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.

In general, comparing the oxidizing strength of different elements can be complex and depends on various factors such as the standard reduction potentials, electronic configurations, and chemical environments. However, as a general trend, bismuth (Bi) tends to exhibit a higher tendency to be reduced compared to antimony (Sb).

In the +5 oxidation state (Bi(V) and Sb(V)), both bismuth and antimony are relatively stable and have a tendency to act as oxidizing agents. However, due to its larger atomic size and lower effective nuclear charge, bismuth tends to have weaker oxidizing properties compared to antimony.

Therefore, in this comparison, antimony (Sb(V)) would typically be considered the stronger oxidizing agent compared to bismuth (Bi(V)).

Red phosphorus is less reactive than white phosphorus due to differences in their molecular structures and arrangements of atoms. White phosphorus consists of tetrahedral P4 molecules, each containing four phosphorus atoms bonded together in a highly strained, reactive structure. These P4 molecules are held together by weak van der Waals forces.

In contrast, red phosphorus has a polymeric structure, with long chains or layers of phosphorus atoms bonded together in a more stable arrangement. This structure makes it less prone to spontaneous combustion and less reactive with other substances compared to white phosphorus.

Additionally, white phosphorus is highly reactive because it readily reacts with oxygen in the air to form phosphorus pentoxide, producing intense heat and light, which can lead to spontaneous ignition. Red phosphorus, on the other hand, is much less reactive with oxygen and requires higher temperatures to ignite.

In H3PO2, also known as hypophosphorous acid, the oxidation number of hydrogen (H) is typically +1.

The sum of the oxidation numbers in a neutral molecule must equal zero. Since there are three hydrogen atoms, each with an oxidation number of +1, their total contribution is +3.

For oxygen (O), the typical oxidation number is -2, except in peroxides and when it's bonded to fluorine. In H3PO2, oxygen's oxidation number is -1.

Given that the overall charge of the molecule is zero, and knowing the oxidation numbers of hydrogen and oxygen, you can calculate the oxidation number of phosphorus (P).

Let's denote the oxidation number of phosphorus as xx:

(+1×3)+(−1×2)+x=0(+1×3)+(−1×2)+x=0

3−2+x=03−2+x=0

1+x=01+x=0

x=−1x=−1

So, in H3PO2, the oxidation number of phosphorus is -1.