Atomic Structure (C 06J) CHEM 0901 PDF

(C 06J) CHEM 0901: ATOMIC STRUCTURE Dr. Mark A. W. Lawrence  REMINDERS 3 Lectures per week Each section of the course is taught by a different person 1 lab session per week (15% of course grade) Lab attire, lab manual, lab submission Failing the lab = failing course...

(C 06J) CHEM 0901: ATOMIC STRUCTURE Dr. Mark A. W. Lawrence  REMINDERS 3 Lectures per week Each section of the course is taught by a different person 1 lab session per week (15% of course grade) Lab attire, lab manual, lab submission Failing the lab = failing course (FP) 1 tutorial each week (starting week 2) Tutorials will be submitted and contributes 5% of the overall grade. 2 course tests (2 x 5%) Final exam (70%) 2 x 2hr written papers OR 3 x 1hr 20 min online papers  LECTURES Check course page for details of the schedule Physical Chemistry Atomic Structure: Dr. Mark Lawrence Gases: Dr. Mark Lawrence Thermochemistry: Dr. Peter Nelson Inorganic Chemistry Organic Chemistry  OPENING REMARKS Physical Chemistry is the physics for chemistry. It therefore involves a fair amount of calculations. DO NOT be afraid or alarmed as CSEC mathematics is sufficient for most of the calculations required at Chem 0900 level. It is very important, indeed essential, that routine calculations are performed by the practitioner, to cement the concepts and simultaneously provide practice and improve the confidence of the practitioner. The concepts will not “jump” from the paper or screen and into the head of the reader! You must take the time to work through every step of the problem questions to fully grasp the subject. The reader should familiarize oneself with units and their conversions eg g → kg, cm → m as well as combined units.  OPENING REMARKS The definitions learned in CSEC physics do form the basis of many concepts e.g. Pressure, work etc. Practice makes it permanent and failing to prepare means you are prepared to fail! If you have had a bad experience with chemistry before, forget about it and let’s start fresh. Note you are only examined on the topics covered and as such you have no excuse should you fail! If you did not understand the mole concept at CSEC; fret not we will cover it under gases and in the lab component of the course. COMMON SI PREFIXES  What is the purpose of having these multiples?  They represent different ways of reporting the same quantity.  SOME PRELIMINARIES Writing very small and large numbers:  100000 = 1.0 x 105 ; 0.00001 = 1.0 x 10-5 Multiples of units (e.g. m): fm = 10-15 m; pm = 10-12 m; nm = 10-9 m; mm = 10-6 m; mm = 10-3 m; cm = 10-2 m; dm = 10-1 m; km = 103 m; Mm = 106 m. A positive index (e.g. 103) implies multiplication (gets larger) where as a negative index (e.g. 10-3) implies division (gets smaller). How can one convert 10 cm to nm?  Since cm (10-2 m) is larger than nm (10-9 m), the apparent quantity will increase (that is the conversion factor must have a positive index. Solution:  Since there is a 107 order of magnitude difference (10-2/10-9)  10 cm ≡ 10cm × 107nm cm-1  10 cm ≡ 10 × 107 nm or 1.0 × 108 nm  SHORT CUT Large unit to small unit = (original value) ×10+y The index is positive indicating multiplication This is so as the apparent quantity will increase Think of changing a single $100 bill into 100 $1 bills; more paper, but same value. Small unit to large unit = (original value) ×10-y The index is negative indicating division This is so as the apparent quantity will decrease (i.e. a fraction of the initial value) Think of changing the 100 $1bills into a single $100 bill; less paper, but same value.  SOME PRELIMINARIES Definitions usually form the basis of units: Work: mechanical work is the amount of energy transferred by a force acting through a distance. i.e. w = F x d. But force F = mass x acceleration. Therefore the S. I. unit of work = kgms-2 × m = kgm2s-2. Understanding indices: 1/a = a-1 Likewise 1/a2 = a-2 etc.  (negative index means fraction) Also √a = a1/2 and ∛a = a1/3  (fractional index represents the nth root) Combinations: a-1/2 = 1/√a etc IMPORTANT CONCEPTS FOR PHYSICAL CHEMISTRY Force Moment and momentum Pressure Energy Kinetic vs Potential Different forms kinetic and potential energy SI units Differentiate amongst hypothesis, theory and (scientific) law Let us begin by looking at the evolution of the “picture” of an atom. We begin by assessing the macroscopic picture, then work our way to the microscopic and finally to the atomic (& sub-atomic) level. Hopefully this will illustrate the systematic nature of science.  NOTE: ATOMIC STRUCTURE YOU DO NOT NEED TO RECALL ANY DATES-THIS IS NOT A HISTORY CLASS!  