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An orbital is a region in space where there is a high probability of finding electron.

Orbitals and Quantum Number:

A wave function represents an electron is the product of two parts, a radial part and an angular part The square of the radial part of the wave function indicate the probability of finding the electron at any distance r from the nucleus.  The square of the angular part of the wave function gives the probability of finding an electron in a particular direction from the nucleus.  The radial dependence and angular dependence of wave function taken together, tell us that a three dimensional standing electron wave (orbital) can be picture to have size, shape, and an orientation of an orbital.
In order to describe the size, shape, orientation of an orbital three quantum number are necessary. 
These quantum number are designed as,
  1. Principal Quantum Number ( n ).
  2. The Orbital angular momentum quantum Number or Azimuthal Quantum Number ( l ).
  3. The Magnetic Quantum Number (ml ).
  4. Spin Quantum Number.

The Principal Quantum Number(n):

    The Principal quantum Number(n) is of primary importance in the determining the size and hence the energy of an electron.
    For hydrogen the energy is fixed by the value of n. In other multi electron atom, the energy of each electron depends on the value of the principal quantum number of the electron. As the value of n increases the radius (nucleus electron separation increases that is the size of the orbital increases).
    The energy also raised, n is always an integer and can assume the value,1,2,3,4.... but not zero.

The orbital angular momentum quantum number or azimuthal quantum number (l):

    The general geometric shapes of an electron wave (Orbital) is described by the Azimuthal Quantum Number. 
    This quantum number related to for the electron in that state.
Thus, l = 0,1,2,3.....(n-1)
Therefore an electron having principal quantum number n, assumed the value l is 0 to (n-1).
Principal Quantum Number(n) Azimuthal Quantum Number(l)
n =1 (K-shall) l=0 (1S)
n =2 (L-shall) l=0 (1S)
l=1 (2P)
n =3 (L-shall) l=0 (3S)
l=1 (3P)

l=2 (3d)

The magnetic Quantum Number:

    The magnetic quantum number associated with the orientation of the electron wave with respect to a given direction, usually that of a strong magnetic fields.
    This quantum number hasn't effect on the shape of orbital or on the energy of an electron.
    For a given Value of l, ml can have any integral value between +1 to -1.
ml = + l, (- 1), (- 2), (- 3) ..... 0 ..... - 1, - 2, - l
Azimuthal Quantum Number(l) Magnetic Quantum Number(ml)
l =0 (S) ml = 0
l = 1 (P) ml = 1, 0, -1
Px, Py, Pz
l = 2 (d) ml = 2, 1, 0, -2, -1
dxy, dxz, dyz, dx2-y2, dz2
    Thus, P - type orbital with three orientation in space describe as PxPy,Pz, and d - type orbital with five orientation in space describe as dxy, dxz, dyz, dx2-y2, dz2.

Spin Quantum Number:

    Besides the three quantum number a fourth quantum number also has forth quantum number namely spin quantum number (s), Was necessary to completely describe an electron in a particular shell.
    The electron itself regarding as a small magnet. A beam of hydrogen atom can be split into two beams by a strong magnetic field. This indicates that there are two kinds of hydrogen atom in which can be differentiates on the basis of their behavior in a magnetic field.
    It has been postulated that each electron spin around it's axis like a lope and they behave like a magnet. A spinning electron can have only two possibilities.
    The electron can either spin clockwise or counterclockwise. The two directions of spin represent as(↑).
    This four quantum number s = 1/2 is independent of other three quantum numbers.
    Two direction of spin is represented as (↑)s can have two possible ms values +1/2 and -1/2 depending on the direction of rotation of the electron about it axis.
    This spin angular momentum is given by,

Shape of The Orbitals:

    The S orbitals penetrate the nucleus most while the P and d orbitals cannot penetrate the nucleus.
    This means that S orbital electron can efficiently screen the nuclear charge from other electrons compared to the other P and d electrons.
    The wave function of the electron in atom is called orbital. The wave function is plotted against distance and the space in three dimensional marked by a curve will gives the shape of the orbitals.
    The probability of finding an electron in space around the nucleus involves two aspects:
  1. Radial Probability.
  2. Angular Probability.
    It is not possible to represent completely in one diagram on paper the directional properties of an electron orbitals. An angular probability distribution must be combined mentally to have an overall shape of the orbital.

