# Solutions 11

- Page ID
- 52477

## Q1

We are given

\[B(I_1) = \dfrac{h}{8 \pi^2 \mu R^2} = 10.5909 \,cm^{-1}\]

We can find \(B(I_2)\) and \(B(I_3)\) by using the relationship

\[B(I_a) = B(I_b) \dfrac{B(I_a)}{B(I_b)}\]

which applies for any pair of rigid rotators. Assuming bond lengths are the same and using the fact that \(B\)

is inversely proportionate to I, we have \(\dfrac{B(I_a)}{B(I_b)} = \dfrac{I_b}{I_a}\). Therefore we have \(B(I_a) = B(I_b)\dfrac{I_b}{I_a}\).## Q2

To arrange the molecules in order of increasing \(\tilde{B}\), we arrange them in order of decreasing moment of inertia, I:

\(\ce{H–C#C–C#C–C#N}\), \(CICN\), \(^{12}C^{18}O\), \(^{13}C^{16}O\), \(^{12}C^{16}O\), \(DF\), \(HF\), \(HD\).

## Q3

Moment of inertia is dependent upon a defined axis of rotation. In the absence of a specified axis, we are free to choose any axis. We can choose the middle carbon atom as the axis of rotation and apply the definition of moment of inertia, \(I\):

\[I = \sum m_i r_i^2\]

where \(m_i\) is the mass of each atom in kg and \(r_i\) is the distance away from the axis of rotation (ie. the bond length) in meters.

We can use \(I\) to find the rotational constant \(B\) via the relation:

\(B = \dfrac{h}{8 \pi^2 I}\)

We can then use this constant in a previously used expression, \(\Delta E = 2Bh(J+1)\), where J refers to the lower quantum number in each transition (ie, 0 and 1), and this energy jump can be used to find the frequency of transitions via: \(E = h\nu\) where \(\nu\) is frequency and \(h\) is Planck's constant.

## Q4

a). \(J\) is the angular momentum quantum number that describes amount of angular momentum the molecule has. \(\tilde{B}\) is the rotational constant giving the proportionality to the second derivative in the Schrodinger equation. Since centrifugal distortion changes the energy, \(\tilde{D}\) is the constant that represents the decrease from this distortion.

b). The selection rule is \( \Delta J = \pm 1 \)

c) If \(E_J = hBJ(J+1) + hDJ^2(J+1)^2 \) then \( \Delta E_{J \rightarrow J+1} = E_{J+1} - E_J \) =

\( 2hB[(J+1)(J+2) - J(J+1)] + hD[J^2(J+1)^2 - (J+1)^2(J+2)^2] \) =

\( 2hB(J+1) + hD(J+1)^2(J^2-J^2-4J-4) \)

= \( 2hB(J+1) -4hD(J+1)^3 \)

Variations of this expression exist for each unit of B.

d) You can take an array of the peak wavenumber values and J values (1, 2, 3, 4) and do a least squares fit to the above expression using a scripting language like Matlab etc. Alternatively we can set any two peak wavenumbers equal to the corresponding J transitions for the situation of two equations to solve for the two unknowns, B and D:

\[\Delta E_{0 \rightarrow 1} = 22.018\, cm^{-1} = 2 \tilde{B}(0+1) - 4\tilde{D}(0+1)^3\]

\[\Delta E_{1 \rightarrow 2} = 44.0218\, cm^{-1} = 2\tilde{B}(1+1) - 4\tilde{D}(1+1)^3\]

etc.

Using the resulting value of B, we can find the bond length finding the reduced mass of HF and using it in the expression:

\[R = \sqrt{\dfrac{h}{8 \pi^2 \mu B}}\]

## Q5

a). We can follow the same procedure in Q4d to the find the constants and bond length of HCl.

b). The rotational constants for DCl are likely to scale with the difference in moment of inertia. We can use the same procedure in Q1. We have one rotational constant \(B(I_{HCl})\) and we are asked to find another \(B(I_{DCl})\) and we know \(I_{HCl}\).

We can calculate \(I_{DCl}\) via the ratio of reduced masses:

\[\dfrac{I_{HCl}}{I_{DCl}} = \dfrac{\mu_{HCl}}{\mu_{DCl}}\]

and use it to find \(B(I_{DCl})\) via the relation:

\[B(I_{DCl}) = B(I_{HCl}) \dfrac{I_{HCl}}{I_{DCl}}\]

## Q6

The \(Ne_2\) ground state electron configuration is

\[ (\sigma_{1s} )^2(\sigma^*_{1s} )^2 (\sigma_{2s} )^2(\sigma^*_{2s} )^2 (\sigma_{2p} )^2 (\pi_{2p} )^4 (\pi^*_{2p} )^4 (\sigma^*_{2p} )^2 \]

With a bond order of 0, it is unstable.

Promoting one electron from an antibonding to bonding orbital results in an electronic configuration of

\[ (\sigma_{1s} )^2(\sigma^*_{1s} )^2 (\sigma_{2s} )^2(\sigma^*_{2s} )^2 (\sigma_{2p} )^2 (\pi_{2p} )^4 (\pi^*_{2p} )^4 (\sigma^*_{2p} )^1 (\sigma_{3p})^1\]

The excited state has a bond order of 1 so it is relatively stable.

Energy would be absorbed as we form excited state \(Ne_2\) molecules and emitted breaking down to separate Ne atoms.