The \(J+1 \leftarrow J\) rotational transitions of \(^{16} \mathrm{O}^{12} \mathrm{C}^{32} \mathrm{S}\) and \(^{16} \mathrm{O}^{12} \mathrm{C}^{34} \mathrm{S}\) occur at the following frequencies \((v / \mathrm{GHz})\) $$\begin{array}{lllll} \hline J & 1 & 2 & 3 & 4 \\ \hline^{16} \mathrm{O}^{12} \mathrm{C}^{32} \mathrm{S} & 24.32592 & 36.48882 & 48.65164 & 60.81408 \\ ^{16} \mathrm{O}^{12} \mathrm{C}^{34} \mathrm{S} & 23.73223 & & 47.46240 & \\ \hline \end{array}$$ Find (a) the rotational constants, (b) the moments of inertia, and (c) the CS and CO bond lengths. Hint. Begin by finding expressions for the moment of inertia \(I\) through \(I=m_{\Lambda} R_{A}^{2}+m_{B} R_{B}^{2}+m_{C} R_{C}^{2},\) where \(R_{X}\) is the distance of atom \(\mathrm{X}\) from the centre of mass. The easiest procedure is to use the result established in Exercise \(10.6,\) which leads to \(I=\left(m_{\mathrm{A}} m_{\mathrm{C}} / m\right)\left(R_{\mathrm{AB}}+R_{\mathrm{BC}}\right)^{2}+\left(m_{\mathrm{B}} / m\right)\left(m_{\mathrm{A}} R_{\mathrm{AB}}^{2}+m_{\mathrm{C}} R_{\mathrm{BC}}^{2}\right)\) The lengths \(R_{\text {AB }}\) and \(R_{\mathrm{BC}}\) may be found only if two values of \(I\) are known. Assume the bond lengths are the same in isotopomeric molecules.

Short Answer

Expert verified
Rotational constants, moments of inertia, and bond lengths can be calculated from the provided transition frequencies, using expressions that relate these quantities to each other. The bond lengths can be found by assuming they are the same in isotopomeric molecules.

Step by step solution

01

Determining the rotational constants B

From the definition, the rotational constant \(B\) is equal to the frequency of the \(J+1 \leftarrow J\) transition divided by \(2J + 1\). In other words, \(B = \frac{v_{J+1,J}}{2J + 1}\), where \(v\) is the transition frequency. For each molecule and each \(J\) value, calculate the respective \(B\) value.
02

Calculating the moments of inertia I

The relationship between the rotational constant \(B\) and the moment of inertia \(I\) is given by \(B=\frac{h}{8 \pi^2 c I}\), where \(h\) is Planck's constant and \(c\) is the speed of light. Solving this equation for \(I\) leads to \(I = \frac{h}{8 \pi^2 c B}\). Substitute the calculated \(B\) values into this equation to find the corresponding \(I\) values.
03

Finding the CS and CO bond lengths

The provided information leads to an expression for \(I\) in terms of the bond lengths \(R_{AB}\) and \(R_{BC}\). Solve this expression for the bond lengths given the \(I\) values calculated in step 2. You might have to solve a system of two equations if two \(I\) values are given.

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