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Chapter 10: Problem 4

Show that the rotational energy levels of a square planar \(\mathrm{AB}_{4}\) molecule may be expressed solely in terms of the rotational constant \(\bar{B}\).

Short Answer

Expert verified
For a symmetric top molecule such as a square planar \(\mathrm{AB}_{4}\) molecule, the energy levels can be expressed using only the rotational constant \(\bar{B}\) which is represented by the formula \(E_{J} = \bar{B}J(J+1)\) where \(J\) is the rotational quantum number.

Step by step solution

01

Understand Rotational Properties

The energy levels of a rotating molecule are quantized which means they can only take on certain discrete values. For a molecule rotating in three dimensions, the energy levels can be described by three rotational constants, generally denoted as \(\bar{A}\), \(\bar{B}\), and \(\bar{C}\). However, if the molecule is symmetric such as in our case of a square planar \(\mathrm{AB}_{4}\) molecule, then two of these constants are equal which we will denote as \(\bar{B}\) and the energy levels can be expressed using only \(\bar{B}\).
02

Express the Rotational Energy Levels

Let's express the rotational energy level \(E_{J}\) of a molecule in terms of \(\bar{B}\) and \(J\). For a symmetric top molecule, which includes the square planar case, the energy level can be expressed as: \(E_{J} = \bar{B}J(J+1)\) , where \(J\) is the rotational quantum number which can take values from 0 to +∞.
03

Understand J values

As explained before, the J values are the quantum number for the rotational energy levels. Each level is associated with a value of \(J\) which starts from 0 and can increase by 1 each time. For example \(J = 0, 1, 2, 3,...\). Each increase corresponds to a different rotational energy level.
04

Proof

Looking at the formula \(E_{J} = \bar{B}J(J+1)\) it is clear that it depends only on \(\bar{B}\) and \(J\). So, we have proved that the rotational energy levels of a square planar \(\mathrm{AB}_{4}\) molecule can be expressed solely in terms of the rotational constant \(\bar{B}\).

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Most popular questions from this chapter

Consider a two-dimensional harmonic oscillator with displacements in the \(x\) - and \(y\) -directions, the force constants being the same for each direction (the two bending modes of \(\mathrm{CO}_{2}\) is an example ). Show that the state resulting from the excitation of the oscillator to its first excited state can be regarded as possessing one unit of angular momentum about the \(z\) -axis. Hint. Show that \(\psi(x) \psi(y) \propto \mathrm{e}^{\mathrm{jp}}\)

\( \mathrm{What}\) is the moment of inertia of (a) a solid disc of mass \(m,\) radius \(R,\) about its axis, (b) a solid sphere of mass \(m,\) radius \(R,\) about its centre?

Find expressions for the moments of inertia of an \(\mathrm{AB}_{3}\) molecule that is (a) planar, (b) trigonal pyramidal.

In \(\mathrm{PCl}_{3}\) the bond length is \(204.3 \mathrm{pm}\) and the CIPCl angle is \(100.1^{\circ} .\) Predict the form of (a) its microwave spectrum, (b) its rotational Raman spectrum, including the general structure of the line intensities. Ignore the effects of nuclear spin statistics. Hint. Establish that \\[ I_{\perp}=m_{\mathrm{B}} R^{2}(1-\cos \theta)+\left(m_{\mathrm{A}} m_{\mathrm{B}} / m\right) R^{2}(1+2 \cos \theta) \text { for } \mathrm{AB}_{3}, \text { with } \\] \(m=m_{A}+3 m_{\mathrm{in}},\) and \(I_{\|}=2 m_{\mathrm{n}} R^{2}(1-\cos \theta) .\) Suppose that the intensities are governed predominantly by the Boltzmann distribution.

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.
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