# What is the difference between the shapes of molecules with different isotopes

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).

• 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). – lemon Oct 4 '14 at 12:17
• see "Quantum Differences between Heavy and Light Water" Physical Review Letters 101, 065502 (2008) journals.aps.org/prl/abstract/10.1103/PhysRevLett.101.065502 – DavePhD Nov 7 '14 at 12:50

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).