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I PUC, Unit 2: Structure of Atom, Chemistry, --------------------------------------------------------------------------------------------------------Wave nature of Electromagnetic Radiation :,  James Maxwell suggested that when electrically charged particle moves under, acceleration, alternating electrical and magnetic fields are produced and transmitted.,  These fields are transmitted in the forms of waves called electromagnetic waves or, electromagnetic radiation (emr).,  These are the radiations associated with electric and magnetic fields., Characteristics of Electromagnetic radiations:, 1. The oscillating electric and magnetic fields are perpendicular to each other and both, are perpendicular to the direction of propagation of the wave., 2. The electromagnetic waves do not require a medium for propagation and can move, in vacuum., 3. All the EM radiations travel through vacuum with a constant speed of 3 x 108 m/s., ------------------------------------------------------------------------------------------------------------- Electromagnetic spectrum: It is a series in which different types of, electromagnetic radiations are arranged in the increasing order of their wavelength, (or decreasing order of frequency).,  The important electromagnetic radiations in the increasing order of wavelength are:,  Cosmic rays, Gamma rays, X-rays, Ultra-violet rays, Visible light, Infra red rays,, Microwaves, Radio waves., , Some terms related to EMR:,  Frequency (ν): It is the number of waves passing through a given point in one, second. The SI unit for frequency is hertz (Hz).,  Wavelength (λ): It is the distance between two adjacent crusts or two adjacent, troughs. Its unit is m or cm.,  Wave number (ῡ): It is the number of wavelengths per unit length. It is the, reciprocal of wavelength. i.e. ῡ = 1/ λ, Its unit is m–1 or cm-1.,  The frequency (ν ), speed of light (c) and the wave length (λ)are related as c= ν λ, --------------------------------------------------------------------------------------------------------------Problem: The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of, 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted, by transmitter. Which part of the electromagnetic spectrum does it belong to?, Solution: The wavelength, λ, is equal to c/ν ,, where c is the speed of electromagnetic radiation in vacuum and ν is the frequency., Substituting the given values, we have, Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 1|P a ge

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λ = c/v = (3.00 x 108 ms-1)/1368 kHz, = (3.00 x 108 ms-1)/(1368 x103 s-1) = 219.3m This is a characteristic radiowave wavelength., ----------------------------------------------------------------------------------------------------------------------------Problem: The wavelength range of the visible spectrum extends from violet (400 nm) to, red (750 nm). Express these wavelengths in frequencies (Hz). (1nm = 10-9 m), Solution : Using equation, frequency of violet light v = c / λ, = (3.00 x 108 m s-1)/(400 x 10-9m), = 7.50 x 1014 Hz, Frequency of red light v = c/λ = (3.00 x 108 m s-1)/(750 x 10-9m) = 4.00 x 1014 Hz, The range of visible spectrum is from 4.0 x 1014 to 7.5 1014 Hz in terms of frequency units., ----------------------------------------------------------------------------------------------------------------------------Problem: Calculate (a) wavenumber and (b) frequency of yellow radiation having, wavelength 5800 Å., Solution : (a) Calculation of wavenumber (, , ), , λ =5800Å =5800 × 10-8 cm = 5800 ×10-10 m, ν = 1/λ = 1/(5800 ×10-10 m) = 1.724 x 10-10m-1 = 1.724×104, (b) Calculation of the frequency (ν ) = c/λ = (3 x 108 ms-1)/(5800×10-10 m)=5.172×1014 s-1, -----------------------------------------------------------------------------------------------------------------------------, , Particle Nature of Electromagnetic Radiation:, ,  Black body radiation: An ideal body which emits and absorbs all frequencies of, radiations is called a black body and the radiation emitted by such a body is called, black body radiation.,  The frequency distribution of radiation emitted from a black body depends only on, its temperature., ------------------------------------------------------------------------------------------------------------- Planck’s Quantum Theory: The phenomenon of black body radiation