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Atoms and Molecules By: Dr. Ashish Kumar LEARNING OBJECTIVES After completing this chapter, you will be able to understand the following: Concept of.

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Presentation on theme: "Atoms and Molecules By: Dr. Ashish Kumar LEARNING OBJECTIVES After completing this chapter, you will be able to understand the following: Concept of."— Presentation transcript:

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2 Atoms and Molecules By: Dr. Ashish Kumar

3 LEARNING OBJECTIVES After completing this chapter, you will be able to understand the following: Concept of the structure of atom as a nucleus with orbital electrons. Progress towards quantum mechanical model of the atom. Quantum mechanical model of an atom and its important features. Application of Schrödinger equation to explain the structure of hydrogen atom. Radial and angular functions of wave equations. Statement of Pauli exclusion principle. The features of quantum mechanical model. Concept of orbitals and quantum numbers. Electronic configuration of atoms. By: Dr. Ashish Kumar

4 Experiments to determine what an atom was J. J. Thomson- used Cathode ray tubes A cathode ray tube with an electric field perpendicular to the direction of the cathode rays and an external magnetic field. The symbols N and S denote the north and south poles of the magnet. The cathode rays will strike the end of the tube at A in the presence of a magnetic field, at C in the presence of an electric field, and at B when there are no external fields present or when the effects of the electric field and magnetic field cancel each other. By: Dr. Ashish Kumar

5 Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

6 Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

7 Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

8 n Passing an electric current makes a beam appear to move from the negative to the positive end Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

9 n Passing an electric current makes a beam appear to move from the negative to the positive end Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

10 n Passing an electric current makes a beam appear to move from the negative to the positive end Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

11 n Passing an electric current makes a beam appear to move from the negative to the positive end Thomson’s Experiment Voltage source +- By: Dr. Ashish Kumar

12 Voltage source Thomson’s Experiment By adding an electric field By: Dr. Ashish Kumar

13 Voltage source Thomson’s Experiment n By adding an electric field + - By: Dr. Ashish Kumar

14 Voltage source Thomson’s Experiment n By adding an electric field + - By: Dr. Ashish Kumar

15 Voltage source Thomson’s Experiment n By adding an electric field + - By: Dr. Ashish Kumar

16 Voltage source Thomson’s Experiment n By adding an electric field + - By: Dr. Ashish Kumar

17 Voltage source Thomson’s Experiment n By adding an electric field + - By: Dr. Ashish Kumar

18 Voltage source Thomson’s Experiment n By adding an electric field he found that the moving pieces were negative + - By: Dr. Ashish Kumar

19 Thomsom’s Model (a) A cathode ray produced in a discharge tube. The ray itself is invisible, but the fluorescence of a zinc sulfide coating on the glass causes it to appear green. (b) The cathode ray is bent in the presence of a magnet. By: Dr. Ashish Kumar

20 Thomsom’s Model Found the electron Couldn’t find positive (for a while) Said the atom was like plum pudding A bunch of positive stuff, with the electrons able to be removed Charge/mass = -1.76 X 10 8 C/g By: Dr. Ashish Kumar

21 Mullikan Charge/mass = -1.76 X 10 8 C/g Found by oil drop method charge on electron= -1.6022 X 10 -19 C By: Dr. Ashish Kumar

22 Rutherford’s Experiment Used uranium to produce alpha particles Aimed alpha particles at gold foil by drilling hole in lead block Since the mass is evenly distributed in gold atoms alpha particles should go straight through. Used gold foil because it could be made atoms thin By: Dr. Ashish Kumar

23 Lead block Uranium Gold Foil Florescent Screen By: Dr. Ashish Kumar

24 What he expected By: Dr. Ashish Kumar

25 Because By: Dr. Ashish Kumar

26 Because, he thought the mass was evenly distributed in the atom By: Dr. Ashish Kumar

27 What he got By: Dr. Ashish Kumar

28 How he explained it + Atom is mostly empty Small dense, positive piece at center Alpha particles are deflected by it if they get close enough By: Dr. Ashish Kumar

29 +

30 Modern View The atom is mostly empty space Two regions Nucleus- protons and neutrons Electron cloud- region where you have a chance of finding an electron By: Dr. Ashish Kumar

31 Modern View By: Dr. Ashish Kumar

32 Properties of Waves Wavelength ( ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1 By: Dr. Ashish Kumar

33 Properties of Waves Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x 7.1 By: Dr. Ashish Kumar

34 Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation x  c 7.1 By: Dr. Ashish Kumar

