Final answer:
Bromine and sodium exchange electrons in a redox reaction due to their significant difference in electronegativity, resulting in sodium being oxidized and bromine being reduced to form sodium bromide.
Step-by-step explanation:
When bromine and sodium react, electrons are indeed exchanged between the two atoms due to the significant difference in their electronegativity values. Bromine has an electronegativity value of 2.96, which is much higher than sodium's value of 0.93. In this chemical reaction, sodium, with a lower electronegativity, tends to lose its valence electron, while bromine, with a higher electronegativity, has a tendency to gain electrons.
Since sodium has only one valence electron and bromine has seven valence electrons, there is a strong drive for an electron from sodium to transfer to bromine. This results in sodium increasing its oxidation number from 0 to +1 as it is oxidized, and bromine decreasing its oxidation number from 0 to -1 as it is reduced. The product formed is sodium bromide (NaBr), which is an ionic compound due to this transfer of electrons. Therefore, this chemical interaction between sodium and bromine is classified as a redox reaction.