was first, explained by Max Planck by his Quantum theory. According to this theory:, 1. Atoms and molecules could emit (or absorb) energy discontinuously in small packets, of energy called quanta or photons., 2. The energy (E ) of a quantum of radiation is proportional to its frequency (ν).,  It is expressed by the equation, E = hν,  Where ‘h’  Planck’s constant and its value is 6.626×10–34 J s (joule-second)., --------------------------------------------------------------------------------------------------------------Photoelectric effect:,  It is the phenomenon of ejection of electrons by certain metals (like potassium,, rubidium, caesium etc.) when light of suitable frequency incident on them.,  The electrons ejected are called photoelectrons.,  This phenomenon was first observed by H.Hertz., , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 2|P a ge

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Characteristics of photoelectric effect:,  The electrons are ejected from the metal surface as soon as the beam of light strikes, the surface.,  The number of electrons ejected is proportional to the intensity or brightness of light.,  For each metal, there is a minimum frequency (known as threshold frequency [ν0]), below which photoelectric effect is not observed.,  The kinetic energy of the ejected electrons is directly proportional to the frequency, of the incident light., Explanation:,  Photoelectric effect was first explained by Einstein using Planck’s Quantum theory.,  According to him, when a photon of sufficient energy strikes the metal surface, it, suddenly transfers its energy to the electron and the electron is ejected without any, time lag.,  A part of the energy is used to eject the electron from the metal surface. i.e. to, overcome the attractive force of the nucleus [this energy is known as work function,, hν0].,  The other part of energy is given to the ejected electron in the form of kinetic energy.,  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.,  Since the striking photon has energy equal to hν and the minimum energy required to, eject the electron is hν0 (also called work function, W0) then the difference in energy, (hν – hν0) is transferred as the kinetic energy of the photoelectron.,  Following the law of conservation of energy principle, the kinetic energy of the, ejected electron is given by,  K.E = hν - hν0,  Or, hν = hν0 + ½ mev2,  Where me is the mass of the electron and v is the velocity of the ejected electron., -------------------------------------------------------------------------------------------------------------Problem: Calculate energy of one mole of photons of radiation whose frequency is 5 ×10 14 Hz., Solution: ν = 5×1014 s-1 (given) w.k.t, h = 6.626 x 10-34 J s, Energy (E) of one photon is given by the expression, E = hν, E = (6.626 x 10-34 J s) (5 x1014 s-1)= 3.313 x 10-19 J, Energy of one mole of photons= (3.313 x 10 -19 J) x (6.022 x 1023 mol-1) = 199.51 kJ mol-1, , ---------------------------------------------------------------------------------------------------------Problem: A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the, number of photons emitted per second by the bulb., Solution: Power of the bulb = 100 watt = 100 J s-1, Energy of one photon E = hν =hc/λ = (6.626 x 10-34 J s x 3 x 108 m s-1)/(400×10-9m)= 4.969 ×10-19 J, Number of photons emitted 100 J s-1/(4.969 x 10-19 J) = 2.012 x1020 s-1, , ----------------------------------------------------------------------------------------------------------Problem: When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium,, electrons are emitted with a kinetic energy of 1.68 ×10 5 J mol-1. What is the minimum energy, needed to remove an electron from sodium? What is the maximum wavelength that will cause a, photoelectron to be emitted?, Solution: The energy (E) of a 300 nm photon is given by, hν = hc/λ = (6.626 x 10-34 J s x 3.0 x 108 m s-1)/(300 x 10-9) = 6.626 × 10-19 J, The energy of one mole of photons = 6.626 ×10 -19 J × 6.022 ×1023 mol-1= 3.99 × 105 J mol-1, Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 3|P a ge

