When writing a possible excited state electron configuration there are two rules, which are?
1)Oxygen atoms have 8 electrons and the shell structure is 2.6. The ground state electron configuration of ground state gaseous neutral oxygen is [He]. 2s2. 2p4 and the term symbol is 3P2.
2)If the element were to become excited, the electron could occupy an infinite number of orbitals. However, in most texts the example will be the next available one. So for oxygen, it might look like this: 1s22s22p33s1 - where the valence electron now occupies the 3s orbital in an excited (i.e. not ground) state
Well ik its not math for this but for the first one if u add
22+22+13+1 u get 58
then second one 1+22+22+2=47
U notice how that most or some are all going up to 50 except the third one
So that means its 1s22s22p24s1
If an element (abbreviated X) has a ground state electron configuration of (Ne)3s23p4, then the formula of the compound that X forms with sodium is Na2X. True
The element with a ground state electron configuration of 1s22s22p63s2 is an alkaline earth metal. True
The element with a ground state electron configuration of (Kr)5s24d105p5 has 15 valence electrons. True
The ground state electron configuration for Cu is (Ar)4s13d10. True
The ground state electron configuration for the undiscovered element 115 is (Rn)7s25f146d107p3. Is not the correct electron configuration for the element moscovium. It is no longer undiscovered but is now well known. The correct electron configuration is Moscovium: [Rn]5f14 6d10 7s2 7p3 OR
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 5f14 6d10 7s2 7p3.
The electron configuration of 1s22s22p63s23p64s13d8 is an excited state electron configuration for Ni.. This statement has to be false because the excited electron must be promoted to a higher energy level. Which was not reflected in the configuration written above.