Question

Identify what is wrong with each electron configuration and write the correct ground state (or lowest energy) based on the number of electrons. Part A Part complete 1s42s42p12 Can only have 2 in s shell and 6 in the p shell: 1s22s22p63s23p64s24p6 Can only have 2 in s shell and 6 in the p shell: 1s22s22p6 Can only have 2 in s shell and 6 in the p shell: 1s22s22p63s23p64s2 Can only have 2 in s shell and 6 in the p shell: 1s22s22p63s23p64s14p1 Previous Answers Correct Part B 1s22s22p63s23p64s2 Use the buttons at the top of the tool to add orbitals. Click within the orbital to add electrons.

Answer 1

s orbital can have maximum of 2 electrons.

p orbital can have maximum 6 electrons.

Part A

(a) 1s⁴ 2s⁴ 2p¹² is wrong electronic configuration. Correct one is: 1s² 2s² 2p⁶

(b)  1s² 2s² 2p⁶3s² 3p⁶ 4s² 4p⁶ is wrong electronic configuration. 3d- orbital has lower energy than 4p orbital. SO, 3d orbital will be filled before 4p orbital. So,Correct electronic configuration is :1s² 2s² 2p⁶3s² 3p⁶ 4s² 3d⁶

(c)1s² 2s² 2p⁶ is correct electronic configuration

(d)  1s² 2s² 2p⁶3s² 3p⁶ 4s¹ 4p¹ is a  is wrong electronic configuration. 4s orbital has lower energy than 4p orbital. Lower energy orbital will be filled first and then higher energy level will be filled. So,Correct electronic configuration is :1s² 2s² 2p⁶3s² 3p⁶ 4s²

Part B