PHYSICAL CHEMISTRY

FUNDAMENTALS OF QUANTUM CHEMISTRY

Dr. Sourav Pal Department of Physical Chemistry

National Chemical Laboratory Dr. Homi Bhabha Road

Pune – 411 008 Email: s.pal@ncl.res.in

CONTENTS Black-body radiation Planck Energy Distribution Formula The Photoelectric Effect The Compton Effect Heat capacities of solids Atomic Spectra Bohr's model of hydrogen atom Wave-Particle Duality The Uncertainty Principle The Schrodinger’s equation Particle in 1-D box The Hydrogen Atom Atomic structure and the different quantum numbers Quantum mechanical models of the chemical bond Multiple bonds between unlike atoms Resonance MO description of homonuclear diatomic molecules Bond Order Some important consequence from molecular orbital theory Molecular orbital theory of heteronuclear diatomic molecules Hydrogen Fluoride Huckel theory: Bonding in polyatomic molecules HMO treatment of benzene To understand the chemical science, it is essential to understand the structure of individual atoms and molecules. In seventeenth century Sir Isaac Newton discovered the laws of classical mechanics. In the beginning of the nineteenth century, the physicists found that the laws of classical mechanics were not adequate to describe the nature of microscopic particles like atoms, molecules, electrons etc. Experiments showed that laws of classical mechanics fail to describe the motion of microscopic particles. These lead to the introduction of the revolutionary theory in physics ‘the quantum mechanics’. In this chapter we will discuss the basics of quantum mechanics as applied to the atomic and molecular structure, chemical bonding and molecular spectroscopy. In the beginning, we discuss some classic

experiments, which showed the failure of classical laws in the regime of atoms and molecules. This will be followed by development of early quantum theory to the description of atoms and molecules within the framework of quantum mechanics. Black-body radiation A black body is defined to be a theoretical object that absorbs 100% of the radiation falling on it. It reflects no radiation and is perfectly black. At the same time, it is also a perfect emitter of radiation. At a particular temperature the black body would emit the maximum possible amount of energy at that temperature. In practice, no material has been found to absorb all incoming radiation. A good approximation of black body is a pin hole of an empty container maintained at a constant temperature, because any radiation leaking out of the hole has been absorbed and re-emitted inside so many times that it has come to thermal equilibrium with the walls. Black Body Radiation Curves

Figure 1: Theoretical black body curve for 5000K Fig 1 is an illustrative curve of energy emitted at every wavelength by a black body at 5000K. It shows that the black body does radiate energy at every wavelength. The curve touches the x-axis at infinite wavelength. It has a maxima at a particular wavelength, at which most of the radiant energy is emitted. Wein (1894) gave an empirical formula to relate the maximum wavelength with temperature, which is known as Wien’s displacement law. According to this law max 2.898T mKλ = (1.1) Stefan-Boltzmann law: This law was given by Stefan experimentally (in 1879) and was explained theoretically by Boltzmann (in1884) using the law of thermodynamics. This law states that the power radiated in a blackbody is proportional to the fourth power of its absolute temperature, that is

4P ATσ= (1.2)

where, P = Power radiated in W (J/s) σ = Stefan's Constant 5.67 x 10-8 W m-2T4

A = Surface area of body (m²) T = Temperature of body (K)

The physicist Lord Rayleigh studied the nature of black body rediation from the classical viewpoint. Along with Jeans he explained the problem considering the radiation within the black body cavity to be made up of a series of standing waves. They proposed that electromagnetic radiation was emitted by oscillating atoms in the walls of the black body and therefore the standing wave between the walls created. Their formula stated:

42 ckTπλ

Ι = (1.3)

where, I is the intensity of the radiation. The Rayleigh Jean equation explains the nature of the blackbody radiation well for large wavelengths but had major problems at shorter wavelengths. Due to having 4λ term in the denominator, when the wavelength tended to zero, the curve would tend to infinity. So according to Rayleigh-Jeans Law there is no maxima in the blackbody radiation curve (which is not the case of the graph). Moreover in this case a very large amount of radiation will emit in the lower wavelength region (ultraviolet wavelength). This absurd behavior of Rayleigh-Jeans formulation is known as the ultraviolet catastrophe.

Figure 2: Comparison of Experimental black body data and Rayleigh-Jeans Law Planck Energy Distribution Formula In the 19th century a major problem for physicists was to predict the intensity of radiation emitted by a black body at a specific wavelength. A major breakthrough was made by Max Planck who came up with a formula that agreed with experimental data, which is

2

5

2

( 1)hc

kT

hcA

eP

λ

π

λ −=

(1.4)

Pλ=Power per m² area per m wavelength h = Planck's constant (6.626 x 10-34 Js) c = Speed of Light (3 x 108 m/s) l = Wavelength (m) k = Boltzmann Constant (1.38 x 10-23 J/K) T = Temperature (K) This equation proved to agree remarkably well with experiment and had come to be known as Planck’s radiation law. However, he had major problems proving this law. In attempting to explain this law Planck made new hypothesis that the radiation can not be emitted or absorbed in any arbitrary amount but only in quanta of energy. This quanta of energy is E = hν, where h is the Planck constant (6.626 x 10-34 Js) and ν is the frequency. The Planck’s equation can explain the nature of blackbody radiation curve both is short as well as large wavelength region.

For short wavelengths, hckTλ is very large and

hckTe λ → ∞ faster than λ → 0 and therefore from

Eq. (1.4), P → 0 as λ → 0.Hence energy density approach to zero as wavelength approaches to zero. For long wavelengths, hc

kTλ 1. So

1 (1 .....) 1hc

kT hchce kT kTλ

λ λ− = + + − ≈

and the Planck;s distribution formula reduces to Rayleigh -Jeans equation.

Again Wein’s law can be obtain from Planck’s distribution law by looking wavelength at which AdPdλ

=0,

the condition for the maximum in the distribution. Making the approximation that the wave length is so short so that hc

λ kT, we get

max 5hckTλ =

Source material for figures: www.egglescliffe.org.uk/physics/astronomy. The Photoelectric Effect Another most well known experiment, which could not be explained by conventional wave picture of light, was photoelectric effect. Albert Einstein in 1905 extended the photon picture to explain - the photoelectric effect. In this effect light is fall on a metal electrons are released. These electrons can be attracted towards a positively charged plate placed at a certain distance below the metal plate and thereby photoelectric current can be produced. One can measure this current conveniently but to measure the stopping potential Vo one require to reduce this current to zero. The maximum kinetic energy of the ejected electrons is related to the stopping potential by eV0 = KE max = 1/2 mν max

2 ...... (2.1) The experimental graph of the stopping potential V0 vs. the frequency ν is plotted in figure 5. This is a straight line, having an x –intercept. This frequency is called cutoff frequency, νc where V0 = 0.

Below this cutoff frequency, νc there is no emission of photoelectrons whatever the applied potential is. This aspect of the photoelectric effect is impossible to understand within the wave picture of light, because within this picture the energy of the light beam which gives the electrons does not depend on the frequency.

Figure 3 Einstein came up with an explanation of the photoelectric effect which built upon Planck's photon hypothesis. In this theory Einstein assumed that Light consists of particles (photons), and the energy of such a particle is proportional to the frequency of the light. E nhν= . (2.2) When these photons hit the metal, they could give up some or all of their energy to an electron. A certain amount of energy would be required to release the electrons from their bonds to the metal - this energy is called the work function, φ, of the metal. Thus the incident light can emit the electrons from the metal surface only when its photons have frequency more than the cutoff frequency of the metal. The remaining energy would appear as kinetic energy of the released electron. Thus, the maximum kinetic energy the electrons could have is KE max = eVo = hf -φ = hc/λ - φ (2.3) And the cutoff frequency, νc = φ/h = c/λ Where, λ is the corresponding cutoff wavelength. The Compton Effect In early 1920, Compton’s experiments prove the wave nature of the light. He described the change in wave length of X-rays produced by scattering and he demonstrated an experiment which is famously known as Compton Effect. Photons have energy, thus from the theory of relativity we can say that they have mass and therefore they obey the law of conservation of momentum.

