0
$\begingroup$

Suppose we have some ion such as $H_2^+$ and we produce some trial electronic wavefunction. For example, we take the simple trial wavefunction made up of a single 1s orbital of the hydrogen atom centred on one of the protons.

How would I know whether such a trial wavefunction would lead to bonding in this ion? In other words what are the conditions on my trial wavefunction to ensure that the ion does or doesn't lead to bonding.

$\endgroup$
5
  • $\begingroup$ lets say simply the 1s orbital of the hydrogen atom centred [sic] on one of the protons That's a really rubbish approximation. $\endgroup$
    – Gert
    Feb 5 at 3:01
  • $\begingroup$ Yes i know, and it also won't lead to bonding but I want to know why. How would i prove that it does not lead to bonding. I know that if i used $c_1 \phi_{1s} + c_2 \phi_{1s} $ i would get bonding but I am unsure how to prove this to myself. $\endgroup$
    – DJA
    Feb 5 at 3:06
  • $\begingroup$ Basically my notes say that the trial wavefunction $\phi_{1s}$ is a poor choice and will not lead to bonding in an $H^+_2$ ion and I would like to prove to myself why this is the case since the notes do not elaborate on this. $\endgroup$
    – DJA
    Feb 5 at 3:12
  • 1
    $\begingroup$ You want to show that there is an energy state with lower energy than you would get for "unbound" $H_2^+$ (unbound $H_2^+$ would pretty much be exactly your ansatz...) $\endgroup$
    – Andrew
    Feb 5 at 3:12
  • 1
    $\begingroup$ Ahh I see, so all I would have to do is show that the expectation value of energy of a wavefunction such as $ c_1 \phi_{1s} + c_2 \phi_{1s}$ is lower than that of $\phi_{1s}$. I would also note that the wavefunction $ c_1 \phi_{1s} + c_2 \phi_{1s}$ does have a bonding and antibonding solutions, so I assume I have to show that $\phi_{1s}$ is higher in energy than the bonding one. $\endgroup$
    – DJA
    Feb 5 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.