ATOMIC STRUCTURE: HISTORY Robert Boyle (1627-1691): Through a study of gases provided clear evidence for atomic make up of matter.  Defined an element as a substance that cannot be chemically broken down further. Joseph Priestly (1733-1804) & Antoine Lavoisier (1743-1794): Preparation of O2 & showed it was essential for combustion. Helped to establish the law of mass conservation; which states:  Mass is neither be created or destroyed in chemical reactions. Joseph Proust (1754-1826): Law of definite proportions:*  Elements combine in specific proportions, not in random proportions.*  ATOMIC STRUCTURE: HISTORY John Dalton (1766-1844): Dalton’s Atomic theory and Law of multiple proportions:  Elements are made of tiny indivisible particles called atoms.  Each element is characterized by the mass of its atoms. Atoms of the same element have the same mass, but atoms of different elements have different masses.  Chemical combination of elements to make different substances occur when atoms join together in small, whole-number ratios.  Chemical reactions only rearrange the way that atoms are combined; the atoms themselves are unchanged. The third point predicts the law of multiple proportions. What is this law of multiple proportions? NO vs NO2 vs NO3 Same atoms (nitrogen and oxygen) Different whole number ratios produces different substances This idea contradicts Joseph Proust’s Law of definite proportions  ATOMIC STRUCTURE: HISTORY We have stated that elements are made up of atoms But what are atoms made up of? Are atoms truly indivisible? Structure of Atoms Dalton’s atomic theory left unanswered the obvious question: What is an atom made of? J.J. Thomson (1856-1940) provided some clues.  THOMSON’S CRT EXPERIMENT Thompson’s experiment (Electron)  Involved the use of cathode-ray tubes (predecessors to CRT displays) flourescent background strip slit (-) (+) cathode anode cathode ray evacuated tube vacuum - + The cathode rays can be deflected by a magnet or electrical plate (i.e. the beam is charged). The beam is produced at the –ve plate and deflected towards the +ve plate; therefore the cathode rays must consist of tiny negatively charged particles (now called electrons).  THOMSON’S EXPERIMENT  Thomson reasoned that the amount of deflection of the electron beam by a nearby magnet or electric field must depend on three factors:  The strength of the magnet or electric field: increasing either of the two increased deflection.  The size of the –ve charge on the electron: higher charge, greater interaction.  The mass of the electron: lighter the particle the greater the deflection.  Thomson was able to calculate the electrons charge : mass ratio, e/m which has a modern value: e/m = 1.758820 x 108 C/g,  where e = magnitude of charge in coulombs (C) and m = mass in grams.  Thomson proposed a plumb-pudding model for the structure of atoms THOMSON’S EXPERIMENT THOMSON’S EXPERIMENT Atoms are electrically neutral Thomson concluded the in order for an atom to remain electrically neutral, electrons are embedded in an positively charged sphere. ‘Plum pudding model’ MILLIKAN’S OIL DROP EXPERIMENT  ATOMIC STRUCTURE: HISTORY  R. A. Millikan (1868 – 1953): determined mass and charge of the electron.  Fine mist of oil sprayed into a box.  The droplets given a negative charge by irradiation with X-rays. By applying a voltage (potential difference) on the plates, it was possible to counteract the fall of the charged drops and keep them suspended.  Millikan was able to show that the charge on the drops was always a small whole number multiple of electron whose modern value is 1.602176 x 10-19 C and substituting into Thomson’s e/m gave a mass of 9.109382 x 10-28 g  e/m = 1.758820 x 108 C/g  Then m  e 1.758820x 108 C/g -19  m= 1.602176x 10 C = 9.109382 x 10-28 g 1.758820x 108 C/g  TESTING THOMSON’S MODEL OF THE ATOM If Thomson’s model is correct, then a metal foil, e.g. gold, can be considered to be a film of positive charge with electrons embedded in it. If a beam of positively charged alpha particles is fired at the foil, then all the particles should be deflected instead of passing through the atoms of the metal foil.  ATOMIC STRUCTURE: HISTORY Ernest Rutherford (1871 – 1937): (Nucleus) Matter is overall electrically neutral. Thomson’s experiment showed that atoms in an electrode can give off negatively charged particles (electrons). This must mean that those same atoms also contain positively charged particles. Rutherford’s work involved the use of α-particles (an emission previously observed to be given off by a number of naturally occurring radioactive elements). It was known that α-particles are ~ 7000 times larger than electrons and had a positive charge 2 x but opposite in sign to the charge on an electron (He2+). RUTHERFORD’S EXPERIMENT  Scintillation (zinc sulfide) screen: phosphoresce when hit by a-particles.  