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The minimum energy needed to remove a mole of electrons from sodium, = (3.99 – 1.68) 105 Jmol-1= 2.31 × 105 J mol-1, The minimum energy for one electron =, (2.31 x 105 J mol-1)/(6.022 x1023 electrons mol-1= 3.84 10-19 J. This corresponds to the wavelength., ∴λ =hc/E =(6.626 x 10-34 J s x 3.0 x 108m s-1)/(3.84 x 10-19J)= 517 nm (This corresponds to green light), , -----------------------------------------------------------------------------------------------------------Problem: The threshold frequency ν0 for a metal is 7.0 ×1014 s-1. Calculate the kinetic energy of an, electron emitted when radiation of frequency ν =1.0 ×10 15 s-1 hits the metal., Solution: According to Einstein’s equation Kinetic energy = mev2=h(ν – ν0 ), = (6.626 ×10-34 J s) (1.0 × 1015 s-1 – 7.0 ×1014 s-1), = (6.626 ×10-34 J s) (10.0 ×1014 s-1 – 7.0 ×1014 s-1), = (6.626 ×10-34 J s) (3.0 ×1014 s-1), = 1.988 ×10-19 J, -----------------------------------------------------------------------------------------------------------------------------, , Dual Behavior of Electromagnetic Radiation, • Electromagnetic radiations possess both particle and wave nature. This is known as, dual nature of Electromagnetic radiation., -------------------------------------------------------------------------------------------------------- ATOMIC SPECTRUM: The spectrum produced by an excited atom or molecule, is called atomic spectrum.,  Atomic spectra are of two types – Emission spectrum and Absorption spectrum.,  The spectrum of radiation emitted by a substance that has absorbed energy is called, an emission spectrum.,  To produce an emission spectrum, energy is supplied to a sample by heating it or, irradiating it and the wavelength (or frequency) of the radiation emitted is recorded.,  The emission spectra of atoms contain some lines with dark spaces between them. So, they are called line spectra or atomic spectra.,  Each element has a unique line emission spectrum. So line emission spectra are also, called finger print of atoms., , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 4|P a ge

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Line spectrum of Hydrogen:, The hydrogen spectrum consists of mainly, five series of lines. They are:, 1. Lyman series, 2.Balmer series, 3.Paschen series, 4.Brackett series, 5.Pfund series.,  Rydberg equation for finding the wave, number of different lines in Hydrogen, spectrum is:, ῡ = 1/ λ =109677 (1/n12 -1/n22) cm-1,  Where n1 = 1, 2, 3,…..and, n2 = n1 + 1, n1 + 2, ……, SERIS, , n1, , n2, , SPECTRAL REGI ON, , Lyman, Balmer, Paschen, Brackett, Pfund, , 1, 2, 3, 4, 5, , 2,3…., 3,4…., 4,5…., 5,6…., 6,7…., , Ultraviolet, Visible, Infrared, Infrared, Infrared, , ------------------------------------------------------------------------------------------------------BOHR’S MODEL FOR HYDROGEN ATOM;, The important postulates are:,  The electron in the hydrogen atom can move around the nucleus in a circular path of, fixed radius and energy.,  These paths are called orbits or stationary states or allowed energy states. These, energy levels are numbered as 1,2,3 etc or designated as K, L, M, N, etc. These, numbers are known as Principal quantum numbers.,  The energy of an electron in the orbit does not change with time. However, when an, electron absorbs energy, it will move away from the nucleus (i.e. to a higher energy, level) and when it loses energy, it will move towards the nucleus (i.e. to a lower, energy level).,  The radius of orbits can be given by the equation: rn = a0 n2 where a0 = 52.9 pm., Thus the radius of the first stationary state is 52.9 pm (called the Bohr radius). As n, increases, the value of r will increase.,  The energy of electron in an orbit is given by the expression: En = -RH (1/n2)., where n = 1,2,3…… and RH is a constant called Rydberg constant., Its value is 2.18x10-18 J.,  The energy of the lowest state (the ground state) is given by E1 = –2.18×10–18J. As, the value of n increases, the energy of the electron also increases.,  The frequency of radiation absorbed or emitted when transition occurs between two, stationary states that differ in energy by ΔE, is given by:, ν = ΔE = E2 – E1, h, h, Where E1 and E2 are the energies of lower and higher energy levels respectively., This expression is commonly known as Bohr’s frequency rule., , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 5|P a ge

