 indicates that there is one less electron than the total number of valence electrons from The neutral atoms:
Total = 41-1=40
The next step is to connect the each of the four fluorines and the oxygen to the central iodine with a single bond, which is two electrons. This uses 10 of the 40 electrons, so 30 remain.
Now determine if there are enough electrons to complete the octets of each F and the O with six more electrons. This would require 30 electron for five atoms, which is the number of electrons left. This leaves iodine with five bonding pairs of electrons and no lone pairs as shown in the attached diagram.
Lastly, if the structure is an ion, put brackets around it toindicate the charge, in this case +1, outside the brackets as shown.
This ion has structural isomers because there are diffferent possible locations for the oxygen in relation to the fluorines in the three dimensional molecule. However, there are no resonance structures. A resonance structure shows a different way to arrange the electrons among the same skeleton structure, for example if a double bond can be placed in more than one location. The only possibility for arranging electrons in this structure is with a single bond to each outer atom.
A structure with 5 pairs of electrons around the central atom is called an sp3d hybrid. The five shared pairs of electrons occupy equivalent orbitals that are an average of three s, a p and a d orbital.
No comments:
Post a Comment