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Sagot :
To determine the correct electron configuration for bromine, let's review some fundamental concepts regarding how electrons are configured around an atom.
1. Atomic Number and Electron Configuration:
- Bromine has an atomic number of 35, which means it has 35 electrons.
2. Order of Filling Orbitals:
- Electrons fill orbitals in an order dictated by the Aufbau principle, where orbitals are filled in order of increasing energy levels.
- The order of filling is typically: \(1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, \ldots\)
3. Filling the Orbitals:
- We will fill the orbitals according to the number of electrons each can hold:
- \(1s\) holds 2 electrons: \(1s^2\)
- \(2s\) holds 2 electrons: \(2s^2\)
- \(2p\) holds 6 electrons: \(2p^6\)
- \(3s\) holds 2 electrons: \(3s^2\)
- \(3p\) holds 6 electrons: \(3p^6\)
- \(4s\) holds 2 electrons: \(4s^2\)
- \(3d\) holds 10 electrons: \(3d^{10}\)
- \(4p\) needs to accommodate the remaining 5 electrons: \(4p^5\)
4. Combining These into the Full Electron Configuration:
- Adding the electrons in the respective orbitals:
- \(1s^2\) : 2 electrons
- \(2s^2\) : 2 electrons
- \(2p^6\) : 6 electrons
- \(3s^2\) : 2 electrons
- \(3p^6\) : 6 electrons
- \(4s^2\) : 2 electrons
- \(3d^{10}\) : 10 electrons
- \(4p^5\) : 5 electrons
- Summing these gives \(2 + 2 + 6 + 2 + 6 + 2 + 10 + 5 = 35\) electrons.
Therefore, 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]
Now, let's match this with the choices given in the question:
1. \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \)
2. \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 \)
3. \( 1s^2 2s^2 2p^6 3s^2 3p^4 4s^2 3d^{10} 4p^5 \)
4. \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9 4p^5 \)
The correct choice is:
[tex]\[ 1. \; 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \][/tex]
1. Atomic Number and Electron Configuration:
- Bromine has an atomic number of 35, which means it has 35 electrons.
2. Order of Filling Orbitals:
- Electrons fill orbitals in an order dictated by the Aufbau principle, where orbitals are filled in order of increasing energy levels.
- The order of filling is typically: \(1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, \ldots\)
3. Filling the Orbitals:
- We will fill the orbitals according to the number of electrons each can hold:
- \(1s\) holds 2 electrons: \(1s^2\)
- \(2s\) holds 2 electrons: \(2s^2\)
- \(2p\) holds 6 electrons: \(2p^6\)
- \(3s\) holds 2 electrons: \(3s^2\)
- \(3p\) holds 6 electrons: \(3p^6\)
- \(4s\) holds 2 electrons: \(4s^2\)
- \(3d\) holds 10 electrons: \(3d^{10}\)
- \(4p\) needs to accommodate the remaining 5 electrons: \(4p^5\)
4. Combining These into the Full Electron Configuration:
- Adding the electrons in the respective orbitals:
- \(1s^2\) : 2 electrons
- \(2s^2\) : 2 electrons
- \(2p^6\) : 6 electrons
- \(3s^2\) : 2 electrons
- \(3p^6\) : 6 electrons
- \(4s^2\) : 2 electrons
- \(3d^{10}\) : 10 electrons
- \(4p^5\) : 5 electrons
- Summing these gives \(2 + 2 + 6 + 2 + 6 + 2 + 10 + 5 = 35\) electrons.
Therefore, 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]
Now, let's match this with the choices given in the question:
1. \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \)
2. \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 \)
3. \( 1s^2 2s^2 2p^6 3s^2 3p^4 4s^2 3d^{10} 4p^5 \)
4. \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9 4p^5 \)
The correct choice is:
[tex]\[ 1. \; 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 \][/tex]
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