Unit 1 Mod 3 Group IV page 1 of 8 a) explain the variation in physical properties of the elements in terms of structure and bonding ( refer to variations in metallic character and electrical conductivity) b) describe the bonding of the tetrachlorides c) explain the reactions of the tetrachlorides with water d) discuss the trends in:- (i) bonding (ii) acid/base character (iii) thermal stability of the oxides of oxidation states II and IV (make reference to Eө values of the elements) e) discuss the uses of ceramics based on silicon(IV) oxide GROUP IV ELEMENTS Group IV elements:C Si Ge Sn Pb Variation in melting points and electrical conductivities of the elements Property m.p (K) C 4003 Si 1683 Ge 1210 Sn 505 Pb 600 The structure of the elements range from macro-molecular non-metals through metalloids to metallic lattices of close-packed ions. Carbon exists either as a diamond formation or as graphite, in either form, there are strong covalent bonds between the carbon atoms which must be broken, resulting in a extremely high melting point. Only in graphite is electrical conductivity observed but as expected it is very low. Both silicon and germanium exist in distorted diamond-like lattices, with germanium’s structure more distorted than silicon’s. The larger the atom, the greater the deviation of its structure from that of diamond. Their melting points are high, however not as high as carbon, but they decrease because of increasing size of the atom, which results in weaker degree of orbital overlap (i.e. weaker bond). Both of them are semiconductors. Unit 1 Mod 3 Group IV page 2 of 8 Tin has 2 allotropes, one stable below 286.2K and another stable above 286.2K, the allotrope stable below 286.2K exists as a structure similar to that of germanium while the allotrope above 286.2K has a metallic structure. It is this transition between states which results in the lower than expected melting point. In its metallic form, it conducts electricity as do all metals, while in its germanium form, it may conduct only in certain conditions like a semi-conductor. Lead exists as a metallic lattice and thus would have a moderately high melting point, and it would conduct electric current. Checkpoint A 1. Using this circle as the size of the silicon atom, draw circles to represent the atomic radii of the elements carbon, germanium, tin and lead. C Ge Sn Pb 2. Using the graph below, sketch the electrical conductivities of the group IV elements 3. Does metallic character of the group IV elements INCREASE or DECREASE down the group? ………………… Unit 1 Mod 3 Group IV page 3 of 8 4. Why does graphite conduct electricity but diamond does not, even though they are both made from carbon atoms? …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… The tetrachlorides The tetrachlorides of group IV elements are covalent, tetrahedral molecules. The molecules exist as discrete units with weak van der Waals’ forces between them and this results in the liquids being quite volatile at room temperature. Within the molecule are strong covalent bonds between the atoms involved. Note that the larger the molecules, the LESS volatile they are Shape of a tetrachloride of a group IV element (tetrahedral) Group IV tetrachlorides are hydrolysed by water (except CCl4) to form the hydroxide (or hydrated oxide) and hydrochloride gas. MCl4(l) + 2H2O(l) M(OH)4(s) or MO2.xH2O + HCl(g) In the case of silicon and all other group IV elements except carbon, hydrolysis of the tetrachloride is accomplished by attack of the water molecule to the electron-deficient metal atom. The group IV metal atom Unit 1 Mod 3 Group IV page 4 of 8 can use a vacant low lying 3d orbital to accommodate the lone pair from the oxygen atom, forming a short- lived 5 co-ordinate intermediate which dissociates to form the hydrogen chloride and the metal trichloride hydroxide, MCl3OH. Repetition of these steps gives the hydrated metal oxide or the metal hydroxide. Carbon cannot accommodate the water molecule as it does not have any empty low-lying 3d orbitals, remember carbon has electronic configuration 1s2 2s2 2p2 therefore no d orbitals present. Therefore the lone pair of electrons from the water molecule cannot be accepted and thus no reaction i.e. no hydrolysis of CCl4 can occur. Checkpoint B 1. Would the rate of hydrolysis of the tetrachlorides increase or decrease down the group? ………………… 2. Suggest a reason for your answer in question 1 above …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… Unit 1 Mod 3 Group IV page 5 of 8 Acid-base nature of the oxides in oxidation state +2 and +4 CO is neutral, CO2 is acidic SiO & GeO do not exist SiO2 is weakly acidic GeO2, SnO, SnO2, PbO and PbO2 are amphoteric Reactions of the oxides in the +2 and +4 oxidation states to show their acidic or basic nature Since CO is neutral, it reacts with neither acids nor bases. CO2 (g) + H2O (l) H2CO3 (aq) CO2 forms an acid when reacted with water SiO2 (s) + OH- (aq) SiO32- (aq) + H2O(l) It react with alkali to form a salt and water General form of group IV monoxide MO General form of group IV dioxide MO2 Reaction of MO with HCl is: Note: Reaction of PbO with HCl forms solid PbCl2 and not (aq) Reaction of MO with NaOH is: Reaction of MO2 with HCl is: Reaction of MO2 with NaOH is: Checkpoint C 1. Write an ionic equation for the reaction of PbO2 with alkali ……………………………………………………………………….. Unit 1 Mod 3 Group IV page 6 of 8 2. What is the name of the acid with the formula H2CO3 ? ………………………………………………. Thermal stability of oxides in +2 and +4 oxidation states Note that +4 oxidation state is more thermally stable than +2 for ALL the group IV elements EXCEPT lead. At lead, the situation reverses and the +2 is MORE thermally stable than the +4 oxidation state. This stability occurs for the aqueous ions as well. Note: There is no +2 oxidation state for Si and Ge Checkpoint D 1. CO can be easily oxidised to CO2. True or False? 2. SnO2 is easily reduced to SnO. True or False? 3. PbO2 is easily reduced to PbO. True or False? Unit 1 Mod 3 Group IV Inert pair effect page 7 of 8 The reason for the reversal in difference in stabilities of the +II and +IV oxidation state for lead is the inert pair effect. On descending group IV, the atoms get larger. The shielding of the nuclear charge from the valence electrons by the core electrons gets more and more ineffective. At lead, the shielding is so poor that the 2 electrons of the s orbital in the valence shell feel the nuclear charge more so than the p electrons of the valence shell and are pulled towards the core electrons. In essence the two electrons are now part of the core electrons and do not take part in bonding and thus are “inert”. Hence lead acts as if it only has 2 valence electrons instead of 4. Thus the term “inert pair effect”. Ceramics These are materials which are fashioned at room temperature from clays or other natural earths and then permanently hardened by heat. Feldspar + carbon dioxide + water clay + silica + potassium carbonate When silica is reacted with other oxides in the process of glazing (the final stage in making ceramics), the important silicate ion (SiO32-) is formed. Properties of ceramics 1. Hard and brittle (unidirectional bonds) 2. Non-corrosive and insulating to heat and electricity (no mobile electrons) 3. Resistant to high and low voltages and high melting point (because of the high strength of the Si—O) Unit 1 Mod 3 Group IV Practice Questions 1. page 8 of 8