Atoms, Molecules & Intermolecular Forces Lecture Notes PDF

ATOMS, MOLECULES AND INTERMOLECULAR FORCES CHE I56  THE ATOMIC NATURE OF MATTER At about 400 B.C., Demotricus, a Greek philosopher proposed the atomic nature of matter-HYPOTHESIS If one piece of matter is divided into smaller and smaller pieces, one would reach an invisibly...

ATOMS, MOLECULES AND INTERMOLECULAR FORCES CHE I56  THE ATOMIC NATURE OF MATTER At about 400 B.C., Demotricus, a Greek philosopher proposed the atomic nature of matter-HYPOTHESIS If one piece of matter is divided into smaller and smaller pieces, one would reach an invisibly smallest particle with all the properties of the parent matter This smallest particle cannot be further sub- divided, it is therefore named atomos which means ‘indivisble’ in Greek  DALTON’S ATOMIC THEORY All matter is made up of tiny, indivisible particles called atoms Atoms can neither be created nor destroyed (law of conservation of mass) Atoms of the same elements are identical and have the same mass and chemical properties (law of constant composition) Atoms of elements combine to form compounds in a small whole number ratio (law of multiple proportion)  DALTON’S ATOMIC MODERN ATOMIC THEORY (1806-1896) THEORY All matter are made An atom can be up of tiny indivisible further sub-divided particles called into sub-atomic atoms particles e.g. electrons, protons, neutrons, positrons, mesons, etc.  DALTON’S ATOMIC MODERN ATOMIC THEORY (1806-1896) THEORY Atoms can neither be Radioactivity- created nor destroyed spontaneous emission from the nuclei of some elements resulting in new atoms  DALTON’S ATOMIC MODERN ATOMIC THEORY (1806-1896) THEORY Atoms of the same Isotopes exist. These element are identical are atoms of the same and have the same element having the mass & chemical same atomic numbers properties and are but different mass different from atoms numbers (16O, 17O and of another element 18O)  DALTON’S ATOMIC MODERN ATOMIC THEORY (1806-1896) THEORY Atoms combine Giant molecules together in simple which contain (& small) whole thousands of atoms do exist. This is number ratio (CO2, referred to as H2O, NH3,etc). CATENATION (e.g. sucrose- C12O22H11)  EXPERIMENTAL EVIDENCE OF ATOMIC STRUCTURE: SUB-ATOMIC PARTICLES THE ELECTRONS William Crooke (1870’s) passed high-voltage electrical discharge through various gases at low pressure inside a sealed tube. He discovered a beam of rays came out from the negative electrode (cathode) and hit the positive electrode (anode) The beam was called cathode rays The cathode ray tube A schematic diagram of the gas discharge tube CharaterisTICS OF THE CATHODE RAYS Cathode rays travel in straight line (they produced shadows of objects placed in the way)  Cathode rays possess momentum and energy (can rotate a light paddle placed in its path)  Cathode rays are deflected by both magnetic and electric fields (They are deflected towards the positive terminal in an electric field)  Other xTICS OF THE CATHODE RAYS They are independent of the material of the electrode or nature of gas in the tube They penetrate small thickness of matter (e.g. aluminium or gold foil) Sir J. J. Thomson’s experiments (1897) B Bev  a negatively charged electron is deflected in an electric field and repelled from the negative plate and attracted towards the positive plate Sir JJ Thomson’s experiments (1897) in a magnetic field, electrons are deflected in a circular arc opposite in direction to that in an electric field Sir JJ Thomson’s experiments (1897) when both fields are present, the forces cancel out Sir JJ Thomson’s experiments (1897) modified cathode ray tube by JJ Thomson  J.J Thomson had balanced the cathode rays between electric and magnetic forces The force (F) on a charged object in an electric field depends on the strength of the electric field (E), multiplied by the charge (e) on the object: The force (F) on a charged object in a magnetic field depends on the strength of the magnetic field (B), multiplied by the charge (e) and velocity (v) on the object:  Since the forces were balanced; The velocity was equal to the electric field strength divided by the magnetic field strength. Since the two field strengths could be measured, then velocity, v, of the cathode ray could then be determined and be used in further calculations  In a magnetic field, an electron moves along the arc of circle of radius, r If the centrifugal force (F) on the electron which has a mass, m, in a magnetic field is given by Recall that in a magnetic field, If we equate the two forces, we have The charge to mass ratio (e:m) will be given by:  Millikan’s oil can including an atomizer to create a mist of oildroplets and a telescope to observe them A cross-sectional view of Millikan’s oil drop apparatus  Millikan’s oil droplet experiMents The oil drop apparatus had two chambers. The upper chamber had an atomizer that dispersed a fine mist of micron-sized oil droplets. Individual droplets would fall through a pinhole into the lower chamber, which consisted of two horizontal plates, held 16 mm above each other. A small window on the side allowed the scientists to observe the droplets through a telescope.  