how the hybridization of IF4+ formed?

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Francesco Paucek · Accepted Answer

The ion IF4^+ has I as the central atom and I has around it five electron pairs, four of which are bonding to F atoms. Therefore, the electron pair geometry is trigonaly bipyramidal, and since one is a lone pair, then the molecular geometry is "see-saw", and the hybridization of iodine is sp3d.

Francesco Paucek · Answer

non-metals on the 3rd energy level and higher have the capability to expand the octet. since flourine is the most electronegative nonmetal AND it is on the 2nd energy level it does not expand its octet in this scenario...there are always exceptions to the rulesCount the number of valence electrons for each atom to get a total of 35 electrons but the +1 charge indicates that 1 electron from the total is missing 35-1= 34 total electrons (17 pairs)iodine is listed first so it becomes the central atom and it can expand its octet, since it is on the 5th energy level it can shift some of the 5s25p5 electrons to the 5d subshell (1 of them to be exact) leaving you with sp3d giving you 5 bonding domains.Iodine gets 4 dashes (NSEW) first then place the flourine atoms on the ends of each dash. This will leave you with 26 total electrons to place on the flourine atoms or 13 pairs left to use.Each flourine atom will get 3 pairs of electrons (total of 24 electrons) leaving a lone pair. This lone pair is placed on the central iodine atom, again because it has expanded its octet to 1 of the d orbital subshells...this will give you the 5 bonding domains around Iodine.

Dr. Nina Hettinger I

how the hybridization of IF4+ formed?

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