Why is Bi(v) a stronger oxidant than Sb(v)?

Bismuth (Bi) in its +5 oxidation state (Bi(V)) is generally considered a stronger oxidizing agent compared to antimony (Sb) in its +5 oxidation state (Sb(V)). This is primarily due to the size and electronic configuration of the atoms.

1. Atomic Size: Bismuth is larger than antimony in the same oxidation state. As you move down a group in the periodic table, the atomic size tends to increase due to the addition of new electron shells. This increased atomic size in bismuth allows it to accommodate more electrons in its outermost shell, leading to a greater dispersion of negative charge over a larger volume. This results in a decreased effective nuclear charge experienced by the valence electrons, making them easier to remove. Consequently, Bi(V) is more willing to accept electrons, making it a stronger oxidizing agent compared to Sb(V).

2. Ionization Energy: Bismuth has lower ionization energy compared to antimony. Ionization energy is the energy required to remove an electron from a neutral atom in the gas phase. Due to bismuth's larger atomic size and the shielding effect of inner electron shells, it requires less energy to remove an electron from Bi(V) compared to Sb(V). Therefore, Bi(V) can more readily accept electrons from other species, making it a stronger oxidizing agent.

3. Electron Configuration: Bismuth and antimony both have the same electron configuration for their +5 oxidation state, which is [Xe]4f^145d^106s^26p^3. However, due to bismuth's larger size, its valence electrons are further away from the nucleus compared to antimony. This results in a weaker attraction between the nucleus and the valence electrons in bismuth, making it easier for Bi(V) to gain electrons and act as a stronger oxidizing agent.

In summary, the larger size and lower ionization energy of bismuth compared to antimony contribute to Bi(V) being a stronger oxidizing agent.