35 7.1 By: Dr. Ashish Kumar

36 x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m Radio wave A photon has a frequency of 6.0 x 10 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? = 5.0 x 10 12 nm 7.1 By: Dr. Ashish Kumar

37 Mystery #1, “Black Body Problem” Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x Planck’s constant (h) h = 6.63 x 10 -34 J s 7.1 By: Dr. Ashish Kumar

38 Light has both: 1.wave nature 2.particle nature h = KE + BE Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 Photon is a “particle” of light KE = h - BE h KE e - 7.2 By: Dr. Ashish Kumar

39 De Broglie (1924) reasoned that e - is both particle and wave. Why is e - energy quantized? 7.4 u = velocity of e- m = mass of e- 2  r = n = h mu By: Dr. Ashish Kumar

40 Let’s consider a base ball mass=0.142 kg and velocity = 42 m/s  = h /mv  = 6.63 x 10 -34 (kg m 2 s -2 )s / 0.142 kg x 42 m s -1 = 1.1 x 10 -31 m --------Undetectably small Why don’t we observe the wave nature of matter in day to day activities ? By: Dr. Ashish Kumar

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43 = h/mu = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kgh in J su in (m/s) 7.4 By: Dr. Ashish Kumar

44 E = h x E = 6.63 x 10 -34 (J s) x 3.00 x 10 8 (m/s) / 0.154 x 10 -9 (m) E = 1.29 x 10 -15 J E = h x c /  7.2 When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. By: Dr. Ashish Kumar

45 Heisenberg’s Uncertainty Principle According to Heisenberg’s uncertainty principle, it is difficult to measure both quantities accurately at the same time. Heisenberg stated that the more precisely we can define the position of an electron, the less certainly we are able to define its velocity, and vice versa. If  x is the uncertainty in defining the position and Δp x (or Δv x ) the uncertainty in the momentum (or velocity), the uncertainty principle may be expressed mathematically as: By: Dr. Ashish Kumar

46 Significance of Uncertainty Principle The uncertainty principle formally limits the precision to which two complementary observables – position and momentum – of a subatomic particle can be measured. It also leads to the conclusion that the properties of a subatomic particle observed are not inde­pendent of the observer. It also leads to the concept that the values of the position and momentum of a particle can take a range of values rather than a single, exact value. Using Bohr’s atomic theory, we could calculate the radius of a one-electron system (such as H, He + and Li 2+ ) and the velocity of the electron in orbit. For objects of large mass, the uncertainty in the position (∆x) and the uncertainty in the velocity (∆v) are insignificant because the value of h/4  is very small and is equal to 5.273  10 −35 J s and its mass is very large compared to the value of h/4 . By: Dr. Ashish Kumar

47 Heisenberg’s principle becomes significant in case of microscopic objects like an electron. The mass of an electron is 9.1  10 −31 kg, then according to Heisenberg’s principle By: Dr. Ashish Kumar

48 QUANTUM MECHANICAL MODEL OF ATOM Quantum mechanics is a theory of atomic structure based on the wave– particle duality of matter. For a system, such as an atom, whose energy does not change with time, the Schrödinger equation is given by is a mathematical operator called Hamiltonian which is constructed based on the total energy of the system. Solutions to the wave equation are called wave functions and given by the symbol . By: Dr. Ashish Kumar

49 Hydrogen Atom and the Schrödinger Wave Equation The application of Schrödinger’s wave equation to multi-electron systems is somewhat more complicated and its solution is not easily obtained. However, computer-aided calculations have shown that the orbitals in multi-electron atoms are similar to that in hydrogen, except that increased nuclear charge leads to their contraction. The probability of finding an electron at a point x, y, z over all space By: Dr. Ashish Kumar

50 Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function (  ) describes: 1. energy of e - with a given  2. probability of finding e - in a volume of space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems. 7.5 By: Dr. Ashish Kumar

51 Important Features of Quantum Mechanical Model The energy of electrons in an atom is quantized, that is, they can have only specific values when bound to nucleus in an atom. All the information about a particle (electron) in a given energy level is contained in the wave function,  which corresponds to the allowed (physically possible) solutions of the Schrödinger wave equation. The probability of finding an electron in an atom is proportional to |  | 2, that is, the square of wave function and is known as probability density. It always has a positive value and the value at different points in atom corresponds to region around the nucleus where the electron is most likely to be found. The wave function and its square |  | 2 have values for all locations about the nucleus. For hydrogen atom, |  | 2 is large near the nucleus and decreases with increase in distance away from the nucleus. The wave function  obtained by solving Schrödinger equation for an electron in an atom corresponds to the atomic orbital. Since the equation can have many solutions, there are various possible wave functions  for an electron and hence many possible atomic orbitals. By: Dr. Ashish Kumar

