Answer:
1[tex]s^{2}[/tex]2[tex]s^{2}[/tex]2[tex]p^{6}[/tex]3[tex]s^{2}[/tex]3[tex]p^{6}[/tex]3[tex]d^{10}[/tex]4[tex]s^{2}[/tex]4[tex]p^{6}[/tex] (alternatively 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6)
Explanation:
The correct configuration for bromide ([tex]Br^{-}[/tex]) is:
1[tex]s^{2}[/tex]2[tex]s^{2}[/tex]2[tex]p^{6}[/tex]3[tex]s^{2}[/tex]3[tex]p^{6}[/tex]3[tex]d^{10}[/tex]4[tex]s^{2}[/tex]4[tex]p^{6}[/tex] (it has one more electron in 4p orbital than bromine)
You can also write it as [Ar] [tex]4s^{2}[/tex] [tex]3d^{10}[/tex] [tex]4p^{6}[/tex], since noble gases are very stable
First can't be correct because there is nothing in 3p orbital and 4p has only 5 electrons here
1s² 2s2 2p6 3s2 3p 4s2 3d10 4p5
Second can't be correct because there is nothing in 2p and 4p orbitales 1s2 2s² 2p 3s2 3p6 4s2 3d10 4p
Third can't be correct because there is nothing in 3p orbital and 4p has only 5 electrons here
1s 2s2 2p6 3s2 3p' 4s2 3d10 4p5
Fourth can't be correct because there is nothing in 2p and 3d orbitales, and 4p has only 5 electrons here
1s2 2s2 2p 3s2 3p6 4s2 3dº 4p5
