Preparation And Properties Of The Halogens (1959)

[Music] the topic of this film is the hallogen we're going to be concerned with three of the four elements which commonly are listed in this group The Elements to be considered are chlorine which is present in the flask on the right bromine in the middle and iodine which is in the bottle on the left in this film we're going to be concerned with the methods of preparation of the free elements from their naturally occurring salts we're going to be concerned with the reactions which the free elements undergo and in the group relationships which these three elements exhibit the fourth hogen Florine is so active that its manipulation in the laboratory is difficult and it will not be considered in this film you've just seen samples of three of the common h chlorine bromine and iodine and mention was made of the fact that we're not going to consider Florine in this film because of its great activity and difficulty of preparation however we do want to spend just a few minutes examining a few of the properties of these elements as a family and discussing their group relationship as you can see when they're arranged in order of increasing atomic weight Florine comes at the top nine at the bottom iodine being many times heavier than Florine atom for atom now this means that the iodine atom contains very many additional electrons as compared to the Florine atom and as a matter of fact each hallogen contains one complete shell of electrons in addition to those possessed by the hallogen above it that is chlorine has one more shell of electrons than Florine and bromine one more than chlorine and so forth now in this we've tabulated the radius of the hogen ion the x minus ion where X may be any hogen and again we see a consistent increase in the size of the ion as we go from Florine to IOD the ions get bigger now since the halogens occur naturally as the ion if we wish to prepare the free hallogen we're going to have to do this by removing an electron from the ion or by oxidizing it and we would expect that it's going to be easiest to remove an electron from an ion where that electron is far out from the nucleus and conversely difficult to remove an electron from an ion where that electron is close to the nucleus and in fact no chemical oxidizing agent is sufficiently attract has sufficient attraction for electrons to remove an electron from the Florine atom on the other hand very many chemical oxidizing agents uh have the ability to oxidize the iodide ion to iodine in this film we're going to examine some of the group Relationships exhibited by these atoms and we're also going to study methods of preparation for each of the halogens and some characteristic chemical properties of each H first we need to develop a test for the presence of chlorine so we can determine when this gas is present or is being produced by a reaction for the production of some chlorine we'll use some potassium permanganate which is a strong oxidizing agent sufficiently strong to liberate chlorine from hydrochloric acid I'll place a few crystals of the permanganate in the test tube now as a source of chlorine we'll use some concentrated hydrochloric acid which of course contains contains the chloride ion which will be oxidized to Elementary chlorine by the potassium permanganate reaction proceeds without the necessity of heating it you can see bubbles of chlorine rising from the reaction mixture in this reaction the permanganate is functioning as the oxidizing agent and is being reduced to the brown manganes dioxide the chlorine is functioning as the reducing agent the chloride ion and is being oxidized to chlorine this tube then contains some chlorine at this point now to develop a test for chlorine we'll use some starch iodide paper this paper contains a mixture of potassium iodide and starch and when it comes in contact with chlorine the iodide is oxidized to iodine and that reacts with the starch to cause a purple color to appear if we leave this paper in the tube too long however the iodine is itself oxidized and the purple card disappears so this test must be judged rather rapidly as I'll illustrate here a second time the paper should just be inserted briefly in the tube and if it turns purple we can conclude that chlorine is present now a second test for the presence of chlorine utilizes litmus paper because chlorine is a bleach I have here a piece of blue litmus paper and as we insert this into the chlorine the color of the litmus paper disappears the paper becomes whitish so the bleaching of litmus paper for the purple color with starch iodide paper each constitutes a good test for the presence of chlorine we produced chlorine in the first experiment by reacting potassium permanganate with concentrated hydrochloric acid the balanced equation for this reaction is on the board the products being water potassium chloride manganas chloride and chlorine gas this equation is a good example of an oxidation reduction equation and uh it would be a good example for you to practice equation balancing with knowing the chlorine gas was given off we then developed a test for the presence of chlorine using starch iodide paper we found the starch iodide paper turns purple when it's uh comes in contact with chlorine gas and the reaction in question is on the lower part of the board chlorine reacts with potassium iodide in the starch iodide paper forming pottassium chloride and iodine the iodine then reacts with the starch in the paper forming the starch iodine complex which has a dark blue or purple color the appearance of this color is a test for any volle oxidizing agent that might convert the iodide ion D in this case the only uh volatile oxidizing agent that might have been present was chlorine so the appearance of this color constitutes a good test for chlorine in each of these test tubes we've placed a sample of an oxygen containing compound this tube contains zinc sulfate this potassium chlorate this lead