Millikan’s oil droplet experiMents The air in this chamber was ionized so that ions or free electrons could be captured on the falling droplets High voltage was applied across the plates, till electric field force balance the force of gravity making the drop to hang suspended in the mid-air Using horizontal lines in the telescope, the speed, size and mass of each droplet could be estimated  When a drop is suspended, its weight (mg) is equal to the electric field force mg=Eq where; m = mass of an oil droplet, g = acceleration due to gravity, E = electric field strength, q= total charge on the oil droplet (q= ne) n= number of electrons e= actual charge on the electron Thus the total charge q could be estimated:  From the values of q obtained, the value of the actual charge, e, was estimated to be : 1.602 x 10-19 Coulombs Recall, if e=1.602 x 10-19 C, Then, This is ≈ 2000 times lighter in weight than H, the first known lightest element This explains the reason behind Crooke’s observation about the penetrating abilities of the electron  PRACTICE QUESTION (pg 37) An experiment to determine the charge-to- mass ratio of the electron was performed twice with the same CRT. In the second experiment, the magnetic field strength was double that used for the first experiment. By what factor must the electric field strength be changed ? Solution Recall that the charge-to-mass ratio is given by: In 1st expt, Let magnetic field strength =B and electric field strength=E; e =E 2........................(i) m nd rB In 2 expt, Let magnetic field strength =2B; and electric field strength= x ) e = x 2 =x 2...........(ii ) m r (2 B ) r 4B Equating (i) and (ii): E 2 =x 2 rB r 4B E=x 4 x = 4 E (If B is doubled, E must be quadrupled i.e. multiplied by a factor of 4)  DISCOVERY OF THE PROTON ❖ Goldstein was used a modified cathode ray tube (CRT), in which the cathode was perforated ❖ He observed a new type of rays that were attracted towards a negatively charged plate, hence they have +ve charge ❖ He called these rays canal rays / anode rays (and later protons) ❖ The e/m ratio were much lower than those of electrons ❖ The e/m ratio also changes when the gas in the CRT was changed ❖ The magnitude of the charge (e) for the canal rays were found to be equal to that of electrons but opposite in charge ❖ The mass of the proton was determined to be 1.672×10-27 kg A modified cathode ray tube  ATOMIC MODELS JJ Thomson had proposed a plum pudding model for the atom negatively charged electrons floating in a sea of positive charge  rutherford’s experiMents To test this model, Ernest Rutherford performed the gold foil scattering experiments A thin sheet of gold foil was bombarded with a beam of α-particles (fast, positive charges) Three different observations ❑ Most of the rays passed through without deflection ❑ Very few were deflected through large angles ❑ Some rebounded Rutherford gold-foil experiment result of Rutherford’s experiment Conclusions from Rutherford’s expt The fact that some of the alpha particles were deflected or reflected meant that the atom had a centre concentrated with positive charges These alpha particles had either hit the positive centre directly or passed by it close enough to be affected by its positive charge. Since many other particles passed through the gold foil, the positive centre would have to be a relatively small size compared to the rest of the atom - meaning that the atom is mostly open space.  Conclusion from Rutherford’s expts Plum pudding model was incorrect The (gold) atom had a concentrated centre of positive charge and relatively large mass The positive centre has to be relatively small size compared to the rest of the atom The atom is mostly open space The concentrated centre is called the nucleus (containing the positive charge and most of the mass of the atom) He proposed the name protons for the positive charge  DISCOVERY OF THE NEUTRON For an atom to be electrically neutral, the number of positive charges must equal the number of negative charges no. of protons = no. of electrons But the mass of the atom is generally found to be twice that of protons (1proton ≡ 1 amu) Rutherford was the first to propose an electrically neutral entity to make up for the residual mass in the nucleus  SIGNIFICANCE OF THE