Answer:
i) 1s²2s²2p⁶3s²3p³
ii) [Ne]3s²3p³
iii) ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑
1s² 2s² 2p⁶ 3s² 3p³
Explanation:
i) The phosphorus, with atomic number (Z) 15, has the following electron configuration:
Z=15 → Number of electrons = 15
1s²2s²2p⁶3s²3p³
ii) The noble gas configuration of the phosphorus is the next:
[Ne]3s²3p³
Neon is the noble gas that precedes phosphorus, its atomic number is 10. Therefore, the first ten electrons in the electronic configuration of P correspond to the noble gas Ne.
iii) The orbital configuration of the P is given by the spins of the electrons in each orbital:
↑↓
1 s²
↑↓ ↑↓ ↑↓ ↑↓
2 s² p⁶
↑↓ ↑ ↑ ↑
3 s² p³
Where ↑ and ↓ represents electron spins.
We can consider every orbital as a box that contains the electrons, and using the Pauli exclusion principle we can fill the boxes with the electrons.
So, following that principle, we have that every box will have a maximum of two electrons, which are represented as ↑↓, when they are paired up, and ↑ when they are not paired up.
In phosphorous case, every s orbital contain 2 electrons represented as ↑↓, and every p orbital has three orbitals: [tex]p_{x}[/tex], [tex]p_{y}[/tex] and [tex]p_{z}[/tex], for a total of 6 electrons (two electrons for each orbital), except for the 3p orbital, which has only 3 electrons, represented as ↑ ↑ ↑.
I hope it helps you!