Shape of S - Orbital:

    The angular probability distribution are greater interest and importance, an S electron has no angular dependence because the relevant wave function is independent of angles θ and Φ.
    There is therefore an equal chance of discovering the electron in any direction of the nucleus.
How to find quantum numbers of an element?
Angular probability distribution of S orbital.
    With the nucleus at the origin of the Cartesian axes, the sphere of the radius r represents the probability of finding the S electron. An S electron has a spherically symmetrical probability distribution.

Shape of P - Orbital:

    The P orbitals are three orientations is represents as, Px, Py and Pz. The orientations of the orbital plane correspond to ml = 1, 0, -1 are mutually at the right angles.
    So the orbitals designated Px, Py and Pz are mutually perpendicular and they are concentrated along the respective coordinate axis X,Y and Z. Unlike the S orbitals the angular part of the P wave function is dependent on θ and Φ.
How to find quantum numbers?
Angular probability distribution of P orbital.
    The angular Probability distribution are shaped like pears along X, Y and Z axis . Thus in Px orbital it is most likely that the electron will be found in the direction of the X axis. There are no probability of its being found along Y and Z axes.
    Similar results are obtained for the Py and Pz orbitals.

Shape of d-Orbital:

    These orbital arises when n =3 (M - shell), that is the orbitals starts with the 3rd main energy level. When l = 2(d - orbital), ml = -2, -1, 0, 2, 1 indicating that d - orbitals have five orientation, that is, dxy, dxz, dyz, dx2 - y2 and dz2.
    All these five d - orbitals, in the absence of magnetic field, are equivalent in energy and are, therefore, said to be five - fold degenerate.
How to find quantum numbers?
Angular probability distribution of d orbital.

The emission spectrum of hydrogen atom was determined experimentally and it was possible to relate the frequencies of different spectral lines by simple equation,
ν = R[1/n12 - 1/n22]
Where n1 and n2 are integers and R is a constant, called Rydberg constant after the name of the discoverer. The existence of discrete spectral emission could hardly be explained by any model at Rydberg's time.

Energy of an Electron in Bohr Orbits:

The Energy associated with the permitted orbits is given by ,
En = - 1/2(e2/r), where r = n2h2/4π2me2
Thus, En = - (2π2me4/n2h2)
Again E1 = - (2π2me4/h2)
En = E1/n2
Thus the values of E2, E3, E4, E5 etc. in terms of E1(- 13.6 eV) are given as,
E2 = -13.6/22 = - 3.4 eV
E3 = -13.6/32 = - 1.51 eV
E4 = -13.6/42 = - 0.85 eV
E5 = -13.6/52 = - 0.54 eV
As n increases the energy becomes less negative and hence the system becomes less stable.

Explanation of the Rydberg Equation:

