Chemistry

5413 s and p – Block ElementsThe bromide ions reduce the sulphuric acid to sulphur dioxide gas. This is a decrease ofoxidation state of the sulphur from +6 in the sulphuric acid to +4 in the sulphur dioxide.You can combine these two half-equations to give the overall ionic equation for the reaction:With Iodide IonsdIodide ions are stronger reducing agents than bromide ions are. They are oxidised to iodineby the concentrated sulphuric acid.The reduction of the sulphuric acid is more complicated than before. The iodide ions arepowerful enough reducing agents to reduce it first to sulphur dioxide (sulphur oxidation state = +4) then to sulphur itself (oxidation state = 0) and all the way to hydrogen sulphide (sulphur oxidation state = -2).The most important of this mixture of reduction products is probably the hydrogen sulphide.The half-equation for its formation is:Combining these last two half-equations gives:Summary of the trend in reducingabilityFluoride and chloride ions won'treduce concentrated sulphuric acid.Bromide ions reduce the sulphuricacid to sulphur dioxide. In the process,the bromide ions are oxidised tobromine.Iodide ions reduce the sulphuric acidto a mixture of products includinghydrogen sulphide. The iodide ionsare oxidised to iodine.Reducing ability of the halide ionsincreases as you go down the Group.Explaining the trend When a halide ion acts as a reducing agent, itgives electrons to something else. That means thatthe halide ion itself has to lose electrons. The bigger the halide ion, the further the outerelectrons are from the nucleus, and the more they arescreened from it by inner electrons. It therefore getseasier for the halide ions to lose electrons as you godown the Group because there is less attractionbetween the outer electrons and the nucleus.