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 The angular momentum of an electron is an integral multiple of h/2π. i.e. mevr =, nh/2π., Where me is the mass of electron, v is the velocity of electron and r is the radius of, Bohr orbit. n = 1,2,3........., Thus an electron can move only in those orbits whose angular momentum is an, integral multiple of h/2π. So only certain fixed orbits are allowed., Significance of negative energy of electron :,  When the electron is free from the influence of nucleus, its energy is taken as zero. In, this situation, the electron is at the orbit with n=∞.,  When the electron is attracted by the nucleus and is present in an orbit n, the energy, is emitted. So it becomes less than zero. That is the reason for the presence of, negative sign in equation., Limitations of Bohr Atom Model:, 1) It could not explain the fine spectrum of hydrogen atom., 2) It could not explain the spectrum of atoms other than hydrogen., 3) It could not explain Stark effect and Zeeman effect., 4) It could not explain the ability of atoms to form molecules by chemical bonds., 5) It did not consider the wave character of matter and Heisenberg’s uncertainty, principle., -------------------------------------------------------------------------------------------------------------Problem: What are the frequency and wavelength of a photon emitted during a transition, from n = 5 state to the n = 2 state in the hydrogen atom?, Solution: Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of, the Balmer series. From equation, ΔE = 2.18 x10-18 J[1/52 - 1/22] = 4.58 x 10-19 J, It is an emission energy The frequency of the photon (taking energy in terms of magnitude) is, given by v = ΔE/h = 4.58×10-19 J/6.626×10-34 Js = 6.91 x 1014 Hz, λ = c/v = 3.0 x 108 ms-1/6.91 x 1014 HZ = 434 nm, ----------------------------------------------------------------------------------------------------------------------------, , Problem: Calculate the energy associated with the first orbit of He+. What is the radius of, this orbit?, Solution: En = -(2.18 x 10-18 J)Z2/n2 atom-1, For He+, n = 1, Z = 2, E1 = (2.18 x 10-18 J)22/12 = -8.72 x 102.18 x 10-18 J, The radius of the orbit is given by equation rn = (0.0529 nm)n2/Z, Since n = 1, and Z = 2, rn = (0.0529 nm)12 = 0.02645 nm, , -------------------------------------------------------------------------------------------------------------Dual Behavior of Matter:,  de Broglie proposed that like radiation, matter also exhibit dual behaviour i.e., both, particle and wave like properties. This means that electrons should also have, momentum as well as wavelength.,  The waves associated with matter is called Matter waves., ----------------------------------------------------------------------------------------------------de Broglie’s equation:,  De Broglie proposed a relation between wavelength (λ) and momentum (p) of a, material particle. This equation is known as de Broglie’s equation.,  The equation is:, λ= h = h, mv p, Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 6|P a ge

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Where m is the mass of the particle, v is the velocity and p is the momentum (p=mv)., This equation is applicable to all material bodies. But for macroscopic bodies the, wavelength is too small to detect., ---------------------------------------------------------------------------------------------------------Problem: What will be wavelength of a ball of mass 0.1 kg moving with a velocity of 10 ms-1 ?, Solution: According to de Brogile equation λ = h/mv = (6.266 x 10 -34 Js)/ (0.1kg)(10 ms-1 ), = 6.626 x 10-34 m (J = kg m2 s-2), -----------------------------------------------------------------------------------------------------------------------Problem :The mass of an electron is 9.1×10-31 kg. If its K.E. is 3.0×10-25 J, calculate its wavelength., Solution: Since K. E. = 1/2 mv2, v = (2K.E./m)1/2 = (2x 3.3x 10-25 kg m2s-2/9.1 x 10-31kg)1/2= 812 m s-1, λ = h/mv = (6.626 x 10-34 Js)/ (9.1 x 10-31kg)(812 ms-1)= 8967 x 10-10 m = 896.7 nm, -----------------------------------------------------------------------------------------------------------------------Problem : Calculate the mass of a photon with wavelength 3.6 Å., Solution: λ = 3.6 Å = 3.6×10−10m, Velocity of photon = velocity of light, m= h/λν = (6.626 x 10-34Js)/(3.6 x10–10m)(3 x 108 ms-1) = 6.135 x 10-29 kg, , -------------------------------------------------------------------------------------------------------------Heisenberg’s Uncertainty Principle:,  It states that “it is impossible to determine simultaneously, the exact position and, exact momentum (or velocity) of a moving microscopic particle like electron”.,  Mathematically, it can be given by the equation: Δx. Δp ≥ h, 4π,  Or, Δx.mΔv ≥ h, 4π,  Or, Δx.