Final answer:
Tellurium generally has a higher ionization energy than indium as it has a higher atomic number, which typically means its valence electrons are more tightly bound to the nucleus. Indium's electron is in a 5p orbital, while tellurium's is in a 6p orbital, which is closer to the nucleus, thereby increasing tellurium's ionization energy.
Step-by-step explanation:
Ionization energy refers to the amount of energy required to remove an electron from an atom. Indium (In) has an atomic number of 49, and Tellurium (Te) has an atomic number of 52. Typically, elements with higher atomic numbers within the same group have higher ionization energies due to increased nuclear charge pulling the electrons in more tightly. However, ionization energy also depends on the electron configuration and the subshell from which the electron is being removed.
Tellurium is higher up in the periodic table than indium, meaning its electrons are more tightly bound to the nucleus, generally leading to a higher ionization energy compared to indium. So, we would expect tellurium to have a higher first ionization energy than indium, largely because indium's valence electron is in a 5p orbital, whereas tellurium's valence electron is in a 6p orbital,
Complicating factors like the penetration of orbitals and electron-electron repulsions do play a role, so these trends are not without their exceptions, but generally, the further away and more shielded from the nucleus the valence electrons are, as in the case with indium, the easier they are to remove. Therefore, tellurium has a higher ionization energy than indium.