NEUTRON They strongly bind protons (protons are unable to bind with each other due to electromagnetic repulsion Neutrons keep the atomic nucleus from flying apart due to its heaviness  The sub-atomic Particles Sub- Location Charge Relative Mass (kg) Relative atomic charge atomic particle mass Proton nucleus +1.6 × 10-19 +1 1.672 1 (p+) ×10-27 (1.0073) Neutron nucleus 0 0 1.674 1 (n) ×10-27 (1.0087) Electron Surround -1.6 × 10-19 -1 9.11× 10-31 0 (e-) -ing the (0.00055) nucleus  Number of protons in the nucleus The number of protons present in the nucleus of an atom was determined by Henry Moseley using X-rays X-radiation (ray) was discovered by Konrad Rőntgen in 1896 When fast-moving electrons (cathode ray) impinge on an object (anode), they loose their kinetic energy. X-rays are the result of excess energy which fast-moving electrons lose when they impinge on objects. He observed a penetrating radiation emitted from the anode of a CRT ❖ It turned wrapped photographic film black ❖ It allows gas to conduct electricity by ionizing them ❖ It makes some substances fluoresce e.g. ZnS ❖ It is not deflected by magnetic or electric field  Henry Moseley’s experiments He used different metals as anode (target for the cathode rays) and analyzed the diffracted radiation of the x-rays produced From his calculations, he observed an integer Z (from the German word zahl -meaning number) for each metal he used was ≈ half the relative atomic mass of the metal He also found out that this no. was equal to the +ve charges (protons) calculated for each metal The atomic number (Z) increases by one from one element to the other as one progress in the periodic table  Atomic and Mass number The atomic number (Z) is equal to the number of protons in the nucleus of an atom Because atoms are electrically neutral, the atomic number must equal the number of electrons surrounding the atom Atomic number (Z ) = no. of protons = no. of electrons The relative atomic mass/weight of an atom (also known as the mass number, A, is equal to number of nucleons (protons and neutrons) Mass number ( A) = no. of protons + no. of neutrons  Atomic notations The atom an element is usually denoted as: The atomic number (Z) increases progressively from one element to the other along the periodic table  Relative atomic masses It is more convenient to quote the mass of atom on a relative scale than the tiny masses of atoms in grams or kilograms On the relative atomic mass scale, carbon-12 (12C) has a mass of exactly 12 That some elements like, chlorine and copper, have non-whole number (fraction) relative atomic masses is due to ISOTOPY The relative abundance of each isotope of an element contributes to its atomic mass Isotopes and relative atomic masses Isotopes are atoms of the same element having different atomic masses For example, there are three isotopes of hydrogen atom namely hydrogen, deuterium and tritium All have the same atomic number but their mass numbers differ Isotopes and relative atomic masses Isotopes are atoms of the same element having different atomic masses For example, there are three isotopes of hydrogen atom namely hydrogen, deuterium and tritium All have the same atomic number but their mass numbers differ  The isotopes of hydrogen Isotopes of H No. of No. of No. of electrons protons neutrons 1 1 H hydrogen 1 1 0 2 1 H deuterium 1 1 1 3 1 H tritium 1 1 2 three different isotopes of hydrogen  Characteristics of isotopes Isotopes of the same element have the same number of protons (and therefore same number of electrons) Isotopes of the same element have different number of neutrons Isotopes of the same element have similar chemical properties because this is determined by the electronic structure of the atom  The mass spectrometer The mass spec (or MS) is used to determine relative atomic/molecular masses of atomic/molecular species It can differentiate between isotopes of the same element and give their relative abundances It measures mass/charge (m/e or m/z) ratio of ionized particles Mass Spectrometry  The operation of a MS There are six important parts in a MS 1. The vaporization chamber The material to be analyzed (element or compound) is injected into the chamber and heated until it vaporizes and changes into the gaseous form 2. Ionization chamber The vaporized sample is bombarded with electrons from a heated filament. Electrons are knocked off from neutral atoms or molecules. Positive ions are formed 3. Accelerating electric field The positive ions formed are accelerated (moves with a very high speed) by an electric field towards the magnetic field 4. Deflecting magnetic field The +ve ions are deflected along a circular path. The lighter the +ve ion, the greater is the deflection