A is incorrect, (1s3?)
I believe bd are all in gs
im not sure about c, but arent all the lone electrons suppose to be moving in one direction? thus the paramagnetism when they are unpaired. in this case, since the electron in the 2py and 2pz orbitals are facing opposite directions, it would cancel each other out (diamagnetism). so i think this is Incorrect.
e is excited, moving from 2p to 3s (higher energy level)
I would say F is incorrect bc i see a 2D subshell, which, unless im mistaken does not exist.

I think only B and D are correct. All the others have at least 1 or 2 problems.
B has n=2, D has n=3. Excited state is anything beside n=1 or ground state so I think both B and D should be the answers.

Therefore, there is no ground state, B and D are excited state, and all the others are incorrect. I bet I am 100 % correct

I think only B and D are correct. All the others have at least 1 or 2 problems.
B has n=2, D has n=3. Excited state is anything beside n=1 or ground state so I think both B and D should be the answers.

Therefore, there is no ground state, B and D are excited state, and all the others are incorrect. I bet I am 100 % correct

Can you please explain why e is incorrect? also, n=2, n=3...that refers to the electron shell right? how did you determine it was 2 and 3? thanx. also, if the OP can post the answers, that would be helpful.

2 and 3 corresponds to the principal quantum number which can be determined by the electronic configuration. What i didnt know was that anything above n=1 was excited. can you please explain this? and also, why is e incorrect, the electron configuration follows all the rules and the filling of the higher energy orbital first indicates excitation. thanx in advance.

Can you please explain why e is incorrect? also, n=2, n=3...that refers to the electron shell right? how did you determine it was 2 and 3? thanx. also, if the OP can post the answers, that would be helpful.

E is incorrect because of 2d
here is the order it is supposed to be
1s2
2s2
2p6
3s2
3p6
4s2
3d( )
4p6

2d doesn't exist.
Kaplan blue book says it doesn't matter if u state 4p before 3d or vice versa.
but anyway, 2d doesn't exist. d starts from 3d.

and if you want more explanation, u should be clear about what's the answer and what's not. because even if i say this, there is no guarantee i am 100% right and i can make things worse if i keep insisting that my wrong answer is correct, if u know what i am saying...

issa, e cannot occur as there is no such thing as 2d. There are only 2s and 2p. Remember the order of filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, etc.

s gets 2, p gets 6, d gets 10, and so on.

e is not an excited state. It's an impossible "state" per the rules.

issa, e cannot occur as there is no such thing as 2d. There are only 2s and 2p. Remember the order of filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, etc.

s gets 2, p gets 6, d gets 10, and so on.

e is not an excited state. It's an impossible "state" per the rules.

lol, but you are wrong regarding question e. e doesn't have 2d, and e needs to fill its 3rd orbital. take a look at the question again and see for yourself. your explanation is correct for question f.

E is incorrect because of 2d
here is the order it is supposed to be
1s2
2s2
2p6
3s2
3p6
4s2
3d( )
4p6

2d doesn't exist.
Kaplan blue book says it doesn't matter if u state 4p before 3d or vice versa.
but anyway, 2d doesn't exist. d starts from 3d.

and if you want more explanation, u should be clear about what's the answer and what's not. because even if i say this, there is no guarantee i am 100% right and i can make things worse if i keep insisting that my wrong answer is correct, if u know what i am saying...

You still havent answered my question regarding e, i know f doesnt make sense bc of 2d orbital. however, why would e be incorrect? i believe e is in its excited state and its filling of its orbital follows every chemical rule. in fact, bc there is a electron moved into a higher energy state leads me to believe it is in its excited state.