V

νc υ

Let us consider a photon striking a static electron which is at rest at the origin relative to the coordinate frame, (thermal motion of the electrons has been neglected). Let us assume that the impact is almost at a grazing angle. So the electron after receiving the impulse begins to move below the X-axis, the photon is deflected above the X-axis

Figure 4 On the basis of the conservation of momentum and energy change in wave length between the original photon and scattered photon at an angle will be

∆ λ = λ / - λ = 0

(1 coshm c

− ө) (3.4)

Therefore the wavelength of scattered photon is always greater than the incident wave length. It formula has been verified with the experimental result. It has been also found that one value of the wavelength shift is observed at a given scattering angle, which implies that the momentum transfer takes place in a discrete manner not continuously. Heat capacities of solids The first experimental studies on heat capacities of solids were performed by French scientist Pierre-Louis Dulong and Alexis-Therese Petit. They determined the heat capacities of a number of mono-atomic solids. On the basis of these experimental results they told that the heat capacity of all the mono-atomic solids are same and it is close to the value 25 J 1 1K mol− − ( in modern units). The Dulong-Petit law can be easily justified in terms of equipartition principle from classical physics at ordinary temperatures. The theorem of equipartition of energy states that molecules in thermal equilibrium have the same average energy associated with each independent degree of freedom of their

motion and that the energy is 12

kT per molecule. According to this principle the mean energy of an atom

vibrate about its mean position of solid is kT for each direction of vibration. As each atom vibrates in three dimensions the total average energy per atom I the solid is 3 kT . So the molar energy is 3N kT , where N is the Avogrado’s number.

So,

3 3NkT RT= =U (4.1)

where R is the universal gas constant.

Therefore the heat capacity of the mono atomic solids is

1 1, 3 24.9m

V mUC R JK molT

− −∂⎛ ⎞= = =⎜ ⎟∂⎝ ⎠ (4.2)

This result agreed with the Dulong-petits law of heat capacity of monoatomic solids.

There is serious deviation of the heat capacity of monoatomic solid given by Dulong –Petit’s law. It was found that the heat capacity decreases and goes to the value zero as 0T → . Einstein accounted the fact by taking the assumptions of Planck’s Quantum hypothesis. Einstein recognized that for a quantum harmonic oscillator at energies less than kT. If the atoms in the solid vibrate with the frequency,ν the statistical distribution of energy in the vibrational states gives average energy:

/ 1h kT

hUe ν

ν=

− (4.3)

There are three degrees of freedom per vibrator, so the total energy is 1

/

31h kT

NhU molee ν

ν −=−

(4.4)

Differentiating the expression we get the following equation, 2 /

1V / 2

3 ( )

( 1)

h kT

h kT

hNk eU kTC moleT e

ν

ν

ν−∂

= =∂ −

(4.5)

In this treatment the appropriate frequency in the expression had to be determined empirically by comparison with experiment for each element. The quantity hυ/k is sometimes called the Einstein temperature. Although the calculated heat capacities of different mono atomic solids using this theory match with experiment reasonably, but it was not exact. Debye advanced the treatment by treating the quantum oscillators as collective modes in the solid which are now called "phonons". Atomic Spectra One of the most important problems to the physicist at that time was to explain the nature of atomic spectra. At that time it was known that every atom emits electromagnetic radiation of characteristic frequencies when it is subjected to high temperatures or electrical discharge. These emission spectra of atoms consist of only certain discrete frequencies and that is why they are called line spectra. The simplest spectrum is the spectrum of hydrogen atom as it is the lightest and simples atom.

Figure 5: The Spectrum of Hydrogen atom The figure represents spectrum of hydrogen atom where a part is in the visible and near ultraviolet region. Since the atomic spectra are the characteristic of the atom involved, it is reasonable to conclude that the spectrum depend on the electronic distribution of the electron on the atom. In 1885 Swiss scientist Johann Balmer showed that the frequencies of the emission lines in the visible region of the spectrum, could be describe by the equation

1

2

48 2 2 0 .2 (1 ) mn

υ −= −

(5.1) where n =3,4,5…….This equation is customarily written in terms of the quantity 1

λ instead of ν. The

reciprocal of wavelength is called a wavenumber. So rearranging the we get the well know Balmer’s formula

12 2

1 110968000( )2

mn

υ −= − (5.2)

where υ = 1λ = c

υ .

Balmer’s formula can successfully account the hydrogen atom pectra in the visible and ultraviolet regions. After this by generalizing the Balmer’s formula Swiss spectroscopist Johannes Rydberg gave a formula which accounted the hydrogen atom spectra in the entire region. This is known as Rydberg’s formula

12 2

1 2

1 1 110968000( )mn n

υλ

−= = − (5.3)

where both n1, n2 are integers in this equation but n2 is always greater than n1. The constant 10968000 m-1 is called Rydberg constant. Bohr's model of hydrogen atom The wave classical electromagnetic theory was unable to explain the discrete nature of atomic spectra. Neils Bohr in1913 developed the theory of the atomic spectra by taking the Planck’s quantum hypothesis.

Bohr made some bold assumptions which violate the law of classical mechanics and electromagnetic theory in his atomic theory. There are following three fundamental postulates in this model. a) The only allowed orbits for the electrons are characterized by values of orbital angular momentum given by L = n ( h / 2.) ............ n = 1, 2, 3, These orbits are called stationary states. n is called the principal quantum number. (b) As long as an electron revolved in a stationary orbit it does not radiate any electromagnetic energy.

Figure 6 (c) During each emission or absorption of radiation, there is the transition of electron from the one stationery orbit to another. The radiation emitted during such transition homogeneous and the frequency of the emitted light is given by the formula i fh E Eυ = − , where h is the Planck constant, iE and fE is the energies in the two stationary states of the atom. If an electron of mass m moving in nth orbit around nucleus then according to Bohr’s atomic model one can derive that the energy of the electron will be

4

2 2 208nmeE

h nε= −

The main success of Bohr's theory was that it could explain the phenomenon of spectral lines of the hydrogen atom. Let's assume that the electron is on the first stationary orbit. After supplying some energy to the atom the electron can "jump" to one of the higher orbits. This electron will come back to the first orbit after emitting the energy and the energy will emit in the form of radiation. The frequency (v) of this radiated light will be according to the following equation: hv= En – E1 where, En is the energy of the electron on the orbit, of which it comes back, E1 is the energy of the electron on the first orbit. Using the above two equations we get:

4

2 3 2 20

1 1( )8 1me

h nυ

ε= −

From this we get the Rydberg constant4

2 308

meRchε

= .

The calculated value of this constant from Bohr's theory has agreed with the experimental value. Similarly one can calculate the emitted energy when the electron goes down from the higher orbit to any lower one. The frequencies of radiation calculated by using Bohr's atomic theory for hydrogen atom spectrum is in agreement with an experimental data. The Bohr's theory describes well the spectrums of the atoms around which the only one electron circulates. Such atoms are: H, He+, Li+2 ( hydrogen-like). The major drawback of this theory is that it cannot describe the spectrums of the atoms around which the two ore more electrons circulate. Soon it was discovered that the spectrum lines of atoms are not homogenous but they consist of several convenient lines. For example for n = 2 there are two such lines, just like there were two electron orbits of almost identical energies. The problem was solved in 1916 by Arnold Sommerfeld. Wave-Particle Duality Now let us go back to the question, ‘what is light’. Is it the electromagnetic wave having an oscillating electric field and a magnetic field perpendicular to each other as described by Maxwell or composites of particles, photon. We have seen that some experiments such as blackbody radiation, the photoelectric effect, where the wave nature of light fails to explained but the photon picture of light gives the proper descriptions. However, some experiments such as diffraction and interference all need the wave picture, as a photon (particle) picture fails. Both pictures are needed in different circumstances; one says that light exhibits a wave-particle duality: Light has a dual nature; in some cases it behaves as a wave, and in other cases it behaves as a photon. de Broglie Waves In 1924 a young physicist, de Broglie, speculated the fact that not only the light has this kind of wave particle duality but every ordinary particle such as electrons, protons, or bowling balls could also exhibit wave nature. For giving a quantitative description, about how this wave nature of matter associated with particle nature he gave the following equation λ=h/mv................... (7.1) Since the momentum of such a particle is p = mv a particle having high momentum has short wavelength. Macroscopic bodies due to having very high momentum (even if when move very slowly) their wavelengths are undetectably small, and wave like properties can not be observed. However, we should emphasize that these two equations have a very different physical content. Tests of this hypothesis would involve demonstrating wave properties of matter at the wavelength given by Eq.(7.1). Relatively straightforward tests are offered by diffraction and interference - if a beam of such ``particles'' were shone at a diffraction grating and a diffraction pattern of a series of light and dark fringes results, then one would be forced to adopt the wave picture for this phenomena. We recall that for a good diffraction pattern to result the size of the diffraction slits should be of the same order as the wavelength of the light used. As we shall see in some examples later, for macroscopic objects such as bowling balls this would require sizes of slits of the order of 10- 34 m or so, which is much outside

present-day technology. However, for electrons the sizes of slits required are of the order of 10- 11 m or so, which are readily available. Thus, it is possible to verify the wave nature of electrons in such diffraction experiments, and indeed this property is the principle behind the relatively common electron microscope. Therefore, Nature seems to be symmetric, in that light and ordinary ``particles'' exhibits this wave-particle duality. The Uncertainty Principle One important consequence of the wave-particle duality of nature was discovered by Heisenberg, and is called the uncertainty principle. To measure the position and the momentum of a particular particle we must ``see'' the particle, and so it should be shined on some light of wavelength λ.. Again there is a limit to the resolving power of the light used to see the particle given by the wavelength of light used. This gives an uncertainty in the particle's position:

∆x λ (8.1) We got these results by considering the light as a wave. However, when we consider it as composites of photon, the light when striking the particle could give up some or all of its momentum to the particle. But we don't know how much it gave up, as we can’t measure the photon's properties, there is an uncertainty in the momentum of the particle; using following Eq. we find ∆p h/λ (8.2) Combining Eqs.(8.1) and (8.2), we find ∆x ∆p .......... (8.3) This equation is independent of the wavelength used, and from this we can say that in principle there is a limitation to measure accurately and simultaneously measure the position and momentum of a particle. If one tries to measure the position more accurately by using light of a shorter wavelength, then the uncertainty in momentum will increase, whereas in order to reduce the uncertainty in momentum if one uses light of a longer wavelength, then the uncertainty in position will be high. Thus one cannot no way reduce both uncertainties to zero simultaneously. This is a direct consequence of the wave-particle duality nature of particle. The arguments used in deriving Eq.(8.3) are somewhat rough. A more refined treatment, developed by Heisenberg, results in the following relation: . ∆x ∆p ≥ h/4π .......... (8.4) For everyday macroscopic objects such as bowling balls the uncertainty principle plays a negligible role in limiting the accuracy of measurements, because in those cases the wave nature of matter is undetectably. However, for microscopic objects such as electrons in atoms the uncertainty principle does become a very important consideration.

The Schrodinger’s equation From de Broglie 's relation we see that microscopic particles due to having very low momentum, have long wave lengths, that is wave nature dominates. For these particles we can suppose that a particle is distributed through space like a wave rather than traveling along a path. The mathematical representation of wave that in quantum mechanics replace the concept of classical trajectory is called a wave function Ψ. In 1926, the Austrian physics Erwin Schrodinger proposed an equation for finding the wave function of any system. The time-independent Schrodinger’s equation for a particle of mass m moving in one dimension (x) with energy E and potential V(x) -(h2/8π2m)d2ψ(x)/dx2 +V(x)ψ (x) = Eψ (x) ..........(9.1) Where V(x) is the potential energy function for the system under consideration.. In the general case the Schrodinger’s equation is written as Hψ =Eψ Where H is the Hamiltonian operator for the system: The wave function, which satisfies the Schrodinger’s Equation represents a state of the system, where measurement of energy gives a definite value, given by E and is considered as a “pure state” of energy. In fact, this provides the significance of the Schrodinger’s Equation in quantum mechanics. However, the state can, in general, may not be an eigen state of H, in which measurement of energy is does not give definite values and only an average value description may be used for such a state. We can get the de Broglie relation for a freely moving particle in the region of constant potential energy V from the above equation (9.1). Substituting V(x) by V in the equation and rearranging it we get d2ψ/dx2 =-(h2/8π2m)(E-V) .........................(9.2) A solution of this equation is ψ =eikx =cos kx + i sin kx k={8π2m(E-V)/h2}1/2 .............(9.3) Now we know that cos(kx), or sin (kx) is a wave of wave length λ = 2π /k, as we can be seen by comparing cos kx with the standard form of harmonic wave, cos(2πx/λ) .The quantity (E-V) is equal to the kinetic energy of the particle, Ek , so k= { 8π2mEk/h2 }1/2 which implies that Ek = k2 h2/8π2m. Because Ek=p2/2m follows that p = kh/2π Therefore, the linear momentum is related to the wave length of the wave function by p = 2π⁄λ × h/2π =h/λ which is de Broglie’s relation.

The Born interpretation of wave function: Quantum, mechanics says that the wave function contains all the dynamical information about the system it describes. The physical interpretation of wave function was given by Max Born. The Born interpretation says that the value of the square of the modulus of the wave function

2ψ at a point is proportional to the probability of finding the particle at that point. The modulus square of the wave function 2ψ is called as the probability density. Thus the probability of finding that particle in the infinitesimal volume dV is P= 2ψ dV In general the wave function itself has no direct physical significance as it may be negative, or complex. The square modulus of the wave function is always a positive quantity and is directly physically significant. Now we see that if ψ is a solution of Schrodinger’s equation then Nψ is also a solution of the solution of that equation where N is a constant .So if we vary the wave function by a constant factor means that it is always possible to find a normalization constant, N, such that the proportionality of the Born interpretation becomes an equality. We see that for a normalized wave function Nψ, the probability of finding the particle in the region dx is equal to (Nψ*)(Nψ )dx. Now the sum over all space of these individual probabilities must be equal to 1. Mathematically we can write N2 ∫ψ*ψ dx = 1 where the integration is over all the space accessible to the particle. From this equation, we can easily find out the value of the normalization constant N. Particle in 1-D box The time-independent Schrodinger’s equation for a particle moving in one dimension can be written as:

2 2

2 ( )2

d V x Em dx

ψ ψ ψ+ =h (10.1)

Where,

h, is Planck's constant m is the mass of the particle ψ is the (complex-valued) wave function that we want to find V(x) is a function describing the potential at each point x and E, is the energy, a real number. Now let us consider a particle in a one dimensional box, in which the particle of mass m is confined

between two infinitely large walls at x=0 and x= L, so that the potential energy of the particle is zero inside the box and infinite outside the box. This is a well studied differential equation and eigen-value problem with a general solution of:

sin( ) cos( )A kx B kxψ = + and energy,

2 2

2kE

m=

h (10.2)

Here, A and B can be any complex numbers, k is any real number because E is real. Now in order find the values for A and B for specific solution of this equation, we must specify the appropriate boundary conditions. As the particle is confined in side the walls the probability of finding the particle outside the box will be zero and hence ψ is equal zero at x = 0 and x = L. ψ(0) = ψ(L) = 0 (10.3) Substituting the general solution from Equation (10.2) into Equation (10.1) and evaluating at x = 0 (ψ = 0), we find that B = 0 (since sin(0) = 0 and cos(0) = 1). It follows that the wave function must be of the form: ψ = A sin(kx) (10.4) and at x = L we find: ψ= A sin(kL) = 0 (10.5) The trivial solution for Equation (10.5) is A = 0, however, it imply that ψ = 0 everywhere (i.e. the particle isn't in the box.) which is absurd. So A ≠ 0 which implies that sin(kL) = 0, which is only true when: kL = n π where n = 1,2,........ k = nπ/L (10.6) (note that n = 0 is possible because then Ψ = 0 everywhere, corresponding to the trivial case where the particle is not in the box. The negative values of n have been also neglected, since they will only change the sign of sin(nx), not the square modulus of the wave function ,that is the probability density will remain same. So both positive and negative values of n will lead to the same solution. Now, in order to find A we must undertake a process called normalization of the wave-function. As the particle must exist somewhere in space and | Ψ |2 is the probability of finding the particle at a particular point in space, so the integral of this value over all x must be equal to 1:

2 2 22

0

1 sin2

L Ldx A kxdx Aψ+∞

−∞

= = =∫ ∫ (10.7)

or

2LA = (10.8)

Thus, A can be any complex number with absolute value √(2/L). But these different values of A will yield the same physical state, so we can choose A = √(2/L). Finally, by substituting the results from Equations 10.7 and 10.8 into Equation 10.2, we obtain the complete set of energy eigen-functions for the one-dimensional particle in a box problem as:

sin( )2nL n x

Lπψ = (10.9)

2 2 2

22nnE

mLπ

=h (10.10) n=1,2,3, . . . . . .

Note that, as mentioned previously, only "quantized" energy levels are possible. Also, since n cannot be zero, the lowest energy from Equation (10.10) is also non-zero. This energy is called zero-point energy and can be explained in terms of the uncertainty principle. Because the particle is bounded within a finite region, the uncertainty in its position is upper bounded. Thus due to the uncertainty principle the uncertainty in the momentum cannot be zero, and hence the particle must contain some amount of energy that increases as the length of the box, L decreases.

Figure 7 Again, since ψ consists of sine waves, for any value of n greater than one, there are regions within the box for which ψ and thus ψ2 both equal zero, which indicates that for these energy levels, there exist nodes in the box where the probability of finding the particle is zero. The Hydrogen Atom Finally, consider the hydrogen atom as a proton fixed at the origin, orbited by an electron of reduced

mass µ. The potential due to electrostatic attraction is

(11.1) in SI units. Here r is the distance between the electron and the nuclei. The kinetic energy term in the Hamiltonian is

(11.2) Thus the total Hamiltonian of hydrogen atom is

( )2 2 2 2

2 2 20

( )2 4

e Ex y z r

ψ ψµ πε

∂ ∂ ∂+ + − =

∂ ∂ ∂h (11.3)

. If we consider the nuclei at the origin of the Cartesian coordinate then, 2 2 2r x y z= + + . How ever, it is more convenient to write this equation in spherical coordinates r, φ, and ө. In terms of spherical coordinate the Schrodinger equation is

(11.4) The advantage of writing the Schrödinger equation in spherical polar coordinate is that the

equation involving three coordinates can be resolved in to three independent equations each involving only one coordinates. Now, we can write the function ψ(r, θ, φ) as a product of three functions: R(r) depending on the radius part r alone; ( )θΘ , which depends only on ө, and ( )ϕΦ , which depends only on ϕ .