Some a-particles are deflected (~1/20,000).  Most a-particles went straight through.  ATOMIC STRUCTURE: HISTORY Rutherford postulated that the metal atom must be almost entirely empty space with its mass concentrated in a tiny central core – the nucleus. The electron move in space at a relatively large distance away from the nucleus. Most α-particles can pass through and is only deflected when it approaches the small but highly positively nucleus (repulsion).  THE RUTHERFORD ATOM –(1911) Electrons  Present in the space around the nucleus.  Electrons occupy this space by repelling electrons of neighbouring atoms. Nucleus  Consists of massive particle, Rutherford identified as protons.  Chadwick later discovered that the nucleus also contains neutrons.  Following on these works (Thomson, Rutherford, and Chadwick) a relatively simple picture of the atom was arrived at.  We now know the atom consists of the electrons (negative) and a nucleus which further consists of protons (positive) and neutrons (neutral).  ATOMIC STRUCTURE: EXPERIMENTS (SUMMARY) Thomson’s CRT expt:  Atoms contained a negatively charged particle within them called electrons  Deduced the charge/mass (e/m) ratio  Proposed plumb pudding model Milliken’s oil drop expt:  Resolved the charge and mass of the electron. Rutherford’s gold foil expt:  Provided evidence for the nucleus which is tiny and contained positively charged particles (protons)  Showed that most of the atom is empty space  ATOMIC STRUCTURE: SUB-ATOMIC PARTICLES Particle Mass / g Charge / C e electron 9.109382 x 10-28 -1.602176 x 10-19 -1 proton 1.672622 x 10-24 +1.602176 x 10-19 +1 neutron 1.674927 x 10-24 0 0 NB: The mass of the proton is slightly less than the neutron and is ~ 1836 times greater than that of an electron. The diameter of the nucleus ~ 10-15 m. The diameter of the atom is of the order 10-10 m.  LECTURE SUMMARY Established a definition for an element Elements consists of atoms Atoms consist of sub-atomic particles: electrons, protons and neutron. Deduced the relative position of the sub-atomic particles. Established the sign & magnitude of the charge on each of the sub- atomic particles. Established the mass of these sub-atomic particles. Relative comparison of the diameter of the nucleus to the diameter of the atom LECTURE 2  ATOMIC NUMBER & MASS NUMBER The nucleus is positive because of the protons and it has a charge of Ze; where Z is the number of protons and e is the elementary charge. Z is also called the atomic number. The atom has the same number of electrons as protons (i.e. Z electrons) and is thus neutral. Elements differ from one another according to the number of protons in their atom (atomic number). In addition to protons, the nuclei of most atoms also contain neutrons. The total number of nucleons: A = Z + N  Where A = mass number; Z = number of protons & N = number of neutrons. The chemistry of an atom depends on the number of electrons it posses (and hence protons). The number of neutrons has no effect! ATOMIC NOTATION Once you know A and Z you can easily calculate N.  ISOTOPES Elements of the same atomic number (i.e. protons & electrons), but different mass number (i.e. neutrons). Recall A = Z + N Isotopes of an element have identical chemical properties and many have identical physical properties except those that depend on the mass of the atom. Some isotopes are radioactive, others are not. Nearly all elements found in nature are mixtures of several isotopes. E.g. 1H  protium, 2 H  deuterium, 3 H  tritium _(radioactive) 1 1 1 12 6C, 13 6C, 146C  carbon_ dating  EXAMPLE Complete the following table. Species Atomic # Proton # Neutron # Mass # charge # of electrons 17O 17 8 8 8 9 0 (A=8+9) p-e = 0 e=p=8 16O2- 8 8 8 16 -2 10 (N=16-8) e =p-(-2) =10 24Mg 12 12 12 24 0 12 (A=12+12) p-e = 0 e=p=12 24Mg2+ 12 12 12 24 +2 10 e =p-(+2) =10 A = Z + N Proton(p) + electron(e) = overall charge where p is positive and e is negative p+(-e) = charge OR p-e = charge  ISOTOPES Contrary to Dalton’s theory, all atoms of a particular element do not always have the same mass. Identify elements based on atomic number.  ATOMIC MASS  Atoms are extremely small and a speck of dust barely visible to the eye contains ≥ 1016 atoms.  