Atomic Structure

ATOMIC STRUCTURE INTRODUCTION The existence of atoms has been propsed by ancient Indian and Greek philosophers (400 B.C.) who believed that all matter consisted of very small indivisible particles called atoms. According to them, atoms were the fundamental building blocks of matter. The continued subdivision of matter would yield atoms, which could not be further subdivided. This means that atom is the fundamental unit of matter. So these ideas remained dominant for a very long time and were reviewed again by scientists in 19th century. In 1808, an English scientist John Dalton formulated a precise definition of the indivisble blocks of matter i.e. atoms. His theory, called Dalton’s atomic theory, regarded atom as the ultimate particle of matter. The postulates are as follows : (a) Every matter is composed of very minute particles called as atoms which take part in chemicals reactions. (b) Atoms can not be further subdivided. (c) The atoms of different elements differ from each other in their properties and masses. while atoms of same elements are identical in all aspects. (d) The atoms of different elements can combine in simple ratios to form compounds. The massed combined elements represents the masses of combined atoms. (e) Atoms can neither be created nor be destroyed. Although Dalton was pretty much successful in explaining the law of conservation of mass, law of constant proportion, law of multiple proportion, however he failed to explain many properties as well. The drawbacks are as follows : (a) It failed to explain the internal structure of atoms and assumed atoms do not have any structure. (b) It could not explain how atoms of different elements differ from one another. (c) It could not explain how atoms of different elements combine with one other. ATTEMPTS FOR DISCOVERING FUNDAMENTAL PARTICLES OF MATTER / ATOMS Earlier Efforts to Reveal Structure of Atom CATHODE RAYS (Discovery of e-) Gas at low pressure (10-4 atm) V > 10,000 volts - + Cathode Anode  vaccum pump (Invisible rays) (Cathode rays) In 1859, Julius plucker started the study of conduction of electricity through gases at low pressure in a discharge tube. When a high voltage of the order 10, 000 volts or more was applied across the electrodes, some sort of invisible rays moved from the –ve electrode to the +ve electrode. Since the –ve electrode is referred to as cathode, these rays were called cathode rays.  Properties of Cathode Rays (1) They travel in straight lines away from cathode with very high velocity (about one - tenth of velocity of light). (2) A shadow of metallic object placed in the path is cast on the wall opposite to the cathode. Production of a shadow of the solid object by cathode rays (3) They produce a green glow when strick the glass wall matter. Light is emitted when they strike the zincsulphide screen. (4) When a small pin wheel is placed in their path, the blades of the wheel are set in motion. Thus the cathode rays consist of material particles which have mass and velocity. Rotation of light paddle wheel by cathode rays (5) They are deflected by the electric and magnetic fields. When the rays are passed between two electrically charged plates, these are deflected towards the positively charged plate. It shows that cathode rays carry -ve charge. These particles carrying negative charge were called negatrons by Thomson. The name negatron was changed to 'electron' by Stoney Deflection of cathode rays towards positive plate of the electric field ATOMIC STRUCTURE (6) They produce heat energy when they collide with the matter. It shows that cathode rays posses Kinetic energy which is converted into heat energy when stopped by matter. (7) These rays affect the photographic plate. (8) Cathode rays can penetrate the thin foil of solid materials. (9) Cathode rays can ionize the gases through which they pass. (10) The nature of cathode rays is independent of (a) The nature of cathode and (b) The gas in discharge tube. MEASUREMENT OF E/M FOR ELECTRON In 1897, J.J. Thomson determined the e/m value (charge/mass) of the electron by studying the deflection of cathode rays in electric & magnetic fields. The value of e/m has been found to be –1.7588 108 coulomb/g.  By performing a series of experiments, Thomson proved that whatever gas be taken in the discharge tube and whatever be the material of the electrodes the value of e/m is always the same.  Electrons are thus common universal constituents of all atoms. DETERMINATION OF THE CHARGE ONAN ELECTRON The absolute value of the charge on an e- was Oil drops measured by R.A. Milikan in 1909 by the Milikan's oil drop experiment.  The apparatus used by him is shown in fig. An oil droplet falls through a hole in the upper plate. The air between the plates is then exposed to X-rays which eject electrons from air molecules. Some of these e- are captured by the oil droplet and it acquires a negative charge.  +++ Spray gun (atomizer) +++ X-rays -----Chargable plates telescope to observe speed of oil drops The metal plates were given an electric charge, and as the electric field between the plates was increased, it was possible to make some of the drops travel upwards at the same speed as they were previously falling. By measuring the speed, and knowing things like the strength of the field and the density of the oil, radius of oil drops, Milikan was able to calculate the magnitude of the charge on the oil drops. He found that the smallest charge to be found on them was approximately 1.59 10–19 C. This was recognised as the charge on an e-. The modern value is 1.602  10–19 C. MASS OF THE ELECTRON Mass of the e- can be calculate from the value of e/m and the value of e m= e 1.602 1019  = 9.1096 10–28 g e / m 17588 108 or = 9.1096  10–31 kg This is termed as the rest mass of the electron i.e. mass of the electron when moving with low speed. The mass of a moving e- may be calculate by applying the following formula. Mass of moving e- = rest mass of e  1   v / c 2 Where v is the velocity of the e- and c is the velocity of light. When v = c  mass of e- =  v > c  mass of e- = imaginary ANODE RAYS OR POSITIVE RAYS - (DISCOVERY OF PROTON)  The first experiment that lead to the discovery of the +ve particle was conducted by 'Goldstein'.  He used a perforated cathode in the modified cathode ray tube. Production of Anode rays or Positive rays  It was observed that when a high potential difference was applied b/w the electrodes, not only cathode rays were produced but also a new type of rays were produced simultaneously from anode moving towards cathode and passed through the holes or canals of the cathode. These rays were termed canal rays since these passed through the canals of the cathode.These were also named anode rays as these originated from anode.  When the properties of these rays were studied by Thomson, he observed that these rays consisted of positively charged particles and named them as positive rays.  The following characteristics of the positive rays we recognised : (i) These rays travel in straight lines and cast a shadow of the object placed in their path. (ii) Like cathode rays, these rays also rotate the wheel placed in their path and also have heating effect. Thus, the rays passess K.E. i.e. mass particles are present. (iii) The rays are deflected by electric and magnetic fields towards the negatively charged plate showing thereby that these rays carry +ve charge. (iv) The rays produce flashes of light on ZnS screen (v) These rays can pass through thin metal foil. (vi) These rays can produce ionisation in gases. (vii) Positive particles in these rays have e/m value much smaller than that of e-. For a small value of e/m, it is definite that positive particles possess high mass. (viii) e/m value is dependent on the nature of the gas taken in the discharge tube, i.e. +ve particles are different in different gases.  Accurate measurements of the charge and the mass of the particles in the discharge tube containing hydrogen, the lightest of all gases, were made by J.J. Thomson in 1906. These particles were found to have the e/m value as +9.579 104 coulomb/g. This was the maximum value of e/m observed for any +ve particle.  