Δv ≥ h, 4πm, Where Δx is the uncertainty in position and, Δp (or, Δv) is the uncertainty in momentum (or velocity) ., -------------------------------------------------------------------------------------------------------------Significance of Uncertainty Principle:,  Heisenberg Uncertainty Principle is significant only for motion of microscopic, objects.,  According to this Principle, we cannot determine the exact position and momentum, of an electron.,  Thus it rules out the existance of definite paths or orbits of electrons.,  We can only say the probability of finding an electron at a given point., ------------------------------------------------------------------------------------------------------------Problem : A microscope using suitable photons is employed to locate an electron in an, atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement, of its velocity?, Solution:, Δx Δp = h/4π or ΔxmΔv=h/4π, Δv = h/4πΔxm, Δv = (6.626 x 10-34Js)/(4×3.14×0.1×10-10m x 9.11 x 10-31kg), = 0.579×107 m s-1 (1J = 1 kg m2 s-2), = 5.79106 m s-1, , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 7|P a ge

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Problem: A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be, measured within accuracy of 2%, calculate the uncertainty in the position., Solution: The uncertainty in the speed is 2%, i.e.,45 × 2/100=0.9ms-1 ., Using the equation ,Δx = h/4πmΔv, = (6.626 x 10-34 Js)/(4×3.14x40gx10-3kg g-1(0.9m s-1)), = 1.46 x 10-33 m, --------------------------------------------------------------------------------------------------------------QUANTUM MECHANICAL MODEL OF ATOM,  Erwin Schrodinger and Werner Heisenberg proposed a new model of atom called, Quantum mechanics., The important features of the quantum mechanical model of atom:,  The energy of electrons in atoms is quantized (i.e., can only have certain specific, values), for example when electrons are bound to the nucleus in atoms.,  The fundamental equation of quantum mechanics was developed by Schrödinger and, is known as Schrödinger equation. The equation is: Ĥ ψ = Eψ, where Ĥ is a mathematical operator called Hamiltonian operator,, E is the total energy of the system (K.E + P.E) and, ψ is called the wave function., Significance of ψ:,  The wave function (ψ) is a mathematical function and is Probability amplitude.,  But ψ2 has some physical significance. It gives the probability of finding an electron, at a point within an atom. So ψ2 is known as probability density.,  From the value of ψ2, it is possible to predict the probability of finding the electron, around the nucleus., ---------------------------------------------------------------------------------------------------------------, , Quantum Numbers,  Quantum Numbers are certain numbers used to explain the size, shape and, orientation of orbitals. Or, Quantum numbers are the address of an electron., There are four quantum numbers. They are:, 1. Principal Quantum number (n), 2. Azimuthal Quantum number (Ɩ), 3. Magnetic Quantum number (m or mƖ), 4. Spin Quantum number (s or ms), ------------------------------------------------------------------------------------------------------Principal Quantum number (n),  It gives the following informations:, 1. The size the orbit., 2. The energy of electron in an orbit., 3. The shell in which the electron is found., 4. The average distance between the electron and the nucleus.,  The possible values of n are 1, 2, 3, 4, etc.,  n= 1 represents K shell,  n = 2 represents L shell,  n = 3 represents M shell etc., ----------------------------------------------------------------------------------------------Azimuthal Quantum Number [Subsidiary Quantum number] (Ɩ), It gives the following informations:, 1. The shape of the orbital., 2. The sub shell or sub level in which the electron is located., 3. The orbital angular momentum of the electron., Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 8|P a ge