The explanation of Rydberg equation is now simple. The most stable orbit must be the one with the lowest energy (ground state).
When n = 1 the energy of an electron is lowest and it the lowest energy state and the states with n 〉 1 are the higher energy exited states.
Provided the right amount of energy is supplied the electron may jump from n = 1 to another higher level. Conversely energy will be released in the form of light of definite frequency when the exited electron returns to its ground state.
Since energy and frequency of the emitted light are connected by Plank Relation
E = hν 
or, ν = E/h
The energy corresponding to a particular line in the emission spectrum of hydrogen atom is the energy difference between the initial state 1 and the final state 2.
So that, E= E2 - E1
= - (2π2me4/n12h2) - [- (2π2me4/n22h2)]
= (2π2me4/h2)[1/n12 - 1/n22]
Then ν corresponding to the energy E is given by,
ν = E/h = (2π2me4/h3)[1/n12 - 1/n22]
ν = R[1/n12 - 1/n22]
Where R is the Rydberg Constant.
This relation tells us that as we go to higher and still higher orbits the energy gap between two immediate orbits continues to decreases.
  • Value of the Rydberg Constant:
Using the value of 9.108 × 10-28 gm for the mass of an electron, the Bohr Theory Predicts the following value of Rydberg Constant.
R = (2π2me4/h3)
= {2 × (3.1416)2 × (9.108 × 10-28 gm) × (4.8 × 10-10 esu)4}/(6.627 × 10-27 erg sec)3 
= 3.2898 × 1015 cycles/sec
Evaluate R in terms of wave number () instead of frequency. Wave number and frequency are connected with,
λν = c and ν = c/λ = c
Thus, = ν/c
where c is the velocity of light.
Recall the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second. Wave number () on the other hand stands for the number of waves connected in unit length that is per centimeter (cm-1) or per metre (m-1).
Thus, R = 3.2898 × 1015 sec-1/2.9979 × 1010 cm sec-1
= 109737 cm-1
= 10973700 m-1
The experimental values of R is 109677 cm-1 (10967700 m-1) showing a remarkable agreement between experiment and Bohr's Theory.

Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom:

Putting n = 1, n = 2, n = 3, etc in Rydberg Equation we get the energies of the different stationary states for the hydrogen electron. The transition energies can be calculated from the Rydberg Equation. The experimental spectra of hydrogen atom also exhibited several series of lines .
Energy Level Diagram for the Hydrogen Spectrum
Hydrogen atom contains only one electron but its spectrum gives a large number of lines - Explain.
This is because any given sample of hydrogen contains almost infinite number of atoms.
Under the normal conditions the electron of the each hydrogen atom remains in the ground state near the nucleus that is n = 1 (K - Shell). 
When energy is supplied to this sample of hydrogen gas by passing an electron discharge through the gas in the discharge tube, individual atom absorbed different amounts of energy. Some of atoms absorbed such energy to shift their electron to third energy level (n = 3) while some others may absorbed large amount of energy to shift their electron to the Fourth (n = 4), Fifth (n = 5), Sixth (n = 6) and Seventh (n = 7) energy level.
The electrons in the higher energy level are relatively unstable and hence drop back to the lower energy level with the emission of energy in the form of line spectrum containing various lines in of particular frequency and wave length.