It was thus assumed that the positive particle given by the hydrogen represents a fundamental particle of +ve charge. This particle was named proton by Rutherford in 1911. Its charge was found to be equal in magnitude but opposite in sign to that of electron. Thus charge on proton = + 1.602  10-19 columb i.e. one unit +ve charge ATOMIC STRUCTURE The mass of the proton, thus can be calculated.  Mass of the proton = e 1.6 0 2  1 0 19  = 1.672  10–24 g e/ m 9.5 79  1 0 4 Mass of proton in amu = 1.6 7 2  1 0 2 4 1.6 6  1 0 2 4 = 1.672  10–27 kg = 1.00757 amu. NEUTRON In 1920, Rutherford suggested that in an atom, there must be present at least a third type of fundamental particles which should be electrically neutral and posses mass nearly equal to that of proton. He proposed the name for such fundamental particles as neutron. In 1932, chadwick bombarded beryllium with a stream of -particles. He observed that penetrating radiations were produced which were not affected by electric & magnetic fields. These radiations consisted of neutral particles, which were called neutrons. The nuclear reaction can be shown as + -particle Charge = +2 Mass = 4 amu [ 42He + Be-atom Atomic No. = 4 Mass = 9 amu 9Be 4  Carbon atom Atomic No. = 6 Mass = 12 amu 12C 6 + Neutron Charge = 0 Mass = 1 amu 1 n] 0 Thus a neutron is a sub atomic particle which has a mass 1.675  10–24 g approximately 1amu, or nearly equal to the mass of proton or hydrogen atom and carrying no electrical charge.  The e/m value of a neutron is zero. ATOMIC STRUCTURE Atom is actually made of 3 fundamental particles 1. Electron 2. Proton 3. Neutron Fundamental Particle Discovered By Charge –19 –1.6 × 10 Electron J.J.Thomson – (e or ) coloumb –10 –4.8 × 10 –1 Unit –19 +1.6 × 10 Proton (P) + (Ionized H atom, H ) Goldstein –10 +4.8 × 10 +1 Unit Charge mass Mass –31 9.1 × 10 kg –28 esu 9.1 × 10 g 0.000548 amu coloumb 1.672 × 10 esu 1.672 × 10 g 1.00757 amu –27 –27 1 (0 n ) James Chadwick 0 –24 8 1.76× 10 C/g kg –24 1.675× 10 Neutron (specific Charge) 4 9.58× 10 C/g kg 1.675× 10 g 1.00893 amu 0 1. 2. esu = electrostatic unit (1 C = 3 × 109 esu) amu = atomic masss unit 1 amu = 1.6 × 10–24 g = 1.6 × 10–27 kg Order of Mass m – 0, the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increase with the increase of frequency of the light used. Photoelectric Effect Equation When a photon of sufficient energy strikes an electron in the atom of the metal, it transfers its energy instantaneously to the electron during the collision and the electron is ejected without any time lag or delay. Greater the energy possessed by the photon, greater will be transfer of energy to the electron and greater the kinetic energy of the ejected electron. In other words, kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation. Since the striking photon has energy equal to h and the minimum energy required to eject the electron is h0 (is also called work function, W0) then the difference in energy (h – h0) is transferred as the kinetic energy of the photoelectron. Following the conservation of energy principle, the kinetic energy of the ejected electron is given by the equation Incident energy = Work Function () + K.E.max Ei =  + (K.E.)max h = h0 + 1 m 2 2 e where me is the mass of the electron and v is the velocity associated with the ejected electron. Ex. The threshold frequency 0 for a metal is 6 × 1014 s–1. Calculate the kinetic energy of an electron emitted when radiation of frequency  = 1.1 × 1015 s–1 hits the metal. Sol.   Ex. Sol. 1 m V2 = h ( – 0) 2 e K.E. = (6.626 × 10–34) (1.1 × 1015 – 6 × 1014) K.E. = (6.626 × 10–34) (5 × 1014) = 3.313 × 10–19 J K.E. = A photon of wavelength 3000 Å strikes a metal surface, the work function of the metal being 2.20 eV. Calculate (i) The energy of the photon in eV (ii) the kinetic energy of the emitted photo electron and (iii) the velocity of the photo electron. (i) Energy of the photon -34 8 -1 hc  6.6 × 10 Js  3 × 10 ms  = = 6.6 × 10–19 J  3 × 10 -7 m 1eV = 1.6 × 10–19 J E = h = Therefore E = (ii) (iii) 6.6 × 10 -19 J = 4.125 eV 1.6 × 10 -19 J / ev Kinetic energy of the emitted photo electron Work function = 2.20 eV Therefore, KE = 2.475 – 2.20 = 1.925 eV = 3.08 × 10–19 J Velocity of the photo electron KE = 1 mv2 = 3.08 × 10–19 J 2 Therefore, velocity (v) = 2 × 3.08 × 10 -19 = 8.22 × 105 ms–1 9.1 × 10 -31 ATOMIC STRUCTURE BOHR’S ATOMIC MODEL It is based on quantum theory of light. It is applicable only for single electron species. Ex. H, He +, Li2+, etc. Assumptions of Bohr’s Model The electron in the hydrogen atom revolves around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states, energy shells, or allowed energy states. these stationary states for electrons are numbered as n = 1, 2, 3, ... or designated as K, L, M, N, ..., etc. shells (Fig.) These integral numbers are known as principal quantum numbers. These orbits are arranged concentricallly around the nucleus.  4 or N 3 or M 2 or L 1 or K + Nucleus Bohr's orbit Electrons revolve only in those orbits where the angular momentum of the electron is quantized. Thus an electron can move only in those orbits for which its angular momentum is an integral multiple of h/2.  mvr = n h 2π V + where n = 1, 2, 3, ..., n ; h is Planck’s constant; m is mass of electron ; v is the velocity of electron; and r is the radius of the orbit. The energy of an electron in the orbit does not change with time. This means that the eneryg of an electron in a particular orbit remains constant; it does not lose or gain energy.  The electron will move from a lower stationary state to a higher stationary state when the required amount of energy is absorbed by the electron. When the electron jumps back to the lower energy level, it emits the same amount of energy. The energy change does not take place in a continuous manner.  The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by E is given by  = ΔE E 2 – E1 = h h where E1 and E2 are the energies of the lower and higher allowed energy states, respectively. This expression is called Bohr’s frequency rule. Mathematical forms of Bohr’s Postulates Calculation of The Radius of The Bohr’s Orbit : Suppose that an electron having mass ‘m’ and charge ‘e’ revolving around the nucleus of charge ‘Ze’ (Z is atomic number & e = charge) with a tangential/linear velocity of ‘v’. Further consider that ‘r’ is the radius of the orbit in which electron is revolving. According to Coulomb’s law, the electrostatic force of attraction (F) between the moving electron and nucleus is – F= KZe 2 r 2 1 where : K = constant = 4 = 9 x 109 Nm2/C2 0 and the centrifugal force F = mv 2 r For the stable orbit of an electron both the forces are balanced. i.e 2 mv 2 KZe = r r2 KZe 2 mr From the postulate of Bohr, nh nh mvr =  v = 2 2mr then v2 = On squaring v2 = ......... (i) n 2h 2 4π 2 m 2 r 2 ........ (ii) From equation (i) and (ii) 2 2 n h KZe 2 = mr 4π 2 m 2 r 2 On solving, we will get n 2h 2 4π 2 mKZe 2 r= On putting the value of e , h , m, the radius of nth Bohr orbit is given by : n2 rn = 0.529 x Å Z n2    r  Z  Ex. Calculate radius ratio for 2nd orbit of He+ ion & 3rd orbit of Be+++ ion. Sol.  22 r1 (radius of 2nd orbit of He+ ion) = 0.529   2  Å    32 r2 (radius of 3rd orbit of Be+++ ion) = 0.529   4 Therefore  Å   r1 0.529× 22 / 2 8 = = r2 0.529×32 / 4 9 Calculation of Velocity of an Electron in Bohr’s Orbit Angular momentum of the revolving electron in nth orbit is given by mvr = v= nh 2π nh 2πmr ......... (iii) put the value of ‘r’ in the equation (iii) then, v= nh ×4π2 mZe2 K 2πmn 2 h 2 v= Ze 2K nh r1 n12 Z2 = × r2 n 22 Z1 ATOMIC STRUCTURE on putting the values of , e-, h and K velocity of electron in nth orbit  vn = 2.18 x 106 x Z m/sec n Z Z v v1 ZZ1 n n 2  v v   = = ×× n n v v 2 ZZ 2 n n1 T, Time period of revolution of an electron in its orbit = Time Period, T = 1.52 × 10–16 × n3  Z2 2πr v v Z ; n3  Z2 v = 2r T1 n13 Z22 = × T2 n 32 Z12 1 T Calculation of Energy of an Electron The total energy of an electron revolving in a particular orbit is T.E. = K.E. + P.E. where : P.E. = Potential energy , K.E. = Kinetic energy , T.E. = Total energy The K.E. of an electron = 1 mv2 2 and the P.E. of an electron = – Hence, T.E. = KZe 2 r 1 KZe 2 mv2 – 2 r we know that, KZe 2 mv 2 = or r r2 mv2 = KZe 2 r  substituting the value of mv2 in the above equation : T.E. = KZe 2 KZe 2 KZe 2 – =– 2r r 2r So, T.E. = – KZe 2  2r T.E. = – K.E. = P.E. 2 substituting the value of ‘r’ in the equation of T.E. Then T.E. = – 4π 2 Ze 2 m 2π 2 Z 2 e 4 m K 2 KZe 2 x = – 2 n 2h 2 n2h2 Thus, the total energy of an electron in nth orbit is given by T.E. = 2 2 2 me 4 k 2  z  En = –  n2  h2   v  1 n substituting the value of 'r' and 'v' we get T f, Frequency of revolution of an electron in its orbit = ; ... (iv) K.E. = 1 KZe 2 2 r Putting the value of m,e,h and  we get the expression of total energy Z2 En = – 13.6 2 eV / atom n as the value of n increases , energy of an electron in the orbit increases. Note : - The P.E. at the infinite = 0 The K.E. at the infinite = 0 KEY POINTS Conclusion from Equation of Energy (a) The negative sign of energy indicates that there is attraction between the negatively charged electron and positively charged nucleus. (b) All the quantities on R.H.S. in the energy equation [Eq. iv] are constant for an element having atomic number Z except ‘n’ which is an integer such as 1,2,3, etc . i.e. the energy of an electron is constant as long as the value of ‘n’ is kept constant. (c) The energy of an electron is inversely proportional to the square of ‘n’ with negative sign. ENERGY LEVEL DIAGRAM (i) Orbit of lowest energy is placed at the bottom, and all other orbits are placed above this. (ii) The gap between two orbits is proportional to the energy difference of the orbits. 0 eV -0.85 eV -1.51eV -3.4eV n= n=4 n=3 n=2 12.1eV -13.6eV 10.2eV n=1 Energy level diagram of H-atom Definition Valid for Single Electron System (i) Ground state Lowest energy state of any atom or ion is called ground state of the atom It is n = 1. Ground state energy of H–atom = – 13.6 ev Ground state energy of He+ on = – 54.4 ev (ii) Excited State States of atom other than the ground state are called excited states : n=2 first excited state n=3 second excited state n=4 third excited state n=n+1 nth excited state Ionisation Energy (IE) Minimum energy required to move an electron from ground state to n =  is called ionisation energy of the atom or ion. (iii) I.E. = Energy of first shell. onisation energy of H–atom = 13.6 ev onisation energy of He+ ion = 54.4 ev ATOMIC STRUCTURE onisation energy of Li+2 ion = 122.4 ev (iv) Ionisation Potential (I.P.) Potential difference through which a free electron must be accelerated from rest, such that its kinetic energy becomes equal to ionisation energy of the atom is called ionisation potential of the atom. .P. of H atom = 13.6 V, .P. of He+ on= 54.4 V (v) Excitation Energy Energy required to move an electron from ground state of the atom to any other state of the atom is called excitation energy of that state. Excitation energy of 2nd state = excitation energy of 1st excited state = 1st excitation energy = 10.2 ev. (vi) Excitation Potential Potential difference through which an electron must be accelerated from rest to so that its kinetic energy become equal to excitation energy of any state is called excitation potential of that state. Excitation potential of third state = excitation potential of second exicited state = second excitation potential = 12.09 V. (vii) Binding Energy ‘or’ Seperation Energy Energy required to move an electron from any state to n =  is called binding energy of that state. Binding energy of ground state = .E. of atom or on. Ex. A single electron system has ionization energy 11180 kJ mol–1 . Find the number of protons in the nucleus of the system. Sol. .E. = Z2 n2 × 21.69 × 10–19 J 11180 ×10 3 Z2 = × 21.69 × 10–19 6.023×10 23 12 Ans. Z = 3 Hydrogen Spectrum Study of Emission and Absorption Spectra An instrument used to separate the radiation of different wavelengths (or frequencies) is called spectroscope or a spectrograph. Photograph (or the pattern) of the emergent radiation recorded on the film is called a spectrogram or simply a spectrum of the given radiation The branch or science dealing with the study of spectra is called spectroscopy. Spectrum Based on Nature Discrete Continuous Band Spectrum Emission Spectra Based on origin Absorption spectrum Emission spectrum Line Spectrum When the radiation emitted from some source e.g. from the sun or by passing electric discharge through a gas at low pressure or by heating some substance to high temperature etc, is passed directly through the prism and then received on the photographic plate, the spectrum obtained is called ‘Emission spectrum’. Depending upon the source of radiation, the emission spectra are mainly of two type : (a) Continuous Spectra When white light from any source such as sun, a bulb or any hot glowing body is analysed by passing through a prism it is observed that it splits up into seven different wide band of colours from violet to red. These colours are so continuous that each of them merges into the next. Hence the spectrum is called continuous spectrum. White light Beam Slit Prism Photographic Plate R O Y G B I V (b) Discrete Spectra : It is of two type (i) Band Spectrum Dark space Band Band spectrum contains colourful continuous bands sepearted by some dark space. Generally molecular spectrum are band spectrum (2) Line Spectrum This is the ordered arrangement of lines of particular wavelength seperated by dark space eg. hydrogen spectrum. Line spectrum can be obtained from atoms. 2. Absorption Spectra When white light from any source is first passed through the solution or vapours of a chemical substance and then analysed by the spectroscope, it is observed that some dark lines are obtained in the continuous spectrum. These dark lines are supposed to result from the fact that when white light (containing radiations of many wavelengths) is passed through the chemical substance, radiations of certain wavelengths are absorbed, depending upon the ATOMIC STRUCTURE nature of the element. White light NaCl Solution Slit Prism Photographic Plate R O Y} G B I V Dark lines in yellow region of continuous spectrum Emission Spectrum of Hydrogen H2 Gas Beam Slit Prism Photographic Plate When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on passing electric discharge is examined with a spectroscope, the spectrum obtained is called the emission spectrum of hydrogen. Line Spectrum of Hydrogen Line spectrum of hydrogen is observed due to excitation or de-excitation of electron from one stationary orbit to another stationary orbit Let electron make transition from n2 to n1 (n2 > n1) in a H-like sample – 13.6 Z n 2 2 2 – 13.6 Z2 n1 2 eV n2 eV n1 photon Energy of emitted photon = (E)n2  n1 = Wavelength of emitted photon = = hc ( E)n2 n1 hc  1 1  13.6Z 2  2 - 2   n1 n 2   1  -13.6Z 2  1  -13.6Z 2   = 13.6Z2  2 - 2  – 2 2 n2  n1   n1 n 2  (13.6)z 2  1 1  1 =  2- 2  hc  n1 n 2  Wave number, 1  1 1  =  = RZ2  2 – 2    n1 n 2  R = Rydberg constant = 1.09678 × 107m–1 ; R ~ 1.1 × 107 m–1 ; R = 8 7 6 5 4 Bracket Series 3 Pfund Series Paschen Series 2 1 13.6eV ; R ch = 13.6 eV hc Balmer Series Lyman Series Ex. Calculate the wavelength of a photon emitted when an electron in H- atom maker a transition from n = 2 to n = 1 Sol.  1 1  1 = RZ2  2 - 2    n1 n 2   1 1  1 = R(1)2  2 - 2   1 2   1 3R 4 = or    3R 4 Spectra Lines of Hydrogen Atom LYMAN SERIES • • • It is first spectral series of H. It was found out in ultraviolet region in 1898 by Lyman. It’s value of n1 = 1 and n2 = 2,3,4 where ‘n1’ is ground state and ‘n2’ is called excited state of electron present in a ATOMIC STRUCTURE H - atom. • 1 1  1 = RH  2 - 2  where n2 > 1 always.  