5
$\begingroup$

I'll explain my question on example of water molecule.

Let us have three water molecules: normal water $H_2 O$, heavy water $D_2 O$ and semiheavy water $HDO$.

  1. Is there any difference between the angles $H-O-H$, $D-O-D$ and $H-O-D$?
  2. Is any differences between the distances $H-O$ and $D-O$?
  3. If we will change the molecule "temperature" (atom vibrations) how it will change that angles and distances?
  4. If we "cool down" the molecules (remove atom vibrations) is that angles and distances will be equal?
  5. What factors can change the angles and the distances?
  6. Can we make the angles and the distances almost equal and how can we do that?

My question is not only about water molecule but for any molecule and any isotops ($NH_4$, $P {Cl}_2$, etc).

$\endgroup$
  • 1
    $\begingroup$ One difference I'm aware of is that isotopes have different zero-point energies due to the different mass. This affects the binding strength (usually stronger (weaker) for larger (smaller) masses). This can certainly have an important effect on the chemistry, e.g. D$_2$O has a slightly larger pH than water, but I imagine the effect on molecular angles and bond lengths is pretty negligible since such things are determined by electronic structure (e.g. hybridisation). $\endgroup$ – lemon Oct 4 '14 at 12:17
  • 1
    $\begingroup$ see "Quantum Differences between Heavy and Light Water" Physical Review Letters 101, 065502 (2008) journals.aps.org/prl/abstract/10.1103/PhysRevLett.101.065502 $\endgroup$ – DavePhD Nov 7 '14 at 12:50
1
$\begingroup$

According to http://www1.lsbu.ac.uk/water/data.html (and with reference to A. G. Császár, G. Czakó, T. Furtenbacher, J. Tennyson, V. Szalay, S. V. Shirin, N. F. Zobov and O. L. Polyansky, On equilibrium structures of the water molecule, J. Chem. Phys. 122 (2005) 214305), the bond angles H-O-H and D-O-D are pretty much the same - 104.50 and 104.49 deg., respectively (gas, 0K, calculated). Please see the data on the O-H (O-D) bond length at the same site (same reference) (again, pretty much the same).

$\endgroup$
1
$\begingroup$

Both isotopes are isovalent so electronically they are identical (in essence your not going to get that much difference in bond angle). However in the asymmetric well approximation since deuterium is heavier the $D-O$ bond is lowered down the well i.e. it has a lower ZPE than the $H-O$ bond and is therefore a stronger bond. This means it has a smaller equilibrium bond length (if you draw a morse potential and just draw two energy levels you can see the lower one is shorter).

Your question in part relates to the kinetic isotope effect and transition state theory where we can get into some heavy thermodynamics about reactivity and reaction coordinates. (if you like I will expand my answer here, but its slightly off topic).

The last consideration is stretching frequencies. If a bond is stronger it will require more energy to break. I think from this you can answer your remaining questions.

$\endgroup$
0
$\begingroup$

Within the Born-Oppenheimer Approximation the different species have the same structure as all nuclei are considered to be infinitely heavy. In fact, you can determine the structure of simple molecules by changing one of the atoms with one of its isotopes (a so-called isotopologue of your initial molecule) and comparing the rotational spectra. Of course the rotational constants change due to the difference in mass (similar to the change in the vibrational constants as mentioned in the comment by lemon) but the distances of the atoms w.r.t. the centre of mass remain the same.

$\endgroup$

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.