The mass of atoms are measured using a unit called an atomic mass unit (u or amu), also called a dalton, Da. 12  * One amu is as exactly 1/12 the mass of an atom of C 6 12  1 amu = mass of one 6 C atom = 1.660539 x 10-24 g 12  Clearly the mass of the proton and neutron is nearly 1 u.  What is the nucleus made up of?  The mass of a given atom (its isotopic mass in u) is close to the atom’s mass number. e.g. 235 92U has a mass of 235.044 u.  The atomic mass of an element quoted on the periodic table is the weighted average of isotopic masses of the element’s naturally occurring isotopes. NB: The modern symbol of atomic mass unit is, u. However some text still uses amu. CALCULATING ATOMIC MASS 35 E.g. Cl has two naturally occurring isotopes: 17 Cl = 34.696 u with 75.77% 37 abundance and 17 Cl = 36.966 u with 24.23%. What is the atomic mass of Cl? Relative atomic mass = ∑(abundance × mass of isotope) 35 Atomic mass = (fraction of 17 Cl x mass of 1735Cl ) + (fraction of 1737Cl x mass of 1737Cl ) = (0.7577 x 34.969 u) + (0.2423 x 36.966 u) = 35.45 u ATOMIC MASS CALCULATION Relative atomic mass = ∑(abundance × mass of isotope) Naturally Percent Mass (u) occurring isotope abundance 24Mg 78.99 23.9850 25Mg 10.00 24.9858 26Mg 11.01 25.9826 Mrel,Mg = (0.7899 × 23.9850u) + (0.1000 × 24.9858u) + (0.1101 × 25.9826u) = 24.3050u i.e. the average mass of one Mg atom = 24.3050 u Try weighing that Or 24.3050 × 1.660 × 10-27 kg = 4.034 × 10-26 kg on a balance!!! ATOMIC STRUCTURE & LINE SPECTRUM LIMITATIONS OF RUTHERFORD’S MODEL OF THE ATOM If the nucleus is positively charged and the electrons are negatively charged. What then stops the electrons from being pulled into the nucleus by electrostatic attraction? According to the laws of classical physics, the electrons orbiting the positive nucleus would gradually lose energy and be pulled into the nucleus. This would effectively cause the collapse of the atom. Why does this not happen?  HOW CAN WE STUDY ATOMS? HOW DO WE SEE THE WORLD AROUND US? How can we study atoms? E.g. Why do group 1 elements display similar chemical behaviours? The interaction of radiant energy with matter have provided immense insight into atomic and molecular structures. Electromagnetic radiation is a spectrum of continuous range of wavelengths and frequencies. Electromagnetic radiation travels at the “speed of light”, c = 2.998 x 108 ms-1. The speed of the wave (transverse) = l x n  (m/s) m s-1  c = l n  c c  l= n or n = l  i.e. there is an inverse relationship between n and l 1 = period, T and 1 = wavenumber ( n ) n l  The intensity is proportional to the square of the amplitude of the wave i.e. I α A2  This is how one quantitatively differentiate between a dull and bright light of the same colour.  PARTICLE-LIKE PROPERTIES OF EMR: THE PLANCK EQUATION In 1900 German physicist Max Planck (1858-1947) proposed a theory to explain blackbody radiation (glow given off by a hot object). Planck concluded that the energy is emitted only in discrete amounts or quanta and not a continuum (stairs vs ramp). The amount of energy (E) associated with each quantum depends on the frequency of the radiation, n: E = hn or E = hc/l (since n = c/l) Where h is fundamental physical constant Planck’s constant h = 6.626 x 10-34 Js  WORKED EXAMPLE What is the energy of a photon of light of wavelength 649 nm?  E = hn = hc/l  = (6.626 x 10-34 Js x 2.998 x 108 ms-1) / 649 x 10-9 m  = 3.06 x 10-19 J  Recall J = work done = F x d (and also recall, F = ma)  F = kgms-2  And work done = kgms-2 x m = J  J = kgm2s-2 Note: higher frequency = higher energy Shorter wavelength = higher energy Because n = c/l (inverse relationship)  ELECTROMAGNETIC RADIATION AND MATTER Each aspect of atomic and molecular energy will correspond to a particular region of the EM spectrum.  E.g. Vibration transitions occurs in the IR region  Valence electrons transition in the Vis-UV region etc This knowledge can allow us to further investigate atomic structure and thus molecular structure. (More at Level I chemistry)  SUMMARY  Atomic number compared to mass number  Defined Isotopes  Defined atomic mass and the atomic mass unit (amu)  Electromagnetic radiation and the electromagnetic spectrum  EM spectrum and Planck’s equation LECTURE 3  LINE SPECTRUM The light from a household bulb or the sun (white light) consists of a continuous distribution of all possible wavelengths spanning the entire visible spectrum. (400 nm – 750 nm) This can be separated by a prism for example. When supplied with energy (for example an electric discharge or heat) atoms become “excited” and give off the energy to return to the “ground state”. This energy is given off as