Final answer:
Only options II. ²⁰⁸Pb²⁺ and III. ²⁰⁷Pb are isotopes of ²⁰⁸Pb because they have the same number of protons as ²⁰⁸Pb. Option I. ²⁰⁸Bi is not an isotope of ²⁰⁸Pb as it is bismuth with a different atomic number.
Step-by-step explanation:
Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. The atomic number (protons) in isotopes of the same element is the same, but their atomic mass number (protons + neutrons) will differ. Given that ²⁰⁸Pb represents an isotope of lead (Pb) with 82 protons and a mass number (A) of 208, its isotopes would have the same number of protons but different numbers of neutrons. For ²⁰⁷Pb, the mass number indicates it has 82 protons (as it is still lead) and one less neutron than ²⁰⁸Pb, making it an isotope. The ²⁰⁸Pb²⁺ is simply the same isotope of lead that has lost two electrons, hence the charge, but the nucleon number remains unchanged. On the contrary, ²⁰⁸Bi is not an isotope of ²⁰⁸Pb as Bi (bismuth) has an atomic number of 83, indicating it has one more proton than lead (Pb). In conclusion, only II. ²⁰⁸Pb²⁺ and III. ²⁰⁷Pb are isotopes of ²⁰⁸Pb. The charge on an ion does not change the isotope, so even though ²⁰⁸Pb²⁺ is positively charged, it is still the same element (lead) with the same atomic and mass numbers. Option I., bismuth, is not an isotope because it pertains to a different element with an additional proton.