A flame test for the detection of boron and the quantum mechanics of multielectron atoms

A reaction involving boric acid is used to create a combustible gas that produces a beautiful green light when burned.

Ingredients: boric acid, methanol, sulfuric acid

Procedure: A complete recipe follows.

1. Prepare a solution of 10.0 mL of methanol and 3.0 mL of glacial sulfuric acid in a test tube.

2. Add 3.0 grams of white boric acid crystals to test tube.

3. Cap test tube with rubber stopper vented through a glass tube.

4. Immerse test tube in a boiling hot water bath and heat for approximately one minute until mixture begins to produce gas.

5. Ignite gas at end of venting tube and observe flame.

Understanding: Boric acid is found to form crystals composed of sheets of molecules, interacting through hydrogen bonding. Boric acid is used as an antiseptic, in eye lotion, as a flame retardant, and in leather tanning. A commonly used pesticide consists of a mixture of boric acid and table sugar. When ingested by an insect, such as the German cockroach, the boric acid is found to be abrasive to the bug's exoskeleton. Not a good way to go.

The reaction between methanol and boric acid is catalyzed by the addition of sulfuric acid

B(OH)3(aq) + 3 CH3OH(aq) → B(OCH3)3(g) + 3 H2O(l)

Our simple reaction is used to create a flamable gas containing boron that is found to emit a unique green flame during combustion.

The flame test is a readily applied tool in the qualitative analysis of gases and liquids. The presense of certain elements in a sample can be demonstrated by exciting electrons in the element and observing the emission. Certain elements emit colors that are readily recognized, even without the use of a spectroscope. Examples include vivid strontium red, sodium yellow, pale barium green, azure copper blue, and potassium violet. A less well known example is the bright greenish-yellow emission of boron.

A wave-mechanical theory for the one-electron atom or ion

From de Broglie's relation we know that the allowed wavelengths of the particle-wave dictate the associated momentum

λde Broglie = h/(mv)

For standing waves formed in a box of length L, the allowed wavelengths are

λn = 2L/n         n = 1,2,3...

Suppose we think of an electron orbiting an atomic nucleus. If the radius of the orbit is r, the length of the orbit is 2πr. The allowed wavelengths for standing waves formed on the circular orbit are

λn = 2πr/n         n = 1,2,3...

Only certain allowed wavelengths are possible, where integer multiples of the wavelength equal the circumference of the orbit.

Using our result for the allowed wavelengths for the electron wave and the de Broglie relation, we find the kinetic energy

T = p2/2m = n2 h2/(8mπ2r2)         n = 1,2,3...

Hey! That looks just like the allowed energies of the particle-in-a-box! (Recall that the total energy of the particle in a box is its kinetic energy.) Increasing n increases the oscillatory character of the particle wave, which increases its kinetic energy.

Note that our condition on the allowed wavelengths leads to Bohr's assumption regarding the allowed values of the angular momentum, as

mvr = nh/2π         n = 1,2,3...

implies that

2πr = nh/(mv) = nλdeBroglie

Bohr's conjecture regarding the quantization of the angular momentum contained within it deBroglie's relation, λdeBroglie=h/p! Bohr had the right mathematics but lacked the deeper conceptual insight into the wave-like nature of the electron, that led de Broglie to his monumental result.

We found that for an orbit of the electron around the nucleus to be stable, there must be a balance between the inward centripetal force and the outward centrifugal force. The force balance leads to the result that the total energy, the sum of the potential energy, U, and the kinetic energy, T, is

E = T + U = U/2 = -T

The potential energy is

U/2 = - Ze2/(8 π ε0 r)

Combining these results through T = -U/2 leads to a set of allowed radii

rn = a0 (n2/Z)         n = 1,2,3...

where Z is the atomic number, the number of protons in the nucleus, and a0 = ε0 h2/ π m e2 = 5.28 x 10-10 m which is exactly Bohr's result for the allowed orbits of the one-electron atom or ion. What is impressive, however, is that we have not had to assume the ad hoc quantization of the angular momentum, as Bohr did. The result follows directly from our wave theory of the electron and the demand that there be standing waves of fixed wavelength.

If we go to the next step of identifying the allowed energies for this model, by inserting the allowed values of the wavelength into the result for the total energy, E=U/2, we find energy

En = - Ry (Z2/n2)         n = 1,2,3...

where Ry = m e4/(8 ε02 h2) = 2.18 x 10-18 J is known as a Rydberg of energy.

This is exactly the result for the Bohr model of the atom, which is coincidently the exact result for the correct wave-mechanical theory for the discrete allowed energies of the one-electron atom or ion. Just as we found for our classical spring, the assumption of a wave-like nature for the electron leads to discrete allowed energy levels when the particle is confined.