52 Difference between orbit and orbitals OrbitsOrbitals They represent planar motion of electron orbits. They represent three-dimensional motion of electron around the nucleus. All orbits are circular or have disk-like shape. Different orbitals have different shapes. They do not have directional characteristics. They have directional characteristics (apart from s orbital). They can accommodate a maximum of 2n 2 electrons. They can accommodate a maximum of two electrons. They are in accordance with classical theory of atom. They are in accordance with wave theory (quantum mechanical model) of atom. By: Dr. Ashish Kumar

53 Schrodinger Wave Equation  fn(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1 n=2 n=3 7.6 distance of e - from the nucleus By: Dr. Ashish Kumar

54 Where 90% of the e - density is found for the 1s orbital swave.gif 7.6 By: Dr. Ashish Kumar http://homepages.ius.edu/kforinas/physlets/q uantum/hydrogen.html

55  = fn(n, l, m l, m s ) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e - occupies l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital Schrodinger Wave Equation 7.6 By: Dr. Ashish Kumar

56 Orbitals and Quantum Numbers The principal quantum number (n) has positive integer values such as 1, 2, 3, etc. Orbitals that have the same value of the principal quantum number form a shell. Orbitals within a shell are divided into subshells that are labeled with the same value of the angular quantum number. The shells are sometimes denoted by the letters K, L, M, N, …, counting outwards from the nucleus. The azimuthal or subsidiary quantum number (also known as orbital angular momentum, l) can have values 0, 1, 2, … (n  1). The magnetic quantum number (m l ) is used to describe the orientation in space of a particular orbital. Value for l012345 Notation for subshell s (spherical)p (dumbbell) d (double dumbbell) fgh By: Dr. Ashish Kumar

57 l = 0 (s orbitals) l = 1 (p orbitals) 7.6 By: Dr. Ashish Kumar

58 l = 2 (d orbitals) 7.6 By: Dr. Ashish Kumar

59  = fn(n, l, m l, m s ) magnetic quantum number m l for a given value of l m l = -l, …., 0, …. +l orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 Schrodinger Wave Equation 7.6 By: Dr. Ashish Kumar

60 m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6 By: Dr. Ashish Kumar

61 Atomic Orbitals Principal quantum number (n) Subsidiary quantum number (l) Magnetic quantum numbers (m l ) Symbol 1001s (one orbital) 2222 0101 0  1, 0, +1 2s (one orbital) 2p (three orbitals) 333333 012012 0  1, 0, +1  2,  1, 0, +1, +2 3s (one orbital) 3p (three orbitals) 3d (five orbitals) 44444444 01230123 0  1, 0, +1  2,  1, 0, +1, +2  3,  2,  1, 0, +1, +2, +3 4s (one orbital) 4p (three orbitals) 4d (five orbitals) 4f (seven orbitals) By: Dr. Ashish Kumar

62 Selection Rules Governing Allowed Combinations of Quantum Numbers The three quantum numbers (n, l and m l ) are all integers. The principal quantum number (n) cannot be zero. The allowed values of n are therefore 1, 2, 3, 4, . The angular quantum number (l) can be any integer between 0 and n  1. If n = 3, for example, l can be either 0, 1 or 2. The magnetic quantum number (m l ) can be any integer between  l and +l. If l = 2, m l can be  2,  1, 0, 1 or 2. By: Dr. Ashish Kumar

63  = fn(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½ By: Dr. Ashish Kumar

64 Principal Quantum Number (n) Azimuthal Quantum Number (l) Magnetic Quantum Number (m l ) Spin Quantum Number (m s ) The main shell in which the electron resides. The number of subshells present in the nth shell. The number of orbitals present in each subshell or the orientation of the subshells. The energy of the orbital.The energy of the orbital in a multi-electron system. Approximate distance of the electron from the nucleus. Shape of the orbitals. The maximum number of electrons present in the shell. For a given value of n, l can have values from 0 to n  1. In each subshell, there are (2l + 1) types of orbitals. For a given value of l, m l =  l to +l. For a particular value of m l, we have m s = +1/2 or  1/2. Represented by integers 1, 2, 3,  or K, L, M, N, etc. l = 0, s subshell l = 1, p subshell l = 2, d subshell l = 3, f subshell For s subshell: s For p subshell: p x, p y, p z For d subshell: d xy, d yz, d xz,, Represented by two arrows pointing in opposite directions,  and . Accounts for the main lines in the atomic spectrum. Accounts for the fine lines in the atomic spectra. Accounts for the splitting of lines in the atomic spectrum. Accounts for magnetic properties of substances. Information obtained from the four quantum numbers By: Dr. Ashish Kumar