Thus, ψ(r, θ, φ)=R(r) ( )θΘ ( )ϕΦ The function ( )θΘ ( )ϕΦ together is written as Yl

m(θ, φ) and called the spherical harmonics. It can be shown that for three different equations for these three functions will be

22

2

1 mϕ∂ Φ

− =Φ ∂

, (11.5)

where m is a constant. 2

2

1 (sin ) ( ) 0sin sin

d d mLd d

θθ θ θ θ

Θ+ − Θ = (11.6)

where L is a constant. and,

2 2 22

2 20

( ) ( ) 02 2 4e e

d dR L er E R rm r dr dr m r rπε

⎡ ⎤− + − − =⎢ ⎥

⎣ ⎦

h (11.7)

which is called the radial equation for the hydrogen atom. Its (messy) solutions are

(11.8) Where 0 ≤ l ≤ n - 1, and a0 is the Bohr radius ε0 h2/πµe2 .The functions Ln+l

2l+1(2r/na0) are the associated Laguerre functions. The hydrogen atom eigenvalues are

(11.9) There are relatively few other interesting problems that can be solved analytically. For molecular systems, one must resort to approximate solutions. Quantum Numbers Now we will solve this set of differential equations. We first solve the equation for ( )ϕΦ and get the values for ‘m’. Then using the value of m the values of L can be obtained from the equation for ( )θΘ . Then using these values of L is used and the values of E, the well behaved solution for R(r) can be obtained. The equation (8.5) ia a differential equation of order 2 and its solution gives the result ( ) imAe ϕϕΦ = (11.10) where, A is a constant. Now Φ is a well behaved function and hence it will be single valued in each points in space. So ( ) ( 2 )ϕ ϕ πΦ = Φ + , or

( 2 )im imAe Aeϕ ϕ π+= (11.11) This is possible only when m takes values zero or positive or negative integers. That is, 0, 1, 2, 3,........m = ± ± ± Hence, m is a quantum number. It is called magnetic quantum number. Now substituting the values of m in equation (11.6) and using the condition that Θ is a well behaved function it can be found that L must be of the form L = l(l+1) where l is an integer equal to or greater than the absolute value of m. This is known as the angular momentum or azimuthal quantum number. From equation (11.9) we see that the energy depends on the number n. This is also an integer and known as principal quantum number.

Hydrogen like wave functions: The complete wave function of any one electronic or hydrogen like atoms is written as the product functions of radial part and the spherical harmonics. Any one-electron function is known as orbital. Hence hydrogenic wave functions are known as orbitals. Now the probability of finding an electron in the infinitesimal volume element 2 sindv r dr d dθ θ ϕ= around a point (r,θ ϕ ) is

2 2 2 2( , , ) ( ) ( ) ( )r R rψ θ ϕ θ ϕ= Θ Φ (11.12) Since the electron must exist somewhere in the space, the integral of the probability density must be equal to one. That means the wave function will be normalized to unity. So,

2

2

0 0 0

( , , ) *( , , ) sin 1nlm nlmr r r d d drπ π

ψ θ ϕ ψ θ ϕ θ θ ϕ∞

=∫ ∫ ∫ (11.13)

Similarly the normalization conditions for the radial part and the spherical harmonics gives

2

0

( ) *( ) 1m m dπ

ϕ ϕ ϕΦ Φ =∫ (11.14a)

{ }2

0

( ) sin 1lm dπ

θ θ θΘ =∫ (11.14b)

{ }2

2

0

( ) 1nlR r r dr∞

=∫ (11.14c)

It can be easily shown that the probability density does not depends on the azimuthal angleϕ . That is for all values of m the probability density is symmetrical around z axis. Angular wave functions The product of the functions ( )θΘ and ( )ϕΦ together is written as Yl

m(θ, φ) and called the spherical harmonics. The analytic expression for ( )lm θΘ are given below

0,l m o= = 0,01( )2

θΘ = (11.15a)

1, 0l m= = 1,06( ) cos

2θ θΘ = (11.15b)

1, 1l m= = ± 1, 13( ) sin

2θ θ±Θ = (11.15c)

2, 0l m= = 22,0

10( ) (3cos 1)4

θ θΘ = − (11.15d)

2, 1l m= = ± 2, 115( ) sin cos2

θ θ θ±Θ = (11.15e)

2, 2l m= = ± 22, 2

15( ) sin4

θ θ±Θ = (11.15f)

l=0 for s orbitals, l=1 for p orbitals and l=2 for d orbitals.

Now we see that we are no longer in the concept that the electron to be moving in the fixed orbit as proposed by Bohr, instead that the motion of electron is describe by the wave function and so this wave function is called the atomic orbital. The analytic expression of the normalized angular wave functions for s, p, and d orbitals are obtained by the product of the corresponding functions for ( )lm θΘ and

12

ime ϕ

π

So for s orbital the normalized angular wave function is 1( )2

Y sπ

=

Now for l=1, m=1, 0, -1.So we will get three p orbitals namely 1 0 1( ), ( ) ( )Y p Y p andY p+ − . Then,

12

1

12

0

12

1

3( ) sin8

3( ) cos4

3( ) sin8

i

i

Y p e

Y p

Y p e

ϕ

ϕ

θπ

θπ

θπ

+

−−

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

(11.16)

So we see that for m>0 the angular wave functions imaginary. It is usually more convenient to use the real functions by taking the linear combinations of these imaginary functions. For p orbitals we take the following combinations.

1 1

0

1 1

( ) ( )( )2

( ) ( )( ) ( )( )

2

x

z

y

Y p Y pY p

Y p Y pY p Y pY p

+ −

+ −

+=

=−

=

(11.17)

By putting the expression for ( )ϕΦ and ( )lm θΘ we will get the expression for the three real p orbitals.

12

12

3( ) sin cos4

3( ) sin sin4

x

y

Y p

Y p

θ ϕπ

θ ϕπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

(11.18)

123( ) cos

4zY p θπ

⎛ ⎞= ⎜ ⎟⎝ ⎠

The Radial wave functions

The solution of the radial part gives the radial wave functions. The radial wave function gives the spatial extent of the orbital. In contrast to the angular part the radial wave function does not only depends on the value of r, but also depends on the quantum numbers n and l.

The analytic expression for the radial wave function for the one electron or hydrogen like atoms are given below

n =1 l=0 1s 0

32

1,00

( ) 2ZraR r e

a

−⎛ ⎞Ζ= ⎜ ⎟

⎝ ⎠

n =2 l=0 2s 0

32

22,0

0 0

1( ) (2 )2 2

ZraZrR r e

a a

−⎛ ⎞Ζ= −⎜ ⎟

⎝ ⎠ (11.19)

n =2 l=1 2p 0

32

22,0

0 0

1( ) ( )2 6

ZraZrR r e

a a

−⎛ ⎞Ζ= ⎜ ⎟

⎝ ⎠

where, 2 0

0 2 0.529a Aeµ

= =h is the radius of first Bohr orbit.

Atomic structure and the different quantum numbers: From Bohr’s theory of atomic structure we saw that the energy of an electron in an atom depends only on principal quantum numbers. These principal quantum numbers are symbolized by n, and denote the major shell in which the electron is located. This is simply called the shell. It can take the value of any integer greater than zero, which is expressed as a series: 1, 2, 3, 4, ... or K, L, M, N, The Bohr’s theory successfully explains the atomic spectra of hydrogen atom. When the spectra of other elements were taken it was found that the lines are divided in to fine structures. Observing the fine structure Sommerfeld realized that two quantum numbers were needed. Then he defined another quantum number known as azimuthal quantum number, denoted by ‘l’ The rule for selecting the proper values of 'l' is as follows:

l = 0, 1, 2…..... (n-1)

‘l’ will always be a whole number and will never be as large as the 'n' value it is associated with.

For n=0, l=0

n=2, l=0, 1

n=3, l=0, 1, 2 etc.

