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IR spectroscopy which has become so useful in identification, estimation, and structure determination of compounds draws its strength from being able to identify the various vibrational modes of a molecule. A complete description of these vibrational normal modes, their properties and their relationship with the molecular structure is the subject of this article.
We are familiar with resolving a translational vector into its three components along the x, y, and z axes. Similarly a rotational motion can also be resolved into its components. Likewise the same is true for vibrational motion. The complex vibration that a molecule is making is really a superposition of a number of much simpler basic vibrations called “normal modes”. Before we take up any further description of “normal modes” it is necessary to discuss the degress of freedom.
Degree of freedom is the number of variables required to describe the motion of a particle completely. For an atom moving in 3dimensional space, three coordinates are adequate so its degree of freedom is three. Its motion is purely translational. If we have a molecule made of N atoms (or ions), the degree of freedom becomes 3N, because each atom has 3 degrees of freedom. Furthermore, since these atoms are bonded together, all motions are not translational; some become rotational, some others vibration. For nonlinear molecules, all rotational motions can be described in terms of rotations around 3 axes, the rotational degree of freedom is 3 and the remaining 3N6 degrees of freedom constitute vibrational motion. For a linear molecule however, rotation around its own axis is no rotation because it leave the molecule unchanged. So there are only 2 rotational degrees of freedom for any linear molecule leaving 3N5 degrees of freedom for vibration.
\(\int\psi_A\psi_B \;dR= 0\) (integration is done over the entire space)
The number of vibrational normal modes can be determined for any molecule from the formula given above. For a diatomic molecule, N = 2 so the number of modes is \(3\times 25 = 1\). For a triatomic linear molecule (CO_{2}), it is \(3 \times 35 = 4\) and triatomic nonlinear molecule (H_{2}O), it is \(3 \times 36 = 3\) and so on.
Example 1: Consider Water 


A linear molecule will have another bend in a different plane that is degenerate or has the same energy. This accounts for the extra vibrational mode.
Example 2: Consider Carbon Dioxide 


Example 3: The Methylene Group 

It is important to note that there are many different kinds of bends, but due to the limits of a 2dimensional surface it is not possible to show the other ones. 
The frequency of these vibrations depend on the inter atomic binding energy which determines the force needed to stretch or compress a bond. We discuss this problem in the next section. The determination of the nature of the relative displacement of each atom with respect to each other is more complicated and beyond the scope of this article. However, such motion can be seen in some common molecules as shown below.
For studying the energetics of molecular vibration we take the simplest example, a diatomic heteronuclear molecule AB. Homonuclear molecules are not IR active so they are not a good example to select. Let the respective masses of atoms A and B be \(m_A\) and \(m_B\). So the reduced mass \(\mu_{AB}\) is given by:
\[ \mu_{AB}=\dfrac{m_A\, m_B}{m_A+m_B} \]
The equilibrium internuclear distance is denoted by \(r_{eq}\). However as a result of molecular vibrations, the internuclear distance is continuously changing; let this distance be called \(r(t)\). Let \(x(t)=r(t)r_{eq}\). When \(x\) is nonzero, a restoring force \(F\) exists which tries to bring the molecule back to \(x=0\), that is equilibrium. For small displacements this force can be taken to be proportional to \(x\).
\[ F=kx \]
where \(k\) is the force constant.
The negative sign arises from the fact that the force acts in the direction opposite to \(x\). This is indeed a case of Simple Harmonic Motion where the following well known relations hold.
\[x(t)= A \sin \left( 2\pi \nu t \right) \]
where
\[ \nu=\dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu_{AB}}}\]
The potential energy is given by \( V=\frac{1}{2}kx^2\). The total energy \(E\) (Kinetic+Potential) is obtained by solving the Schrödinger equation:
\[\dfrac{h^2}{8\pi^2\mu_{AB}} \dfrac{d^2\psi}{dx^2}+\dfrac{1}{2} kx^2\psi = E\psi\]
A set of wave functions \(\psi_n\)) and the corresponding Eigenvalues \(E_n\) are obtained. \(E_n=(n+(1/2))hv\) where \(n\) is an integer (1,0,1,2 etc.). The energy is quantized, the levels are equally spaced, the lowest energy is \((1/2)hv\), and the spacing between adjacent levels is \(hv\).
As show above, the energy difference between adjacent vibrational energy levels is hv_{vibration}. On the other hand, the photon energy is hv_{photon}. Energy conservation requires that the first condition for photon absorption be,
Hv_{vibration} = hv_{photon} or v_{vibration} = v_{photon}.
Such photons are in IR region of the electromagnetic spectrum. In addition, two more conditions must be met.
Spectroscopy in the IR region can determine the frequency and intensity of absorption. These frequencies are generally specific for a specific bonds such as cc, c(double bond)c, c(triple bond)c, co, c(double bond)o, etc. So the IR absorption data is very useful in structure determination. The intensity depends on the concentration of the resposble spec. So it is useful for quantitative estimation and for identification.
1.) NH_{3} – 6
C_{6}H_{6} –30
C_{10}H_{8} –48
CH_{4} –9
C_{2}H_{2} – 7
2.) N_{2} – IR inactive
C0 – active
C0_{2} (stretching) – inactive
HCl – active
3.) m_{AB} = m_{A}xm_{B}/(m_{A}+m_{B}) = 11.395x10^{27}
v = (1/2pi)(k/m_{AB})^{.5 }= 2143.3 cm^{1}
4.) Energy of the mode for n = 0
E_{0} = (1/2)hv = 2.13x10^{20}J
Energy per mole = 2.13x10^{20}x6.022x10^{23} = 12.8KJ/mole
5.) v for D_{2}O will be lower because v is inversely proportional to 1/(m^{.5}), where m is the reduced mass.
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