Final answer:
The correct order of ionization enthalpy among the given species, considering periodic trends and electronic configurations, is likely Cs < Ba²⁺ < I⁻ < Te₂⁻. This takes into account the factors of atomic size, nuclear charge, and electron configuration.
Step-by-step explanation:
The question revolves around the comparison of ionization enthalpy for different chemical species. Ionization enthalpy refers to the energy required to remove an electron from a gaseous atom or ion. The order of ionization enthalpy given in the question is Ba₂⁺ < Cs < I⁻ < Te₂⁻. To analyze this order, we need to consider several factors such as atomic size, nuclear charge, and electron configuration.
Generally, ionization enthalpy decreases down a group and increases across a period on the periodic table. For metals such as Ba and Cs, which belong to the s-block, the values of ionization enthalpy are relatively low. Given that Ba is above Cs in the periodic table, Ba should indeed have a higher ionization enthalpy than Cs, contradicting the order provided. For nonmetals like I and Te, which are in the p-block, the additional electron in the anion form (I⁻ and Te₂⁻) increases the repulsion between electrons and hence, lowers the ionization enthalpy as compared to their neutral atoms.
Considering the periodic trends and the electronic configurations, a more likely order of increasing ionization enthalpy for these species would, therefore, be: Cs < Ba²⁺ < I⁻ < Te₂⁻, since Cs has the lowest ionization enthalpy being the largest and most metallic element among the given, whereas Te₂⁻, being a heavier nonmetallic anion, would have higher ionization enthalpy.