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  GR9277 #58
Problem
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\prob{58}
The ground state configuration of a neutral sodium atom (Z=11) is

  1. $1s^22s^22p^53s^2$
  2. $1s^22s^32p^63$
  3. $1s^22s^22p^63s$
  4. $1s^22s^22p^63p$
  5. $1s^22s^2sp^5$
Atomic}Orbitals

Eliminate (E) immediately since the superscripts do not add to 11. Each superscript stands for an electron.

Eliminate (B) because the s orbital can only carry 2 electrons.

Ground state means none of the electrons are promoted, and there are no states with unfilled gaps in them.

Eliminate (A) since it promotes the 2p electron to 3s, leaving a unfilled orbital of lower energy.

Eliminate (D) since it promotes the 3s electron to 3p, leaving an empty orbital of lower energy.

Choice (C) is it.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
ETScustomer
2017-09-23 18:07:55		For those who would rather not use the method of elimination:\r\n\r\nFirst use up two (two because spin up and spin down are different states) electrons to fill the (n,l)=(1,0). These two electrons give us the 1s^2. Then, use up another two electrons to get the 2s^2 for (n,l)=(2,0). Then, use up six more electrons to fill the (n,l)=(2,1), and that gives the 2p^6. We\'ve used up now ten electrons, so place the final electron in the (n,l)=(3,0) shell to get a 3s^1. Altogether, this state is notated as 1s^22s^22p^63s^1.\r\n\r\nFor higher atomic number atoms, the way of filling the shells to get the ground state is a little more elaborate than the pattern that I used here might suggest. See https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule for more about this.Alternate Solution - Unverified
Comments
deneb
2018-10-13 23:23:14		If you know enough about orbitals to eliminate 4 answers, it would be way quicker to just figure out the correct configuration for 11 electrons and pick CNEC
ETScustomer
2017-09-23 18:07:55		For those who would rather not use the method of elimination:\r\n\r\nFirst use up two (two because spin up and spin down are different states) electrons to fill the (n,l)=(1,0). These two electrons give us the 1s^2. Then, use up another two electrons to get the 2s^2 for (n,l)=(2,0). Then, use up six more electrons to fill the (n,l)=(2,1), and that gives the 2p^6. We\'ve used up now ten electrons, so place the final electron in the (n,l)=(3,0) shell to get a 3s^1. Altogether, this state is notated as 1s^22s^22p^63s^1.\r\n\r\nFor higher atomic number atoms, the way of filling the shells to get the ground state is a little more elaborate than the pattern that I used here might suggest. See https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule for more about this.Alternate Solution - Unverified
diggydax
2015-03-23 12:27:14		I'm a little confused about notation in chemistry. In Quantum Mechanics, the notation of (2s+1)^L_(j) is used where L is a place holder for the total angular momentum quantum number (l=0 ->S, l=1->P, l=2 ->D, etc). How does 2s+1 relate to the proton number, in this case Z=11. The solution says to sum up the superscripts to see which add up to 11 but why?Help
blackerester
2012-04-16 22:24:54		This is learned in chemistry isn't it? I'm an astrophysics major and chemistry is not a requirement so all the atomic orbital and spectroscopic notation questions are foreign to me.
mpdude8
2012-04-20 00:00:06		 It's definitely taught in basic chemistry, but it's also very relevant to Quantum Mechanics and Electromagnetism. I remember discussing this topic in my first E&M class when going over paramagnetism/ferromagnetism.

In fact, this atomic and spectroscopic stuff actually originates from Quantum Mechanics (spin, angular momentum, etc.)... introductory college chemistry courses just simplify it tremendously.
NEC
hoyas08
2008-06-17 15:48:49		Choice (B) should read: 1s^22s^32p^6Typo Alert!
neutrino
2007-11-02 10:27:37		should we read;

3s=3s^(1)
?

Because else choice C does not add to 11 electrons (2+2+6=10)
neutrino
2007-11-02 10:29:06		 oeps, latex does not catch my brackets:

3s^1

it should be...
evanb
2008-06-23 19:40:06		 Yes, 3s should be read as 3s^1. Otherwise, you wouldn't write it at all, or you would specify 3s^2.
NEC

Post A Comment!
You are replying to:
For those who would rather not use the method of elimination:\r\n\r\nFirst use up two (two because spin up and spin down are different states) electrons to fill the (n,l)=(1,0). These two electrons give us the 1s^2. Then, use up another two electrons to get the 2s^2 for (n,l)=(2,0). Then, use up six more electrons to fill the (n,l)=(2,1), and that gives the 2p^6. We\'ve used up now ten electrons, so place the final electron in the (n,l)=(3,0) shell to get a 3s^1. Altogether, this state is notated as 1s^22s^22p^63s^1.\r\n\r\nFor higher atomic number atoms, the way of filling the shells to get the ground state is a little more elaborate than the pattern that I used here might suggest. See https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule for more about this.

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