In this paper, it is mentioned:

Furthermore, since the energy of $H_2^+$in the ground state must be lower than that of an H atom in the ground state,the negative (attractive) forces in the $H^+ – H$ interaction must play the dominant role

How do we intuitively expect the ground energy of hydrogen molecule ion $H_2^+$ should be less than that of hydrogen atom without doing exact quantum mechanical calculation.


I had a look at the paper, and I think the author means that the energy for the reaction:

$$ H + H^+ \rightarrow H_2^+ $$

is negative i.e. the ground state energy of $H_2^+$ is less than the sum of the ground state energies of $H$ and $H^+$. The reason for this is simply the observation that the $H_2^+$ ion is stable. If the energy of $H_2^+$ were higher than $H$ and $H^+$ the ion would dissociate.

Response to comment:

If you mixed $H^+$ ions and $H$ atoms then they would react to form $H_2^+$. But in nature we don't find such mixtures because the hydrogen is normally neutral and not ionised. If you mix hydrogen atoms you get neutral $H_2$ not $H_2^+$ because the reaction:

$$ H_2^+ + e \rightarrow H_2 $$

is exothermic.

  • $\begingroup$ does it mean that we get more amount of $H_2^+$ than $H$ in nature, because former is more stable? $\endgroup$
    – DurgaDatta
    Apr 21 '13 at 8:30

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