Question

How do you write the complete and abbreviated ground state electron configuration for sulfur?

Answer 1

Hint: To compose the Sulfur electron arrangement, we first need to know the quantity of electrons for the S iota (there are 16 electrons). At the point when we compose the design we'll place every one of the 16 electrons in orbitals around the core of the Sulfur particle.

Complete answer:
At. Number of Sulphur = 16
No. of electron in 1st orbit = 2 (Max. Capacity is 2)
No. of electrons in 2nf orbit = 8 (Max. Capacity is 8)
No. of electrons in 3rd orbit = 16-(2+8) = 6 (Max. Capacity is 8)
1st orbit has only 1s orbital. So, electronic configuration is
The 2nd orbit has 2s and 2p orbitals. As we know s orbital can hold max. 2 electrons and p orbital can hold max. 6 electrons, so the configuration is $2{p^6}$.
The 3rd orbit has 3s, 3p and 3d orbitals. As Energy of 3s < Energy of 3p < Energy of 3d So, 3s is filled at first followed by 3p and 3d as per Aufbau's rule. So, electronic configuration is \[3{p^4}\]. As all the electrons are accommodated within 3p so, 3d will remain empty. Moreover, in a 3p orbital all the 3 degenerate orbitals get 1 electron at first and then pairing occurs according to Hunds' rule.

Note: Recorded as a hard copy the electron arrangement for Sulfur the initial two electrons will go during the 1s orbital. Since 1s can just hold two electrons the following 2 electrons for sulfur go during the 2s orbital. The following six electrons will go in the 2p orbital.