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.

Short Answer

Expert verified
The microwave spectrum of PCl3 will display complex rotational transitions due to the non-equivalence of the moment of inertia along different axes. Similarly, the rotational Raman spectrum will reveal complex line structures. The complexity arises from the fact that PCl3 has a trigonal pyramidal shape which does not give a symmetric top and there is a dissimilarity between the inertia along different axes.

Step by step solution

01

Identification of the Molecular Shape

Identify the molecular shape of PCl3. Using the VSEPR theory, we know that when a molecule's central atom has 5 outermost electrons and is bonded with 3 other atoms, the molecule forms a Trigonal Pyramidal shape, similar to the AB3 form provided in the formula.
02

Calculation of Moment of Inertia

Calculate the moment of Inertia for PCl3 using the provided formula. Here, mb represents the mass of chlorine, R is the bond length, θ is the CIPCl angle, mA represents the mass of the phosphorus atom and, m represents the whole mass of the molecule. So, calculate I_⊥ using the formula I_⊥=m_{B} R^{2}(1−cos θ) + (m_{A} m_{B} / m) R^{2}(1+2cos θ) and I_|| using the formula I_||=2m_{B} R^{2}(1−cos θ). By substituting the given values, we find the moments of inertia.
03

Prediction of Microwave and Raman Spectra

After calculation, note that for an AB3 type molecule, as there is a difference in the inertia along the different axes, it does not follow the rules for linear molecules (I_⊥ ≠ I_||). Also, due to its shape, there is no symmetric top, so we can predict that the microwave spectrum will exhibit complex rotational transitions. For the same reasons, the rotational Raman spectrum will exhibit more complex line structures.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine all of the symmetry species spanned by the normal modes of chlorofluoromethane.

At very high values of the angular momentum \(J\), the rotational wavenumbers of a linear rotor can go through a maximum due to the presence of centrifugal distortion. Find the value of \(J\) for \(\mathrm{HCl}\left(\bar{B}=10.4400 \mathrm{cm}^{-1} \text {and } \bar{D}=\right.\) \(0.0004 \mathrm{cm}^{-1}\) ) where \(\bar{F}\) is a maximum. Hint: You will need to find a root of a cubic equation.

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.

Identify the conditions for the existence and locations of heads in the \(P\) - and \(R\) -branches of a diatomic molecule.

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

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free