Associate Professor

Chemistry

Sri Tika Ram Kanya Mahavidyalaya, Aligarh

 Uttar Pradesh, India 

The Schrodinger wave equation finds application in many chemical systems. All aromatic compounds show lowering of energy as compared to their non aromatic counterparts. Such compounds show an amazing bond strength which comes from conjugation and delocalization.

Conjugation in several systems finds support in the theory. It becomes important to understand the theory, as applied to simple chemical systems and extrapolate it to more complicated ones. The example of preparation of phenylazo β-naphthol dye from aniline and β-naphthol is a classic example of understanding the coloured product from mostly colourless compounds. It gives an easy insight into why the product is coloured to an undergraduate student.

Although, this example is that of a 2-d box problem, we can simplify it for the undergraduate student by explaining the 1-d box problem.

Writing the time independent Schrodinger Wave Equation, we have,

(-h²/8 π²m) (d²Ψ /dx²+ d²Ψ /dy²+ d²Ψ /dz²)+U Ψ=E Ψ

Writing the same for a 1-d box, we write the equation for 1 dimension.

Particle in 1-d box problem

Let us write the Schrodinger wave equation in one dimension. Such a particle stays in one dimension only, never leaving it. It can be imagined as a small particle moving along a thin wire.

(-h²/8π²m). d²Ψ /dx²+UΨ=E Ψ……    (1)

Boundary conditions:

 If we consider an electron of mass m in a box of 1- dimension whose length is ‘a.’The particle can exist anywhere between x=0 to x=a.Let the potential energy of the particle be zero inside this potential energy well and let the potential energy at x=0 and x=a as well as outside the box be ∞.

Thus, the particle is unable to cross this energy barrier and go outside the box.

We can write the Schrodinger wave equation outside the box:

(-h²/8π²m). d²Ψ /dx² +∞Ψ=E Ψ……(2)

 Multiplying this equation (2) by -8π²m/h², we get

                 d²Ψ /dx² - 8π²m/h² ∞Ψ= -8π²m/h² E Ψ

This equation can be rearranged as

                d²Ψ /dx² = 8π²m/h² ∞Ψ -8π²m/h² E Ψ

= 8π²m/h² (∞-E) Ψ……….                              (3)

For finite values of energy, equation (3) can be written as

                             d²Ψ /dx²= ∞ Ψ……       (4)

The left- hand side of this equation must be finite (Ψ is a well- behaved wave function), so the right- hand side must be finite which is possible only if Ψ=0.

Thus Ψ=0 for all points outside the box and the particle cannot exist outside the box at all.

When the particle is inside the box, the potential energy is 0. Therefore, the

Schrodinger wave equation inside the box is:

(-h²/8π²m). d²Ψ /dx² =E Ψ……..              (5)

Or, d²Ψ /dx²+ (8π²mE Ψ)/h²=0………      (6)

                  Let        k²=8π²mE/h²………..  (7)

        Thus, rewriting we have

                             d²Ψ /dx²+ k²Ψ=0 ……   (8)

 

This is a second order differential equation whose solution is of the form      

                   Ψ=A.sinkx + B.coskx …         (9)

    where A and B are arbitrary constants.

  The values of these constants can be calculated using the boundary conditions.

Since the wave function is 0 outside the box, it must also be 0 at the walls of the box as there must be a continuity in the values of Ψ at the walls of the box.

Applying the boundary conditions,

             Ψ must be 0 at x=0 and x=a.

Thus, at x=0 the equation (9) becomes:

          0= Asink.0+Bcosk.0         since sin0=0

          0=Bcosk.0

           Since cosk.0=1, therefore B=0

      Thus equation (9) is reduced to

Ψ=Asinkx ………..(10)

At the point x=a, equation (9) becomes:

           0=Asinka

For this to be true, either A=0 or sinka=0

If A=0, the wave function will become zero everywhere inside the box which is not acceptable as the particle does exist inside the box, so sinka=0

Since sinka can be zero for all values of sinnπ, where n is an integer,

                       Therefore sinka= sinnπ

                                       Or ka= nπ

                                        Or k= nπ/a ……..      (11)

Where n is an integer having values 0,1,2,3…

 Finally, the wave function for the particle inside the box becomes

                     Ψ=A sinnπx/a ………          (12)

 Using equation (7) and (11), we get

                                     (nπ/a)²=8π²mE/h²

Or,             n²π²/a² = 8π²mE/h²

or E_n =n²h²/8ma² ……….         (13)

The value of A is calculated by normalizing the wave function. The wave

function becomes:

                  Ψ=(2/a)^1/2 sinnπx/a

      For a particle moving between two points, the energy is quantized.

No such discrete levels are expected from classical mechanics.

Although n=0 is permitted but it is not acceptable as it would make the wave function 0 everywhere inside the box.

Thus, lowest energy is obtained by substituting n=1 in equation (13).

This energy is known as the Zero- point energy.

                    E _{zero point} = h² /8ma²

The salient features of the particle in a box problem are summarized below.

1) The particle is not at rest even at 0 Kelvin.

Therefore, the position of the particle cannot be precisely known.

In such a situation, only the mean value of the kinetic energy can be known.

Therefore, the momentum of the particle is also not known precisely.

The occurrence of zero- point energy is in accordance with Heisenberg’s Uncertainty Principle.

