GREPhysics.NET: GR9277 #58

GR9277 #58

Problem

GREPhysics.NET Official Solution

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

$1s^22s^22p^53s^2$
$1s^22s^32p^63$
$1s^22s^22p^63s$
$1s^22s^22p^63p$
$1s^22s^2sp^5$

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

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Comments

hoyas08
2008-06-17 15:48:49

Choice (B) should read: 1 $s^2$ 2 $s^3$ 2 $p^6$

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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 $^2$ .

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$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
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$\vec{E}$	$\vec{E}$
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