Para and Ortho Hydrogen Statements Analysis

The question asks to identify the correct statements regarding the nuclear spin isomers of hydrogen: para-hydrogen and ortho-hydrogen.

Statement A: Total spin of ortho hydrogen is zero.

Ortho hydrogen has nuclear spins aligned parallel. For two protons (each with spin $s=1/2$), the total nuclear spin quantum number ($I$) can be $1/2 + 1/2 = 1$ (parallel alignment, triplet state). Therefore, the total spin is not zero. This statement is incorrect.

Statement B: Total spin of para hydrogen is zero.

Para hydrogen has nuclear spins aligned anti-parallel. For two protons, the total nuclear spin quantum number ($I$) is $1/2 - 1/2 = 0$ (anti-parallel alignment, singlet state). Thus, the total spin is zero. This statement is correct.

Statement C: The physical properties of two forms are different.

Para and ortho hydrogen differ in their nuclear spin states, which affects their rotational energy levels. These differences lead to variations in their physical properties, such as boiling points, melting points, and specific heats. This statement is correct.

Statement D: The chemical properties of two forms are different.

Chemical properties are primarily determined by the electronic structure of molecules. Since both para and ortho hydrogen have the same electronic configuration, their chemical reactivity and properties are virtually identical. This statement is incorrect.

Statement E: At higher temperature ortho is more stable.

Hydrogen exists as an equilibrium mixture of para and ortho forms. At room temperature, para-hydrogen is slightly more stable. However, as the temperature increases, the equilibrium shifts towards the ortho form. At higher temperatures, ortho-hydrogen becomes thermodynamically more stable due to the population of higher energy states. This statement is correct.

Conclusion on Correct Statements

Based on the analysis, the correct statements are B, C, and E.

The correct option combining these statements is 3.