Chapter 29

The Giancoli onLine Tutor

Molecules and Solids.

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I. Molecular Bonding

The following movie goes with the text below. It is meant to illustrate the ionic bond.
If you don't have the Quicktime plugin you will get a broken image . You can view a GIF animation version of the movie here.



The Ionic Bond - NaCl

Interactive Activity

The electron affinity of Cl is -3.61 eV (this means it gives up this much energy when it acquires an extra electron and becomes a negative ion). The energy required to remove the least bound electron in Na, the ionization energy, is 5. 14 eV. The horizontal line in the movie repesents the zero of energy corresponding to the two neutral atoms. The following questions refer to different steps in the movie. It is best to use the buttons on the right side of the movie controller to step back and forth through the different frames. After answering each question, click in the feedback field beneath the questions to get the results!


1) What is the energy (in eV) at frame A (upper right hand corner)?

2) What is the energy (in eV) at B?

3) What is the energy (in eV) at C?

4) What is r (the separation distance) at C?

5) At the potential minimum the separation is given as .24 nm, using the Coulomb potential to estimate the minimum in energy (in eV) ?



The reason the PE starts to rise abruptly when the minimum is past in the movie is that the electron 'clouds' surrounding each atom start to overlap and, because of the Pauli Exclusion Principle, the electrons are forced up into higher and higher energy levels.

Note: at the end of the movie the atoms are shown oscillating around their minimum position. This is true for all molcules but is especially easy to understand for diatomic molecules. The bottom of the potential energy curve resembles closely a parabola kx2/2, where x=(r-ro) and ro is the separation at the minimum. The potential energy kx2/2 is the potential energy of a spring, hence, vibration! From Quantum Mechanics we know that the possible vibrational energies are quantized and are given by
En = (n+1/2)hf, n = 0, 1, 2, ... .
The value of f depends upon the shape of the curve but even with n = 0, the molecule is still vibating (with energy hf/2 ), as shown.

The Covalent Bond

The ionic bond is not the only type of bond. A 'cartoon' of the Hydrogen Molecular Ion, H+ , is shown at the left. Notice that the protons equally share the lone electron. The bonding occurs because the electron spends much of its time between the two protons, thus partially shielding the positive charges from each other while at the same time providing attraction for each. Such a bond is a covalent bond. It, along with the ionic bond, are the major forms of molecular binding.


II. Moleculer Energy Levels

II. Vibrational Levels

A typical vibrational structure is shown at the left.
Like the ionic bond, the covalent bond also produces vibrational energy levels. They are evenly spaced and transistions between them are limited by the selection rule Dn = +/- 1. Generally the radiation produced is in the infrared, for example, for NaCl it is 1.5x1013 Hz and for diatomic hydrogen, H2, it is 1.3x1014 Hz. Answer the following questions :

Interactive Example

Remembering the selection rule above, what is the frequency, f, of vibration of the NaCl molecule?

What is its lowest vibrational energy (in eV) , that is, its ground state energy? Hint: use h= 4.14x10-15 eV/s?

What is the lowest vibrational energy (in eV) of H2?


Rotational Levels


Besides vibating, diatomic or polyatomic molecules can also rotate. Classically a dumbell will rotate about its middle with an energy

Note: h/2p is usually shortened to h with a slash through the riser and is called hbar.


As shown at the left unlike the vibrational energy levels, the rotational energy levels increase in size as L increases.


Calclating Rotational Energy Levels

STEP 1: Calculating I for H2
For diatomic hydrogen I = mR2/2
m=1.08 u = 1.08 u*1.66x10-27 kg/u = 1.793x10-27 kg
R = 0.074 nm
I = 4.9x10-48 kgm2


STEP 2: Calculating Rotational levels for H2
hbar2=(1.055x10-34)2= 1.11x10-68
hbar2/2I = 1.13x10-21 J = .007 eV
  • E0 = 0*.007= 0 eV
  • E1 = 2*.007=.014 eV
  • E2 = 3E1= .042 eV
  • E3= 6E1= .084 eV
  • E4 = 10E1= .14 eV


STEP 3: Calculating I for NaCl
For a diatomic molecules in which the masses are not the same, I= mR2, where R is theseparation distance and m, the reduced mass is given by m1m2/(m1+m2). There are two isotopes present in nature for Cl we will assume we are dealing with the most common for which the mass is very close to 35 u and wewill use 23 u for Na.
m=23*35/(23+35) u = 13.9 u = 2.3x10-26 kg
R = 0.24 nm
I = 1.34x10-45 kgm2


STEP 4: Calculating Rotational levels for NaCl
hbar2=(1.055x10-34)2= 1.11x10-68
hbar2/2I = 4.18x10-24 J = 26x10-6 eV
  • E0 = 0*26x10-6= 0 eV
  • E1 = 2*26x10-6=52x10-6 eV
  • E2 = 3E1= 156x10-6 eV
  • E3 = 6E1= 312x10-6 eV
  • E4 = 10E1=520x10-6 eV


Interactive Example

What type of Electromagnetic waves would result from the lowest rotational transistion in H2? (See figure 22-10 in Giancoli)

What type of Electromagnetic waves would result from the lowest rotational transistion in NaCl?

PART II



© 1999 Carl Adler mailto:Carl@Image-ination.com