dioxide and this one manganese dioxide we're going to test or treat each of these compounds with concentrated hydrochloric acid to see which of these oxygen containing compounds are sufficiently active oxidizing agents to liberate chlorine from hydrochloric acid we now test each tube for the presence of chlorine first we'll test the tube in which we placed the zinc sufate and the hydrochloric acid with a moistened piece of starch iodide paper and you notice that nothing happens no chlorine then is produced and no reaction is taking place I'll move this paper over to the next tube which contains the potassium chlorate hydrochloric acid mixture and here we see the paper immediately darkens indicating that plenty of chlorine is present the third tube contains the lead dioxide hydrochloric acid mixture and here we get a strong positive test for the presence of chlorine the fourth tube contains manganes dioxide and hydrochloric acid and strong positive test is obtained here also so three of our four compounds potassium chlorate lead dioxide and manganous dioxide were active enough oxidizing agents to liberate chlorine from hydrochloric acid while the fourth substance zinc sulfate was not a strong enough oxidizing agent on the board are equations for several of the reactions which we've just seen in each case we used hydrochloric acid as our source of chlorine and we found at least three other oxidizing agents that are sufficiently active to liberate chlorine from hydrochloric acid these are potassium chlorate lead dioxide and manganese dioxide the equations for these three reactions appear rather similar we obtain the salt of the metal used chlorine and water and then to demonstrate that not all oxygen containing compounds are good enough oxidizing agents to liberate chlorine from hydrochloric acid we added some of the acid to sodium sulfate and found that no chlorine was evolved because the sulfate ion under these conditions is not a strong enough oxidizing agent to liberate chlorine from hydrochloric acid the purpose of this experiment is to demonstrate that virtually any chloride salt when it's mixed with a strong oxidizer agent and in acid medium will react to liberate chlorine in these three test tubes we've placed samples of sodium chloride calcium chloride and potassium chloride respectively and then to each sample of a chloride we've added some manganes dioxide which we saw earlier is an active enough oxidizing agent to release chlorine from hydrochloric acid now to each tube We'll add some sulfuric acid now we'll warm each tube wait a few minutes and then test the gas in each tube for the presence of chlorine the contents of each tube have been heated for a few moments and reaction appears to be taking place in each first we'll test the tube containing the mixture of sodium chloride manganese dioxide and sulfuric acid with starch iodi paper positive test for the presence of chlorine is obtained next we'll test the tube containing the calcium chloride manganes di oxide and sulfuric acid a positive test is obtained here also finally we'll test the tube containing a mixture of potassium chloride manganese dioxide and sulfuric acid and a positive test is obtained here so in all three cases when we've mixed a chloride salt with manganese dioxide and sulfuric acid chlorine has been liberated by the reaction and this uh method of producing chlorine is frequently used in the laboratory in order to study the chemical properties of chlorine we need to to set up a generator and produce several bottles of the gas in this generator bottle we've placed some potassium permanganate crystals and a small quantity of water to cover the bottom of the thistle tube through the thistle tube we'll then add some hydropic acid we've already established that the reaction between permanganate and hydrochloric acid liberates chlorine more hydrochloric acid can be added from time to time to keep the reaction going smoothly the chlorine is passing through this tube and down into the bottle notice that the tube into the bottle extends almost to the bottom since chlorine is heavier than air you can introduce the chlorine at the bottom of the bottle and it will rise in the bottle displacing air from the top since chlorine is rather soluble in water we can't collect chlorine by the displacement of water as we have with other gases a total of four bottles of chlorine will be collected and the chemical properties of chlorine studied using this gas made from the generator we will first illustrate the reaction of chlorine with Metals the tip of the spatula we have a small quantity of finely powdered antimony metal the bottle is one that we filled with chlorine from our generator I'll remove the watch glass from the bottle and let small particles of antimony metal fall into the Container you can probably see little flashes of fire at the bottom of the the bottle the bottle becomes filled with a heavy white material which is antimon Tri chloride from this experiment then we can see that chlorine reacts very rapidly with most metal secondly we will illustrate the reaction between chlorine and a compound of carbon and hydrogen for the hydrocarbon we've chosen turpentine which has the formula C10 h16 I've Place several drops of turpentine on this small piece of filter paper and we'll insert the filter paper and the turpentine into the bottle of chlorine you can see that after a few moments the reaction takes place very rapidly a puff of flame is produced and the inside of the bottle becomes filled with soot so obviously one product of the reaction is carbon the other product is hydrochloric acid thirdly we will discuss the reaction of chlorine with another compound hydrogen sulfide in this bottle we have a solution of hydrogen sulfide in water I'll place a few milliliters of this solution in a small bottle of chlorine and then agitate you'll notice the formation of a dense milky material in the solution this is free Elementary sulfur which has been