Sure, let's analyze each reaction based on the activity series:
1. [tex]\( CuI_2 + Br_2 \rightarrow \)[/tex]
- According to the activity series [tex]\( F > Cl > Br > I \)[/tex], bromine (Br) is more reactive than iodine (I).
- However, since Cu is part of the compound as CuI2, we need to compare the reactivities of Br and I directly.
- Since Br is higher in the activity series than I, theoretically Br can displace I.
- But considering the actual reactivity of the compound CuI2, where Cu and I have formed a stable compound, Br2 will not react with CuI2.
- So, no reaction occurs here.
2. [tex]\( Cl_2 + AlF_3 \rightarrow \)[/tex]
- Chlorine (Cl) is more reactive than fluoride (F) as indicated by the activity series.
- But in the compound AlF3, fluorine is already in a highly stable state bounded with aluminum.
- Thus, chlorine will not displace fluorine from this stable compound.
- Therefore, no reaction occurs here.
3. [tex]\( Br_2 + Na \rightarrow \)[/tex]
- Sodium (Na) is highly reactive, much more than bromine (Br).
- When sodium is introduced to bromine, it will react to form sodium bromide (NaBr).
- Therefore, a reaction will occur producing NaBr.
4. [tex]\( CuF_2 + I_2 \rightarrow \)[/tex]
- Fluorine (F) is the most reactive halogen according to the activity series.
- Iodine (I) is the least reactive; it will not be able to displace fluorine from its compounds.
- Thus, iodine will not displace fluorine from CuF2.
- Hence, no reaction takes place here.
Summary of reactions:
1. [tex]\( CuI_2 + Br_2 \rightarrow \)[/tex] No reaction.
2. [tex]\( Cl_2 + AlF_3 \rightarrow \)[/tex] No reaction.
3. [tex]\( Br_2 + Na \rightarrow \)[/tex] Reaction occurs, forming NaBr.
4. [tex]\( CuF_2 + I_2 \rightarrow \)[/tex] No reaction.
Therefore, the answer is: [tex]\( (False, False, True, False) \)[/tex].
