It is known that Para-Hydrogen has opposite spin and Ortho-Hydrogen has same spin. Therefore, the energy of Para-Hydrogen is lower than that of Ortho-Hydrogen. Therefore, the stability order should be: Para-Hydrogen > Ortho-Hydrogen.

Is this true or is the stability order the opposite of what I predicted?. What is the reason for this stability order?


At room temperature, about $75$% of $\ce{H2}$ is $\ce{o-H2}$ and $25$% is $\ce{p-H2}$. As the temperature drops, the relative amounts of $\ce{p-H2}$ increases. The two forms are in a temperature-dependent equilibrium $\ce{o-H2<=> p-H2}$, $\Delta H = \pu{-1.66 kJ/mol}$.

Ref. M. Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, New York $1994$, 1202 p. ISBN 0-89925-4547-8.

  • $\begingroup$ This fails to answer OP's question: why? $\endgroup$ – Karl Knechtel Oct 18 '20 at 0:46
  • $\begingroup$ So, which one is more stable? $\endgroup$ – Param Budhadev Oct 18 '20 at 1:28
  • $\begingroup$ Both are more stable. It is dependent on temperature and their ratio. $\endgroup$ – Poutnik Oct 18 '20 at 13:22

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