The orbital with same value of n but different values of l are said to form a subshell in a given shell. These subshells are generally referred by the letter.

l = 0 1 2 3 4 5 …………….

s p d f g h

This azimuthal quantum number represents the shape of the orbital. From the solution of Schrodinger’s equation of hydrogen atom it was found that the shape of s orbital is spherical, The energy of the subshells increases in the following order: 1s<2s<2p<3s< 3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d…. When an atom is placed in a strong magnetic field, the lines of spectrum are split up; this effect was discovered by Zeeman and is known as Zeeman’s effect. This indicates that the energy levels in atom are divided in presence of magnetic field. An additional quantum number is therefore needed to account these energy sub-levels. This quantum number is called ‘magnetic quantum number’ and denoted by ‘ml’. The rule for selecting m is as follows:

ml = -l, -l+1, -l+2, …0, +1, _+2, .…… +l So, for each value of l there are (2l+1) values of m. The magnetic quantum number represents the orientation of the subshell. The energy of the orbital does not depend on the orientation of the orbital, and hence orbitals correspond to different values of magnetic quantum number but having same azimuthal quantum number will be same in energy. The orbitals having same energies are called degenerate orbitals. For example, in case of 2p orbital, l=1 so ml = -1, 0, +1 Therefore we get three degenerate 2p orbitals. They are denoted as 2 , 2 2x y zp p and p There are mainly three governing principle of filling of atomic orbitals within the energy levels. They are the Aufbau principle, the Pauli Exclusion Principle, and Hund’s rule.

The Aufbau principle: This principle was given by Neils Bohr. According to this, electrons enter orbitals of lowest energy first. The various orbitals within a sublevel of a principal energy level are always of equal energy. Yet the range of energy within a principal energy level can overlap the energy levels of a nearby principal energy level. As a result, the filling of orbitals does not follow a simple pattern beyond the second energy level. For example, the 4s orbital is lower in energy than the 3d. So, electron will fill up the 4s orbital before 3d orbital.

The Pauli Exclusion Principle: This principle says that an atomic orbital may only hold two electrons. When two electrons occupy the same orbital, they must have opposite spin. That is the spin quantum numbers will be different for each electron in the same orbital.

Hund’s rule: Electrons fill degenerate orbitals according to rules first stated by Friedrich Hund. Hund's rules can be summarized as follows.

(a) One electron is added to each of the degenerate orbitals in a subshell before two electrons are added to any orbital in the subshell.

(b) Electrons are added to a subshell with the same value of the spin quantum number until each orbital in the subshell has at least one electron.

When one electron enters in the 2p subshell, it can enter to any of the degenerate orbitals. This choice is arbitrary. The second electron enters the same subshell with same spin. The wave-function contains states, which are equal linear combinations within the degenerate space.

So, for example the electronic configuration in carbon atom is C (Z = 6) and consists of equal linear combinations of 1s2 2s2 2px

1 2py1, 1s2 2s2 2pz

1 2py1 and 1s2 2s2 2px1 2pz

1

Both the electrons will have same spin quantum number, and number can be shown by representing an electron for which s = +1/2 with an arrow pointing up and an electron for which s = -1/2 with an arrow pointing down.

The electrons in the 2p orbitals on carbon atom for one of the above combinations can therefore be represented as follows.

Similarly in case of nitrogen, 2 2 31 2 2s s p

The Electron Configurations of the Elements

atomic Number Symbol Electron Configuration 1 H 1s1 2 He 1s2 = [He] 3 Li [He] 2s1 4 Be [He] 2s2 5 B [He] 2s2 2p1 6 C [He] 2s2 2p2 7 N [He] 2s2 2p3 8 O [He] 2s2 2p4 9 F [He] 2s2 2p5 10 Ne [He] 2s2 2p6 = [Ne]

11 Na [Ne] 3s1 12 Mg [Ne] 3s2 13 Al [Ne] 3s2 3p1 14 Si [Ne] 3s2 3p2 15 P [Ne] 3s2 3p3 16 S [Ne] 3s2 3p4 17 Cl [Ne] 3s2 3p5 18 Ar [Ne] 3s2 3p6 = [Ar] 19 K [Ar] 4s1 20 Ca [Ar] 4s2 21 Sc [Ar] 4s2 3d1 22 Ti [Ar] 4s2 3d2 23 V [Ar] 4s2 3d3 24 Cr [Ar] 4s1 3d5 25 Mn [Ar] 4s2 3d5 26 Fe [Ar] 4s2 3d6 27 Co [Ar] 4s2 3d7 28 Ni [Ar] 4s2 3d8 29 Cu [Ar] 4s1 3d10 30 Zn [Ar] 4s2 3d10 31 Ga [Ar] 4s2 3d10 4p1 32 Ge [Ar] 4s2 3d10 4p2 33 As [Ar] 4s2 3d10 4p3 34 Se [Ar] 4s2 3d10 4p4 35 Br [Ar] 4s2 3d10 4p5 36 Kr [Ar] 4s2 3d10 4p6 = [Kr

We see that there are some exceptions also.

predicted electron configurations: Cr (Z = 24): [Ar] 4s2 3d4 Cu (Z = 29): [Ar] 4s2 3d9

But the experimentally found configuration is

Cr (Z = 24): [Ar] 4s1 3d5

Cu (Z = 29): [Ar] 4s1 3d10

In each case, one electron has been transferred from the 4s orbital to a 3d orbital, even though the 3d orbitals are supposed to be at a higher level than the 4s orbital. The reason for this kind of configuration is that if a subshell gets fulfilled or half filled it gains extra stabilization.

Quantum mechanical models of the chemical bond According to the description of quantum mechanics the electron can not be located precisely. We can only find out the probability of finding a particle in a location. The shape of the orbital is defined by the volume of the space where the probability of finding the electron is maximum. Thus the shape of the orbital is defined by a mathematical function that relates the probability to the (x,y,z) coordinates of the molecule. In quantum models of chemical bonding, one attempt to show how more electrons can be simultaneously close to more nuclei. The Valence-bond theory In 1927 Heitler and London first applied wave mechanics to valency problems in attempt to describe the stability of the covalently bonded hydrogen molecule. This theory is known as valence bond (VB) theory. This theory considers the involvement of electrons in terms of atomic orbital. The covalent bond is formed by pairing and resultant neutralization of opposite spins electrons from the different atoms. To form a covalent bond an atom must have one or more electrons which can pair with those of another atom by canceling their spins. Thus the valency is determined by the number of unpaired electrons. The paired electrons in atoms generally do not take part in the formation of the chemical bond. They, however, can be unpaired with the expenditure of small energy and there by can take part in bond formation. And this is possible when the impairing of the electrons occur within the same principal quantum number. Let us consider the bonding of hydrogen molecule. The two hydrogen atoms have two 1s orbital in their ground state. Let us denote the wave function of these two hydrogen atom as A BandΨ Ψ . If the atoms are sufficiently isolated so that they can not interact with each other the wave function for the system of two atoms will be

A (1) (2)BΨ = Ψ Ψ ( 13.1) where A and B are the designation for atoms and the number 1 and 2 for electrons. When the atoms will be in the bonding distance there will be change of the wave function of the system. Now since the electrons are indistinguishable they can not be labeled as above. Moreover now electron 1 can be on B atoms and electron 2 can be on A atom. Hence the proper description of the wave function will be

A A(1) (2) (2) (1)B BΨ = Ψ Ψ +Ψ Ψ (13.2) Taking the above wave function the Schrödinger equation for molecule is solved. This is the

fundamentals behind the valence bond theory. The improvement of valence bond theory has been done by considering the fact that there is a

possibility of finding both the electrons of hydrogen molecule on the same atom. It is commonly call the influence of ionic structure on the overall wavefunction:

H H H H H H+ − − +− ↔ ↔ The total wave function now becomes

(13.3)

A A A B(1) (2) (2) (1) (1) (2) (2) (1)B B A Bλ λΨ = Ψ Ψ +Ψ Ψ + Ψ Ψ + Ψ Ψ

The first two terms represent the covalent structure and the last two terms for the ionic structure of the bond. The constant λ accounts the ionic character of the corresponding bond. Because electrons will repel each other the possibility of finding both the electron on the same electron will be small and hence the value of λ is always less than one. If the electronegativity difference of the bonded atoms is high λ value will be more and correspondingly the bond will be more ionic. The hybrid orbital model In 1931 Linus Pauling first developed the quantum-based model of chemical bonding. According to this model, the atomic s, p, and d orbitals which are occupied by the valence electrons of adjacent atoms are combined in a suitable way, form the hybrid orbitals. The character and directional properties of these hybrid orbilats are consistent with the bonding pattern of the molecule. The rules for bringing about these combinations turn out to be remarkably simple, so once they were worked out it became possible to use this model to predict the bonding behavior in a wide variety of molecules. The hybrid orbital model is most usefully applied to the p-block elements the first two rows of the periodic table, and is especially important in organic chemistry. Hybrid types and multiple bonds Up to now we have dealt with molecules in which the orbitals on a single central atom are considered to be hybridized. Now let us discuss how the various models of hybridization can help us understand bonding between pairs of such atoms Let us discuss the bonding properties in hydrocarbon compounds for convenience. The most well known problem was the structure of the methane molecule, which consists of four hydrogen atoms and one carbon atom. The valence shell electronic configuration of carbon atom is 2 1 12 2 2x ys p p , that is it has only two unpaired electrons. So according to VB theory carbon is capable to form only two bonds not four. To form four bonds, one electron will unpaired from 2s orbital and will go to 12 zp orbital. But 12 zp orbital energy is higher than 2s orbital energy and hence the four bonds will not be identical and also bond angles will differ. But in practical this is not the case. In methane molecule all the C-H bonds are identical and the four bond angles are exactly equal. In account to solve this problem, in 1931 Linus Pauling first developed the hybrid orbital model of chemical bonding. According to this model, the atomic s, p, and d orbitals which are occupied by the valence electrons of adjacent atoms are combined in a suitable way, form the hybrid orbitals. The character and directional properties of these hybrid orbilats are consistent with the bonding pattern of the molecule. So according to this model due to mixing or linear combination of pure one 2s and three 2p orbitals of carbon atom form four hybrid orbital are formed which all are exactly identical. As three p orbital make hybridization with one s orbital this type of hybridization is called 3sp hybridization. The linear combinations that give rise four equivalent hybrid orbitals are