1 n 2  • The wavelength of marginal line = • st line of lyman series  2  1 1 n12 for all series. So for lyman series  = R . RH H nd line of lyman series = 3  1 Last line of lyman series =   1 [10.2 eV  (E) lyman  13.6 eV] 12400   lyman  12400 Aº 13.6 10.2 12400 ( E)min • Longest Line : longest wavelength line  longest or  max. = • Shortest Line : shortest wavelength line  shortest or  min = • • 12400 ( E)max First line of any spectral series is the longest ( max) line. Last line of any spectral series is the shortest ( min) line. Series Limit t is the last line of any spectral series. Wave no of st line of Lyman series = 1 1  1 =  = R × 12  2 - 2   1 2   4 -1  2 = R×1  4    =  R × 3 3R = 4 4 4   λ = 3R    Wave no of last line of Lyman series 1   1  = R × 12  2  2  1     =R For Lyman series,  longest = 12400 , shortest (ΔE) 2-1 = 12400  ΔE   1 BALMER SERIES 125 • • • It’s value of n1 = 2 and n2 = 3,4,5,............. • The wavelength of marginal line of Balmer series = •  1 1  1  = RH  2  2  where n2 > 2 always. n2   2 It is the second series of H-spectrum. It was found out in 1892 in visible region by Balmer. n12 4 22 = = RH RH RH 1.9  (E)balmer  3.4 eV.. All the lines of balmer series in H spectrum are not in the visible range. nfact only st 4 lines belongs to visible range. 12400 12400 Å Aº   balmer  3 .4 1. 9 3648 Å   balmer  6536 Å Lines of balmer series (for H atom) lies in the visible range. Ist line of balmer series = 3  2 last line of balmer series =   2 1  5R  1 (  ) 1st line = R ×1  2  2  = 3  36 2 1  R  1 (  ) last line = R  2  2  =   4 2 PASCHEN SERIES (a) It is the third series of H - spectrum. (b) It was found out in infrared region by Paschen. (c) It’s value of n1 = 3 and n2 = 4,5,6 ........ (d) The wavelength of marginal line of Paschen series = (e) 1 = RH   1 1  2  2  where n2 > 3 always. n 2   3 BRACKETT SERIES (a) It is fourth series of H - spectrum. (b) It was found out in infrared region by Brackett. (c) It’s value of n1 = 4 and n2 = 5,6,7 .............. n12 9 32 = = . RH RH RH ATOMIC STRUCTURE (d) The wavelength of marginal line of brackett series = (e) 1 = RH  n12 42 16 = = RH RH RH  1 1  2  2  where n > 4 always. 2 n2   4 PFUND SERIES (a) It is fifth series of H- spectrum. (b) It was found out in infrared region by Pfund. (c) It’s value of n1 = 5 and n2 = 6,7,8 ............... where n1 is ground state and n2 is excited state. (d) The wavelength of marginal line of Pfund series = (e) 1 = RH  n12 52 25 = = RH RH R H  1 1  2  2  where n > 5 always. 2 n2   5 HUMPHRY SERIES (a) It is the sixth series of H - spectrum. (b) It was found out in infrared region by Humphry. (c) It’s value of n1 = 6 and n2 = 7 , 8 , 9 ................... (d) The wavelength of marginal line of Humphry series = (e) 1 = RH  n12 62 36 = = RH R H RH  1 1  2  2  where n > 6. 2 n2   6 Ex. Calculate wavelength for 2nd line of Balmer series of He+ ion Sol. 1 1 1  R(2)2  2  2    n1 n2  n1 = 2 n2 = 4 1 1  1  R( 2 2 )  2  2   4  2 1 3R   4 = 4 Ans. 3R No. of Photons Emitted by a Sample of H Atom If an electron is in any higher state n = n and makes a transition to ground state, then total no. of different photons emitted is equal to n  (n  1) . 2 If an electron is in any higher state n = n2 and makes a transition to another excited state n = n1, then total no. of different photons emitted is equal to (n 2 – n1 )(n 2 – n1 +1) 2 Note : In case of single isolated atom if electron make transition from n th state to the ground state then max. number of spectral lines observed = (n–1) Ex. If electron make transition from 7th excited state to 2nd state in H atom sample find the max. number of spectral lines observed. n = 8 – 2 = 6 Sol.  6  1 7  = 6× spectral lines = 6  = 21  2  2 SOMMERFELD EXTENSION OF THE BOHR'S MODEL  According to sommerfeld electron revolve around the nucleus in the Elliptical Orbits.  Circular orbit is a special case of elliptical orbit when the length of major axis becomes equal to the length of minor axis then the shape of orbit will be circular.  1. If electrons revolve in elliptical orbit then its angular momentum shows two components Radial Component nrh 2 where nr = radial quantum number. Jr = 2. [nr = (n – 1).................0] n = Shell number Azimuthal Components h 2 n = Azimuthal quantum number J= n [ n = 1, 2, 3, 4.............n] n = Shell number So total Angular momentum = Jn Jn = Jr + J nh 2π n = = nr h h + n 2 2π nr + n where n = principal quantum number ATOMIC STRUCTURE Ex. Let Then n=4 n 4 4 = = = nr + n 3 +1 2 +2 4 4 = = 1 +3 0 +4  The length of major axis indicates by nr + ni.e. n and length of minor axis indicates by n  The path of electron K= n r  n n = Length of major axis n = n  Length of minor axis n varies is from 1 to n.  If n = 4 then n= 1, 2, 3, 4 K = 4 4 , , 1 2 4 , 3 4 4 3 Elliptical path  If n = 5 K =   circular path then n= 1, 2, 3, 4, 5 5 , 1 5 5 , , 3 2 4 Elliptical path If n = 1 Then Elliptical path Circulars path In nth Orbit : 5 5 Circular path 5 , 4 = ( n – 1) = ( 1 – 1) = 0 = 1 Number of elliptical path = (n – 1) Number of circular path = 1 In every atom, 1st orbit is always circular. Failures / Limitations of Bohr’s Theory (a) He could not explain the line spectra of atoms containing more than one electron. (b) He also could not explain the presence of multiple spectral lines. (c) He was unable to explain the splitting of spectral lines in magnetic field (Zeeman effect) and in electric field (Stark effect) (d) No conclusion was given for the principle of quantisation of angular momentum. (e) He was unable to explain the de-Broglie’s concept of dual nature of matter. (f) He could not explain Heisenberg’s uncertainty principle. THE DUALNATURE OF MATTER (THE WAVE NATURE OF ELECTRON) 1. In 1924. a French physicist, Louis De Broglie suggested that if the nature of light is both that of a particle and of a wave, then this dual behavior should be true for the matter also. 2. According to De Broglie, the wavelength of an electron is inversely proportional to its momentum p.  1 p or  1 mv  h p Here h = Planck's constant p = momentum of electron  Momentum (p) = Mass (m) × Velocity (c)   h = mv h 2m(K.E.) From the de-Broglie equation it follows that wavelength of a particle decrease with increase in velocity of the particle. Moreover, lighter particles would have longer wavelength than heavier particles, provided velocity is equal.  If a charged particle Q is accelerated through potential difference V from rest then de-broglie wavelength is λ= h 2mQV for an electron  = 150 12.25 Å= Å v v  de-Broglie concept is more significant for microscopic or sub-microscopic particles whose wavelength can be measured.  The circumference of the nth orbit is equal to n times the De-broglie wavelength of the electron revolving around the nucleus. 2 rn  n  where n = no. of shells or no. of waves. Ex. Two particles X and Y are in motion. If the wavelength associated with particle X is 4 × 10–8 m, calculate the wavelength associated with particle Y if its momentum is half of X. Sol. According to de Broglie equation x = h h and y = p px y p x  y y px But py = ½ px (given) x 1 / 2p x  =½ y px B = 2A = 2 × 4 × 10–8m = 8 × 10–8m Ex. Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving with a speed of 30 ms–1. Sol. = h 6.6 × 10 -34 = mv 0.1 × 30  = 2.2 × 10–34 m This is apparent that this wavelength is too small for ordinary observation. Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles. Since, we come across macroscopic objects in our everyday life, de Broglie relationship has no significance in everyday life. HEISENBERG UNCERTAINTY PRINCIPLE ATOMIC STRUCTURE Bohr's theory considers an electron as a material particle. Its position and momentum can be determined with accuracy. But, when an electron is considered in the form of wave as suggested by de-Broglie, it is not possible to ascertain simultaneously the exact position and velocity of the electron more precisely at a given instant since the wave is extending throughout a region of space. In 1927, Werner Heisenberg presented a principle known as Heisenberg uncertainty principle which states as : "It is impossible to measure simultaneously the exact position and exact momentum of a body as small as an electron." The uncertainty of measurement of position, x, and the uncertainty of momentum p or mv, are related by Heisenberg's relationship as : ( p = mv, p = mv) h 4 where h is Planck's constant. x . p > or x . mv > h 4 or x.v  h 4 m x v= uncertainty product For an electron of mass m (9.10 × 10–28 g), the product of uncertainty is quite large. -27 x . v > 6.624 × 10 4πm 6.624 × 10 -27 > 4 × 3.14 × 9.10 ×10 -28 = 0.57 erg sec per gram approximately When x = 0, v= and vice-versa. In the case of bigger particles (having considerable mass), the value of uncertainty product is negligible. If the position is known quite accurately, i.e., x is very small, vbecomes large and vice-versa. In terms of uncertainty in energy E, and uncertainty in time t, this principle is written as,  E.t  h 4 Heisenberg replaced the concept of definite orbits by the concept of probability. According to Heisenberg we can only define the probability of finding electrons around the nucleus.  Ex. Why electron cannot exist inside the nucleus according to Heisenberg's uncertainty principle ? Sol. Diameter of the atomic nucleus is of the order of 10–15 m The maximum uncertainty in the position of electron is 10–15 m. Mass of electron = 9.1 × 10–31 kg. h 4 x × (m.v) = h/4 x. p = v = 1 h 1 6.63 × 10 -34  = × -15 22 4  x.m 10 × 9.1 × 10 -31 4× 7 v = 5.80 × 1010 ms–1 This value is much higher than the velocity of light and hence not possible. DE BROGLIE RELATIONSHIP & HEISENBERG'S UNCERTAINTY PRINCIPLE Ex. The mass of a particle is 1 mg and its velocity is 4.5 × 105 cm per second. What should be the wavelength of this particle if h = 6.652 × 10–27 erg second. (1) 1.4722 × 10–24 cm Sol. (2) 1.4722 × 10–29 cm (3) 1.4722 × 10–32 cm (4) 1.4722 × 10–34 cm Given that m = 1 mg = 1 × 10–3 g c = 4.5 × 105 cm/sec. h = 6.652 × 10–27 erg sec.  Ex. Sol.  6.652 × 10 -27 6.652 × 10 -29 h = = cm = 1.4722 × 10–29 cm mc 1× 10 -3 × 4.5 × 10 5 4.5 Which of the following should be the wavelength of an electron if its mass is 9.1× 10 –31 kg and its velocity is 1/10 of that of light and the value of h is 6.6252× 10–24 joule second? (1) 2.446 × 10–7 metre (2) 2.246 × 10–9 metre (3) 2.246 × 10–11 metre (4) 2.246 × 10–13 metre Given that m = 9.1 × 10–31 kg c 1 of velocity of light 10 c or 1 × 3 × 108 metre/second i.e. 3 × 107 metre/second 10 h = 6.6252 × 10–34 joule second  6.6252 × 10 -34 6.6252 × 10 -34 h = = mc 9.1× 10 -31 × 3 × 10 7 27.3 × 10 -24 0.2426 × 10–10 metre 2.426 × 10–11 metre or or Ex. Sol. What should be the momentum (in gram cm per second) of a particle if its De Broglie wavelength is 1 Å and the value of h is 6.6252 × 10–27 erg second ? (1) 6.6252 × 10–19 (2) 6.6252 × 10–21 (3) 6.6252 × 10–24 (4) 6.6252 × 10––27 Given that = 1 Å = 1 × 10–8 cm h = 6.6252 × 10–27 erg second p= or 6.6252 × 10 -27 = 6.6252 × 10–19 gram cm/sec. 1× 10 -8 Ex. What should be the mass of the sodium photon if its wavelength is 5894Å, the velocity of light is 3 × 10 8 metre/ second and the value of h is 6.6252 × 10–34 kg m2/sec.? (1) 3.746 × 10–26 (2) 3.746 × 10–30 (3) 3.746 × 10–34 (4) 3.746 × 10–36 Sol. = h m×c  m= (= 5894Å = 5894 × 10–10 m) m= or h c 6.652 × 10 -34 6.652 × 10 -32 = 0.0003746 × 10–32 -8 -10 or 17682 3 × 10 × 5894 × 10 3.746 × 10–36 kg ATOMIC STRUCTURE Ex. Sol. What should be the uncertainty in the velocity of an electron if the uncertainty in its position is 0.005 nm, the mass of electron is 9.109 × 10–31 kg and the value of h is 6.6252 × 10–34 joule/second? (1) 2.316 × 105 (2) 1.158 × 107 (3) 2.316 × 109 (4) 2.316 × 1011 Uncertainty in position (x) = 0.005 nm = 0.005 × 10–9 m = 5 × 10–12 m Mass of electron (m) = 9.109 × 10–31 kg. v =  Ex. 6.6252 × 10 -34 4 × 3.14 × 5 × 9.109 × 10 -43 or v =  v = 1.15816 × 107 m/sec. What should be the uncertainty in velocity of a particle of 1 kg mass if uncertainty in position is 1Å and the value of h is 6.6252 × 10–34 Joule sec.? (1) 1.055 × 10–22 Sol. -34 6.6252 × 10 h = m/sec. 4 m  x 4 × 3.14 × 5 × 10 -12 × 9.109 × 10 -31 (2) 1.055 × 1022 (3) 5.25 × 10–25 (4) 1.055 × 1024 Given that x = 1Å = 1 × 10–10 m m = 1 kg h = 6.6252 × 10–34 Joule sec. v  6.6252 × 10 -34 h = 4   m  x 4 × 3.14 × 1× 10 -10 Δv = or 6.6252 × 10 -34 m/sec. 12.56 × 10 -10 = 0.52525 × 10–24 m/sec. = 5.25 × 10–25 m/sec. Ex. What should be the uncertainty in position if uncertainty in momentum is 1 × 10 –2 g cm/sec. and value of h is 6.6252 × 10–34 Joule sec. ? (1) 1.054 × 10–22 m Sol. (2) 1.054 × 10–25 m (3) 0.525 × 10–27 m (4) 1.054 × 10–32 m Given that p = 1 × 10–2 g cm/sec. = 1 × 10–7 kg m/sec. h = 6.6252 × 10–34 Joule sec. Ex. x × p = h 4 or Δx = 6.6252 ×10 -34 = 0.525 × 10–27 m 4 × 3.14 ×10 -7 x = h 4   p A ball weighs 25 g moves with a velocity of 6.6 × 104 cm/sec then find out the De Broglie associated with it. Sol.  =  = h mv 6.6 ×10-34 ×107 10 3 8 ergsec 1  = 25 25 × 6.6 ×10 4 cm / sec = 0.04 × 10–38 × 107 = 0.04 × 10–31 cm = 4 × 10–33 cm Ex. Which of the following has least De Broglie if they have same velocity. 1. e– Sol. 2. p  = 3. CO2 4. SO2 h mv mass of SO2 is greater than the mass of e– , p, CO2 = h constant = v Same least will be SO2   1 m Ex. If uncertainty in position of an e– is same as the x of He atom. If p of e– is 32 ×105 then find p in He atom. Sol. x × p = h 4 Since x is same for both. therefore p will be same by Δx e × Δp e ³ Δx (He) × ΔPHe  h 4π h ³ 4π  Pe 1 PHe Pe  PHe 32 × 105 = 32 × 105 PHe= 32 × 105 Ex. Calculate the uncertainty in the position of a particle when the uncertainty in momentum is (a) 1 × 10–3g cm sec–1 Sol. Given p = 1 × 10–3 g cm sec–1 h = 6.62 × 10–27 erg sec. = 3.142 According to uncertainty principle (b) Zero. ATOMIC STRUCTURE x. p  x  So,  h 4 h 1 . 4  p 6 .62  10 2 7 1  3 4  3.142 10 = 0.527 × 10–24 cm Ex. Sol. (b) When the value of p = 0, the value of x will be infinity. The uncertainty in position and velocity of a particle are 10–10 m and 5.27 × 10–24 ms–1 respectively. Calculate the mass of the particle (h = 6.625 × 10–34 Joule Sec.) According to Heisenberg's uncertainty principle, x.m v  = h 4 or m= h 4πΔx.Δv 6.625 × 10 -34 4 × 3.143 × 10 -10 × 5.27 × 10 -24 = 0.099 kg Ex. Calculate the uncertainty in velocity of a cricket ball of mass 150 g if the uncertainty in its position is of the order of 1Å (h= 6.6 × 10–34 kg m2 s–1). Sol. x . m v = h 4 h 4πΔx.m 6.6 × 10 -34 = 4 × 3.143 × 10 -10 × 0.150        v = = 3.499 × 10–24 ms–1 Orbital An orbital may be defined as the region of space around the nucleus where the probability of finding an electron is maximum (90% to 95%) Orbitals do not define a definite path for the electron, rather they define only the probability of the electron being in various regions of space around the nucles. less dots represent low probability of finding an electron Difference Between Orbit and Orbitals Maximum dots represent high probability of finding an electron S. No. 1 2 3 4 Orbit It is well defined circular path followed by revolving electrons around the nucleus Orbitals It is the region of space around the nucleus where electron is most likely to be found It represents planar motion of electron It represen ts 3 dimensional motion of an electron around the nucleus. The maximum no. of electron in an orbits is 2n where n stands for no. of orbit. Orbits are circular in shape. 2 Orbitals can not accomodate more than 2 electrons. Orbitals have differen t shape e.g. s-orbital is spherical, p - orbital is dumb- bell shaped. 5 Orbit are non directional in character. Orbitals (except s-orbital) have directional character. Hence, they cannot explain shape of molecules Hence, they can account for the shape of molecules. 