GREPhysics.NET: GR8677 #30

GR8677 #30

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Atomic $\Rightarrow$ }Orbitals

Recall that:

$\begin{eqnarray} s&=&0\\ p&=&1\\ d&=&2\\ f&=&3\\ g&=&4...(etc.),<br /> \end{eqnarray}$

where the right side of the equal sign is the angular momentum $l$ .

And then there's also the formula $2(2l+1)$ , which determines the number of electrons in each subshell.

(A) The configuration does not specify anything about $3d$ . This is the first row of transitional metals.

(B) From the formula above, one sees that the $s$ subshell needs $2$ electrons to be filled. The $4s$ subshell is thus not filled.

(C) $l=4$ corresponds to $g$ . The configuration does not specify $g$ .

(D) Because the problem states that potassium is in the ground state, the atomic number is the same as the number of electrons in the configuration. The sum of the superscripts is 19.

(E) The valance shell is $4s^2$ . Many-electron atoms have wave functions that take after the hydrogen atom, and one recalls that $\psi_{40m}$ looks like a spherically symmetrical blob of a ball. This is it.

Alternate Solutions

a_coiled_atom
2011-07-10 12:58:10

Another way of thinking about it that might be helpful is that , in general, $\ell=0\rightarrow$ no orbital angular momentum $\rightarrow$ spherical symmetry.

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Comments

joshuaprice153
2019-08-09 02:49:54

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a_coiled_atom
2011-07-10 12:58:10

Another way of thinking about it that might be helpful is that , in general, $\ell=0\rightarrow$ no orbital angular momentum $\rightarrow$ spherical symmetry.

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suvas
2010-09-09 16:23:08

I would choose E. The electronic charge distribution of atom is determined mainly by valence electron.

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wittensdog
2009-07-25 17:54:02

ETS is good at thinking of traps. I almost wanted to go for C. I saw the 4s1 electron being the least tightly bound, and in my mind a connection was made with the number 4. Clearly, l = 4 is supposed to trick you, when of course the electron actually has n = 4, not l = 4. It's value for orbital angular momentum is l = 0, of course.

jmason86
2009-08-16 16:36:12

I must have fallen for a trap too. I chose (A) immediately and without hesitation.
I was thinking that the n=3 shell only means the s-subshell. So when I saw that we were given $3s^2$ , I assumed that (A) was correct. However, the n=3 is a total shell consisting of the s, p, and d subshells... If (A) were true then we would be talking about Zinc with [Ar] $4s^23d^1^0$ .

neon37
2010-11-01 09:56:59

same here @ wittendog. Ofcourse reading E, changed made me realize my mistake.

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kyros
2007-10-30 21:42:24

Shouldn't that be Sigma_4,0,0, since m can only range from -l to +l anyway?

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simpsoxe
2006-11-30 21:20:14

well the valence shell is $4s^1$ , but my question is about the other electrons in the lower shells. don't we have to consider them as part of the electron distribution? or do we assume they all have spherically symmetric distributions? (and why would that be?)

Blake7
2007-07-23 18:26:38

The lower shells all appear to be completely filled

2 2 6 2 6

if I'm not mistaken that will make them all symmetric

kronotsky
2018-10-23 03:51:23

Filled subshells are the singlet component of the direct product of the component orbitals, so they have zero orbital angular momentum. So does 4s. The rotation operator(s) is an exponent of the orbital angular momentum operator, so it is equal to 1 (i.e. rotational symmetry).

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Another way of thinking about it that might be helpful is that , in general, $\ell=0\rightarrow$ no orbital angular momentum $\rightarrow$ spherical symmetry.

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