and the smaller the radius of the circular path ⇨ mass (m) The higher the charge of the +ve ions, the greater the deflection ⇨ charge (e) Ions with low m/e ratio are deflected more than those with high m/e 5. Ion detector The detector converts the m/e ratio into electric current The magnitude of the current is proportional to the number of ions present 6. Recorder The ions detected are traced out by a recorder which shows the m/e ratio and corresponding intensity of the ions. The position of the peak gives the m/e ratio of an ion The height of the peak gives the relative abundance of the ion (usually in %) The chart is called a mass spectrum Typical mass spectra A typical mass spectrum of chlorine Interpretation of the MS of Cl m/e ratio Ions 35 35Cl+ 37 37Cl+ 70 35Cl- 35Cl+ 72 35Cl-37Cl+ 74 37Cl-37Cl+ A typical mass spectrum of chlorine for example, Chlorine has two isotopes: 35Cl & 37Cl If in a sample of chlorine, 75% is 35Cl, while the remaining 25% is 37Cl The relative atomic mass will be (75 × 35) + (25 × 37) = 3550 100 3550 = 35.5 100 The relative atomic mass of chlorine is 35.5 Calculating relative atomic masses PRACTICE QUESTIONS 1. `A sample of neon is found to consist of 20Ne, 21Ne, 22Ne in the following percentages 20Ne 90.92% 21Ne 0.26% 22Ne 8.82% Calculate the relative atomic mass of neon. Solution The rel. mass of neon= (20  90.92) + (21 0.26) + (22  8.82) 2017.9 = = 20.18 90.92 + 0.26 + 8.82 100 Question 2 The relative atomic mass of potassium is 39.1. Calculate the relative abundances of its two isotopes, 39K and 41K in a sample of potassium  Σ ( mass of each isotope × its rel. abundance ) R.a.m = Σ ( relative abundances of all isotopes ) Let the rel. abundance of 39K be x and 41K be y. 39 x + 41y. x= 1.9 y = 19 y 39.1 = x+ y 0.1 39.1( x + y ) = 39 x + 41y but x + y = 100 39.1x + 39.1y = 39 x + 41y 19 y + y = 100 20 y = 100 39.1x − 39 x = 41y − 39.1y y =5 0.1x = 1.9 y If y = 5, then x = 95  Mass defect It has been observed that the mass of a nucleus of an atom is always less than the sum of the masses of the protons and neutrons it contains The difference between the actual mass and the observed mass is referred to as mass defect (∆m) This is due to the effect of the binding energy that the nucleons exert towards one another  Mass defect and Nuclear Binding energy ∆m = {(mass of p ×no. of p)+(mass of n ×no. of n)} – {actual mass of the nucleus}  Example Find the mass defect of a nucleus if the actual mass of 63Cu is 62.91367 amu. Solution: The atomic number of Cu is 29, therefore the number of protons (p) is 29 number of neutrons (n) in 63Cu is 63 – 29 = 34 (Recall that the mass of proton is 1.00728 amu and mass of neutron is 1.00867 amu) m = (1.00728  29 ) + (1.00867  34 )  amu − 62.91367  amu = 0.59223amu  NUCLEAR BINDING ENERGY (EB) This is the energy required to breakdown a nucleus into its components (nucleons) It is expressed in Joules per nucleus It is given by the formula: EB= ∆mc2 where ∆m is the mass defect in kg (1 amu = 1.6606×10-27 kg) c is the velocity of light =3.0×108 m/s For 63Cu with ∆m=0.59223, EB is equal to: ( 0.592231.6606 10 −27 ) kg ( 3.0 10 ) 8 2 m2 s −2 −11 2 −2 = 8.839 10 J (kgm s  J )  Exercise Calculate the mass defect and the nuclear binding energies of the following nuclides: NUCLIDE ATOMIC NO ACTUAL ATOMIC MASS (amu) U-238 92 238.050784 Ni-58 28 57.935346 O-16 8 15.994915  The electronic structure of the atom The electronic structure is a description of the arrangement and distribution of the electrons around the nucleus of an atom Different models were postulated to account for the actual position of the electrons around the nucleus of an atom ❖ Stationary model: electrons cannot be stationary ⇨ they would be pulled into the positively charged nucleus ❖Continuous motion model: If they were to continue moving in a circular orbit, they would radiate energy and slow down continuously until they spiral into the nucleus  Electrons & light: Spectral analysis When an electrical discharge is passed through a gas at low pressure in a CRT, an electromagnetic radiation is produced An electromagnetic radiation is Energy that travels as a wave through space It is observed as a spectrum consisting of narrow lines of bright colours. This is referred to as a ‘Line spectrum’ (Can be seen when passed through a prism) Each line corresponds to light of a specific wavelength  Each element (gas) gives its own (characteristic) line spectrum i.e. every element has a unique pattern (colours) WHEN AN ELECTRON ABSORBS ENERGY, IT GIVES OFF LIGHT  Electromagnetic radiations They consist of waves which have electrical and magnetic properties A wave conveys energy from a vibrating object (source) to a distant place Waves are