electromagnetic radiation. If we examine the atom’s emission spectrum, we will see discrete lines separated by dark areas—a line spectrum. Each atom produces a unique spectrum. LINE SPECTRUM  CONTINUUM VS LINE SPECTRUM A continuum refers to a function where in a particular domain, there is no discontinuity. f(x) f(x) is a continuous function between x1 and x2. x1 x2 x f(x) f(x) is a discontinuous function between x1 and x2. Based on Rutherford model, one would expect a continuous spectrum for atoms! x1 x2 x  IMPLICATIONS OF A LINE SPECTRUM The fact that when one excite atoms, and they are allowed to relax, the same spectrum is always observed. This says a lot about atomic structure. The electrons can only have certain well defined energies AND NOT ANY ARBITRIARY ENERGY PREDICTED BY CLASSICAL MECHAINCS. The energy separations correspond to the energy of the photons emitted when the atom “relaxes” from its excited state(s). Spectroscopy (specifically atomic) was one of the earliest tools to hint at the inappropriateness of models proposed by Thomson, Rutherford or Millikan. It also hinted to the inappropriateness of trying to apply classical mechanics to solve or explain atomic structure.  BALMER-RYDBERG EQUATION J.J. Balmer (1825-1898) was the first to discover a pattern in the atomic spectrum of hydrogen. H2 produces four visible lines: Red (656.3 nm), blue-green (486.1 nm), blue (434 nm) & indigo (410.1 nm). By thought & trial & error, Balmer discovered that the 4 lines can be expressed by the equation: 1 1 1 1 1  R 2  2  l 2 n  or v  R.c 2  2  2 n  Where R is a constant now called the Rydberg constant R = 1.097 x 10-2 nm-1 or 1.097 x 107 m-1 Red, n = 3 ; blue-green, n = 4; blue, n = 5; indigo, n = ? Subsequent to Balmer’s discovery H2 was found to contain many lines in the non- visible regions of the electromagnetic spectrum.  BALMER-RYDBERG EQUATION Swedish physicist J. Rydberg was able to show a more generalized expression, now called Balmer-Rydberg equation: 1  1 1  1 1  RH  2  2  v  RH.c 2  2  l m n  or m n  RH = 1.097 x 10-2 nm-1 or 1.097 x 107 m-1 Where m and n are integers with n > m. If m = 1 UV and n = 2, 3….. Lyman series  m=2 visible and n = 3, 4….. Balmer “  m=3 IR and n = 4, 5….. Paschen “  m=4 IR and n = 5, 6….. Brackett “  m=5 IR and n = 6, 7….. Pfund “  BALMER-RYDBERG (B-R) EQUATION Recall E = hc/l Thus the B-R expression can also be written as:  1 1  1 1 E  hcRH  2  2  E   RH  2  2  m n  m n  RH  1.097107 m1 RH  2.1781018 J  The negative sign is introduced as the energy levels lie below 0. (Zero energy is when the electron is free from the influence of the nuclei.) The equation shows that the each energy level is of the form: RH En   2 n The final B-R expression results from a difference of two terms and series are the result of transitions between energy levels. IMPLICATIONS OF THE BALMER-RYDBERG EQUATION Suggesting a shell arrangement of the electron in the H atom (and other atoms).  QUESTIONS 1. What is the longest wavelength line (nm) in the Lyman series of the hydrogen spectrum? (Ans.= 121.5 nm) 2. If an electron in a H atom relaxes from n = 5 to n = 2 level, what is the wavelength (nm), frequency (Hz), and energy of light emitted? (Ans = 434.1 nm; freq = 6.911 x 1014 Hz; E = 4.579 x 10-19J)  ANSWER (Q1) 1  1 1  R 2  2  l  n f ni    longest wavelength, lowest energy, i.e. between n f  1 and n i  2 1 1 1  1  R  2  2   1.097 102 nm1 1    8.228 103 nm1 l 1 2   4 1 l 3 1  121.5nm 8.228 10 nm  CONTINUUM VS LINE SPECTRUM Now that we have the elementary mathematics out the way, what is the importance of a line spectrum versus a continuum with regards to atomic structure? As implied in the Balmer-Rydberg equation, since m and n can only be integers, it means that the energy levels are quantized. Now having an appreciation that the energy levels are quantized, how do we attempt to solve for the energies of electrons as they “oscillate” around the nucleus? This is a problem that needed to be solved. LECTURE 4  BOHR’S THEORY Rutherford proposed a model with the electrons disordered in energy and distribution. From classical physics this model lack stability because –ve & +ve attracts; hence the electron will not remain stationary. The electron must orbit around the nucleus (since it smaller, like the earth around the sun). But in orbiting it would radiate electromagnetic waves and lose energy and spiral to the nucleus in ~10-10 s. Thus the atom would be very unstable and collapse almost immediately. But this is not observed! In fact hydrogen the simplest atom is very stable, as are most elements.  