0
$\begingroup$

Do I understand correctly?

In the case of the $\mathrm{H}_2$ molecule, the overall state is $S = \Psi(r)\chi(s)$.....(Spatial and spin aspects)

  • $\Psi(r)$ is inherently anti-symmetric under exchange since we're dealing with Fermions (electrons in this case)

  • In order for there to be bonding S must be symmetric under exchange to experience an "attractive exchange force"

  • Therefore $\chi(s)$ must be anti-symmetric under exchange in order to make S symmetric under exchange

ConclusionAnd for these reasons the singlet state (which is anti-symmetric) is bonding, and triplet state (which is symmetric) is anti-bonding

$\endgroup$
0
$\begingroup$

Correct conclusion but incorrect argument. The full wave function - not only its spatial part - must be antisymmetric. The spatial part can be symmetric if the spin part is antisymmetric.

In fact in the case of $H_2$ the bonding is for a symmetric spatial wavefunction (see here). Indeed “singlet” usually refers to spin-singlet, which is $S=0$ and antisymmetric, thus forcing the spatial part to be symmetric under permutation.

$\endgroup$
  • $\begingroup$ Thank you for your response, let me see if I understand correctly 1) S = Psi(r)Chi(s) must be anti-symmetric for fermions 2) Psi(r) being symmetric leads to bonding. Psi(r) being anti-symmetric leads to anti-bonding. 3) Chi(s) can be either symmetric or anti-symmetric Conclusion 1: Chi(s) being symmetric leads to anti-bonding because it forces Psi(r) to be anti-symmetric in order to satisfy 1) $\endgroup$ – Omar Azami Aug 5 '18 at 20:20
  • $\begingroup$ Conclusion 2: Chi(s) being anti-symmetric leads to bonding because it forces Psi(r) to be symmetric in order to satisfy 1) Conclusion 3: Triplet states of H2 are symmetric and are therefore anti-bonding. Conclusion 4: Singlet states of H2 are anti-symmetric and are therefore bonding. $\endgroup$ – Omar Azami Aug 5 '18 at 20:30

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.