Tin

Tin ! One of the longest known elements ! Symbol Sn from Latin stannum ! Bronze items from ca. 3500 B.C.E. contain 10-15% Sn alloyed with Cu. ! Cassiterite, mineral SnO2, found in Britain, was an early source. ! Malaysia, Russia, and Bolivia are current principal sources. ! About 40% of tin used in plating to provide a non-toxic coating for sheet metal and “tin” cans. ! Important component of many alloys. Solder 50% Sn, 50% Pb Aluminum solder 86% Sn, 9% Zn, 5% Al Pewter 85% Sn, 6.8% Cu, 6% Bi, 1.7% Sb Bronze 80% Cu, 15% Sn, 5% Zn Bell metal 78% Cu, 22% Sn Babbitt 90% Sn, 7% Sb, 3% Cu ! Tin has more stable isotopes than any other element. Tin Chemistry ! Tin is stable toward water and oxygen at 25 oC, but reacts with steam or when heated in oxygen to give SnO2. Sn(s) + 2H2O(g) ÷ SnO2(s) + 2H2(g) Sn(s) + O2(g) ÷ SnO2(s) ! Tin shows little reaction with dilute HCl or H2SO4, but with dilute HNO3 it forms Sn(NO3)2 and NH4NO3. ! Hot conc. HCl gives SnCl2 and H2, and hot conc. H2SO4 gives SnSO4 and SO2. ! In hot KOH(aq), Sn dissolves to give K2[Sn(OH)6] and H2. K Note that both +2 and +4 Sn compounds form, showing that both states have comparable stability. K Which oxidation state forms depends upon the thermodynamics of ionization versus bond stability. State Trends in Group 14 ! All group 14 elements have the valence configuration ns2np2. M(ns2np2) ÷ M2+(ns2np0) ÷ M4+(ns0np0) ! Stable carbon and silicon compounds have the group 14 element in the +4 state. K Going down the group, the +4 state becomes less stable and the +2 state becomes more stable. • The increasing stability of the lower state (+2) has been called the “inert pair effect” for the tendency of the ns2 configuration to be retained. • This has nothing to do with inherent “inertness” of the ns2 configuration, but rather thermodynamics. ! A similar "inert pair" effect is found in groups 13, 14, 15. Ionization Energies and Bond Enthalpies kJ/mol Group 14 Ionization Energies 20000 18000 16000 14000 12000 10000 8000 6000 4000 2000 0 M(II) M(IV) C Si Ge Sn Pb Group 14 MX4 Halide Bond Enthalpies 600 500 400 kJ/mol M-F M-Cl 300 M-Br M-I 200 100 0 C Si Ge Sn Pb Explanation of Inert Pair Effect ! Although energy “cost” of forming M(IV) is high for the lighter elements, the “pay-back” of bond formation is high, too. ! Ionization energy drops dramatically after C, declining slowly through the rest of the group. ! Bond strengths peak at Si, slowly decline through Sn, and then drop off significantly at Pb. ! At Ge and Sn, both M(II) and M(IV) states are stable. ! At Pb, the bond strength is too low to compensate for the slightly higher ionization energy requirement of the Pb(IV) state in many cases. Hence, the +2 state is favored. L In all these compounds, the oxidation state is merely a formalism, because all are molecular (not ionic) compounds. Group 14 Tetrahalides ! All are known, except PbI4, whose Pb-I bonds are too weak to support the higher oxidation state. ! Although C–X bonds are strong, CI4 has significant steric crowding and decomposes on heating or in u.v. light to C2I4, whose I–C–I angle of 114.2o reduces the strain. 2CI4 ÷ I2C=CI2 + 2I2 ! Except for SnF4 and PbF4, all are very volatile, suggesting covalent bonding with weak van der Waals forces between molecules. ! SnX4 compounds can act as Lewis acids in the presence of excess halide ion. SnX4 + 2X– ÷ SnX62– • Lewis acid strength is in the order SnF4 > SnCl4 > SnBr4 > SnI4. • SnI4 can also undergo redox with I– to give SnI2. SnI4 + I– ÷ SnI2 + I3– The resulting solution is brown in polar solvents. ! Except for CX4, all hydrolyze in water to give various hydrated oxides, although it is possible to isolate a few hydrates of Sn(IV). ! SnX4 compounds hydrolyze to give hydrated SnO2 and HX. SnX4 + 2H2O ÷ SnO2 + 4HX • Similar reaction occurs with alcohols. Synthesis of SnI4 ! Direct reaction of tin metal with iodine in methylene chloride yields SnI4 as the principal product, with formation of SnI2 as a side reaction. Sn + 2I2 ÷ SnI4 Sn + I2 ÷ SnI2 ! SnI4 is a red-orange solid; SnI2 is a yellow-red solid. • Color due to polarizability of I–, probably involving charge transfer (I÷Sn). ! SnI2 is polar, and SnI4 is non-polar. • As a result, SnI2 is slightly more soluble in CH2Cl2. ! SnI2 is somewhat more ionic and has a higher m.p. SnI2 m.p. . 330 oC SnI4 m.p. = 143-144 oC