formed by the oxidation of the hydrogen sulfide by the chlorine in the bottle the last reaction of chlorine to be studied is its activity as a bleaching agent in my hand I have have a small piece of colored cloth the Dy used on this cloth is of the type that is quickly affected by chlorine I'll place this cloth in the bottle of chlorine and you can see that almost instantly purple color of the cloth has disappeared as the Dy was attacked by the vigorous bleaching action of the chlorine the equations on the board represent several of the reactions of chlorine which we've just seen you will recall that to our bottle of chlorine gas we added some finely divided animony metal that the animony glowed uh with the heat of the reaction and formed a quite powdery substance anamon Tri chloride we then put several drops of turentine which has the formula C NH h16 on a piece of filler paper and plac this in our bottle of chlorine and a rather violent reaction occurred producing a puff of soot and hydrochloric acid again or hydrogen chloride we see that the chlorine reacted with the turpentine by removing the hydrogen from its combination with the carbon forming hydrogen chloride and releasing the free Carbon when we added some hydrogen sulfide water to our chlorine the solution became milky due to the presence of free sulfur which is produced when chlorine acts as an oxidizing agent on the hydrogen sulfide then we used chlorine as a bleach on a piece of moist claw the reactions that take place there are more complex but probably start out by having the chlorine react with the water to form mixture of hypochlorous and hydrochloric acids a hypochlorous acid is a very strong oxidizing agent and probably is the active bleaching ingredient when chlorine attacks the dye such as it did in this experiment bromine also occurs naturally in the form of bromide salts and in order to prepare the free element from these salts an oxidation will be necess necessary since bromine is more easily oxidized than chlorine any of the oxidizing agents that were found suitable for the production of chlorine will also be suitable for the production of bromine in this retort we've placed some crystals of potassium bromide as a source of bromine and manganese dioxide and oxidizing agent that we found suitable for the production of chlorine the manganes dioxide is black and imparts a black color to the entire mass in order to cause the reaction to start We'll add some dilute sulfuric acid and then warm with the bunson burner shortly you'll be able to see the appearance of the redish brown fumes of the element bromine in the retort we'll heat the mixture until bromine begins to distill over and down the neck of the retort it will then be condensed by the cold water in the beaker and drops of liquid bromine should be seen falling into the test tube we wish first to examine the reaction of bromine with an active metal such as magnesium in the large test tube I've placed a quantity of finely divided magnesium in the small test tube we have some of the bromine water which we just prepared I'll now add the bromine water to the test tube containing the Magnesium you should note that a gas is obviously being produced since bubbling is taking place and that the red color of the bromine is rapidly disappearing actually several different reactions are occurring in the tube which we'll discuss later we'll wait for a few moments and then examine the tube after the reaction has gone to completion after a few minutes the reaction is over and you should note that the red color of the elementary bromine has entirely disappeared the next reaction to be studied is that between bromine water and six normal sodium hydroxide solution I'll now add some of the sodium hydroxide to the bromine water in the tube you should note the almost instant disappearance of the characteristic red color of the element bromine the solution now has a light yellow color a useful test for the presence of Elementary bromine is given by the extraction of bromine with carbon tetrachloride and the formation of a characteristic color in the carbon tetrachloride layer in this test tube we've placed a few two drops of roaming water which we then glut with considerable distilled water and to this test tube We'll add a small quantity of carbon tetrachloride I'll then remove the tube from the stand and Shake to cause the two liquids to come into contact then replace the tube in the stand and permit the carbon tetrachloride to settle and the bright orange color in the carbon tetrachloride bead serves as a useful and characteristic test for the presence of the element bromine in the retort we prepared bromine by reacting a mixture of cassium bromide manganese dioxide and sulfuric acid actually the reaction probably takes place in two steps with the first step being the reaction of the sulfuric acid with the potassium bromide to produce some hydrogen bromide and then a reaction of the hydrogen bromide with the manganes dioxide to produce bromine however the uh overall equation which we would get by adding those two equations together is given on the board the products being bromine potassium hydrogen sulfate mangana sulfate and water the bromine distilled over was collected in water and then subjected to seral reactions first we saw that groing water reacts rather vigorously with powdered magnesium actually several reactions are going on at once there's the direct Union reaction between the bromine and the Magnesium to produce magnesium bromide however the bromine also reacts with the water forming hypobromous acid and hydr bromic acid and the hydrogen ions from the ionizations of these acids react with the Magnesium producing hydrogen and magnesium I and you saw this hydrogen bubbling from the solution so all of these reactions were taking place simultaneously when magnesium was placed in bromine water you also saw that the color of the elementary bromine disappears rapidly when the bromine water is made