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issa

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Circle g.s. [ground state]; e.s. [excited state]; or i [incorrect]


Which one of these in the excited state and why?
 

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tom_servo_dds

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issa said:
Circle g.s. [ground state]; e.s. [excited state]; or i [incorrect]


Which one of these in the excited state and why?

a) i - no 1s3
b) g
c) i - two problems
d) g
e) e - not the correct filling order
f) i - should fill all d subshells first
 

issa

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whiskeysour said:
wrong, it is the correct filling order.

P has a 5 not 6 in part e. is e in the excited state or not? and anyone have the correct explanation?
 

Envision

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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.
 

joonkimdds

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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 :)
 

creestoL

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joonkimdds said:
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.
 
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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.
 

joonkimdds

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creestoL said:
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...

plz tell me what the answer is.
 

issa

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i don't have the answers for these questions. is e in excited mode or what?
 

aranjuez

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f 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.

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

aranjuez

NOTE: editted after it was brought to my attention that my explanation was regarding (f) and not (e). My mistake. Corrected.
 

joonkimdds

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aranjuez said:
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.

aranjuez


Yay~ so i was correct!
Community College Rules!
Woo HOo~~!
 

issa

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aranjuez said:
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.

aranjuez

you are talking about f not e. f is incorrect. can you give me an example of an excited state configuartion?
 

issa

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joonkimdds said:
Yay~ so i was correct!
Community College Rules!
Woo HOo~~!

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.
 

joonkimdds

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aranjuez said:
Wupps! Sorry about that. You're right. It was for f.

aranjuez


same here~ lol but it's becuz u linked the photo so whenever i click it the entire window goes to that link and i have to press back and so on.

if i quote the problem, there wasn't this mistake :laugh:

anyway, I still believe that anything beside n=1 is excited state becuz n=1 is called ground state.

at least that's what i read from the book.
 

dat_student

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joonkimdds said:
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...

plz tell me what the answer is.

issa said:
Circle g.s. [ground state]; e.s. [excited state]; or i [incorrect]


Which one of these in the excited state and why?

e is the correct answer without a doubt :)

Ground state:
1s2 2s2 2p6 3s1

Excited state:
1s2 2s2 2p5 3s2
(one of the electrons in one of the p suborbitals is excited and moved to the next shell (i.e. shell #3, orbital S))
 

Envision

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joonkimdds said:
same here~ lol but it's becuz u linked the photo so whenever i click it the entire window goes to that link and i have to press back and so on.

if i quote the problem, there wasn't this mistake :laugh:

anyway, I still believe that anything beside n=1 is excited state becuz n=1 is called ground state.

at least that's what i read from the book.

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.
 

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dat_student said:
e is the correct answer without a doubt :)

Ground state:
1s2 2s2 2p6 3s1

Excited state:
1s2 2s2 2p5 3s2
(one of the electrons in one of the p suborbitals is excited and moved to the next shell (i.e. shell #3, orbital S))

i agree. DAT student, what about b and d? i was under the impression that they are in gs. what do you think?
 

R.L.H

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Envision said:
i agree. DAT student, what about b and d? i was under the impression that they are in gs. what do you think?

I agree, B & D are 100% gs. The answer is E.
 
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