write down the electronic configuration of Ca and Fe
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Answered by
4
Ca - 2,8,8,2
And
Fe - 2,8,8,6
And
Fe - 2,8,8,6
Answered by
4
Iron is on the fourth row of the periodic table, sixth column of the transition metals, atomic number 26.
What we have is:
Its core orbitals are the 1s, 2s, 2p's, 3s, and 3p's.
Its valence orbitals are the 4s and 3d's.
Writing the electron configuration, you really only need the valence orbitals, and you can omit the core orbitals by notating it via the noble gas shortcut.
So, [Ar] can be written instead of 1s22s22p63s23p6.
Note that the total number of electrons in the neutral atom adds up to the atomic number, so 2+2+6+2+6=18, which is the atomic number of Ar.
Finally, the 4s orbitals are higher in energy than the 3d orbitals by about 3.75 eV when they are partially filled like this, so we write the 3d before the 4s.
Therefore, we get:
[Ar]3d64s2v
CALCIUM...
In writing the electron configuration for Calcium the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Calcium go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the next six electrons. We now shift to the 4s orbital where we place the remaining two electrons. Therefore the Calcium electron configuration will be 1s22s22p63s23p64s2.
What we have is:
Its core orbitals are the 1s, 2s, 2p's, 3s, and 3p's.
Its valence orbitals are the 4s and 3d's.
Writing the electron configuration, you really only need the valence orbitals, and you can omit the core orbitals by notating it via the noble gas shortcut.
So, [Ar] can be written instead of 1s22s22p63s23p6.
Note that the total number of electrons in the neutral atom adds up to the atomic number, so 2+2+6+2+6=18, which is the atomic number of Ar.
Finally, the 4s orbitals are higher in energy than the 3d orbitals by about 3.75 eV when they are partially filled like this, so we write the 3d before the 4s.
Therefore, we get:
[Ar]3d64s2v
CALCIUM...
In writing the electron configuration for Calcium the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Calcium go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the next six electrons. We now shift to the 4s orbital where we place the remaining two electrons. Therefore the Calcium electron configuration will be 1s22s22p63s23p64s2.
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