To determine which pair of compounds is isoelectronic and isostructural, we need to analyze each option.$\\$Isoelectronic species have the same number of electrons. Isostructural species have the same geometry.$\\$Let's analyze each option:$\\$Option 1: $$TeI$_2$ and $$XeF$_2 \\$• TeI$_2:$Tellurium (Te) has 6 valence electrons, and each iodine (I) has 7.$\\$ Total electrons = 6 + 2(7) = 20.$\\$• XeF$_2:$Xenon (Xe) has 8 valence electrons, and each fluorine (F) has 7.$\\$ Total electrons = 8 + 2(7) = 22.$\\$ These are not isoelectronic.$\\$Option 2: $$IBr$_2^{-}$ and $$XeF$_2 \\$• IBr$_2^{-}:$Iodine (I) has 7 valence electrons, each bromine (Br) has 7, and the negative charge adds 1 electron.$\\$ Total electrons = 7 + 2(7) + 1 = 22.$\\$• XeF$_2:$Total electrons = 8 + 2(7) = 22.$\\$ These are isoelectronic.$\\$ Both have a linear structure, so they are isostructural.$\\$Option 3: $$IF$_3$ and $$XeF$_2 \\$• IF$_3:$Iodine (I) has 7 valence electrons, and each fluorine (F) has 7.$\\$ Total electrons = 7 + 3(7) = 28.$\\$• XeF$_2:$Total electrons = 22.$\\$ These are not isoelectronic.$\\$Option 4: $$BeCl$_2$ and $$XeF$_2 \\$• BeCl$_2:$Beryllium (Be) has 2 valence electrons, and each chlorine (Cl) has 7.$\\$ Total electrons = 2 + 2(7) = 16.$\\$• XeF$_2:$Total electrons = 22.$\\$ These are not isoelectronic.$\\$Therefore, the correct option is Option 2: $$IBr$_2^{-}$ and $$XeF$_2$ are isoelectronic and isostructural.