Problem 58
Question
(a) Use the following reactions to prepare an activity series for the halogens: $$ \begin{aligned} \mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) & \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\ \mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) & \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q) \end{aligned} $$ (b) Relate the positions of the halogens in the periodic table with their locations in this activity series. (c) Predict whether a reaction occurs when the following reagents are mixed: \(\mathrm{Cl}_{2}(a q)\) and \(\mathrm{KI}(a q) ; \operatorname{Br}_{2}(a q)\) and \(\operatorname{LiCl}(a q)\)
Step-by-Step Solution
Verified Answer
In summary, an activity series for the given halogens is: Cl₂ > Br₂ > I₂. The reactivity decreases as we move down Group 17 in the periodic table. A reaction will occur between Cl₂ and KI, forming KCl and I₂. However, no reaction will occur between Br₂ and LiCl as Br₂ is less reactive than Cl₂.
1Step 1: Analyze the given reactions
The reactions are as follows:
\(
\begin{aligned}
\mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) & \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\
\mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) & \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q)
\end{aligned}
\)
In reaction 1, bromine (Br₂) displaces iodine (I₂) from its salt, sodium iodide (NaI), resulting in the formation of sodium bromide (NaBr) and iodine (I₂).
In reaction 2, chlorine (Cl₂) displaces bromine (Br₂) from its salt, sodium bromide (NaBr), resulting in the formation of sodium chloride (NaCl) and bromine (Br₂).
2Step 2: Prepare the halogen activity series
Based on the given reactions, we can arrange the halogens in decreasing order of their reactivity as:
1. Chlorine (Cl₂)
2. Bromine (Br₂)
3. Iodine (I₂)
This activity series tells us that chlorine is the most reactive among the given halogens, followed by bromine, and then iodine.
3Step 3: Relate halogen positions in the periodic table and the activity series
In the periodic table, all halogens belong to Group 17 (or Group VIIA). Their positions in the periodic table are as follows:
1. Chlorine (Cl) - Period 3
2. Bromine (Br) - Period 4
3. Iodine (I) - Period 5
When comparing their locations in the periodic table with their positions in the activity series, we can notice that the reactivity of halogens decreases as we move down the group. This is due to the decrease in the electron affinity and the increase in atomic size, which make it more difficult for the halogens to attract an additional electron to complete their octets.
4Step 4: Predict the outcome of reactions based on the activity series
We have two reactions to predict:
1. \( \mathrm{Cl}_{2}(a q) + \mathrm{KI}(a q)\)
2. \( \mathrm{Br}_{2}(a q) + \mathrm{LiCl}(a q)\)
(a) Since chlorine (Cl₂) is more reactive than iodine (I₂), it will displace iodine from potassium iodide (KI) to form potassium chloride (KCl) and iodine (I₂). Thus, a reaction will occur, and the products will be KCl and I₂.
(b) Bromine (Br₂) is less reactive than chlorine (Cl₂), so it cannot displace chlorine from lithium chloride (LiCl). Therefore, no reaction will occur in this case.
Key Concepts
Activity SeriesPeriodic Table TrendsHalogen Displacement Reactions
Activity Series
The activity series is a list that ranks elements according to their reactivity. When it comes to halogens, each has a different level of reactivity and they can be ranked accordingly. This ranking is very helpful when predicting the outcomes of halogen displacement reactions.
In the activity series for halogens, chlorine (Cl₂) is at the top, followed by bromine (Br₂), and finally iodine (I₂) at the bottom. This means that chlorine is the most reactive, and iodine is the least reactive of the three.
The activity series is crucial for predicting if one halogen can displace another from a compound. For instance:
- Chlorine can displace bromine and iodine because it is more reactive than both.
- Bromine can only displace iodine due to its lower reactivity compared to chlorine.
- Iodine cannot displace bromine or chlorine as it is the least reactive.
Periodic Table Trends
In the periodic table, halogens are found in Group 17, moving from top to bottom, they are fluorine (not shown in this problem), chlorine, bromine, and iodine.
A key trend in the periodic table is that as you move down a group, the atomic size of elements increases, and so does the number of electron shells. Halogens demonstrate a decrease in reactivity as you move down the group.
This trend is attributed to several factors:
- An increase in atomic size means that the outermost electrons are farther from the nucleus.
- The inherent attraction of the nucleus for additional electrons decreases, reducing electron affinity.
- More electron shells add to the electron shielding effect, further decreasing the effective nuclear charge felt by the valence electrons.
Halogen Displacement Reactions
Halogen displacement reactions occur when a more reactive halogen displaces a less reactive halogen from its compound. These reactions are useful for determining the relative reactivity of different halogens.
Such reactions are two-part processes: a more reactive halogen takes the place of a less reactive halogen in a compound, leading to two new products.
Let's explore how this works in practice:
- When chlorine (Cl₂) is added to potassium iodide (KI), it displaces iodine, forming potassium chloride (KCl) and iodine (I₂).
- No reaction occurs when bromine (Br₂) is mixed with lithium chloride (LiCl), since chlorine is more reactive than bromine, and the latter cannot displace it.
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