Answer :
To identify the Lewis acid in the balanced equation:
[tex]\[ \text{SnCl}_4 + 2 \text{Cl}^- \rightarrow \text{SnCl}_6^{2-} \][/tex]
we need to understand the definitions and roles of Lewis acids and bases. According to the Lewis definition:
- A Lewis acid is a substance that can accept an electron pair.
- A Lewis base is a substance that can donate an electron pair.
Now let's analyze the equation step by step:
1. Identify the species involved:
- [tex]\(\text{SnCl}_4\)[/tex]
- [tex]\(\text{Cl}^-\)[/tex]
- [tex]\(\text{SnCl}_6^{2-}\)[/tex]
2. Determine what happens in the reaction:
- [tex]\(\text{SnCl}_4\)[/tex] reacts with 2 [tex]\(\text{Cl}^-\)[/tex] ions to form [tex]\(\text{SnCl}_6^{2-}\)[/tex].
3. Analyze the electron movement:
- [tex]\(\text{Cl}^-\)[/tex] ions each have an extra electron pair that they can donate.
- [tex]\(\text{SnCl}_4\)[/tex] will accept these electron pairs to form coordination bonds with the chloride ions, resulting in the formation of [tex]\(\text{SnCl}_6^{2-}\)[/tex].
4. Classify the species:
- Since [tex]\(\text{Cl}^-\)[/tex] ions donate electron pairs, they act as Lewis bases.
- Since [tex]\(\text{SnCl}_4\)[/tex] accepts electron pairs from the chloride ions, it acts as a Lewis acid.
Therefore, given the balanced equation, [tex]\(\text{SnCl}_4\)[/tex] is the Lewis acid as it accepts electron pairs from the chloride ions, [tex]\(\text{Cl}^-\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{\text{SnCl}_4} \][/tex]
[tex]\[ \text{SnCl}_4 + 2 \text{Cl}^- \rightarrow \text{SnCl}_6^{2-} \][/tex]
we need to understand the definitions and roles of Lewis acids and bases. According to the Lewis definition:
- A Lewis acid is a substance that can accept an electron pair.
- A Lewis base is a substance that can donate an electron pair.
Now let's analyze the equation step by step:
1. Identify the species involved:
- [tex]\(\text{SnCl}_4\)[/tex]
- [tex]\(\text{Cl}^-\)[/tex]
- [tex]\(\text{SnCl}_6^{2-}\)[/tex]
2. Determine what happens in the reaction:
- [tex]\(\text{SnCl}_4\)[/tex] reacts with 2 [tex]\(\text{Cl}^-\)[/tex] ions to form [tex]\(\text{SnCl}_6^{2-}\)[/tex].
3. Analyze the electron movement:
- [tex]\(\text{Cl}^-\)[/tex] ions each have an extra electron pair that they can donate.
- [tex]\(\text{SnCl}_4\)[/tex] will accept these electron pairs to form coordination bonds with the chloride ions, resulting in the formation of [tex]\(\text{SnCl}_6^{2-}\)[/tex].
4. Classify the species:
- Since [tex]\(\text{Cl}^-\)[/tex] ions donate electron pairs, they act as Lewis bases.
- Since [tex]\(\text{SnCl}_4\)[/tex] accepts electron pairs from the chloride ions, it acts as a Lewis acid.
Therefore, given the balanced equation, [tex]\(\text{SnCl}_4\)[/tex] is the Lewis acid as it accepts electron pairs from the chloride ions, [tex]\(\text{Cl}^-\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{\text{SnCl}_4} \][/tex]