1

2

x y z

x y z

s p p p

s p p p

ϕ

ϕ

= + + +

= − − + 3

4

x y z

x y z

s p p p

s p p p

ϕ

ϕ

= − + −

= + − − (13.4)

Now electrons from these four identical hybrid orbital of carbon paired with 1s orbital of hydrogen form each C-H bond of methane molecule. The wave function for the bond of the methane molecule is thus

1 1(1) (2) (2) (1)A Aϕ ϕΨ = + , A represent the wave function for 1s orbital of hydrogen molecule. Thus methane is tetrahedral and all H-C-H bond angles are identical (109° 28´)

Figure 8 Reference: http://en.wikipedia.org We have already seen how 3sp hybridization in carbon leads to its combining power of four in the methane molecule. To form the molecule ethane C2H6, two such tetrahedrally coordinated carbons can link up together. In this molecule, each carbon is bonded in the same way as the other; each is linked to four other atoms, three hydrogen and one carbon atoms. The ability of carbon-to-carbon linkages to extend themselves indefinitely and through all coordination positions accounts for the millions of organic molecules that are known.

Figure 9 Trigonal hybridization in carbon: the double bond In case of ethylene (ethene), each carbon atom is joined to only three other atoms. Here, carbon is regarded as being trivalent. This trivalence can be explained by supposing that only two of the three p orbitals of carbon mix with the 2s orbital to form hybrids orbital and the remaining p-orbital, let, 2pz orbital, remains unhybridized. As two p orbitals of carbon mix with one s orbital this kind of hybridization is called orbital sp2 hybridization. So, in this case each carbon is bonded to three other

atoms in the same kind of plane trigonal and the bond angles around each carbon are all 120°.

Figure 10 Now the unhybridised 2pz orbital on each carbon atom from another type of bond by overlapping and this type of bond is called Π bond.

Figure 11

This figure is the pictorial representation of the structure ethylene. (a) the backbone structure with sigma bond forming by overlapping the sp2 hybrid orbital of carbon with s orbital of hydrogen. Triple bonds In case of alkynes for example acetylene each carbon atom is bonded with only one hydrogen atom and one carbon atom. Here, 2s and one 2p orbital mixes and form sp hybridization and the other two 2p orbital remains unhybridized and these two 2p orbital from each carbon atom form two Π bond by lateral overlapping. Thus this type of compounds is linear in structure. www.chem1.com/acad/webtext/chembond/cb07.html Multiple bonds between unlike atoms One should not think that the formation of multiple bonds occur only between`n the similar atoms.

Figure 12 There are several examples such as carbon dioxide, hydrogen cyanide etc, where we find multiple bond

formation between dissimilar atoms. In carbon dioxide there are two unhybridized atomic p orbitals in the carbon atom and one p orbital available on each oxygen atom. The sp hybridized orbital of carbon atom form σ bond with two oxygen atoms and the unhybridized orbital forms Π bond with the p orbital of oxygen. As two p orbital on oxygen are perpendicular, the two C-O Π bonds are mutually perpendicular. Resonance In the valence bond theory more than one acceptable structure can be drawn to describe the bonding of some molecules, or in more precise way to tell that more than one wave function can written for that molecule. These structures are called the resonance structure. For example it is seen in the hydrogen molecule that, H-H and H+H- both are acceptable structure for the molecule. Since the ionic term having higher energy compared to the covalent term it will contribute less in the total wave function. Due to addition of this term the total energy of hydrogen molecule decreases. This is case of ionic-covalent resonance. Now let us consider the bonding structure of carbonate ion. The valence bond theory suggests that there will be three σ bonds and one Π bond. Since there are three oxygen atoms in the ion, the Π bond can be three different positions. So, we will get three equivalent structures. Each structure can be describe by a wave function, 1Ψ , 2 3orΨ Ψ .

The actual structure in fact is none of these three but a resonance hybride formed by linear combination of the three canonical structures: 1 2 3a b cΨ = Ψ + Ψ + Ψ (13.5) No simple structure is possible to describe the actual bonding, but structure will be similar to the following

D

It has been found that structure D is more stable compared to A, B and C. The energy difference between the structure D and A is called the resonance energy of carbonate ion. This arises due to the delocalization of lone pair of electrons.Since in the case of carbonate ion all the three resonance structure are equivalent they will contribute equally (a=b=c) and hence the hybride will be exactly the

intermediate of these three. Rules for drawing contributing resonance structures: Some rules must be considered when drawing resonance structures.

Rule 1: All resonance structures must have the same number of valence electrons.

Rule 2: The octet rule must be obeyed. For example hydrogen atom can not have more than one bond, carbon atom will have maximum four bond etc. The structure violating the octate rule in any atom will not be a contributing structure.

Resonance structure G is acceptable. Structure H is not acceptable because the carbon has ten valance electrons.

Rule 3: Nuclei do not change positions in space between resonance structures. This rule eliminates the possibilities of tautomers as resonance structures. Resonance structures differ only in the arrangement of valence electrons.

Structures K and L are both acceptable Lewis structures, but they are not related by resonance because the circled hydrogen atom has changed position in space. Rule 4: The propose resonance structure should have maximum number of bonds, consistence, of course, with other rules. The more and more resonance structures possible in molecule resonance energy will be more and hence the molecule will be more stable. Limitations of the valence bond theory The valence bond theory can successfully explain the structure and bonding of the molecules. However this theory is inadequate in many cases. According to this theory, two electrons from two different atoms

paired to form a covalent bond. Every unpaired electron will form the covalent bonds. Therefore, in the oxygen molecule, all sixteen electrons will pair up and the molecule will be diamagnetic. But, normally oxygen exhibits strong paramagnetic moment indicating the presence of unpaired electrons. One major limitation of the valence-bond theory arises from extension to large molecules. The wave function in this theory is built up from atomic orbitals, based on different atoms. The set of the above orbitals is non-orthogonal in nature and this non-orthogonality leads to serious problem in extension to lare molecules. Molecular orbital theory The molecular orbital theory is quite different from the valence bond theory and is more popular in its implementation. In the valence bond theory, the wave function is constructed from the overlap of orbitals of different atoms to form covalent and ionic bonds. In the molecular orbital theory (MO), instead of treating atomic orbital, electrons are said to be present in the molecular orbital. The molecular orbital is the volume of space where the probability of electron is maximum in a molecule as a whole and electros are fed in to these molecular orbital according to ‘aufbau’ principle. The basis principles of the MO theory are thus same as those relevant to atoms. Now, the problem is how these MO are to be formed. The most reasonable approximation is the linear combination of the atomic orbitals from different atoms to form the molecular orbitals. This method is known as LCAO (linear combination of atomic orbitals) method. Let us illustrate the method by taking a general case of AB molecule. If A Bandψ ψ are the atomic orbital, the molecular orbital will be

b A A B B

a A A B B

c cc c

ψ ψ ψψ ψ ψ

= += − (13.6)

where cA and cB are the coefficients. Among these two molecular orbitals bψ has lower energy than that of atomic orbital from which they formed. This orbital leads to formation of a stable molecule and is termed as bonding orbital. On the other hand aψ has the energy lower than the atomic orbitals and is termed as antibonding orbital. In case of bonding orbital an electron is occupied in the region of space between two nuclei and therefore exhibits a mutual attraction from the two positive centers, leading to a net binding effect. Conversely, if the electron is off to one side, in an anti-binding region, it actually adds to the repulsion between the two nuclei. One way to visualize the formation of the bonding and the antibonding MO from the atomic orbitals is the interference of light. We know that two waves can interact with each other in two ways. When two waves are in same phase, they combine together forming stronger wave. This is called constructive interference. Other hand if they are out of phase there occur destructive interference. In molecular orbital theory bonding MO are formed due to constructive interference and antibonding MO are formed due to destructive interference.

