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Ch.42 notes – P.1 Ch.42 – Group IV Elements A. Introduction Group IV elements include__________,__________, ________________, _______ and _________. Electronic configurations: Carbon Silicon Germanium [Kr]4d105s25p2 Tin [Xe]4f145d106s26p2 Lead Carbon is a ________________, silicon and germanium are _____________, while tin and lead are typical ____________. B. Structure and bonding 1. Carbon Carbon exists in two important allotropic forms, ___________ and___________. They possess _______________________________: Diamond Graphite Diamond is a __________;graphite is used for making ____________and _______________. Ch.42 notes – P.2 2. Silicon and germanium Silicon and germanium exist as a ________________________ in which the atoms are covalently bonded to one another. They are used as ______________________ in transistors. 3. Tin and lead Tin has two__________________: white tin (more stable) and grey tin. White tin has ______________________ and grey tin has a ________________ structure which is similar to that of diamond. Lead exists as a typical ______________________. Tin is used to plate iron to prevent it from rusting; lead is used as a protective shield from radioactivity and used in accumulators. Ex. 1. Describe the trend of electrical conductivities of the group IV elements. Ch.42 notes – P.3 C. Variation in physical properties 1. Melting point The melting points show a general ______________ on going down the group. Diamond has giant covalent structure, it needs a great amount of energy to break the _________________________. From carbon to germanium, the bond lengths increases and the ____________________ decrease. Tin and lead have metallic structures and with these, there is ___________ __________ strong metallic bonds to bring about melting. Also, in both of these metals, only two of the four valence electrons are delocalized to form metallic bonds (_________________). Ch.42 notes – P.4 2. Boiling point The general trend and explanation of the variation in boiling point are similar to those for melting point. Germanium has an abnormally high boiling point. The probable explanation is that germanium changes to _____________________________________ in the liquid state and each atom has four valence electrons participate in the formation of metallic bonds. Ch.42 notes – P.5 D. Chlorides of Group IV elements All group IV elements form ___________________ and they are ___________ at room temperature. They are all _____________________________with a tetrahedral shape. Although each M-Cl bond is polar, the molecule as a whole has ____________________________ because it has a symmetrical shape. Only germanium, tin and lead form _______________, GeCl2, SnCl2 and PbCl2. They exist as _____________________ at room temperature and pressure. Germanium(II) chloride and tin(II) chloride are ______________________ while lead(II) chloride is ________________________. A general trend is observed when going down the group. The relative stability of the ______oxidation state______________, and the _____ oxidation state becomes more______________. Ch.42 notes – P.6 Ex. 1. Refer to the electronegativities of carbon, silicon and germanium, predict which of their M-Cl bond has the highest ionic character. 2. SiCl4 is readily hydrolyzed in water while CCl4 is stable to water. Give an explanation. E. Oxides of Group IV elements There are two series of oxides: ______________ (MO) and ____________ (MO2). All group IV elements except silicon from the monoxides at normal conditions. CO is a __________________ compound, GeO is a black solid, SnO and PbO are predominantly__________ solids. In going down the group, there is a relative ____________________________ of the monoxides(oxidation state of +2) relative to the dioxides (oxidation state of +4). Ch.42 notes – P.7 Monoxide Formula Bond type Stability Dioxide Formula Bond type Stability Carbon Silicon Germanium Tin Lead Ex. 1. Carbon dioxide exists as a gas while the dioxides of other group IV elements are crystalline solids at room temperature and pressure. Explain. Ch.42 notes – P.8 2. Explain why Lead(IV) oxide is an oxidizing agent. 3. Given: Bonding Enthalpy (kJmol-1) C-C 348 Si-Si 176 C-O 360 Si-O 374 Explain why carbon catenates to form chains and rings, which are stable in air while no compounds containing Si-Si bonds are found in nature. Ch.42 notes – P.9 F. Silicon Silicon is the______________ most abundant element found in the Earth’s crust. It is commonly found as__________________________ in a variety of forms such as sand, quartz and flint. It is also found as____________ in rocks and clay. Silicon can be obtained from silica by _____________ ____________________ in an electric furnace. SiO2(s) + C(s) Extremely pure silicon an be obtained by the reaction silicon(IV) chloride with hydrogen, followed by _________________________ of the resultant silicon. SiCl4(s) + 2 H2(g) Silicon is the basic material for making ___________________________. Other uses include making steel and aluminium _____________. Ch.42 notes – P.10 G. Silicates 1. Basic unit of silicates Silicates are compounds of silicon, oxygen and one or more metals. Silicates are the largest and the most complicated class of minerals. About 30% of all minerals are silicates. The basic chemical unit of silicates is the ____________ anion, which has a __________________ shape. A few silicate minerals contain SiO44- as discrete ions, and they are known as _____________________, e.g. Zircon, ZiSiO4. 2. Structures of silicates The SiO44- tetrahedral can be joined up to form chain, sheet sheet or network silicates by sharing oxygen atoms, e.g. Si2O76-. Ch.42 notes – P.11 a) Chain silicates When two oxygen atoms of an SiO44- tetrahedron with other SiO44- tetrahedra, the silicates form a __________ or an _____________________. Chain silicates tend to have ________________________________________ because the individual chains of tetrahedral can be separated much more easily than the chains themselves that can be broken. In some ______________, the fibrous nature is resulted from the long-chain silicate anions present in the mineral. b) Sheet silicates When each SiO44- tetrahedron _______________________________ atoms with neighbouring SiO44- tetrahedra, sheet silicates are formed. Ch.42 notes – P.12 Sheet silicate anions are found in ___________ and ___________. Since only weak van der Waals’ forces exist between the sheets of the SiO44anions, mica and clay readily cleave into_______________. c) Network silicates The mineral quartz consists of SiO44- tetrahedral in which_____________ __________________ with adjacent SiO44- tetrahedra. Another group of network silicates is the ______________ group. Every oxygen atom is shared between SiO44- tetrahedra, but some of the tetrahedra have____________ at their centers instead of silicon. Aluminium has one electron less to share than silicon atom. This allows the __________________ ______________ to atoms of sodium, potassium or calcium.