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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 58a

The following reactions (note that the arrows are pointing only one direction) can be used to prepare an activity series for the halogens: Br21aq2 + 2 NaI1aq2¡2 NaBr1aq2 + I21aq2 Cl21aq2 + 2 NaBr1aq2¡2 NaCl1aq2 + Br21aq2 (a) Which elemental halogen would you predict is the most stable, upon mixing with other halides?

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Welcome back everyone. The activity series of allergens can be illustrated by the following reactions where one mole of chlorine reacts with two moles of sodium bromide to form two moles of sodium chloride and one mole of. Then we have bromine reacting with two moles of sodium. I'd I to form two moles of sodium bromide and one mole of iodine. Based on these reactions, predict the most stable halogen upon mixing with other hey leads, we want to recognize that in our given reactions the more reactive element were, the more reactive compound is most likely to undergo oxidation and displaces the less reactive adam from it's compound. We want to recall the trend on our periodic table that the more electro negative and adam will correspond to a greater reactivity of that atom recall our trend on our periodic table where electro negativity increases towards the top right of our periodic table. And because we know our halogen Zarin Group seven a of our periodic table recalled. That consists of flooring underneath flooring, we have chlorine. Then we have roaming iodine and turpentine at the bottom. And based on our first reaction between chlorine and sodium bromide, we see that chlorine has a greater electro negativity than browning, which has a lower electro negativity. And so we would say that chlorine is more reactive and we can see that in our product sodium chloride chlorine displaces bromine from sodium bromide. Since we can outline that chlorine is more reactive, it's going to be less stable than bromine. Whereas in our second reaction we have roaming now reacting with sodium iodide. So we're comparing it to the hay lied ein. And on our periodic table we see that the electro negative activity of bromine because it is Across period four and higher than iodine, which is in period five of our periodic table than Aydin has a lower election. Or sorry, I'm writing this backwards. So the election negativity of iodine is going to be less than the election negativity of blooming, which has a greater electro negativity since again, it is across period four and higher on the periodic table than I did. And so because bromine is more electro negative than I did, then in this case we have bromine which is more reactive and less stable than I did. And so we can therefore determine the following order in which chlorine has a greater reactivity than grooming, which browning has a greater reactivity than hiding. And so iodine is going to be the least stable halogen and chlorine is going to be the most stable halogen and if I'd seen is the least stable halogen, that means it's the least prone to oxidation. Whereas chlorine being the most stable halogen is the most prone to oxidation. So chlorine is going to be our final answer as our most stable halogen corresponding to choice C in the multiple choice. So I hope this made sense. And let us know if you have any questions