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Chapter 10: Problem 2

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

Short Answer

Expert verified
The concept of moment of inertia is critical in understanding the rotation of molecules. For an AB3 type molecule, the moments of inertia depend on the geometry of the molecule. In a planar arrangement, IA=0 and IB=\(m*r^2\). As for a trigonal pyramidal configuration, IA=0 as well, while IB is obtained by summing the moments of inertia about an axis through B and perpendicular to the AB bond.

Step by step solution

01

Figure Out the Geometry of the AB3 Molecule

To find the moments of inertia, it's critical first to conceptualize the geometry of the molecule. In its planar form, all four atoms in the AB3 molecule lie in the same plane forming a triangular shape. The molecule takes on a trigonal pyramidal shape with three atoms forming a base in a plane and the fourth one perpendicular to this plane in the trigonal pyramidal framework.
02

Calculate the Moment of Inertia - Planar

In the planar framework, A lies at the center with the three B atoms at each vertex of an equilateral triangle. Therefore, all dAB (the distance between A and B) values are identical. The moment of inertia (IA) about an axis perpendicular to the plane of the molecule through A is zero as there is no mass at a distance from the axis. The moment of inertia (IB) about each B atom is \(IB = m*r^2\), where m is the mass of atom B and r is the distance AB.
03

Calculate the Moment of Inertia - Trigonal Pyramidal

In the trigonal pyramidal arrangement, atom A is at the apex while the three B atoms form the base of the pyramid. The moment of inertia (IA) about an axis through A and perpendicular to the base of the pyramid will again be zero, as all the B atoms lie in the plane perpendicular to this axis. To calculate the moment of inertia (IB) about each B atom, take the sum of the moments of inertia about an axis through B and perpendicular to the AB bond. This gives the total moment of inertia.

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Most popular questions from this chapter

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

The rotational constant of \(^{1} \mathrm{H}^{35} \mathrm{Cl}\) is \(10.4400 \mathrm{cm}^{-1}\) in the ground vibrational state and \(10.1366 \mathrm{cm}^{-1}\) in the state \(v=1 .\) Plot the wavenumbers of the \(\mathrm{P}\) -, \(\mathrm{Q}\) -, and \(\mathrm{R}\) -branches against \(J\) as a representation of the structure of the \(1-0\) transition. Take \(\tilde{v}=2990.95 \mathrm{cm}^{-1}\) and neglect anharmonicity. (The Q-branch is not observed.)

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