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GR9277 #58
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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$\Rightarrow$}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.

Alternate Solutions
 ETScustomer2017-09-23 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.Reply to this comment
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 C
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.
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?
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.
 mpdude82012-04-20 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.
hoyas08
2008-06-17 15:48:49
Choice (B) should read: 1$s^2$2$s^3$2$p^6$
neutrino
2007-11-02 10:27:37

$3s=3s^(1)$
?

Because else choice C does not add to 11 electrons (2+2+6=10)
 neutrino2007-11-02 10:29:06 oeps, latex does not catch my brackets: $3s^1$ it should be...
 evanb2008-06-23 19:40:06 Yes, 3s should be read as 3s$^1$. Otherwise, you wouldn't write it at all, or you would specify 3s$^2$.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$