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Atomic Structure, Bonding and the Periodic Table. Revision questions.

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´╗┐Task 1: Atomic Structure ________________ 1. Copy and complete this table: Relative Mass Relative Charge Location Proton 1 1 Nucleus Neutron 1 0 Nucleus Electron 1/1836 -1 shells / orbital?s 2. Describe how 12C and 13C are different and how they are the same. You should include the different types of sub-atomic particles in your answer. What term is used to describe how these two atoms are related? 12C and 13C are isotopes of carbon, like most elements carbon has several isotopes. It is the number of protons (atomic number Z) which defines a specific element, and so carbon-12 and carbon-13 are still distinct as the element carbon, as they each have 6 protons (atomic number Z = 6). Both the 12C and 13C isotopes also still have the same electronic configuration, with 6 electrons in total, meaning that they are identical in how they react chemically. The reason for the categorisation as isotopes of carbon is because their atomic mass is different, with carbon-12 possessing 6 neutrons and carbon-13 possessing 7 neutrons. 3. For each of the following species, give a full electronic configuration in the format 1s2, 2s2, a) ...read more.


We can predict that the molecular geometry of NCl3 will be trigonal planar. b. H2S ? Hydrogen sulphide (Sulphane). From the Lewis structure of hydrogen sulphide we can see that again, there are four pairs of electrons around the central sulphur atom, only this time there are two loan pairs present. This means that the hydrogen-sulphur bonds assume two points of a tetrahedron but are even further compressed because of the two loan pair, resulting angular in a (bent) molecular geometry. c. CCl4 ? Carbon tetrachloride (Tetrachloromethane). Here the central carbon is using all four of its valence electrons to bond with chlorine atoms. As there are no loan pair electrons carbon tetrachloride will have a tetrahedral shape with bond angles of 109.5o. 5. Use your knowledge of bonding and intermolecular forces to explain the following observations: a. Lithium iodide is a solid at room temperature. The bonds which form between lithium and iodine are mostly ionic, arising from the electrostatic attraction between the positive lithium cation and the negative iodine anion. As lithium?s charge density (charge to volume ratio) is high the bonds do exhibit some covalent characteristics, such as dissolving in organic solvents. ...read more.


As this minimises the repulsive force between the bonds. Ethane which is more complex should be considered as two trigonal planar molecules joined double bond in the middle. This gives bond angles of 120o around each carbon atom. With molecules such as H2O where the central oxygen atom has two loan pairs in its valence shell, the shape of the molecule is affected. Instead of a linear geometry like CO2, the molecular orbital forms a tetrahedron, including the bond electrons and loan pairs. As such the hydrogen-oxygen bonds take and angular (bent) geometry. However due to loan pair electrons exerting a greater repulsive force than bonded ones, the bond angle is not 109.5o, but 104.5o as a result of the compression this causes. VSEPR theory also works on polyatomic ions, such as ammonium (NH4+). This is because the bonds between the nitrogen and hydrogen atoms are covalent, thus have a fixed position and angle. In the ammonium ion the central nitrogen has a full octet, with all four electron pairs involved in a bond. Even though one of the covalent bonds is dative the repulsion it exerts on the other bond is the same as a single or double bond; so has no effect on the geometry. This means four equal pairs, giving a tetrahedral geometry with 109.5o bond angels. ...read more.

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