Consider the H2 molecule. Now you replace one H atom with a deuterium atom ... H2  HD. What is the effect on the vibrational energies? (A) The vibrational energies decrease, because the force constant K changes. (B) The vibrational energies increase, because the force constant K changes. (C) The vibrational energies increase, because the reduced mass  changes. (D) The vibrational energies decrease, because the reduced mass  changes. (E) The vibrational energies stay the same, because the changes in  and K cancel each other.

Consider the H2 molecule. Now you replace one H atom with a deuterium atom ... H2  HD. What is the effect on the vibrational energies? (A) The vibrational energies decrease, because the force constant K changes. (B) The vibrational energies increase, because the force constant K changes. Both (A) and (B) are wrong. The molecular bonding is caused by the electrons. Since H and D both have the same electron configuration, the chemical bonding properties of deuterium are the same as for H. Therefore, K will not change. (C) The vibrational energies increase, because the reduced mass  changes. ... wrong, see (D) (D) The vibrational energies decrease, because the reduced mass  changes. ... correct: since the reduced mass increases and , the vibrational energies decrease. (E) The vibrational energies stay the same, because the changes in  and K cancel each other. ... wrong, see all above.

What problems do we encounter with the HO model, if we want to describe real molecules for higher energies? (A) there is no dissociation in HO model (B) the potential curve in most real molecules becomes steeper for large distances and flatter for short distances (C) the potential curve in most real molecules flattens out for large distances and becomes steeper for short distances (D) A and B (E) A and C

What problems do we encounter with the HO model, if we want to describe real molecules for higher energies? (A) there is no dissociation in HO model (B) the potential curve in most real molecules becomes steeper for large distances and flatter for short distances (C) the potential curve in most real molecules flattens out for large distances and becomes steeper for short distances (D) A and B (E) A and C V(x) x