What is the most likely binary compound that would form between Se and Br
The most likely binary compound that would form between Se and Br is selenium bromide, which has the chemical formula SeBr2.
To determine the binary compound that would form between selenium (Se) and bromine (Br), we need to consider the charges of each element.
Selenium belongs to Group 16 (also known as Group VIA) on the periodic table, so it has a valence electron configuration of 2-8-6. It tends to gain two electrons to achieve a stable noble gas configuration. Therefore, selenium will have a charge of -2 in a compound.
Bromine belongs to Group 17 (also known as Group VIIA), so it has a valence electron configuration of 2-8-7. It tends to gain one electron to achieve a stable noble gas configuration. Therefore, bromine will have a charge of -1 in a compound.
When determining the formula for the compound formed, we need to balance the charges so that the overall compound is electrically neutral. Since selenium has a charge of -2 and bromine has a charge of -1, we need two bromine atoms to balance the charge of one selenium atom.
Therefore, the most likely binary compound that would form between selenium (Se) and bromine (Br) is selenium dibromide, which has the chemical formula SeBr2.
To determine the most likely binary compound that would form between Selenium (Se) and Bromine (Br), we need to consider their respective charges or oxidation states.
Selenium can have variable oxidation states, including -2, +2, +4, +6, while Bromine usually has an oxidation state of -1.
To form a compound, the total charge of the compound must be zero. This means that the sum of the charges of the atoms involved should equal zero.
Considering this, the most likely binary compound that would form between Se and Br would be Selenium Bromide with the chemical formula SeBr2.
Here is the reasoning behind it:
- Since Bromine has an oxidation state of -1 and we have only one Bromine atom, the total charge contributed by Bromine is -1.
- To achieve a total charge of zero in the compound, Selenium must have an oxidation state of +2 because there are two Bromine atoms in SeBr2. The total charge contributed by Selenium is +4 (2 x +2).
- The sum of the charges of the atoms in SeBr2 (4 + (-1) + (-1)) equals zero, satisfying the charge balance requirement.
Therefore, the most likely binary compound that would form between Selenium (Se) and Bromine (Br) is Selenium Bromide (SeBr2).