The xenon halides and oxides are isoelectronic with many other compounds and ions containing halogens. Give a molecule or ion in which iodine is the central atom that is isoelectronic with each of the following. $\begin{array}{ll}{\text { a. xenon tetroxide }} & {\text { d. xenon tetrafluoride }} \\ {\text { b. xenon trioxide }} & {\text { e. xenon hexafluoride }} \\ {\text {c. xenon difluoride}}\end{array}$

The known xenon compounds are given below, and their corresponding total electrons are calculated. a. Xenon tetroxide (XeO4): Xe(54 electrons) + 4 * O(8 electrons each) = 54 + 4 * 8 = 86 electrons. b. Xenon trioxide (XeO3): Xe(54 electrons) + 3 * O(8 electrons each) = 54 + 3 * 8 = 78 electrons. c. Xenon difluoride (XeF2): Xe(54 electrons) + 2 * F(9 electrons each) = 54 + 2 * 9 = 72 electrons. d. Xenon tetrafluoride (XeF4): Xe(54 electrons) + 4 * F(9 electrons each) = 54 + 4 * 9 = 90 electrons. e. Xenon hexafluoride (XeF6): Xe(54 electrons) + 6 * F(9 electrons each) = 54 + 6 * 9 = 108 electrons.

To find isoelectronic iodine compounds, we will identify the species with the required number of electrons by adjusting the number and type of ligands (atoms surrounding iodine). a. For 86 electrons, we need iodine (53 electrons) plus an additional 33 electrons from ligands. Four oxygen atoms can provide this (4 * 8 = 32 electrons), and we have a negative charge: \(\text{IO}_4^-\). b. For 78 electrons, we need iodine (53 electrons) plus an additional 25 electrons from ligands. Three oxygen atoms can provide this (3 * 8 = 24 electrons), and additionally we have a negative charge: \(\text{IO}_3^-\). c. For 72 electrons, we need iodine (53 electrons) plus an additional 19 electrons from ligands. Two fluorine atoms can provide this (2 * 9 = 18 electrons), and additionally we have a positive charge: \(\text{IF}_2^+\). d. For 90 electrons, we need iodine (53 electrons) plus an additional 37 electrons from ligands. Four fluorine atoms can provide this (4 * 9 = 36 electrons), and we have a positive charge: \(\text{IF}_4^+\). e. For 108 electrons, we need iodine (53 electrons) plus an additional 55 electrons from ligands. Six fluorine atoms can provide this (6 * 9 = 54 electrons), and we have a positive charge: \(\text{IF}_6^+\).

The iodine-containing species isoelectronic with the given xenon compounds are as follows: a. \(\text{IO}_4^-\). b. \(\text{IO}_3^-\). c. \(\text{IF}_2^+\). d. \(\text{IF}_4^+\). e. \(\text{IF}_6^+\).