What is the electron configuration for Oxygen?
Answer
586.5k+ views
Hint: Electron configuration system is the arrangement of electrons of a molecule or an atom in their orbitals, shells, subshells. It gives us an idea about the number of valence electrons that particular atom or molecule has.
Complete answer:
To deal with this question we have to consider the electron configuration system and the rules governing it.
So, basically there are three main and common rules that govern this system and they are – Aufbau principle, Pauli-exclusion principle and Hund’s rule .
Principal quantum number is given to shells.
And, Azimuthal quantum number \[(n)\] is assigned to the subshells \[(l)\]\[ = \] \[(s,p,d,f)\]
\[
n = 0 = s \\
n = 1 = p \\
n = 2 = d...and{\text{ so on }} \\
\]
This arrangement is in increasing order of their energies and the orbitals are filled in the increasing order of their energies only , such as first electron will go in the s orbital them in p then d and so on.
\[1s < 2s < 2p < 3s < 3p < 4s\]
So, in case of oxygen
The atomic number of oxygen is \[8\]
So, its electron configuration will be, \[O(8) = 1{s^2}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{4}}}\]
The electron in the last shell of a particular atom is the valence electron, these are the electrons that jump out first on excitation.
Note:
Energies of the orbitals are calculated by adding their azimuthal quantum number \[(n)\] and subshells\[(l)\]. Like, for \[1s\] energy will be \[(1 + 0 = 1)\] as s orbital has value. For 2p orbital our energy will be \[(2 + 1 = 3)\] as the value of p is \[1\] . Like this way the energies are counted for each orbital and electrons are filled accordingly.
Complete answer:
To deal with this question we have to consider the electron configuration system and the rules governing it.
So, basically there are three main and common rules that govern this system and they are – Aufbau principle, Pauli-exclusion principle and Hund’s rule .
Principal quantum number is given to shells.
And, Azimuthal quantum number \[(n)\] is assigned to the subshells \[(l)\]\[ = \] \[(s,p,d,f)\]
\[
n = 0 = s \\
n = 1 = p \\
n = 2 = d...and{\text{ so on }} \\
\]
This arrangement is in increasing order of their energies and the orbitals are filled in the increasing order of their energies only , such as first electron will go in the s orbital them in p then d and so on.
\[1s < 2s < 2p < 3s < 3p < 4s\]
So, in case of oxygen
The atomic number of oxygen is \[8\]
So, its electron configuration will be, \[O(8) = 1{s^2}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{4}}}\]
The electron in the last shell of a particular atom is the valence electron, these are the electrons that jump out first on excitation.
Note:
Energies of the orbitals are calculated by adding their azimuthal quantum number \[(n)\] and subshells\[(l)\]. Like, for \[1s\] energy will be \[(1 + 0 = 1)\] as s orbital has value. For 2p orbital our energy will be \[(2 + 1 = 3)\] as the value of p is \[1\] . Like this way the energies are counted for each orbital and electrons are filled accordingly.
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