The following reactions (note that the arrows are pointing only one direction) can be used to prepare an activity series for the halogens: Br 2 (aq) + 2 NaI(aq) → 2 NaBr(aq) + I 2 (aq) Cl 2 (aq) + 2 NaBr(aq) → 2 NaCl(aq) + Br 2 (aq) (a) Which elemental halogen would you predict is the most stable, upon mixing with other halides?

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

Ch.4 - Reactions in Aqueous Solution

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Brown 15th Edition

Ch.4 - Reactions in Aqueous Solution

Problem 58a

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:
Br₂(aq) + 2 NaI(aq) → 2 NaBr(aq) + I₂(aq)
Cl₂(aq) + 2 NaBr(aq) → 2 NaCl(aq) + Br₂(aq)
(a) Which elemental halogen would you predict is the most stable, upon mixing with other halides?

Verified step by step guidance

Identify the reactions given: \( \text{Br}_2 + 2 \text{NaI} \rightarrow 2 \text{NaBr} + \text{I}_2 \) and \( \text{Cl}_2 + 2 \text{NaBr} \rightarrow 2 \text{NaCl} + \text{Br}_2 \).

Recognize that these reactions are single displacement reactions where a more reactive halogen displaces a less reactive halogen from its compound.

In the first reaction, \( \text{Br}_2 \) displaces \( \text{I}_2 \), indicating that \( \text{Br}_2 \) is more reactive than \( \text{I}_2 \).

In the second reaction, \( \text{Cl}_2 \) displaces \( \text{Br}_2 \), indicating that \( \text{Cl}_2 \) is more reactive than \( \text{Br}_2 \).

Conclude that \( \text{Cl}_2 \) is the most reactive and therefore the least stable when mixed with other halides, while \( \text{I}_2 \) is the least reactive and most stable.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Activity Series

The activity series is a list of elements organized by their ability to displace other elements in reactions. In the context of halogens, it ranks them based on their reactivity, with more reactive halogens able to displace less reactive ones from their compounds. Understanding this series helps predict the outcomes of reactions involving halogens.

Halogen Reactivity

Halogens are a group of elements in Group 17 of the periodic table, known for their high reactivity. Their reactivity decreases down the group, meaning fluorine is the most reactive, while iodine is the least. This trend is crucial for predicting which halogen will be more stable when mixed with other halides.

Stability of Halides

The stability of halides refers to how likely a halogen is to remain in its ionic or molecular form when mixed with other halides. More reactive halogens tend to form less stable compounds with other halides, while less reactive halogens are more stable. This concept is essential for determining which halogen would be the most stable in a given reaction.