Figure 13 Molecular orbital diagrams Now let us illustrate the MO theory by taking the consideration of hydrogen molecule. In this case there are only two electrons from hydrogen atoms. So within LCAO approximation the simplest way to form molecular orbitals is by combining linearly the 1s orbital from two hydrogen atoms to form one bonding MO and one antibonding MO. The bonding MO is called σ MO and the antibonding MO is called σ* . Both electrons will enter the bonding orbital, following ‘aufbau’ principle. Thus the MOs in terms of 1s orbitals of atom A (hydrogen atom) and B ( the other hydrogen atom) are Ψg = N1 (ψ A1s + ψ B1s ) Ψu = N2 ( ψ A1s - ψ B1s )

(13.7)

The subscript g and u represent the symmetry of the molecular orbital. The letter ‘g’ is for ‘garade’ symmetry, which means that the orbital does not change sign on inversion through the center of symmetry at the mid point between two nuclei, whereas ‘u’ is for ‘ungerade’ symmetry, which means it will change sign on inversion through center of symmetry. N1 and N2 are normalization constants for the molecular orbitals, which are orthogonal to each other. N1 and N2 can be found out from the overlap of two atomic orbitals ψ A1s and ψ B1s . The following diagram is the pictorial representation of molecular orbitals of hydrogen molecule. These types of diagrams are known as MO diagram.

Figure 14 In this simple description of MO for hydrogen molecule, the wave function obtained can be shown to have covalent and ionic configurations in equal proportions in the “Valence-Bond” sense. This is unphysical and thus the simple MO represents overweighting of ionic terms. This can, however, be corrected by more elaborate theoretical formulations, which is outside the scope of this chapter. The

extra weightage of ionic terms in simple MO theory results in wrong dissociation components, for example, the simple MO theory will lead to H+ H- as dissociation products in case of H2 fragmentation. Sigma and pi orbitals In case of second period elements there are p orbitals which also participate in the molecular orbital formation. The s orbitals form usual bonding σ and antibonding σ* MO. But in case of p orbitals, they are not spherical, like s orbitals, but are elongated, and thus possess definite directional properties. According to directions of Cartesian space, the three p orbitals are designated as px, py, and pz, indicating the axis along which the orbital is aligned. If we take the molecular axes as x axes, then two px orbital will make head on overlap and form bonding σ and antibonding σ* molecular orbital. The remaining py, and pz orbitals from each atom are now perpendicular to the bond axes. Thus they will overlap laterally and form two types of molecular orbital, known as Π and Π* orbital. As there are four such atomic orbitals in two atoms there will be two Π MO and they are degerate i.e. having same energy. Π*

Figure 15

MO description of homonuclear diatomic molecules In this section will discuss the molecular orbital description of homonuclear diatomic molecules. In

the first approximation only the atomic orbitals having similar energy will combine to form the MO. In this approximation the simplest way to form molecular orbital is to combine the corresponding orbitals on two atoms (i.e. 1s + 1s, 2s + 2s, etc.). The appropriate combinations are

*

*

*

(1 ) 1 1

(1 ) 1 1( 2 ) 2 2

( 2 ) 2 2( 2 ) 2 2

( 2 ) 2 2

A B

A B

A B

A B

x A x B

x A x B

s s s

s s ss s s

s s sp p p

p p p

σ

σσ

σσ

σ

= +

= −= +

= −= +

= −

*

*

(2 ) 2 2

(2 ) 2 2

(2 ) 2 2

(2 ) 2 2

y yA yB

z zA zB

y yA yB

z zA zB

p p p

p p p

p p p

p p p

Π = +

Π = +

Π = −

Π = −

( 13.8)

According to this approximation the 1s orbitals from two different atom will form σg(1s)* and σu(1s)* just like the MO’s of hydrogen molecule described in the section 13.8. MO. In the similar way 2s orbitals from two different atoms will form σg(2s)* and σu(1s)* MO. These two molecular orbital will look like σg(2s) and σu(2s)* . Since an atomic 2s orbital has higher energy than atomic 1s orbital σg(2s) and σu(2s)* MO will have higher energy in comparison to σg(2s) and σu(2s)* MO respectively. So the ordering of the molecular orbitals discussed so far will be σg(2s) < σu(2s)* < σg(2s) < σu(2s)* .

Now let us discuss about the combination of the 2p orbitals. Except hydrogen atom 2p orbital energy is higher than the 2s orbital energy. As a result the bonding and antibonding molecular orbital formed from the combination of 2p orbital will have higher energy than σg(2s) and σu(2s)* respectively. If we consider the x axis as molecular axis then then these two orbitals will be denoted as σg(2p) and σu(2p)*. The remaining 2p orbitals i.e. 2py, and 2pz will form П molecular orbitals by lateral overlapping. The bonding and antibonding molecular orbital forming by overlapping 2py orbitals are denoted Пu(2py) and Пg(2py)* .These two orbital will be directed towards Y axis. The 2pz will combine in the similar manner will result bonding and antibonding П MO but these will be directed along the Z axis. As discussed in the previous section the bonding П orbitals arising from the combination of two 2py atomic orbitals and two 2pz will be degenerate and similarly the antibonding П* MOs.

Now to determine the electronic configuration of molecule by placing the electrons in these MOs in accord to the Pauli’s exclusion principle and Hund’s rule just like multi-electronic atoms, we need to know the energy ordering of these molecular orbitals. The energy of the molecular orbitals depends on the atomic number (atomic charge) on the nuclei. As the atomic number increases from lithium to fluorine, the energy of the σg(2p) and energy of Пu(2py) and Пu(2pz) orbitals approaches to each other and interchange order in going from N2 and O2 .

In fig18 a simplified molecular orbital energy labels for diatomic molecules of elements in the second period is shown. σg(1s)* and σu(1s)* MO are not shown here as they are independent from the rest and formed from 1s AO. No mixing between 2s and 2p orbital is considered here. This type of diagram is for O2, F2 and Ne2.

Fig16

But in case of N2 molecule the MO diagram the energy of the σg(2p) orbital will be less than that of Пu(2py) and Пu(2pz) orbitals . The following diagram gives the electronic configuration from Be2 to N2 molecules.

Fig17

Bond Order The net bonding in a diatomic molecule is define by a quantity called bond order, b;

12 ( *)b n n= − ( 13.9)

Where n is the number of electron in the bonding orbital and n* is the number of electrons in the antibonding orbital. This is very useful quantity for describing the characteristics of bonds,

because it correlates with bond length and bond strength. If the bond order between atoms of a given pair of elements is higher the bond length will be shorter and consequently the bond will be stronger. The bond strength is measured by bond dissociation energy which is the energy required to separate the atoms to infinity.

Single bond s have bond order one; double bonds have two; and so on. The bond order zero indicates that there is no bond between the given pair of atoms. For example bond order for He2 is zero, because there are two electrons in both bonding and antibonding orbital. For this reason He2 molecule does not exist.

Some important consequence from molecular orbital theory

Prediction whether a molecule exists or not: From MO theory we can predict whether a diatomic molecule exists or not by simply calculating the bond order. If the value is greather than zero the molecule will exist. For example the ground state electronic configuration of He2

+ is σ(1s) 2σ(1s)*2 , and the bond order is 0.5 . So He2

+ ion exists but since the bond order of the He2 molecule is zero it will not exist.

Lithium and beryllium molecules: The six electrons from two lithium atoms will fill in the molecular orbital according to aufbau principle. Four will fill in the σg(1s) and σg(1s)* with no bonding. The last two electrons will enter in the σg(2s) orbital. Hence the bond order in the lithium molecule will be one and the electronic configuration will be

KK σg(2s)2 where K stands for the K (1s) shell. The electronic configuration of beryllium molecule will be KK σ(2s)2 σ(2s)*2 Since the number of electrons in the bonding and anti-bonding molecular orbital is equal the bond

order will be zero. Hence like dihelium molecule Be2 molecule does not exist. Experimentally it is found that lithium is diatomic and beryllium is mono-atomic in the gas phage.

Paramagnetic property of oxygen molecule: One of the most impressive successes of the molecular orbital theory is the prediction of correct electronic configuration of oxygen molecule. The experimental result shows that the oxygen molecule is paramagnetic, the net spin of oxygen molecule correspond to two unpaired electron having same spin. According to molecular theory the ground state of oxygen molecule is KK σ(2s)2 σ(2s)*2 σ(2px)2 П(2py)2П (2pz)2 П(2py)1П(2pz)1. According to Hund’s rule two electrons having parallel spin will occupy two degenerate П (2py) and П (2pz) orbital. This explains the magnetic behavior of oxygen molecule.