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 the possible value of Ɩ are 0, 1, 2, 3, .......... (n-1).,  When n = 1, Ɩ= 0. i.e. K shell contains only one sub shell - s sub shell.,  when n = 2, Ɩ = 0 and1. i.e. L shell contains two sub shells - s and p.,  when n = 3, Ɩ = 0, 1 and 2. i.e. M shell contains three sub shells – s, p and d.,  when n = 4, Ɩ = 0, 1, 2 and 3. i.e. N shell contains four sub shells – s, p,d and f., ----------------------------------------------------------------------------------------------Magnetic Quantum Number (m),  It gives the orientation of orbitals in space.,  For a given ‘Ɩ’ value, there are (2Ɩ+1) possible values for m.,  The possible values for m are:– Ɩ to 0 to + Ɩ.,  When Ɩ = 0, mƖ = 0. i.e. s-sub shell contains only one orbital called s orbital.,  For Ɩ = 1, mƖ = –1, 0 and +1. i.e. p subshell contains three orbitals called p, orbitals (px, py and pz)., For Ɩ = 2, mƖ = –2, –1, 0, +1 and +2. i.e. d subshell contains five orbitals called d, orbitals (dxy, dxz, dyz, dx2- y2 and dz2), ------------------------------------------------------------------------------------------------------Spin Quantum Number (s or ms),  It is the only experimental Quantum number.,  It gives the spin orientation of electrons.,  It has 2 values: +½ or -½.,  +½ represents clock-wise spin,  -½ represents anticlock-wise spin., ------------------------------------------------------------------------------------------------------Problem : What is the total number of orbitals associated with principal quantum number n = 3 ?, Solution: For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0, and ml = 0); there are three 3p orbitals (n = 3, l = 1 and ml = -1, 0, +1); there are five 3d orbitals, (n = 3, l = 2 and ml = -2, -1, 0, +1+, +2)., Therefore, the total number of orbitals is 1+3+5 = 9. The same value can also be obtained by using, the relation; number of orbitals = n2, i.e. 32 = 9., ----------------------------------------------------------------------------------------------------------------------------Problem : Using s, p, d, f notations, describe the orbital with the following quantum, numbers (a) n = 2, l = 1, (b) n = 4, l = 0, (c) n = 5, l = 3, (d) n = 3, l = 2, Solution:, n, , l, , orbital, , a), , 2, , 1, , 2p, , b), , 4, , 0, , 4s, , c), , 5, , 3, , 5f, , d), , 3, , 2, , 3d, , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 9|P a ge

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Shapes of orbitals:,  s-orbital,  For s-orbitals, Ɩ = 0 and hence m=0. So there is only one possible orientation, for s orbitals.,  They are spherically symmetrical.,  All s-orbitals have same shape but they have different size.,  The plots of probability density (ψ2) against distance from the nucleus (r) for, 1s and 2s atomic orbitals are as follows:, ,  For 1s orbital the probability density is maximum around the nucleus and it, decreases with increase in r.,  But for 2s orbital the ψ2 first decreases sharply to zero and again starts, increasing.,  After reaching a small maximum it decreases again and approaches zero as the, value of r increases.,  Node:,  The region where the probability density (ψ 2) reduces to zero is called node.,  There are two types of Nodes: Radial node and angular node.,  Number of radial nodes = n - Ɩ – 1,  Number of angular nodes = Ɩ,  Total number of nodes = n-1,  The boundary surface diagrams for 1s, 2sand 3s orbitals are:, , ------------------------------------------------------------------------------------------------------Shape of p-orbitals:,  For p-orbitals, Ɩ = 1 and mƖ = -1, 0, +1.,  i.e., there are three possible orientations for p orbitals.,  So there are 3 types of p-orbitals – px, py and pz.,  Each p orbital consists of two lobes.,  They have dumb-bell shapes.,  The boundary surface diagrams for three 2p orbitals are as follows:, Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 10 | P a g e

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 The no. of radial nodes for p-orbitals = n–2.,  i.e. for 2p orbital, no. of radial nodes =1, for 3p orbital, it is 2 and so on.,  The probability density functions for the p-orbitals are zero at the plane, passing through the origin (angular node).,  For example, in the case of p z orbital, xy-plane is a nodal plane.,  The number of angular nodes for p-orbitals = 1., ------------------------------------------------------------------------------------------------------Shape of d-orbitals:,  For d-orbitals, Ɩ = 2 and m = -2, -1,, 0, +1 and +2.,  i.e., there are 5 types of d-orbitals., They are dxy, dxz, dyz, dx2-y2 and, dz2.,  The shapes of the first four dorbitals are double dumb-bell.,  The shape of dz2 orbital is dumbbell having a circular collar in