6 Concept of well defined orbit is against Heisenberg’s uncentainty principle. Concept of orbitals is in accordance with Heisenberg’s principle Shape of The Orbitals Shape of the orbitals are related to the solutions of Schrodinger wave equation, and gives the space in which the probability of finding an electron is maximum. s- Orbital : Shape  spherical Y Z X s- orbital is non directional and it is closest to the nucleus, having lowest energy. s-orbital can accomodate maximum no. of two electrons. p-Orbital : Shape  dumb bell Dumb bell shape consists of two lobes which are separated by a region of zero probability called node. Nodal Plane – The imaginary plane where probability of finding an electron is zero. Y Y Z X Px nodal plane – yz Y Z X Py nodal plane – xz Z Px nodal plane – xy X ATOMIC STRUCTURE p - subshell can accomodate maximum of six electrons. d - Orbital : Shape  double dumb bell Y Z Z 45° Y X dxy dyz Z Y Z Y X X dxz d x -y 2 2 ring X dz 2 d - subshell can accomodate maximum of 10 electrons. f - Orbital : Shape  leaf like or Complex f - orbital can accomodate maximum no. of 14 electrons. Quantum Numbers The set of four numbers required to define the possible location of an electron in an atom are called quantum numbers. The first three have been derived from Schrodinger wave equation. (i) Principal Quantum Number (n) : (Proposed by Bohr) It describes the size of the electron wave and the total energy of the electron. It has integral values 1, 2, 3, 4 ...., etc., and is denoted by K, L, M, N. ..., etc.  Number of subshell present in nth shell = n n subshell 1 2 3 4 s s, p s, p, d s, p, d, f  Number of orbitals present in nth shell = n2 .  The maximum number of electrons which can be present in a principal energy shell is equal to 2n2. No energy shell in the atoms of known elements possesses more than 32 electrons. nh 2 (ii) Azimuthal Quantum Number () : (Proposed by Sommerfield)  Angular momentum of any orbit = It describes the shape of electron cloud and the number of subshells in a shell.  It can have values from 0 to (n – 1). Each value of  represents a subshell.  value of  subshell 0 s 1 p 2 d 3 f  Number of orbitals in a subshell = 2 + 1  Maximum number of electrons in particular subshell = 2 × (2 + 1)  Orbital angular momentum L = i.e. Orbital angular momentum of s orbital = 0, Orbital angular momentum of p orbital = h 2 (1) =  Orbital angular momentum of d orbital = 3 (  1) h     2     2 h , 2 h 2 (iii) Magnetic Quantum Number (m) : (Proposed by Linde) It describes the orientations of the subshells. It can have values from – to +  including zero, i.e., total (2 + 1) values. Each value corresponds to an orbital. s-subshell has one orbital, p-subshell three orbitals (p x, py and pz), d-subshell five orbitals (d xy , d yz , d zx , d x 2  y 2 , dz 2 ) and f-subshell has seven orbitals. The total number of orbitals present in a main energy level is ‘n2’. (iv) Spin Quantum Number (s) : (Proposed by Samuel Goudsmit & Uhlenbeck) It describes the spin of the electron. It has values +1/2 and –1/2. signifies clockwise spinning and anticlockwise spinning. ATOMIC STRUCTURE (i) Spin magnetic moment s = eh 2 mc s( s  1) or  = n (n  2) B.M. (n = no. of unpaired electrons) (ii) It represents the value of spin angular momentum which is equal to (iii) Maximum spin of atom = h 2 s( s  1) 1 x No. of unpaired electron. 2 Rules for Filling of Orbitals 1. Aufbau Principle Aufbau is a German word and its meaning 'Building up' Aufbau principle gives a sequence in which various subshell are filled up depending on the relative order of the Energies of various subshell.   Principle :The subshell with minimum energy is filled up first and when this subshell obtained maximum quota of electrons then the next subshell of higher energy starts filling. The sequence in which various subshell are filled is the following. 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 7p 3d 4d 5d 6d 4f 5f 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6 2. (n +) Rule According to it the sequence in which various subshell are filled up can also be determined with the help of ( n +) value for a given subshell. PRINCIPLE OF (n +) RULE The subshell with lowest( n +) value is filled up first, when two or more subshell have same (n+) value then the subshell with lowest value of n is filled up first. Sub Shell n  n+ 1s 2s 1 2 0 0 1 2 2p 2 1 3 3s 3 0 3     (1) (2) 3p 3 1 4 4s 4 0 4 3d 3 2 5  (1) 4p 4 1 5 (2) 5s 5 0 5 4d 4 2 6 5p 5 1 6 6s 6 0 6            (1) (2) (3) (1) (2) (3) KEY POINTS For H atom the energy of orbital depends only on the value of n. i.e. 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f.................. 3. Pauli's Exclusion Principle In 1925 Pauli stated that no two electron in an atom can have same values of all four quantum numbers.  4. An orbital can accomodates maximum 2 electrons with opposite spin. Hund's Maximum Multiplicity Rule (Multiplicity : Many of the same kind)  According to Hund's rule electrons are distributed among the orbitals of subshell in such a way as to give maximum number of unpaired electron with parallel spin. i.e. in a subshell pairing of electron will not start until and unless all the orbitals of that subshell will get one electron each with same spin. SPIN MULTIPLICITY It is given by 2S + 1 where S is the total spin. (a) For (a), S =  (b) 1 1  =0 2 2 Spin multiplicity = 2S + 1 = 0 + 1 = 1 (singlet) For (b), S =  1 1  =1 2 2 Spin multiplicity = 2S + 1 = 2 × 1 + 1 = 3 (triplet) Ex. Find out the angular momentum of an electron in (a) 4s orbital (b) 3p orbital (c) 4th orbital ATOMIC STRUCTURE Sol. Angular momentum in an orbital = (a)  = 0 for 4s orbital, hence orbital angular momentum = 0 (b)  = 1 for 3p orbital  Angular momentum = (c) Angular momentum in 4th orbit = Ex. Sol. h l l  1  2 h 2 1  1   1  h 2 nh 4 h 2 h   2 2  Given below are the sets of quantum numbers for given orbitals. Name these orbitals. (i) n = 4,  = 2, m = 0 (ii) n = 3,  = 1, m = ±1 (iii) n = 4,  = 0, m = 0 (iv) n = 3,  = 2, m = ±2 (i) 4dz2 (ii) 3px or 3py (iii) 4s (iv) 3d x2  y2 or 3dxy ELECTRONIC CONFIGURATION OF ELEMENTS Based on the rules, we can easily determine the electronic configurations of most element. We just need to know the atomic number of an element, the order in which orbitals are to be filled and the maximum number of electrons in a shell, sub-shell or orbital. The configurtion so obtained can be represented in two ways. As an illustration, let us consider fluorine (Z = 9) : F(Z = 9) = 1s2, 2s2, 2px2, 2py2, 2pz1 or 1s 2s 2px 2py 2pz Importance of knowing the exact electronic configuration of an element lies in the fact that the chemical properties of an element are dependent on the behaviour and relative arrangement of its electrons. Electronic configurations of heavier elements (beyond Z = 56) deviate a little from the order mentioned previously. These are listed below : Lanthanides La (Z = 57) : [Xe]6s25d1 (not 4f1) Ce (Z = 58) : [Xe]6s25d14f1 Pr (Z = 59) : [Xe]6s25d14f2 Actinides Ac (Z = 89) : [Rn]7s26d1 (not 5f1) Th (Z = 90) : [Rn]7s26d15f1 Pa (Z = 91) : [Rn]7s26d15f2 Beyond Z = 103 Z = 104 : [Rn]5f146d27s2 Z = 105 : [Rn]5f146d37s2 Z = 106 : [Rn]5f146d47s2 Z = 112 : [Rn]5f146d107s2 Ex. Write the electronic configuration and find the no. of unpaired electrons as well as total spin for the following atoms (1) C 6 (2) O 8 (3) P 15 (4) Sc 21 Sol. (5) (6) (i) Fe Ne 10 C  1s2, 2s2, 2p2 6 26 No. of unpaired electrons  2. Total spin = (ii) 8 1s 2 2 or 2 2 O  1s2, 2s2, 2p4 2s 2p No. of unpaired electrons = 2  Total spin = 2 2 or 2 2 (iii) 15 P 1s2, 2s2, 2p6, 3s2, 3p3 3s  3p No. of unpaired electrons = 3 Total spin = 3 3 or 2 2 (iv) 21 Sc  1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d1 or [Ar] 4s2 3d1 3d [Ar] 3d 4s 1 2  No. of unpaired electrons = 1  Total spin = (v) or Fe  1s2, 2s2 2p6, 3s2, 3p6, 4s2 3d6 [Ar] 4s2, 3d6 1 1 or 2 2 26 4s 3d No. of unpaired electrons = 4  Total spin = 4 4 or 2 2 4s ATOMIC STRUCTURE (vi) 10 Ne  1s2, 2s2 2p6 No. of unpaired electrons = 0 Total spin = 0 Ex. Sol. Write down the four quantum numbers for fifth and sixth electrons of carbon atom. 2 2 2 6C : 1s , 2s 2p fifth electron : n = 2  = 1 m = – 1 or +1 s=+ 1 1 or – 2 2 sixth electron : n=2 =1 m=0 s=+ 1 1 or – 2 2 Ex. Calculate total spin, magnetic moment for the atoms having at. no. 7, 24 and 36. Sol. The electronic configuration are : 1s2, 2s2 2p3 7N : 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1 24Cr : 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 36Kr  unpaired electron = 3 unpaired electron = 6 unpaired electron = 0 Total spin for an atom = ± 1/2 × no. of