We have learned that the reactivity of a given element is a sensitive function of the energy require to ionize an atom of that element. The first ionization energy of the hydrogen atom is defined as the energy

H → H+ + e-         ΔE = I.E.

Our exact result for the allowed energies of the hydrogen atom allow us to exactly predict the ionization energy

I.E. = En=∞ - En=1 = Ry = 13.6 eV

We have an exact theory for the emission spectra and ionization energy of the hydrogen atom. How can we understand and interpret the ionization energies and emission spectra of multi-electron atoms and ions?

Estimating ionization energies of a many electron atoms or ions

Suppose that we want to understand the trend in the first ionization energy for the elements of the Periodic Table. How can we extend our quantum mechanical theory for the one-electron atom to treat the case of many-electron atoms?

Let's see how far we can get with a few basic assumptions. We assume that the most weakly bound electron will be the outermost valence electron in the atom. That electron is bound to the positively charged ion by the electron attachment energy

H+ + e- → H         ΔE = E. A. = -I.E.

Let's assume that we can model that attachment by assuming that the most weakly bound valence electron is attracted to the nucleus with an effective nuclear charge, Zeffe, which is less than the total nuclear charge, Ze. The effective charge takes into account the fact that there are other electrons in the atom or ion, and that the negative charge of those electrons will partially shield the electron from the full charge of the nucleus.

For example, think of the Li atom. There are three electrons. Two core electrons and one valence electron. For the outer valence electron, looking back at the nucleus, it would not see the full +3e charge of the three protons in the nucleus, because swarming around the nucleus are the two negatively charged core electrons. The valence electron does occasionally find itself close to the nucleus, with the core electrons a bit outside of it. So the effective nuclear charge that the valence electron sees is less than +3e, but greater than +e.

Based on that observation, we can model the mutli-electron atom as a one-electron atom where the electron is bound to a nucleus, with an effective nuclear charge, Zeffe. We can estimate the allowed energies for that beast using our result for the one-electron atom or ion

En = - Ry (Zeff2/n2)         n = 1,2,3...

where we have simply substituted Zeff for Z. Using our definition of the ionization we find

I.E. = En=∞ - En = Zeff2Ry /n2

where n is the principal quantum number of the outermost valence electron. That gives us a direct way to determine Zeff! Let's try that out. Experiments show that I.E.(H) = Ry = 13.6 eV and that I.E.(Li) = 5.39 eV. The outermost valence electron in Li has n=2. That means that Zeff(Li) = n(I.E.(Li)/Ry)½=2(5.39/13.6)½ = 1.26. The outermost valence electron in Li sees an effective nuclear charge that is somewhat greater than +e and less than +3e, just as we'd reasoned.

Emission and absorption spectroscopy of the many-electron atom

We understand the absorption of light in terms of the conservation of energy

Eparticle before + Ephoton = Eparticle after

It follows that the frequency of the absorbed light is

ν = (Efinal - Einitial)/h = Ry Zeff2 [ 1/ninitial2 - 1/nfinal2 ]

The energy of the particle increases when light is absorbed, and nf > ni.

For the emission of light, the conservation of energy dictates that

Eparticle before = Ephoton + Eparticle after

so that the frequency of the emitted light is

ν = -(Efinal - Einitial)/h = Ry Zeff2 [ 1/nfinal2 - 1/ninitial2 ]

The energy of the particle decreases when light is emitted, and nf < ni.

Ionization energy and effective nuclear charge

Question: Using experimental data for first ionization energies, compute the value of Zeff for the elements H through Ar. Plot that value versus the atomic number, Z.

You can check your answers here.

Space oddity

Question: He+ and H exist in interstellar atmospheres. Calculate all possible frequencies of light emitted as a result of transition in He+ where the initial state of the He+ ion is the third excited state (n=4).

Can light emitted by He+, initially prepared in the third excited state, excite H, initially in its ground state (n=1)? Support your answer.

You can check your answers here.

Modeling electronic transitions in multielectron atoms

Question: The atomic number of copper is 29. Compute the energy of a photon that can excite a Cu+28 ion in its ground state (n=1) to its first excited state (n=2).

In the x-ray spectrum of copper atoms, radiation of wavelength 154. x 10-12 m is emitted by an electron in the copper atom that undergoes a transition from the first excited (n=2) state to its ground (n=1) state. What is the energy of the photon that is emitted (h = 6.626 x 10-34 J s and c = 2.998 x 108 m/s)?

Compare your two results derived above. You should find that the transition energy of the one electron ion Cu+28 ion is greater than the transition energy in the multielectron Cu atom. Explain the difference between the two energies.

You can check your answers here.