basic with sodium hydroxide and this this case the reaction is between the bromine molecule and two hydroxide ions forming the hypo bromide ion the bromide ion and a molecule of water and since neither of these ions are colored the dark red color of the original bromine rapidly disappeared iodine like chlorine and bromine occurs naturally in the form of iodide salts so that an oxidation reaction is necessary in order to produce the free element in this Beaker we' placed some crystals of potassium iodide and some manganes dioxide which will again function as the oxidizing agent in the evaporating dish we place some cold water which will serve to condense the vapors of the iodine which are formed in this bottle we have phosphoric acid which we will add to the beaker in order to cause the reaction to start I'll then replace the evaporating dish and warm the reaction mixture with the flam of the bunson burner in a few moments you should begin to see the purplish Vapors of Elementary iodine in the beaker we've removed the evaporating dish from the beaker and have inverted it here on the wire gauze I'll rotate the dish slowly so that you can see the flashes of light from the faces of the crystals of iodine which are formed on the bottom of the ditch notice that the iodine vapor condensed directly to the solid rather than condensing into liquid first this is moderately unusual behavior when the reverse process happens that is when a solid changes directly to a gas without melting it is of course called sublimation this is a useful method for the purification of iodine we'll scrape some of these crystals from this dish and use them to test several chemical reactions of iodine we'll now examine the solubility of iodine in several different solvents in each of these three test tubes we've placed several crystals of iodine to the tube on the right I'll add a few milliliters of water to the center tube We'll add an approximately equal quantity of potassium iodide solution and to the tube on the left we'll add a quantity of ethyl alcohol we now permit the iodine in each tube to dissolve for a few moments we've stirred each of the uh mixtures for a few moments and you should notice that the tube on the right which iodine is in contact with water has assumed a faint brownish color indicating that some iodine has gone into solution but the tube in the middle containing potassium iodide solution and the tube on the left containing alcohol have both become highly colored this indicates that iodine is much more soluble in either of these two solvents than it is in pure water we will now apply the carbon tetrachloride extraction test to the very dilute solution of iodine which we've just prepared using pure water as the solvent I'll remove this tube from the holder and pour a few milliliters of this solution into the empe then to this solution We'll add a few two drops of carbon tetrachloride and Shake as before this time the purplish pink color in the bead serves as a characteristic test for the presence of iodite the reaction of iodine with starch produces a blue color which also serves as a useful and characteristic test for the presence of iodine in this tube we've placed a few milliliters of dilute starch solution from this bottle and to the tube We'll add a few milliliters of our iodine solution in water the deep blue color which is produced immediately serves as a good test for the presence of iodine we'll now study the activity of the halogens by examining the reaction of chlorine and bromine water with Solutions of potassium bromide and potassium iodide salts in these three test tubes I've placed on the left a sample of potassium bromide solution and in the two tubes on the right samples of potassium iodide solution now to each tube we'll add a few drops then to the potassium bromide we'll add several drops of chlorine water we'll also add some chlorine water to the first tube of potassium iodide solution and then finally we'll add a few drops of bromine water to the second potassium iodide solution and we'll now shake each two in this tube we observe the characteristic orange color of free bromine which has been displaced by the chlorine and here the iodine color which has been displaced by the chlorine so obviously chlorine is more active than both bromine and iodine in this tube we see the iodin color caused by the reaction of potassium iodide with bromine so bromine must be more active than iodine in the preparation of iodine we used the same type of reaction that we used before for Bromine we took a mixture of potassium iodide manganes dioxide and phosphoric acid in this occasion and obtained iodine potassium hydrogen phosphate manganas hydrogen phosphate and water the substitution of phosphoric acid for the sulfuric acid that we used previously uh was for a good reason because iodides are able to reduce sulfuric acid to hydrogen sulfide and uh if we use sulfuric acid in this preparation the iodine that we obtained would be contaminated to some extent with reduction products of sulfuric acid later we studied the activity of the three halogens chlorine bromine and iodine with respect to displacement reactions of one with another and we found that chlorine was sufficiently active to react with both bromides and iodides producing the free hallogen in each case and that bromine was active enough to react with iodide producing bromide and iodine these reactions permit us to rank the three hallogen chlorine bromine and iodine in order of activity it's obvious from these three equations the chlorine is the most active since it replaces the other two bromine is more active than iodine since it replaces iodine and therefore iodine must be the least active of these three halogens as oxidizing agents when we consider the ions as reducing agents however exactly the reverse order takes place and we find that the iodide ion is the most active reducing agent and the chloride ion of the three the least active reducing ache [Music]