Which of the following pairs of compounds is isoelectronic and isostructural?

For polyatomic species to be considered isoelectronic in this context, they must have the same number of valence electrons. For them to be isostructural, they must possess the same geometry.

• $\text{XeF}_2$: The total number of valence electrons is $8 \text{ (from Xe)} + 2 \times 7 \text{ (from F)} = 22$. The central Xenon atom forms 2 bond pairs and has 3 lone pairs (steric number = 5, $sp^3d$ hybridisation). According to VSEPR theory, the 3 lone pairs occupy equatorial positions to minimise repulsion, resulting in a linear shape.
• $\text{IBr}_2^-$: The total number of valence electrons is $7 \text{ (from I)} + 2 \times 7 \text{ (from Br)} + 1 \text{ (from negative charge)} = 22$. Similar to $\text{XeF}_2$, the central Iodine atom forms 2 bond pairs and has 3 lone pairs (steric number = 5, $sp^3d$ hybridisation), which also gives it a linear shape. Because both species have 22 valence electrons and a linear structure, they are isoelectronic and isostructural.