In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.
Let's consider the following balanced redox reaction.
2 Sc + 3 Br₂ ⟶ 2 ScBr₃
We can identify both half-reactions.
Oxidation: 2 Sc ⟶ 2 Sc⁺³ + 6 e⁻
Reduction: 6 e⁻ + 3 Br₂ ⟶ 6 Br⁻
As we can see, 6 electrons are involved in the formation of 2 formula units of ScBr₃. Thus, 3 electrons are involved in the formation of 1 formula unit of ScBr₃.
In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.
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