BOHR’S THEORY Likewise if the electron had a random orbit , the atomic spectra of the elements should be a continuum like a light bulb and not discrete lines as observed. In 1913 Niels Bohr (1885-1962) proposed that the orbitals and the energies of the hydrogen atom are quantized. Therefore only certain discrete orbits and energies are allowed. Thomson’s plum- pudding model. Bohr’s Model used at CSEC  BOHR’S POSTULATES The quantized nature of atomic orbitals implied that classical mechanics cannot be applied at the atomic level and new laws were needed.  When an electron is in one of the quantized orbitals, it does not emit radiation (a stationary state). The electron can make a discontinuous transition or quantum jump from one stationary state to another and emit radiation during this transition. This is consistent with discrete nature (line spectrum) of atomic spectra.  The laws of classical mechanics can be applied to the orbital motion of the electron in a stationary state, but these laws do not apply during the transition from one to another.  When an electron makes a transition from one stationary state to another the energy difference E is released as a single photon of frequency n = E/h. Bohr’s model, though important historically, because of its suggestion of the quantized nature of orbitals, fails beyond hydrogen.  BOHR’S POSTULATES Successes: Provided a better picture of the atom: electrons move in a quantized orbit around the atom Explained the emission and absorption of discrete wavelengths as well as the stability of atoms. Line spectra and ionization energy for H and 1 electron ions that are predicted, are in excellent agreement with experimental data. Failures: Could not predict the line spectra for more complex atoms—not even neutral He which has only two electrons. The model did not explain why energy levels are quantized or why atoms make transitions. Could not explain why some emission lines consisted of two or more closely spaced lines (fine structure of atoms), or why some lines are brighter than others. Could not explain bonding. The model failed because it mixed classical and quantum ideas.  PARTICLE-LIKE PROPERTIES OF EMR: THE PLANCK EQUATION CONSEQUENCES  The idea that electromagnetic energy is quantized rather than continuous, was further supported in 1905 when Albert Einstein (1879-1955) used Planck’s idea to explain the photoelectric effect.  Einstein explained the photoelectric effect by assuming that a beam of light behaves as if it were composed of a stream of small particles called photons of energy, E, related to their frequency by the Planck equation, E = hn.  Planck’s and Einstein’s works showed that electromagnetic radiation is more complex than previously thought.  In addition to behaving as waves, electromagnetic radiation can also behave as small particles— wave-particle duality.  WAVELIKE PROPERTIES OF MATTER: THE DE BROGLIE EQUATION In 1924 French physicist L. de Broglie (1892-1987) showed that matter also exhibited wave-particle duality. de Broglie used Einstein famous equation E = mc2 to show : m = E/c2 and since E = hc/l m = h/lc and for a particle c = v m = h/lv or l = h/mv or l = h/P; where P = mv, the momentum. The de Broglie equation: m = mass of the particle and v = the velocity of the particle. Experimental evidence of the de Broglie equation came soon after: C. Davisson & L. Germer showed the diffraction of electrons. Note diffraction is only observed by waves! G. P. Thompson also observed the diffraction of electrons by a thin film of celluloid. So we have established the particle-wave duality of energy & matter. How does this relate to atomic structure?  QUESTION Calculate the wavelength associated with an electron (mass = 9.109x10-28 g) traveling at 40.0% the speed of light. Solution: v = (2.998 x 108 m/s)(0.40)=1.20 x 108 m/s l = h/mv l = (6.626 x 10-34 J s) = 6.06 x 10-12 m (9.11x10-31 kg)(1.20 x108 m/s) Remember 1J = 1(kg)(m)2/s2  SUMMARY Continuum compared to a line spectrum Balmer-Rydberg equation B-R equation implications on atomic structure The Planck equation The de Broglie equation Wave-particle duality  QUANTUM MECHANICS AND THE HEISENBERG UNCERTAINTY PRINCIPLE The breakthrough in understanding atomic structure came in 1926, when Austrian physicist E. Schrodinger (1887-1961) proposed what is now called the quantum mechanical model of the atom.  The primary idea is the abandonment of the particle view of electron orbiting the nucleus in a defined path and concentrate only on the electron’s wavelike properties. Using the de Broglie equation, the l of the electron in a hydrogen atom is greater than the diameter of the atom.  