unpaired electron For 7N, it is = ± 3/2 ; For 24Cr, it is = ± 3 ; Also magnetic moment = n (n  2) For 7N, it is = 15 ; For 24Cr, it is = 48 ; For 36Kr, it is = 0 For 36Kr, it is = 0 EXCEPTIONALCONFIGURATIONS Stability of Half Filled and Completely Filled Orbitals Cu has 29 electrons. its expected electronic configuration is 1s2,2s2, 2p6, 3s2, 3p6, 4s2, 3d9. But a shift of one electron from lower energy 4s orbital to higher energy 3d orbital will make the distribution of electron symmetrical and hence will impart more stability. Thus the electronic configuration of Cu is 1s2, 2s2, 2p6, 3s2, 2p6, 4s1, d10 Fully filled and half filled orbital are more stable. Ex. We know that fully filled and half filled orbital are more stable. Can you write the electronic configuration of Cr(Z = 24) ? Sol. Cr (Z = 24) 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5. Since half filled orbital is more stable, one 4s electron is shifted to 3d orbital. Ex. A compound of vanadium has a magnetic moment of 1.73 BM work out the electronic configuration of the vanadium in the compound. Sol. Magnetic moment = n n  2  Where n is number of unpaired electrons 1.73 =  n  n  2  or (1.73)2 = n2 + 2n, n = 1 Vanadium atom must have the unpaired electron and thus its configuration is : 23V 4+ : 1s22s22p63s23p63d1 WAVE MECHANICALMODEL OFATOM Nodes Schrodinger Wave Equation General wave equation y = A sin t where, y = displacement A = amplitude t = time Developed by schrodinger, this model is based on the particle and wave nature of electron is known as WAVE MECHANICAL MODEL of atom. The motion of electron around nucleus is round motion and may be considered to be analogous to the STANDING WAVES, the waves which are generated by plucking the stretched string. The amplitude of the standing wave is independent of time and is a function of the distance from one fixed end. The derived eq. by schrodinger is Schrodinger wave equation  where 2  2  2  8 2 m    (E  V)  0 x2 y2 z2 h2  = Amplitude of e– wave (or wave function) m = mass of e– E = Total energy V = Potential energy or 2   8 2 m (E  V)  0 h2      where 2 = Laplacian operator =  2    2  x   y        z2   h2 2+ (E – V)  = 0 8 2 m  h 2 2   2   V   = E  8 m  H   E  H h 2 2  V 8 2 m SCHRODINGER EQUATION IN CARTESIAN COORDINATE Hemiltonion operator ATOMIC STRUCTURE Z = rcos y = rsinsin x = rsincos The schrodinger equation can be written in terms of cartesian coordinates (x, y, z) or in terms of spherical polar coordinates (r, , ). However for most calculations it is simpler to solve the wave equation in polar coordinates. When Schrodinger wave equation in polar coordinates is solved for hydrogen atom the solution obtained can be factorized into 2 separate parts, one being the function of r and other the function of  and .  (r, ) = R(r) f (, ) R(r) = Radial function f (, ) = Angular function SIGNIFICANCE OF () The wave function may be regarded as amplitude of electron wave expressed in terms of coordinates (x, y, z) or (r, , ). The wave function may have +ve or –ve values depending on the values of coordinates. As such there is no physical significance of . SIGNIFICANCE OF ()2 In classical theory of electromagnetic radiation, the square of amplitude is proportional to the intensity of light. A very similar concept was suggested by MAX BORN in QUANTUM MECHANICS according to which the square of function  at any point is proportional to the probability of finding an electron at that point 2 is known as PROBABILITY DENSITY and is always +ve. The region of space in which there is maximum probability of finding an electron (say 90%) is termed as an orbital. QUANTUM NUMBERS In an atom, a large no of orbitals are permissible. These orbitals are designated by a set of 3 numbers known as QUANTUM NUMBERS (principle, azimuthal, magnetic) which arise as a natural consequence in the solution of schrondinger wave equation. These quantum numbers describe energies of electron in an atom, information about shapes and orientation of orbitals. In order to designate the electron an additional quantum number called as SPIN QUANTUM NUMBER is needed to specify spin of the electron. GRAPHICAL REPRESENTATION OF   = Amplitude of wave 2 = probability density  It gives us probability of finding an electron at a point or per unit volume. Probability  2  dv  =  (x, y, z) =  (r, , )  = R(r) ~ () R2 = radial probability density, r = radius R2 = radial probability density of finding an electron at a distance R from the nucleus in any direction. R (i) r Hyperbola  2r  1 R(r)1s  2   e 2 where   na0  a0  a0 = 0.529 Å n = principal quantum number 3 r  1 2  R(r)1 s  2   e a 0  a0  R 2 Graph between R2 and r will be same hyperbolic. r 3 (ii)   1  1 2 R(r)2s  (2   )e 2   2 2  a0  After reaching 2a0 now curve would start decreasing and again goes up to cut at . R(r)2s R2(r)2s  2a0 r r (iii) R(r)3s vs r R(r)3s  R (r)3s 2 r r RADIAL NODES : Points at which the probability of finding an electron is zero is known as radial nodes. No. of radial nodes = n –  – 1 2a0 Electron cannot be present at 2a0 distance from nucleus. If we join all 2a0 points to form a sphere, we can say that electron cannot be present on surface of sphere, 2a0 2a ATOMIC STRUCTURE however it may be present inside or outside. At 2a0 distance probability of presence of electron is zero. Since a sphere is formed radial nodes are also called SPHERICAL NODES. Radial nodes are spherical in shape also known as spherical nodes or nodal sphere. GRAPHS FOR P-ORBITALS 3   1  1 2 2 R(r)2 p   a  e 2 6 0 2p R(r)2p r 3p R(r)3p R2(r)3p r r 4p R(r)4p 3d R(r)3d r r RADIAL PROBABLITY DISTRIbUTION FUNCTION dr is very very less 4 3 3 volume = π((r + dr) - r ) shell 3 = dr r 4 π((r 3 + 3r 2 dr + 3r dr 2 + dr 3 - r 3 ) 3 dr2 and dr3 are neglected 4 (3r 2 dr) = 4r2dr 3 Radial probability in given shell = 4r2drR2 = Radial probability function = 4r2R2 The probability of finding an electron at a distance r from the nucleus in all the direction is called radial probability function (RPF). GRAPHS BETWEEN R PFAND r R (r)1s = Ce  2 42R2r2 = 4r2C2 e = C'r2 e  4 R r = C'r e 2 2 2 2 2r a0   2 4p2R2r2 1s r(max) 2r a0 For s-orbital, R and R2 at nucleus is not zero but probability is almost equal to zero because of very small size. R2 dv  0 as dv  0 for nucleus rmax  distance at which the probability of finding an electron is max. 1  R1 s  2    a0  2 2 1s P  4 R r 3/ 2 r e a0 now differentiate eq. by putting value of R1s 3 1  P = 4r2 × 4   e  a0  2 r 2 r  2 r a0 2 r dP  C(2 r e a 0  r 2 e a 0 ) dr a P = Cr 2 e 0 dP  Cr e dr 2 r a0 2r    2    0 a0   r  a 0 Maxima Peaks of the curves are increasing Here 1st peak is smaller than 2nd and 2nd smaller than 3rd. ANTINODE POINT - Point at which probability of finding an electron is max. COMPARISON OF rmax. & ravg FOR DIFFERENT ORBITALS Note : ravg is always greater than rmax Case I - when  is same but n is different. As value of n increases rmax increases. Peaks are numbered according to value of (n – ). As n increases ravg increases if ravg is more, electron will be more away from nucleus. PENETRATION POWER : Penetration power of orbital is a measure of its closeness to the nucleus. 1s > 2s > 3s Case II - When 'n' is same but '' different. As value of '' increases, r increases ATOMIC STRUCTURE As value of '' increases, rmax decreases   rmax.    ravg.  Closeness to nucleus - 3s > 3p > 3d ENERGY COMPARISON For energy comparison Aufbau rule should be used. But for hydrogen atom, subshell belonging to particular shell possess equal energy. 2p = 2s, 3s = 3p = 3d ANGULAR FUNCTION It gives us an idea about the shape, orientation of an orbital Ex.  = 1  m = –1, 0, +1 Py Px For s orbital, angular part is independent of  and .  