Answer :
To determine the electron configuration for bromine, we need to know its atomic number, which is 35. This means bromine has 35 electrons. We will fill the electron orbitals according to the Aufbau principle, which states that electrons occupy the lowest energy orbitals first.
Let’s go through each option:
1. Option 1: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^5 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 10 (3d) + 5 (4p) = 35
This adds up to 35, the correct number of electrons for bromine. Configuration also follows the correct order of filling orbitals.
2. Option 2: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 10 (3d) + 6 (4p) = 36
This configuration has 36 electrons, which is one more than the atomic number of bromine.
3. Option 3: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4 \, 4s^2 \, 3d^{10} \, 4p^5 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 4 (3p) + 2 (4s) + 10 (3d) + 5 (4p) = 33
This configuration has only 33 electrons, which is two fewer than needed for bromine.
4. Option 4: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^9 \, 4p^5 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 9 (3d) + 5 (4p) = 34
This configuration has only 34 electrons, which is one fewer than needed for bromine.
Hence, the correct electron configuration for bromine is:
[tex]\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^5 \][/tex]
So, the correct answer is the first option.
Let’s go through each option:
1. Option 1: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^5 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 10 (3d) + 5 (4p) = 35
This adds up to 35, the correct number of electrons for bromine. Configuration also follows the correct order of filling orbitals.
2. Option 2: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 10 (3d) + 6 (4p) = 36
This configuration has 36 electrons, which is one more than the atomic number of bromine.
3. Option 3: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4 \, 4s^2 \, 3d^{10} \, 4p^5 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 4 (3p) + 2 (4s) + 10 (3d) + 5 (4p) = 33
This configuration has only 33 electrons, which is two fewer than needed for bromine.
4. Option 4: [tex]\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^9 \, 4p^5 \)[/tex]
Total number of electrons = 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 6 (3p) + 2 (4s) + 9 (3d) + 5 (4p) = 34
This configuration has only 34 electrons, which is one fewer than needed for bromine.
Hence, the correct electron configuration for bromine is:
[tex]\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^5 \][/tex]
So, the correct answer is the first option.