the, xy-plane., , ------------------------------------------------------------------------------------------------------Shape of f-orbitals:,  For f-orbitals, Ɩ = 3 and m = -3, -2, -1, 0, +1, +2 and +3.,  i.e., there are seven possible orientations for f orbitals.,  So there are 7 types of f-orbitals, [fx3, fy3, fz3, fxyz, fx(y2-z2), fy(z2-x2), fz(x2-y2)].,  They have diffused shapes., ----------------------------------------------------------------------------------------------Energies of Orbitals, The energy of an electron in a hydrogen atom is determined solely by the, principal quantum number. Thus the energy of the orbitals increases as follows, : 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f, , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 11 | P a g e

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ORBITAL, , VALUE OF, , n, , VALUE OF, , l, , VALUE OF (n+l), , 1s, 1, 0, 1+0=1, 2s, 2, 0, 2+0=2, 2p, 2, 1, 2+1=3, 2p(n=2) has lower energy than, 3s, 3, 0, 3+0=3, 3s(n=3), 3p, 3, 1, 3+1=4, 3p(n=3) has lower energy than, 4s, 4, 0, 4+0=4, 4s(n=4), 3d, 3, 2, 3+2=5, 3d(n=3) has lower energy than, 4p, 4, 1, 4+1=5, 4p(n=4), Arrangement of Orbitals with Increasing Energy on the Basis of (n+l ) Rule, , ------------------------------------------------------------------------------------------------------Rules for Filling of electrons in various orbitals :,  Various orbitals are filled according to the 3 rules:, 1. Aufbau principle, 2. Pauli’s exclusion principle and, 3. Hund’s rule of maximum multiplicity., ------------------------------------------------------------------------------------------------------Aufbau principle:,  The German word aufbau means, ‘build up’.,  It states that the orbitals are filled, in order of their increasing, energies.,  This rule has two sub rules:, i. The various orbitals are filled in the, increasing order of their (n+Ɩ) value., ii. If two orbitals have the same (n+Ɩ), values, the orbital with the lower n, value is filled first., The increasing order of energy for various orbitals are:, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ……, ------------------------------------------------------------------------------------------------------Pauli’s exclusion principle :,  It states that no two electrons in an atom can have the same set of four, quantum numbers.,  i.e. an orbital can accommodate a maximum of only 2 electrons with opposite, spin.,  If 2 electrons have same values for n, Ɩ and m, they should have different, values for s.,  i.e. if s = +½ for the first electron, it should be -½ for the second electron., ------------------------------------------------------------------------------------------------------Hund’s rule of maximum multiplicity:,  It states that electron pairing takes place only after partially filling all the, degenerate orbitals.,  For example the electronic configuration of N is 1s 2 2s2 2px1py1pz1 and not, 1s2 2s2 2px2py1.,  Orbitals having same energies are called degenerate orbitals., ------------------------------------------------------------------------------------------------------Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 12 | P a g e

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------------------------------------------------------------------------------------------------------ The electronic configuration can be simplified by using noble gas, configuration as follows:, Noble Gas, , Atomic, Number, , Sub shell filled, after this confgn., , He, , 2, , 2s, , Ne, , 10, , 3s, , Ar, , 18, , 4s, , Kr, , 36, , 5s, , Xe, , 54, , 6s, , Rn, , 86, , 7s, , (ii) Orbital diagram :, , ------------------------------------------------------------------------------------------------------Extra Stability of Half Filled and Completely Filled Subshells :,  Atoms having half filled or completely filled electronic configurations have, extra stability.,  This is due to their symmetrical distribution of electrons and greater exchange, energy.,  The electronic configuration of Cr is [Ar] 3d54s1 and not 3d44s2.,  This is because of the extra stability of half filled d5 configuration.,  Similarly for Cu the electronic configuration is [Ar] 3d 104s1 and not 3d94s2.,  This is due to the extra stability of completely filled d 10 configuration., ============================================================, , Dept. Of Chemistry, Holy Rosary Composite PU College, Honnavar, , 14 | P a g e