The previous models did not predict this and could not explain this! In 1927 W. Heisenberg (1901-1976) showed that it is impossible to know precisely where an electron is, and what path it follow simultaneously a statement called the Heisenberg Uncertainty Principle.  QUANTUM MECHANICS AND THE HEISENBERG UNCERTAINTY PRINCIPLE (x)(mv) ≥ h/4p the HUP; where x is the position. We cannot know both the position and velocity of an electron (or any object) beyond a certain level of precision. If the velocity is known to a high degree of certainty (mv is small) then the position must be uncertain (x must be large). Therefore the electron will always appear as a blur when we attempt to measure its position and velocity. Why? For us to “see” the electron, photons of appropriate frequency would have to interact with and bounce off the electron. This would transfer energy to the electron and increase the energy of the electron, causing it to move faster. Recall the photon (that we use to see the electron) transfers energy to the electron for you to observe it.  QUANTUM MECHANICS AND THE HEISENBERG UNCERTAINTY PRINCIPLE Thus the very act of determining the electron’s position would change its position and speed! For particles in daily life, this uncertainty is too small to have any serious consequence. So we can tell the position and velocity of a football or cricket ball with fair accuracy. We will investigate this further in the tutorials. LECTURE 5  WAVE FUNCTIONS & QUANTUM NUMBERS Schrodinger’s quantum mechanical model of atomic structure is framed in wave mechanics, a mathematical equation similar in form to that used to describe the motion of ordinary waves in fluids. wave function probability of finding the wave solve electron in a region of space or obital  (similar to equation Amplitude) 2 (similar to I = A2) A wave function contains 4 variables: n, l, ml and ms which describe the energy level of the orbital, the 3 dimensional shape of the region of space occupied by a given electron and its relative direction of travel.  QUANTUM NUMBERS  The principal quantum number (n):  A positive integer with values n = 1,2,3,4…..∞, on which the size and energy level of the orbital primarily depend.  As the value of n increases, the number of allowed orbitals increases, and the orbitals become larger, allowing the electron to be farther from the nucleus.  As n increases, the orbital energy increases.  The angular-momentum quantum number (l):  Defines the 3 dimensional shape of the orbital.  For an orbital with principal quantum number n, the angular momentum quantum number (l) can have any integral value from 0 to n-1.  Thus if n = 1 then l = 0  if n = 2 then l = 0 or 1  if n = 3 then l = 0, 1 or 2 etc  So if n is considered the shell l would be the sub-shell of n.  QUANTUM NUMBERS Different sub-shells are usually referred to by a letter rather than a number  Quantum # l: 0 1 2 3 4 ….  sub-shell notation: s p d f g …. The magnetic quantum number (ml): Defines the spatial orientation of the orbital along a standard set of coordinate axes. For an orbital of angular-momentum quantum number l, the magnetic quantum number ml can have integral value from –l to +l. Thus, within each sub-shell (orbitals of the same shape), there are 2l +1 different spatial orientations for those orbitals. If l = 0 then ml = 0  If l = 1 then ml = -1, 0 or +1 therefore p-sub-shell has 3 orbtials px, py, pz.  If l = 2 then ml = -2, -1, 0, +1 or +2 therefore d- sub-shell has 5 orbitals etc.  QUANTUM NUMBERS The spin quantum number (ms): Refers to the spin of an electron and the orientation of the magnetic field produced by this spin. For every ml value, ms can have the value of +½ or -½; i.e. ms = ±½. The values of n, l and ml describes a particular atomic orbital. Each atomic orbital can accommodate no more than 2 electrons, one with ms +½ and another ms -½. This is summarized in Pauli Exclusion Principle: No 2 electron in an atom can have the same four quantum numbers. The result is the s-orbitals accommodates 2 electrons, p-orbitals: 6 electrons, d- orbitals 10 electrons. How similar is this to the model used at CSEC? THE ANGULAR-MOMENTUM QUANTUM NUMBER, L THE ELECTRON SPIN QUANTUM NUMBER (MS) s-Orbitals Spherical Node: 0 amplitude; therefore probability of finding electron is zero. NB: They y-axis represents the probability of finding the electron Radial wave functions* * You will learn more on this at Chem1900 P-ORBITALS Dumbbell shaped and the electron density is concentrated in identical lobes on either side. Radial wave function D-ORBITALS f-orbital  ENERGY LEVELS OF DIFFERENT ORBITALS Note hydrogen is the only atom/element in which the energy levels of different orbitals only depend on n. LECTURE 6  ENERGY LEVELS OF DIFFERENT ORBITALS Note hydrogen is the only atom/element in which the energy levels of different orbitals only depend on n. ELECTRONIC CONFIGURATIONS OF MULTIELECTRON ATOMS Having established the modern/quantum mechanical picture of the atom, let’s now turn our attention to how this principle relates to the groups and period of the periodic table. 