Atomic Theory & Atomic Structure Lecture 1 PDF

14/2/2023 Dr. Wafa Hatim Balla Atomic Theory & Atomic Structure Lecture 1 ❖ History of Atomic Theory of Matter: Democritus (Greek philosopher) believed that there was a smalles...

14/2/2023 Dr. Wafa Hatim Balla Atomic Theory & Atomic Structure Lecture 1 ❖ History of Atomic Theory of Matter: Democritus (Greek philosopher) believed that there was a smallest particle— “Atomos” (uncuttable)—that made up all of nature. Experimental evidence for the atomic nature of matter was realized in the 18th century. In the mid ‘80’s a tool was developed which for the first time allowed scientists to actually “see” individual atoms and molecules. John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. This theory explains several concepts that are relevant in our world today. Before discussing the Atomic Theory, we will explain the theories that Dalton used as a basis for his theory: the law of conservation of mass and the law of constant composition. Law of Conservation of Mass: 1775 - Lavoisier “Father of Modern Chemistry”: states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; thus, mass is conserved. Example (1): Potassium chlorate (KClO3) decomposes to potassium chloride (KCl) and oxygen (O2) when heated. In one experiment 100.0 g of KClO3 generated 36.9g of O2 and 57.3 g of KCl. What mass of KClO3 remained unreacted? Solution: Mass of KClO3 before reaction = mass of KCl + mass of O2 + mass of unreacted KClO3 100.0 g of KClO3 = 57.3 g KCl + 36.9g O2 + g unreacted KClO3 g unreacted KClO3 = 100.0 g - 57.3 g - 36.9 g = 5.8 g Law of constant composition “Joseph Proust”: ✓ Also known as Law of Definite Proportions: state that in a given chemical compound, the proportions by mass of the elements that compose it are fixed, regardless of the source of the compound. ✓ The ratio of elements in a compound is fixed regardless of the source of the compound. ✓ For example: Water is made up of 11.1% by mass of hydrogen and 88.9% oxygen. Example (2): In a set of experiments very pure tin (Sn) was combined with bromine (Br) forming tin tetrabromide (SnBr4). Using the data below, confirm the law of definite proportions by calculating the % of tin in each sample of SnBr4. Grams of Sn reacted Grams of SnBr4 formed 2.8445 10.4914 3.0125 11.1086 4.5236 16.6752 Solution: 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑺𝒏 𝒓𝒆𝒂𝒄𝒕𝒆𝒅 Need to determine = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑩𝒓 𝒓𝒆𝒂𝒄𝒕𝒆𝒅 Mass of Br reacted = Mass of SnBr4 formed - mass of Sn reacted Grams of Sn reacted Grams of Br reacted 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑆𝑛 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = Grams of SnBr4 formed - 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵𝑟 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 Grams of Sn reacted 2.8445 10.4914 - 2.8445 = 7.6469 0.3721 3.0125 11.1086 - 3.0125 = 8.0961 0.3721 4.5236 16.6752 - 4.5236 = 12.1516 0.3721 ❖ Dalton's Atomic Theory: Experiments in the eighteenth and nineteenth centuries led to an organized atomic theory by John Dalton. Page 2 of 6 ❖ Discovery of electron: J.J. Thomson’s Model: Thompson provided the first hint that atoms were made up of even smaller particles In 1897, J.J. Thomson used a cathode ray tube to deduce the presence of a negatively charged particle: the electron Thompson proposed the plum pudding model of an atom o He stated that the electrons negatively (-) charged float in material that is positively (+) charge o The negatively charged electrons in an atom are like the pieces of fruit in a plum pudding, while the positively charged material is like the batter. ❖ Discovery of the Nuclear Model: Rutherford’s Model Ernest Rutherford (1871-1937): Rutherford shot a beam of alpha particles at a thin foil of gold. Around the gold foil Rutherford placed sheets of zinc sulfide. These sheets produced a flash of light when struck by an alpha particle. Rutherford predicted that particles in an alpha beam would pass through matter unaffected by a tiny amount of particles slightly deflected. The particles would only be deflected if they happen to come into contact with electrons. Page 3 of 6  However, this experiment produced results that contradicted Rutherford's hypothesis. According to the result of experiment Rutherford discovered the nucleus. He stated that atoms have a small, dense, positively (+) charged center called a nucleus. ❖ The Bohr Model of Atomic Structure ✓ The electrons are in continues motion round the nucleus in closed orbits of defined energy levels, This orbits are named as K, L, M, N, … or numbered as 1, 2, 3 ,4 … from the nucleus. ✓ Electrons can jump from a path on one level to a path on another level. ❖ Modern theory: ✓ Electrons travel in regions called “electron clouds” ✓ You cannot predict exactly where an electron will be found Page 4 of 6 ❖ Atoms: are made of three particles: 1. Proton (+) ✓ Located in the nucleus ✓ Have a positive charge ✓ Have a mass of one 2. Neutron (N): ✓ Located in the nucleus ✓ Have no charge ✓ Have a mass of one –similar to the proton 3. Electron (e-) : ✓ Have a negative charge ✓ Orbit the nucleus of the atom ✓ Are very small (have basically NO mass) In a neutral atom (the net charge is zero), there are the same number of protons and electrons. Each element has a unique number of protons. ❖ Atomic Number & Mass Number Atomic Number: the number of protons in an atom Mass Number: the number of protons and neutrons in an atom Mass number = (number of protons) + (number of neutrons) Mass number = atomic number + number of neutrons The accepted way to denote the atomic number and mass number of an atom of element X is as follows: Example (3) Give the number of protons, neutrons and electrons in each of the following species: a) 𝟏𝟔𝟖𝑶 b) 𝟏𝟗𝟗 𝟖𝟎𝑯𝒈 Solution: Page 5 of 6 ❖ Isotopes: ⚫ Atoms with the same number of protons, but different numbers of neutrons. ⚫ Atoms of the same element (same atomic number) with different mass numbers, e.g. Isotopes of chlorine: Chlorine - 35 Chlorine - 37 𝟑𝟓 𝟑𝟕 𝟏𝟕𝑪𝒍 𝟏𝟕𝑪𝒍 Example (4): Naturally occurring carbon consists of three isotopes, 𝟏𝟐𝟔𝑪 𝟏𝟑𝟔𝑪 𝟏𝟒𝟔𝑪. State the number of protons, neutrons, and electrons in each of these carbon atoms. Solution: ❖ Atomic weight: ✓ Can be defined as the average weight of an element with respect to all its isotopes and their relative abundances ✓ Atomic weight = fractionA mA + fractionB mB +... ✓ Atomic weight = Σ[(isotope mass)×(fractional isotope abundance)] Average atomic mass is known as the atomic weight. Example (5): Boron has two naturally occurring isotopes. In a sample of boron, 20% of the atoms are B-10, which is an isotope of boron with 5 neutrons and a mass of 10 amu. The other 80% of the atoms are B-11, which is an isotope of boron with 6 neutrons and a mass of 11 amu. What is the atomic mass of boron? Solution: Boron has two isotopes: Atomic mass = fraction1 m1 + fraction2 m2 Atomic mass = (0.20) (10) + (0.80) (11) Atomic mass = 10.8 amu The mass of an average boron atom, and thus boron’s atomic mass, is 10.8 amu. Page 6 of 6