  • Lyman Series:
Transition to the ground state (n = 1) from n = 2, n = 3, n = 4, etc exited states constitute the Lyman Series in the ultraviolet region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = R[1/n12 - 1/n22]
For Lyman Series, we have n1 = 1 and n2 = 2, 3, 4 ....
Thus, (i) When n2 = 2,
1/λ = 109677{1 - (1/4)}
or, 1/λ = {(109677 × 3)/4} cm-1
λ = {4/(109677 × 3)}
= 1215 × 10-8 cm
= 1215 Å
Thus, (i) When n2 = ∞,
1/λ = 109677{1 - (1/∞2)}
or, 1/λ = 109677 cm-1
λ = (1/(109677)
= 912 × 10-8 cm
= 912 Å
  • Balmer Series:
Thus the transition occurs to the ground state (n = 1) from n =2.
Thus the transition occurs to the ground state (n = 1) from n =2.
Transition to the n = 2 level from n = 3, n = 4, n = 5, etc exited states constitute the Balmer Series in the visible region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/22) - (1/n22)]
  • Pschen Series:
Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Pschen Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/32) - (1/n22)]
  • Brackett Series:
Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Brackett Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/42) - (1/n22)]
  • Pfund Series:
Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
= 1/λ = 109677[(1/52) - (1/n22)]
Series of Lines n1 n2 Spectral Region Wavelength
Lyman Series 1 2,3,4..etc Ultraviolet ㄑ4000 Å
Balmer Series 2 3,4,5..etc Visible 4000 Å to 7000 Å 
Paschen Series 3 4,5,6..etc Near Infrared 〉7000 Å
Brackett Series 4 5,6,7..etc Far Infrared 〉7000 Å
Pfund Series 5 6,7,8..etc Far Infrared 〉7000 Å
The wave number of the experimental spectra of hydrogen atom in Lyman Series is 82200 cm-1. Show that the transition occur to the Ground state (n =1) from n = 2.
(R = 109600 cm-1)
We know that, Wave number,
= R[1/n12 - 1/n22]
For Lyman series n = 1 and 
Wave Number () = 82200 cm-1
Thus, 82200 = 109600(1/12 - 1/n22)
or, 1/n22 = 1 - (822/1096)
= 274/1096 
∴ n2 = 2
Which of the Balmer lines fall in the visible region of the spectrum wave length 4000 to 7000 Å ? (Given  R = 109737 cm⁻¹)
The Balmer Series of the hydrogen spectrum comprise the transition from n 〉2 levels to the n = 2 level.
4000 Å = 4000 × 10-8 cm = 4 × 10-5 cm and 7000 Å = 7 × 10-5 cm
Therefore wave number,
= 1/λ = (1/4) × 105 to (1/7) × 105 cm⁻¹
Transition energy of the Balmer lines,
= 1/λ = 109677[(1/22) - (1/n22)]
Taking () = (1/4) × 105 cm-1
We have,
(1/4) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/4 ⋍ 1.1{(1/4) - (1/n22)}
or, n2 ⋍ 44
So n2 = 7 ( nearest whole number).
Taking () = (1/7) × 105 cm-1
We have,
 (1/7) × 105 cm-1 = 109737{(1/4) - (1/n22)}
= 1.1 × 105{(1/4) - (1/n22)} cm-1
∴ 1/7 ⋍ 1.1 {(1/4) - (1/n22)}
or, n2 ⋍ 3
So n2 = 3 ( nearest whole number).
So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.
Calculate the wave length of Hɑ and Hβ of the Balmer Series.
We know that, for the Balmer Series,
= 1/λ = 109677[(1/22) - (1/n22)]
Again for Hɑ line n2 = 3 and for Hβ line n2 = 4
Thus the wave length for Hɑ - line,
1/λ = 109677 {(1/22) - (1/32)}
= 109677 × (5/36)
λ = 36/(5 × 109677)
= 6.564 × 10-5 cm
= 6564 Å
Again the wave length for Hβ - line,
1/λ = 109677 {(1/22) - (1/42)}
= 109677 × (3/16)
λ = 16/(3 × 109677)
= 4.863 × 10-5 cm
= 4863 Å