18 group periodic table. ELECTRONIC CONFIGURATIONS AND ORBITAL FILLING DIAGRAMS OF MULTIELECTRON ATOMS Aufbau principle (used to determine ground state configurations):  Lower energy orbitals fill before higher energy orbitals.  An orbital can only hold 2 electrons which must have opposite spins.  If two or more degenerate orbitals are available, one electron goes into each orbital, until all are ½ filled (Hund’s rule). Only then does a 2nd electron fill one of the orbitals. Each of the singly occupied orbitals must have the same spin quantum number. H: 1s1 Be: 1s2 2s2 He: 1s2 Energy 2s Energy 2s 1s Li: 1s2 2s1 1s ELECTRONIC CONFIGURATIONS OF MULTIELECTRON ATOMS 2p x y z Recall for p-subshell B: 1s2 2s2 2p1 l =1 and ml = -1,0,+1 Energy 2s 1s O: 1s2 2s2 2p4 C: 1s2 2s2 2p2 2p x y z 2p x y z Energy Energy 2s 2s 1s 1s F: ? N: 1s2 2s2 2p3 2p x y z Ne: ? Energy 2s 1s  Now how do these quantum numbers relate to the different orbtials described previously? Why can an s-orbital only accommodate only two electrons? Hint include the ms quantum number and think of Pauli’s Exclusion principle. ELECTRONIC CONFIGURATIONS OF MULTIELECTRON ATOMS Across the period an electron is added each time until a nobel gas configuration (full shell) is achieved. Going down the group the principal quantum number n increases. Notice how the periodic table is also arranged in blocks: s-block, p-block, d-block, f-block ELECTRONIC CONFIGURATIONS OF MULTIELECTRON ATOMS Notice how the atomic number is equivalent to the number electrons as discussed previously. Also the group number is the same as the number of electrons in the valence-shell (i.e. same valence-shell electron configurations). Neon configuration E.g. Na (atomic # 11): 1s22s22p6 3s1 or [Ne] 3s1 Mg (Z = 12): 1s22s22p63s2 or [Ne] 3s2 Al (Z = 13): [Ne] 3s23p1 Si: [Ne]3s23p2 P: [Ne] 3s23p3 S: [Ne] 3s23p4 Cl: [Ne] 3s23p5 Ar: [Ne] 3s23p6 SHORT CUT USING THE PERIODIC TABLE Determine the electronic configuration of K and Si: Strategy: Identify the noble element in the row immediately before the species of interest Identify the row the element/ion is located in. This gives the q.n. of the shell being filled. Identify the block the element K= [Ar]4s1 is located within Si = [Ne]3s23p2 Then the group number within the block tells how many electrons are/is in the subshell. ELECTRONIC CONFIGURATIONS OF MULTIELECTRON ATOMS: TRANSITION ELEMENTS Now notice the relative energies or the orbitals After the 3p orbitals are filled, which orbital(s) is filled next? Recall that the lowest energy orbitals are filled first. TRANSITION ELEMENTS  Called the d-block elements  Notice at even higher values of n the f-shells are filled before all the d-orbitals.  These are shown at the bottom as the lanthanide and actinide series. ELECTRONIC CONFIGURATIONS OF MULTIELECTRON ATOMS: TRANSITION ELEMENTS Starting from the s-block element Ca: [Ar] 4s2 to the first transition element Sc: Sc: [Ar]4s23d1 Ti: [Ar]4s23d2 V: [Ar]4s23d3 Zn:[Ar]4s23d10 Notice Cr: [Ar]4s13d5 and Cu: [Ar]4s13d10 are anomalous (i.e. ½ filled s-orbital) Expect Cr: [Ar]4s23d4 and Cu: [Ar]4s23d9 This anomalous behaviour will be explained later-Inorganic Chem. These configurations based on Aufbau principle are called ground-state configurations. The ground state is the electronic configuration of the lowest energy. These states can be excited (as seen with hydrogen and the Balmer series) which will relax and emit the energy as photons.  QUESTIONS Write the electronic configuration of the following: Na [Ne]3s1 Na+ [He]2s22p6 Mg [Ne]3s2 Mg2+ [He]2s22p6 Cl [Ne]3s23p5 Cl- [Ne]3s23p6 Cu [Ar]4s23d9 ?? [Ar]4s13d10 Cu2+ [Ar]3d9 V3+ V=[Ar]4s23d3 but V3+=[Ar]3d2 Note: When forming cations from transition metals, the 4s electrons are lost before any of the 3d electrons.  SUMMARY Be able to deduce the electronic configuration of atoms and ions Understand the order in which orbitals are filled. Use periodic table to assist in figuring out the electronic configuration. Understand the order in which electrons are lost from transition metal species.