characterized by wavelength ( λ) and frequency (ν) λ is measured in (Å)angstrom or nanometers 1 Å = 0.1nm (1nm = 1x10-9m) ν is measured in Hertz (Hz)  Why did electrons emit energy as light of only certain wavelength? Atoms of a particular element could only radiate energy within certain definite values because 1. Electron moves in a certain allowed orbit around the central nucleus 2. A definite amount of energy (quantum) is associated with the electron in each allowed orbit (i.e. the electronic energy is quantized) 3. The electron does not radiate (lose) energy when it is in these orbits 4. These orbits are arranged around the nucleus in increasing order of energy 5. The electron only radiates energy when it undergoes transition from one orbit to another of a lower energy  Bohr’s theory & Planck’s hypothesis Max Planck was the one who suggested the theory of quantization of energy ⇨ Quantum Chemistry This theory suggests that electromagnetic radiation consists of minute packets of energy called photons (or quanta) Niel Bohr’s theory states that in the atom, electrons exist in quantum energy levels (n levels) Electrons have the lowest energy when closest to the nucleus Under resting conditions, the electrons are in the lowest possible energy level called ground state The allowed energy states associated with individual electrons in an atom can be described by a set of quantum numbers  Bohr’s Model Nucleus Electron Orbit Energy Levels  When an electron moves from one energy level to the other, it absorbs or emits energy The difference in energy between the two energy level is given by: ΔE = hν (where ΔE = Efinal state- Einitial state ) h is the Planck’s constant 6.626 x10 -34 Js ν is the frequency of the emitted radiation The lowest electronic energy state is called the ground state Any state with greater energy than the ground state is called the excited state  Energy Levels in an atom The allowed energy levels within which an electron has definite energy values are designated by four quantum numbers 1. Principal quantum number (n) and also by letters K to P quantum no (n) = 1 2 3 4 5 6 shell =K L M N O P The maximum number of electrons in a particular shell, is equal to 2n2 For example, within the L-shell (n=2), the maximum no of e- = 2 (2)2 =8 electrons 2. Angular momentum/azimuthal quantum number (l ) divides the shells into subshell which are further divided into orbitals It describes the shape of the orbitals in which electrons are found The minimum value of l is 0, while the maximum value is n-1(l = 0,1,2,…., n-1) Each value of l corresponds to a particular subshell s, p, d, f,___, respectively. The maximum number of electron that is allowed in a subshell is given by 2(2l + 1) The s subshell (with l =0 )can therefore contain 2(2×0+1) electrons = 2 electrons The f shell (with l =3 ) will have a maximum of 2(2×3+1) electrons=14 electrons 3. Magnetic quantum number (ml or m) This quantum number arises to differentiate the orbitals that are available in a subshell When these orbitals are exposed to strong magnetic field, orbitals which are degenerate i.e. have the same energy level become non-degenerate i.e have different energies ml ranges from –l , ___, 0, ___, +l The ml of the s subshell (l =0) is 0 (it has only one orbital) For p subshell (l =1), ml is -1, 0, or +1 (has three orbitals) For d subshell (l =2) has five orbitals, ml = -2, -1, 0, +1 or +2 4. Spin quantum number (ms or s ) Describes the orbital angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum Because angular momentum is a vector, the Spin quantum number (s) has both a magnitude (½) and direction (+ or -)  Each atomic orbital can admit only two identical electrons with opposite spin Pauli exclusion principle The numbers of electrons that can occupy each shell and each sub-shell arise from equations of quantum mechanics One of such is Pauli exclusion principle It states that no two electrons in the same atom can have the same values for the four quantum numbers pg 18  How many subshells, orbitals and electrons are contained within the principal shell with n=4? Solution A. Given n=4, l can have values from 0 to n-1 l=0,1,2 and 3 (four subshells ≡ s, p, d, f) B. l =0, ml =0: 1 orbital l =1, ml = -1, 0, +1: 3 orbitals l =2, ml = -2, -1, 0, +1, +2: 5 orbitals l =3, ml = -3, -2, -1, 0,+1,+2,+3: 7 orbitals Total = 16 orbitals Total number of electrons = 2n2 =2(4)2 =2x16=32 electrons  Filling electrons into subshells and orbitals According to Aufbau principle, (German word meaning construction or building up), electrons orbiting an atom fill subshell/orbitals in order of increasing orbital energy The lowest energy subshell are filled before electrons are placed in higher energy subshell: s