There would be zero angular nodes for s-orbital. No.of angular nodes for any orbital = l SHAPE OFANGULAR NODE For Px : Angular Node or Nodal plane  For dxy : Angular node or nodal plane Px  yz plane Py  xz plane Pz  xy plane Pz y x dxy  xz and yz dyz  xz and xy dzx  xy and yz  45° d x 2 – y2 We cannot predict the designation of angular nodes but can be said that at an angle of 45° with axis. Nuclear Chemistry Spontaneous disintegration of nuclei due to emission of radiations like , ,  is called radioactivity. Radioactivity is a nuclei phenomenon. Radioactivity is not depend on external conditions like temperature, pressure etc. Radioactivity of a substance is independent to its physical state. x(s), x(l), x(g), (x)+(g), (x)–(g) in all form, x is radioactive. CO2, 146C(s), 146C(g) is radioactive. Radiations  : 2He4 ( 42He2+) (nucleus of He-atom) or – : –1e0 (fast moving electron emitted from nucleus) 0  : 0 (electromagnetic radiation (waves) of high frequency) speed : >> penetrating power : >> ionisation power : >> 14 1. Emission of rays Usual condition Effect  Z > 83 n ratio increases p Process representation / example XA  Z– 2X’A – 4 + 2He4 Z 92 2.  n If p ratio is high. n ratio decreases p U238  90Th234 + 2He4 YA  Z+ 1Y’A – 4 + –1e0 Z ATOMIC STRUCTURE Ex. n 6 6 C12 (stable) p = 6 6 C14 (radioactive) p = (high) 6 6 n Ex. n 11 Na23 (stable) p = 11 11 Na22 p = ( ratio low) 11 p 11 0 n 7 p =7 n1  1p1 + –1e0 (from nucleus) 12 n If nucleus energy level is high  n 8 p =6 13 Na24 (radioactive) p = (high) 11 n 3. 8 11 n C14  7N14 + –1e0 nucleus energy level decreases 43 Tc99  43Tc99 +  high low nucleus nucleus energy energy (metastable) 4. (a) Positron emission n If p ratio is low (+1e0) Na22  10Ne22 + +1e0 p1  0n1 + +1e0 (from nucleus) 1 n If p ratio is low (EC) or K-shell electron capture n ratio increases ZX´A + 1e0  Z – 1X´´A p K  shell 80  1e 0  0 x1 (K  shell ) X-ray emission occurs K-shell electron capture L K YA  Z – 1Y´A + +1e0 Z 11 (b) Electron capture 1 1p n ratio increases p + Nuclear Stability Belt Hg197 + –1e0  79Au197 120 80 bond high n/p ratio stable low n/p ratio 60 n/p ratio is 1 100 No. of n 40 20 0 20 40 60 80 100 No. of p -Emission p No. of n n 0 Z (Z + 1) No. of p     Z n1  1p1 + –1e0 (Z + 1) Z upto 20 : nuclei stable with n/p ratio nearly 1 : 1 Z > 20 : n/p ratio increases with Z in stable nuclie region. More number of neutrons are required to reduce repulsion between protons. Bi209 : Stable with largest n/p ratio 83 n 1.52 = p 1 Even - Odd Rule : (Out of syllabus) no. of n no. of p no. of stable nucleic even even 155 (max) even odd 55 odd even 50 odd odd 5 (min) * Expected pairing of nucleus Magic Numbers Nuclei in which nucleons have magic no. (2, 8, 20, 28, 50 ....) are more stable. Ex. He4 , 8O16 2 Group Displacement Law : (Given by Soddy and Fajan)  When 1 emission takes place from a nuclie, new formed nuclie occupy two position left in periodic table.  When 1 emission takes place from a nuclie, new formed nuclie occupy one position right in periodic table. Due to emission of 1 particle; isobars are formed. Due to emission of 1 particle; isodiaphers are formed. ATOMIC STRUCTURE Due to emission of 1 and 2 ; isotopes are formed. Isotopes : Isobars : Isotones : same number of proton same mass number same number of neutron eg. 6C14 and 6C12 eg. 6C14 and 7N14 eg. 2He4 and 1H3 Isodiaphers : Same (n – p) difference Ex. 9F19 and 19K39 ; (n – p) = 10 Isosters : Same number of atoms and electrons Ex. N2 and CO N2O and CO2 Artifical Nuclear Reaction Ex.  specific nuclei + stricking particle  New nuclei + emitted particle 1. (, p type) 7 N14 + 2 He4  8 O17 + (s.p.) p1 (or 1H1) (e.p.) 2. (n,  type) 11 N23 + 1 n0  3. (D,p type) 13 Al27 + 1 H2  4. (p,  type) 3 + 1 H1  Li7 1 11 Na24 +  13 Al28 + 1 H1 He4 + 2 He4 2 Nuclear Fission and Nuclear Fusion In both processes, large amount of heat evolved due to conversion of some mass into energy. Nuclear Fission : Is a process where heavy nuclei splits into large nuclei. 92 U235 + 0n1  92 U*236 Ex. atom bomb is based on fission. Nuclear Fusion Is a process where light nuclei fused together to form heavy nuclei. 1 H2 + 1H3  2He4 + 0n1 1 H2 + 1H2  2 He4 Hydrogen bomb is based on fusion. Very high temperature is required in this process. For Objective Questions Total time (T) = no. of halves (n) × Half life (t1/2) T  n  t1/ 2 No. of half life 1 2 3 Int. wt. w w/2 w/4 Final weight w/2 w/4 = w/22 w/8 = w/23 Amount of substance left in n halves (wt / mole) = Initial amount (wt / mole) 2n Radioactive Disintegration Series A series of continued disintegrations starting from an unstable nucleus (radioactive elements) and ending at a stable nucleus, is known as radioactive disintegration series. Mainly Radioactive Disintegration Series are of Four Type (1) Thorium series (4n series) : 23 2 Th (Starting element) 90    –6,–4 20 8 Pb (last element) 82    –7,–4 209 Bi (last element) 83 –8,–6 206 Pb (last element) 82 –7,–4 207 Pb 82 (2) Neptunium series (4n + 1 series): 237 93Np (Starting element) (3) Uranium Series (4n + 2 series) : 23 8 U 92 (Starting element)    (4) Actinium Series (4n + 3 series) : 235 U 9 2 (Starting element)    (last element) * Thorium series (4n series), Uranium Series (4n + 2 series), Actinium Series (4n + 3 series) are naturally series. But Neptunium series (4n + 1 series) is artificial series because Neptunium–237 is prepared artificial and the only member of this series found in nature is the stable end product Bi–209. 24 11 Na Ex. 23 Sol. n/p ratio of 24Na is 13/11 and thus greater than one. It will therefore decay following -emission. Na is the most stable isotope of Na. Find out the process by which 24 24 11 Na  12 Mg Ex. 0 1e The number of -particle emitted during the change ca X  (A) Sol. + ab 4 c X  b Y + m 4 He a d 2  c = b + 4m and a = d + 2m – n by (i) & (ii)  c b  ab (B) d +  2  + c   + n 01e n = d +  2  – a. Ans. (C)   ......(i) ......(ii) can undergo radioactive decay..  c b b is : dY (C) d +  2  – a    ab (D) d +  2  – c   ATOMIC STRUCTURE Ex. The decay product of 13 7 N is : (A) 13 8 O + 0 1e (C) 13 6 C + K electron capture Sol. 13 7 N is positron emitter ; p ratio is low.. Ex. A radioactive element X has an atomic numbers of 100. It decays directly into an element Y which decays directly n (B) 13 6 C + 0 1e (D) 9 5 Be + 24 He Ans. (B) into an element Z. In both processes a charged particle is emitted. Which of the following statement would be true? (A) Y has an atomic number of 102. (B) Z has an atomic number of 101. (C) Z has an atomic number of 97. (D) Z has an atomic number of 99. Sol. X and Y can decay one  each or one  each or X-decays, 1 , Y-decays 1  or X-decays 1 or Y-decays 1 . In either case (a), (b) and (c) cannot be true. Ans. (D) TIPS & FORMULAS c  1. Frequency, 2. Energy/photon, E = h = Also, E = 3. = hc  1 23 75 eV , if  is in Å  Electronic energy change during transition, E = E n 2  E n1 n2 > n1, emission spectra if electron jumps from n2 to n1 shell and absorption spectra if electron excites from n1 to n2 shell. 4. Radius of nth Bohr orbit of H atom, rn = n2h2 (where K = 9 × 109) 4  2 me 2 K r1 for H = 0.529 Å ; rn for H like atom rn = 0.529 × 5. Velocity of electron in nth Bohr orbit of H atom, v = v = 2.18 × 108 6. n2 Å Z 2 KZe 2 nh Z cm / sec . n Energy of electron in nth Bohr orbit of H atom, E = 2  2 mZ2 e 4 K2 n2h2 where n = 1, 2, 3.......... [E = –13.6 × 7. Z2 kcal/mole (1 cal = 4.18 J)] n2 E1 for H = – 21.72 × 10–12 erg = – 13.6 eV, E1 for H like atom = E1 for H × Z2 Wavelength emitted during transition in H atom, 1  RH  2 4 1 1  2  me  2  2  = ch 3  n1 n2  1 1   2  2  (in C.G.S.) n n  1 2  1 mu 2 2 8. Photoelectric effect hv = w + 9. Possible transitions for a jump from n2 to n1 = 10. Angular momentum of electron in an orbit = n. (h/2) 11. Angular momentum of electron in an orbital = (nh/2) l  l  1   or  (n hv = I.E. + K.E. 2  n1 ) ATOMIC STRUCTURE 12. 1  Total spin = ±   n  ; where n is no. of unpaired electrons. 2  13. Magnetic moment of an atom 14. Nodal planes : Radial nodes = n –  – 1, 15. de Broglie equation :  = h  mu n  n  2  B.M.; where n is no. of unpaired electrons. Angular nodes = 1, Total nodes = (n – ) h2 2  K.E.  m where  is wavelength, m is mass and u is velocity of particle. 16. Heisenberg uncertainty principle : p.x  h 4 u.x  h 4 m where p, u and x are uncertainties in momentum, velocity and position respectively. Planck's constant is h and m is mass of subatomic particle.