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Atomic structure Part 1 2.1 The nuclear atom • Essential idea: The mass of an atom is concentrated in its minute, positively charged nucleus Understandings • Atoms contain a positively charged dense nucleus composed of protons and neutrons (nucleons). • Negatively charged electrons occupy the space outside the nucleus. • The mass spectrometer is used to determine the relative atomic mass of an element from its isotopic composition. Application and skills • Use of the nuclear symbol notation to deduce the number of protons, neutrons and electrons in atoms and ions. • Calculations involving non-integer relative atomic masses and abundance of isotopes from given data, including mass spectra. John Dalton (1766 – 1844) John Dalton (1766 – 1844) • Dalton’s atomic theory 1. All matter is made of atoms. Atoms are indivisible and indestructible. 2. All atoms of a given element are identical in mass and properties. Atoms of different elements are different 3. Compounds are formed by a combination of two or more different kinds of atoms in specific whole number ratios. 4. A chemical reaction is a rearrangement of atoms, not a change in the atoms themselves. Sir Joseph John “J. J.” Thomson (1856 – 1940) Sir Joseph John “J. J.” Thomson (1856 – 1940) • Challenged Dalton’s notion that atoms are indestructible • First suggestion of electrons • “Negatively charged electrons scattered in a positively charged sponge-like substance” Deflecting “cathode ray”, now known as electron beam When a high voltage was applied to the metal plate, a “cathode ray” is released. The cathode ray’s path can be deflected towards a positively charged metal plate, suggesting that the cathode ray is negatively charged. Sir Joseph John “J. J.” Thomson (1856 – 1940) Ernest Rutherford (1871 – 1937) Gold foil experiment Discovery of subatomic particles • If Thomson’s “plum pudding model” was correct, the alpha particles (positively charged) should either pass straight through or get stuck in the positive “sponge” Discovery of subatomic particles • Results: – Most alpha particles passed straight through and get detected at the expected location on the screen – A very small number were repelled and bounced back Discovery of subatomic particles • Conclusion – The atom is mainly empty space – Deflection occurred when the alpha particles hit and were repelled by a dense, positively charged center called the “nucleus” Subatomic particles Niels Bohr (1885 – 1962) Bohr model of the hydrogen atom • A small solar system with an electron moving in an orbit • Most of the atom is empty space • Protons? • Neutrons? • Electrons? Ions 40Ca2+ Number of subatomic particles • Protons? • Neutrons? • Electrons? 127I- Number of subatomic particles • Protons? • Neutrons? • Electrons? How should these three symbols be interpreted? Isotopes Isotopes: Atoms of the same element with different number of neutrons Isotopes and Atomic mass • https://phet.colorado.edu/en/simulation/isot opes-and-atomic-mass Take a look at the data booklet 6. Periodic table with relative atomic mass • The values are expressed to 2 decimal places • Calculated as the weighed, relative average mass based on the percentage abundance of different isotopes Question type 1. Calculate the relative atomic mass from the information provided below Relative atomic mass (Ar) = 34.97 x 75.77% + 36.97 x 24.23% = 35.45 Question type 2: Mass spectra interpretation Question type 3: • Relative atomic mass of Chlorine is 35.5 • The two isotopes are 35Cl and 37Cl • Calculate the relative abundance of each isotope Question type 3: • Relative atomic mass of Chlorine is 35.5 • The two isotopes are 35Cl and 37Cl • Calculate the relative abundance of each isotope • Let – x = relative abundance of 35Cl – y = relative abundance of 37Cl x+y=1 y=1–x 35x + 37y = 35.5 x = relative abundance of 35Cl y = relative abundance of 37Cl x+y=1 y=1–x 35x + 37y = 35.5 35x + 37 (1-x) = 35.5 35x + 37 – 37x = 35.5 -2x = -1.5 x = 0.75 y = 1 – x = 0.25 x = relative abundance of 35Cl = 0.75 or 75% y = relative abundance of 37Cl = 0.25 or 25% x+y=1 y=1–x 35x + 37y = 35.5 35x + 37 (1-x) = 35.5 35x + 37 – 37x = 35.5 -2x = -1.5 x = 0.75 y = 1 – x = 0.25 Practice: Magnesium has three stable isotopes – 24Mg, 25Mg, and 26Mg. • The lightest isotope has an abundance of 78.90%. Calculate the percentage abundance of the other isotopes. The Quantum Mechanical model of the atom … continued in Part 2