Constituent Particles of an Atom:

Dalton’s Model to Modern Structure Of Atom:
  1. All the matter is made of atoms are indivisible and indestructible.
  2. All the atoms of a given element are identical mass and properties.
  3. Compounds are formed by combination of two or more same or different kind of atoms.
  4. A chemical reaction is rearrangement of atoms.
Rutherford has remarked that it is not in the nature of things for any one man to make sudden violet discovery. Science goes step by step and every man depends on the work of his predecessor.
The journey from Dalton model of the atom to modern structure of atom was long and arduous one. At turn this century many valuable information were being compiled.
This clearly indicates that Dalton’s atomic model no longer enjoyed the exalted position to grant it. Today an atom is considered to made up of a tiny nucleus carrying neutrons and protons.
This tiny nucleus has around itself a certain number of negatively charged particle carrying negligible mass, called electrons, arranged in a definite order.
Discussion on the Fundamental Particles of Atom:

Cathode rays - Discovery of Electron:

Gases at low pressures, when subjected to high potential, becomes conducting and various luminous effect were observed.
When the pressure is quite low(0.01 mm), the tube remains dark (Crooks dark space) but a streak of rays, named cathode rays, emanate from the cathode.
This is confirmed from the fact that a fluorescence is produced on the opposite wall where the rays impinge.
The cathode rays have been very carefully studied for their many definite characteristics.
Atomic Structure
Deflection of Cathode Rays on Electric Field
Characteristics of Cathode Rays:
  1. Cathode rays excite fluorescence on the glass walls where they impinge.
  2. The rays have travel in straight lines, confirmed by the shadows of objects placed in their path. 
  3. The rays have penetrating power and can pass through thin metal foils. 
  4. They also possess considerable momentum, small paddle wheels placed in their path rotate from the impact with the rays.
  5. The cathode rays are deflected from their path by the application of magnetic or electrostatic field. From the direction of deflection, the charge accompanying the rays is a negative one.
  6. When the rays impinge on a metal target, called anti cathode, and placed on the path , a different type of radiation, the X-rays is produced. these new radiation not deflected in an electric or magnetic field. X-rays are really electromagnetic radiation of very short wave length.
  • Charge of an Electron:
The electron carrying negatively charged and the charge of an electron(e),
4.8 × 10-10 esu = 1.60 × 10-19 coulomb.
  • Mass of an Electron:
Let the mass of an electron = m and charge = e,
then e/m = 1.76 × 108 Coulomb/gram.
∴ Mass of an electron,
= (1.60 × 10-19)/(1.76 × 108) gram
= 9.11 × 10-28 gram
  • Coulometric Determination of the Electronic Charge:
Electrode position of silver from an aqueous solution of a silver salt is a suitable experiment for the determination of the electronic charge.
Faraday's Low are readily interpreted by reference to the electrolysis of Silver Nitrate. The change at the cathode requires one electron for every Silver ion reduced.
Ag+ + e Ag
If the electrons consumed at this electrode is equal to Avogadro Number (6.023 × 1023 mol-1), 1 mole of Silver metal (107.9 gm) is produced. At the same time, 1 mole of electrons is removed from the anode and 1 mole of nitrate ions is discharged.
Thus, 96500 coulomb of electricity which is necessary to produce 1 equivalent mass of a substance at the electrode will be total charge carried by 1 mole of electrons.
Hence Charge carried by each electron is given by,
 e = (96500 C mol-1)/(6.023 × 1023 mol-1)
= 1.60 × 10-19 C

Positive Ray - Discovery of Protons:

Since the Electrons contribute negligible to the total mass of the atom and the atom is electrically neutral. That the nucleus must carry particles which will account both for the mass and positive charge of the atom.
We have so far deliberately restricted the discussion on the discharge phenomena at low gas pressure. The production of cathode rays in discharge tubes inspired physicists to look for the oppositely charged ions, namely positive ions.
Goldstein added a new feature to the discharge tubes by using holes in the cathode. With this modification it is observed that on operating such discharge tubes there appeared not only cathode rays travelling from the cathode to anode but also a beam of positively charged ions travelling from around anode to cathode. Some of the positively charged particles passed through the hole in the cathode and produces a spot on the far end of the discharge tube.
Structure of Atom and Constituent Particles of Atom (Electron, Proton and Neutron)
Goldstein Experiment.
The nature of these positive rays is extensively investigated by Thomson. It proved much more difficult to analyse the beam of the positive rays than to analyse a beam of electrons. On deflection by a magnetic and electric field the positive ray beam produced a large defuse spot indicating that the e/m ratio of the constituents of the beam was not same and that the particles moved with different velocities.
Thomson further demonstrated that each different gas placed in the apparatus gave a different assortment of e/m. Since a H+ is produced from a hydrogen atom by the loss of one electron, which has but a negligible mass, it follows that the mass of a H+ is same as that of the hydrogen atom(=1). The particle represented by H+ is called a Proton and is considered a fundamental constituent that accounts for the positive charge of the nucleus.
Charge of a Proton:
The Proton carrying Positively charged and the charge of a proton(p),
4.8 × 10-10 esu = 1.60 × 10-19 coulomb
Mass of an Proton:
Let the mass of an Proton = m and charge = e,
then e/m = 9.3 × 104 Coulomb/gram.
Mass of an Proton 
= 1.6725 × 10-24 gm

Discovery Of Neutron:

Attempts were now directed towards a correlation of atomic mass number ( = integer nearest to the atomic weight) and nuclear charge (= atomic number).
If A stands for the mass number and Z for the nuclear charge of an element, then Z units of nuclear charge means Z number of proton inside the nucleus. But Z protons can contribute only Z mass units. The shortfall of the (A - Z) mass units bothered chemists and physicists for the quite same time.
Rutherford then suggested this shortfall must be made up by another fundamental particle. This particle has electrically neutral, and mass equal to that of the proton, namely 1.
He named the particle in advance as neutron. The glory of discovering the neutron went to Chadwick, one of Rutherford Students.
An interpretation of the atomic nuclei on the basis of neutrons and protons is now a simple affair. Taking oxygen of mass number 16, for example and recalling that the atomic number of the element is 8, we have an atomic nucleus composed of 8 protons and 8 neutrons. Since neutrons contribute only to the mass of the element but does nothing towards charge it follows that there may exist species with the same number of protons but varying numbers of neutrons inside the nucleus. Such species must belong to the same element and must vary only in their mass numbers. They are called isotopes.
Thus 1H1, 1H2, 1H3 are three isotopes of hydrogen, and 8O16, 8O17 and 8O18 are three isotopes of oxygen.

Rutherford and his students describes some alpha particle scattering experiments. The alpha particles were already established by the group as He+2 from their behavior in electric and magnetic fields. A beam of alpha particles obtained from spontaneously disintegrating Polonium, were directed on to a very thin foil of Platinum or Gold.
With the help of fluorescent zinc sulphide screen around the Platinum or Gold foil, any deflection of alpha particle was observed.
Scattering of Alpha Particles from a Metal Foil
The vast majority of the alpha particles passed straight line through the foil. But a very limited few were found to be deflected back from the foil, some even appearing on the side of incidence. 
Rutherford concluded that since most of the alpha particles passed straight through there must be a very large volume of empty space in the atom of the platinum or gold.
A very small part of the platinum or gold atom must be responsible for the large scattering of the few alpha particles, central part was called atomic nucleus.

Conclusions of Rutherford experiment:

The following Conclusions and an atomic model emerged from the Rutherford's experiment.
  1. All the positively charged and almost entirely mass of the atom was concentrated in very small part of the atom and these central core called atomic nucleus.
  2. The large deflection of alpha particle from its original path was due to Coulombic repulsion between the Alpha particle and Positive Nucleus of an atom. Simple impact between the two such massive particles can lead to a Scattering of the order of only 10. 
  3. An alpha particle suffers little deflection while passing by an electron.
  4. The Radius of atomic nucleus is ∼ 10-13 being the same as that of an electron. Since the radius of an atom is ∼ 10-8 it is obviously that an atom must have a very empty structure. From the above Conclusions an atomic structure is proposed by Rutherford.

Rutherford's Atomic Model:

According to Rutherford's Model the entire mass of the atom was concentrated in a tiny, positively charged nucleus around which the extra nuclear electrons were moving in a circular orbits. 
Rutherford divided the atom into two part,
  1. Centre of Atom or The Nucleus:
    Almost the entire mass of the atom is concentrated in a very small, central core called the atomic Nucleus.
    Since the extra nuclear electrons contribute negligible to the total mass of the atom ans since the atom is electrically neutral it follows that the nucleus must carry particles which will account both for the mass and positive charge of the atom.
  2. The Extra Nuclear Electrons:
    A very small positive nucleus was considered surrounded by electrons. Such a system cannot be stable if the election were in rest.
    Therefore it is proposed that the electron moving in circular orbits around the nucleus so that the Coulombic attraction between the nucleus and the electron was equal to the centrifugal force of attraction.
Defects of Rutherford's Model:
  1. The Rutherford model is however, not in conformity with the classical model of electromagnetic radiation. A moving charged particle will emit radiation, will then loss kinetic energy and eventually will hit the nucleus.
  2. If the electrons lose energy continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other.

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