Molecular orbital theory of heteronuclear diatomic molecules Since in this case the energies of the atomic orbital on the two atoms from which the molecular orbital are formed are different, the combination of atomic orbital to form molecular orbital will be different from that in the case of homo-nuclear diatomic molecules. In order to develop the molecular orbital description of hetero nuclear diatomic molecule we need to consider the fact that different types of atoms have different capacities to attract the electrons. So basically the electronegativity of the bonded atoms plays vital role in the treatment of hetero nuclear bond. The bonding electron will be more stable

in the presence of the nuclei of the atom having greater electronegativity, i.e. the atom having lower energy. Probability of finding the bonded electrons will be more near that nucleus. The electron cloud will be distorted towards that nucleus and hence the MO will resemble that AO more than the AO on the less electronegative atom. This description can easily describe the polarity of the heteronuclear molecules. Hydrogen Fluoride Hydrogen Fluoride is a simple heteronuclear diatomic molecule. Since the electronegativity of Fluorine atom is much higher than that of hydrogen atom the energy of 1s orbital of hydrogen atom is much higher than the atomic orbital of fluorine atom. The molecular orbital will be formed due the combination of hydrogen 1s orbital with 1s, 2s and 2p orbitals of fluorine.The pictorial representation of the MO diagram is given fig18. There are eight valence electrons which occupy four molecular orbitals. The two highest energy MO's are degenerate, are π-type formed onmly from two 2p orbitals of fluorine atom and hence have no electron density associated with the hydrogen atom, ie. they are Non-Bonding Orbitals (NBO).In Lewis Theory these two electrons are represented as two "Lone Pairs". We can see from the MO diagram that the electron density is not equally distributed about the molecule. There is a much greater electron density around the fluorine atom because of extremely electronegativity of fluorine atom, and in each bonding molecular orbital, fluorine will take a greater share of the electron density.

fig.18 Reference of the diagram is http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.php Huckel theory: Bonding in polyatomic molecules In case of polyatomic molecules the combinations among the atomic orbitals are complicated. One widely used approximated theory to describe the molecular orbitals in polyatomic molecule is Huckel

molecular theory (HMO). Huckel MO theory is based on the treatment of Π electrons in conjugated molecules. Here Π orbitals are treated separately from the σ orbitals which gives the framework of the molecule. The Π electron Hamiltonian is simply written as a sum of one-electron terms.

( 13.10) It follows that the total energy is the sum of one-electron energies. The sum is over all electrons. Since each molecular orbital is doubly occupied (for a normal closed shell hydrocarbon, which is the class we shall restrict ourselves to for now), this is twice the sum of energy terms for each molecular orbital.

( 13.11) Each molecular orbital i is now described as a linear combination of atomic orbitals, µ

( 13.12) Now the next task is to calculate the coefficient of the atomic orbitals to find out the molecular orbitals and their energies. Let us illustrate the theory with the example of ethene:

The number of atoms (i.e. C atoms) (n) is 2, so the molecular orbitals are of the form:

( 13.13) Since there are two atomic orbitals, two molecular orbitals, will formed due to the combination of the orbitals, one will be bonding and other will be antibonding. The electrons will in the bonding orbital of the system. Note that each C atom always contributes one electron to the system for a neutral hydrocarbon, so the number of electrons is then equal to the number of atoms, and the number of occupied orbitals (m) is n/2. The best molecular orbitals are those which give the minimized the total energy. This is achieved by imposing the condition:

( 13.14) for all µ. The details of the procedure is beyond the scope of the book, the reader can read the book Quantum Chemistry by I.N. Levine for it , but you may wish to just accept the result. The energy and coefficients satisfy the following equations:

( 13.15)

The H and S terms are simple integrals over the atomic orbitals, with H including the one electron Hamiltonian. Since we have not specified the form of the one-electron Hamiltonian, we will not make any attempt to evaluate these integrals We know that the Hamiltonian must include the kinetic energy term, the Coulombic attraction to all the nuclei in the molecule (H as well as C), a repulsion term to all the electrons and some average repulsion to the other electrons. In order to evaluate the above integrals Hückel made some approximation called the Hückel approximations:

1. The diagonal element of the matrix H, i.e. Hµµ, depends on a single C atomic orbitals are all equal, in spite of the fact that the environment may differ. We just label all Hµµ terms as . Note that in our example of ethene, the two terms must be identical due to symmetry.

2. The off diagonal terms Hµv will be zero if the two atoms are not neighbors. In our example, we have no atoms that are not neighbors. All the Hµv terms for orbitals of neighboring atoms are all equal and this term is denoted as .

3. All the atomic orbitals are orthonormal. This means that they are scaled is such a way that the diagonal elements of S are unity and the off diagonal elements are assumed to be zero

In summary, the Hückel approximations are:

This makes the equations for the energy and coefficients:

These simultaneous equations are called homogeneous equations. The non-trivial solutions to these equations occur when:

This is called the secular equation. This determinant can be easily multiplied out to give:

We obtain two values of , which is reasonable since we expect to find two molecular orbitals. For these values of , non-zero values of the C terms can be found: For i = 1, ie.

For i = 2, ie.

We scale these C terms to normalize the molecular orbitals. This gives:

The result is:

If we were to evaluate and , they would both be negative quantities as they are describing the binding of the electron to the atom or molecule. Thus 1 lay in energy a value of below the energy of the electron if there was no bonding (i.e. ). HMO treatment of benzene Benzene is an interesting molecule in chemistry in bonding respect. It has conjugated six carbons each of having one Π electron. The secular equation in this case will be a 6×6 determinate. So we will get six roots for this equation which are

1 2ε α β= + 2ε α β= + 3ε α β= + 4ε α β= − 5ε α β= − 6 2ε α β= + We can see that there are two nondegenerate and two doubly degenerate Huckel levels. In the HMO ground state electronic configuration all the six electrons will be in the bonding orbitals.

The concept of HOMO and LUMO: This is very useful concept in chemistry. The molecular orbital which is highest in energy among the occupied orbital is called HOMO (highest occupied molecular orbital) and the energetically lowest of the unoccupied molecular orbital is called LUMO (lowest unoccupied molecular orbital). If we want to remove an electron from a molecule it will be removed from the HOMO orbital just like in case of atom the electron remove from the outer most atomic orbital. Hence the energy of the HOMO is the measurement of the ionization potential of the molecule. Similarly if we want to add an electron to a molecule it will go to the LUMO and hence its measures the electron affinity of molecule. The HOMO and LUMO energy can give the account of the molecular stability and the chemical reactivity of the molecule.

Greater the energy difference between HOMO and LUMO energies greater will be the stability of the molecule. If the energy difference between HOMO and LUMO is small an electron from HOMO can easily go to the LUMO and there by decreasing the bond order of the molecule and hence the stability will be lower. Lower the stability of the molecule higher will be the reactivity of the molecule. Hence, lower the energy gap between HOMO and LUMO, greater will be the reactivity of the molecule. Exercises:

1. If the work function for sodium metal is 1.82eV, then calculate the threshold frequency for sodium.

Fig19. Huckel MOs for benzene

2. Calculate the de Broglie wavelength for a moving ball having of mass 600mg. 3. Calculate the de Broglie wavelength of an electron traveling with velocity 2% of velocity of

light. 4. Calculate the ionization energy of hydrogen atom. 5. What will be the uncertainty of momentum if one wants to locate an electron within an atom?

Given that the uncertainty of position is 55 pm. 6. A line in the Lyman series of hydrogen has a wavelength of 1.0228×10-7. Find the original

energy level of the electron. 7. Derive the Bohr formula for wavelength for a nucleus of atomic number Z. 8. Which of the following functions are normalized in the given interval? Normalize those which

are not normalized.

a) 2 2( )x ye− +

00

xy

≤ ≤ ∞≤ ≤ ∞

b) ( )x ye− + 00

xy

≤ ≤ ∞≤ ≤ ∞

c) 4( )sin sinx yab a b

π π 00

xy

≤ ≤ ∞≤ ≤ ∞

9. A particle of mass 3.00×10-27 g is in a one dimensional box of length 9.00nm. Find the frequency and the wavelength of the photon emitted when this particle goes from n=4 to n=3 level.

10. Write the electronic configuration of Ni2+ ion and Ni2+ and calculate their total spin quantum number.

11. Find out the normalization constants N1 and N2 in Eq (13.7). 12. Judge whether N2

+ will have greater or smaller dissociation energy than N2 ?

13. Write the MO electronic configuration for the NO- ion a) What is the bond order? b) How many unpaired electron will be present? c) Will the bond length be shorter or longer than in NO?

14. Write the electronic configuration of O2+, O2

-, O2+2 and O2

-2 in the light of MO theory. a) Calculate the bond order of the ions. b) Arrange the ions in the order of increasing bond length and decreasing dissociation

energy.

15. Solve the secular equation for benzene in Huckel theory and write down the molecular orbital. Which functions will be HOMO and LUMO? 16. Calculate the HMO energies of butadiene molecule and write the HMO orbitals.