65 Existence (and energy) of electron in atom is described by its unique wave function . Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation  = fn(n, l, m l, m s ) 7.6 By: Dr. Ashish Kumar

66 7.6 By: Dr. Ashish Kumar

67 Schrodinger Wave Equation  = fn(n, l, m l, m s ) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½  = (n, l, m l, ½ ) or  = (n, l, m l, - ½ ) An orbital can hold 2 electrons 7.6 By: Dr. Ashish Kumar

68 How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e - 7.6 By: Dr. Ashish Kumar

69 Energy of orbitals in a single electron atom Energy only depends on principal quantum number n E n = -R H ( ) 1 n2n2 n=1 n=2 n=3 7.7 By: Dr. Ashish Kumar

70 Energy of orbitals in a multi-electron atom Energy depends on n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2 7.7 By: Dr. Ashish Kumar

71 ARRANGEMENT OF THE ELEMENTS IN GROUPS IN THE PERIODIC TABLE If the elements are arranged in groups which have the same outer electronic arrangement, then elements within a group should show similarities in chemical and physical properties. Elements with one s electron in their outer shell are called Group 1 (the alkali metals) and elements with two s electrons in their outer shell are called Group 2 (the alkaline earth metals). These two groups are known as the s-block elements, because their properties result from the presence of s electrons. Elements with three electrons in their outer shell (two s electrons and one p electron) are called Group 13, and similarly Group 14 elements have four outer electrons, Group 15 elements have five outer electrons. Group 16 elements have six outer electrons and Group 17 elements have seven outer electrons. Group 18 elements have a full outer shell of electrons. Groups 13–18 all have p orbitals filled and because their properties are dependent on the presence of p electrons, they are called jointly the p-block elements. By: Dr. Ashish Kumar

72 In a similar way, elements where d orbitals are being filled are called the d- block, or transition, elements. In these, d electrons are being added to the penultimate shell. For example, the element scandium Sc is the first transition element, and follows immediately after the element calcium Ca, which is in Group 2. Finally, elements where f orbitals are filling are called the f-block, and here the f electrons are entering the antepenultimate (or second from the outside) shell. By: Dr. Ashish Kumar

73 Table 6 Periodic table By: Dr. Ashish Kumar

74 Electronic Configuration Pauli Exclusion Principle The Pauli’s exclusion principle states that no two electrons in one atom can have all four quantum numbers the same. By permutating the quantum numbers, the maximum number of electrons which can be contained in each main energy level can be calculated Quantum numbers, the permissible number of electrons and the shape of the periodic table. By: Dr. Ashish Kumar

75 “Fill up” electrons in lowest energy orbitals (Aufbau principle) H 1 electron H 1s 1 He 2 electrons He 1s 2 Li 3 electrons Li 1s 2 2s 1 Be 4 electrons Be 1s 2 2s 2 B 5 electrons B 1s 2 2s 2 2p 1 C 6 electrons ?? 7.9 By: Dr. Ashish Kumar

76 C 6 electrons The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). C 1s 2 2s 2 2p 2 N 7 electrons N 1s 2 2s 2 2p 3 O 8 electrons O 1s 2 2s 2 2p 4 F 9 electrons F 1s 2 2s 2 2p 5 Ne 10 electrons Ne 1s 2 2s 2 2p 6 7.7 By: Dr. Ashish Kumar

77 Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7 By: Dr. Ashish Kumar

78 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1 7.8 By: Dr. Ashish Kumar

79 What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 2 + 2 + 6 + 2 = 12 electrons 7.8 Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p 5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3l = 1m l = -1, 0, or +1m s = ½ or -½ By: Dr. Ashish Kumar

80 Outermost subshell being filled with electrons 7.8 By: Dr. Ashish Kumar

81 7.8 By: Dr. Ashish Kumar

82 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8 By: Dr. Ashish Kumar


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