2) The allowed integral values of n come naturally as a consequence of the solution and not as an arbitrary postulate as given by Bohr.  ‘n’ is called a quantum number.

3) The energies of the electron are quantized. The only permitted values are as given in the table.

4) Plots of Ψ and Ψ 2 for different values of n are as shown.

 

The plots of Ψ and Ψ ² for different values of n are as shown. The appearance of nodes and antinodes in the wave function is another striking feature of this problem.

 The plots of Ψ versus x show that there are n-1 nodes (regions of zero amplitude and zero probability) in each wave function. The antinodes are regions of high probability e.g., at x=a/2, in case of Ψ₁ and at x=a/4

and 3a/4 in case of Ψ₂.

There are nodal points in between positions other than x=0 and x=a

5)  The probability density Ψ ² has the same number of maxima as the quantum number ‘n.’

For n=2, the probability of finding the particle at the centre of the box is zero, which is quite different from the classical result.

6) As we go to higher energy levels with more nodes, the maxima, and minima of probability curves come closer together and the variations in probability along the 1-d box become undetectable.

For higher quantum numbers, we approach the results of uniform probability density.

This agrees with ‘Bohr Correspondence Principle.’ According to this principle, the quantum mechanical result must go over to classical mechanics when the quantum number describing the system becomes very large.

7) The energy expression E_n =n²h²/8ma² showsthat energy is inversely proportional to a² i.e., square of the length of the box. Longer the box, lower will be its energy.

More localized the electron, higher will be its energy.

In chemical systems, larger the extent of delocalization, more stable is the system energetically. (For example, benzene and other conjugated systems.)

8) At a first glance, the energy expression is inversely proportional to the mass of the particle. It seems to contradict the fact that kinetic energy is proportional to mass of the particle.

However, if we understand that submicroscopic particles travel close to the speeds of light, we can understand this contradiction.

The energy expression suggests that the lighter particles will have velocities close to the velocity of light and heavier particles will have lower velocities.

This would suggest that β-rays would have higher velocities than α- rays.

 

The phenyl azo β-naphthol Dye preparation

When benzene diazonium chloride, obtained by the diazotization of aniline at 0-4⁰ C, is made to undergo coupling reaction with β-naphthol in the presence of sodium hydroxide, phenyl azo β-naphthol is obtained. It is strange, aniline, being pale yellow in colour, and β-naphthol, being light grey in colour, produce a brilliant orange red dye. We shall try to understand the reaction and the colour of the product formed in light of the problem of the 1-d box problem.

Reagents:

Aniline 5 mL

Conc. HCl 16 mL

β-naphthol 7.8 mL

10% NaOH solution 50 mL

Sodium nitrite 4 g

Procedure of preparation of dye:

In a 250 mL beaker dissolve 5 mL aniline in 16 mL conc. HCl and dilute the solution with 20 mL water. Place the beaker in ice bath. Dissolve 4 g sodium nitrite in 20mL of water in a test tube and place in ice bath. Now stirring the aniline solution add the sodium nitrite solution, temperature being maintained between 0-4⁰C.

In another 250 mL beaker dissolve 7.8 g in 50 mL sodium hydroxide solution and place in ice bath, stirring it till the solution thickens and develops a grey shine. Now add the diazotized solution of aniline, stirring continuously. Red crystals of phenyl azo β-naphthol. Allow the mixture to stand for 15 minutes and filter. Dry the precipitate and recrystallize with alcohol.

Particle in 1-d box and the colour of the dye phenyl azo β-naphthol

Looking at the reaction and structures of the reactants and products, it becomes easy to understand the lowering of energy due to conjugation. It can be seen that there is already delocalization in aniline and β-naphthol. The process of diazotization and subsequent coupling of the two molecules gives rise to additional conjugation. From two otherwise nearly colourless compounds we get a brilliant-coloured dye. The additional delocalization lowers the energy of the product considerably, leading to a lowering of energy from UV to visible region, resulting in red colour.

With increase in ‘length of the box,’ the energy lowering takes place, leading to an absorption in the visible region, making the dye a brilliant red.

This is an example of a 2-d box problem. Its energy can be written as follows:

E_nx,ny= (n_x²+ n_y²)h²/8ma²where n_x and n_y are the two quantum numbers and the length of the box is equal in both x and y directions, namely, a.

References

1.     Quantum Chemistry, B.K Sen, Tata Mc Graw Hill, 1992

2.     https://everyscience.com/Chemistry/Physical/Quantum_Mechanics/b.1129.php

3.     https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_ Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_ Theoretical_Chemistry)/Quantum_Mechanics/05.5%3A_Particle_in _Boxes/Particle_in_a_2-Dimensional_Box

4.     https://labmonk.com/to-prepare-and-submit-1-phenylazo-2-naphthol-from-aniline

5.     https://www.google.com/url?sa=i&url=https%3A%2F%2F

6.     www.cei.washington.edu%2Fwp-content%2Fuploads%2F2018%2F11%2 FNanoparticle-Lab-for-Chem-162.pdf&psig=AOvVaw1proDyDuebKeahkzambfgP &ust=1665577521247000&source=images&cd=vfe&ved=0CNcBEK-JA2o XChMIsLHv9ZXY-gIVAAAAAB0AAAAAEAM