4
$\begingroup$

How do I find the set of quantum numbers for a specific electron in an element? For example,

Calculate the set of quantum numbers for the 19th electron in chromium.

The electronic configuration of the 19th electron is 1s2 2s2 2p6 3s2 3p6 4s1. From here on, how do I decide what would the principle quantum number be?

$\endgroup$
  • $\begingroup$ This sounds like a homework problem. Could you please write out your attempt at solving this question? For example, what are the quantum numbers and how do they relate to atomic 'structure'? One example of this would be that the principal quantum number, n, is the electron shell of an atom. $\endgroup$ – LordStryker Apr 29 '14 at 14:03
  • $\begingroup$ the electronic configuration of the 19th electron is 1s2 2s2 2p6 3s2 3p6 4s1. from here on, how do i decide what would the principle quantum number be? @LordStryker it isn't a homework problem.. i'm just trying to clear up my concept which is pretty much messed up :/ i put in the example to make it more clear as to what direction would i like the answer to proceed.. $\endgroup$ – krazkat Apr 29 '14 at 16:48
  • $\begingroup$ Okay so, from your example (1s2 2s2 2p6 3s2 3p6 4s1), the principle quantum number for the 19th electron is what? You have it written right there in your provided electron configuration. N refers to something. How about the angular momentum number (L)? This number refers to the shape of the orbital. S orbitals have different shapes from p orbitals (sphere vs. dumbell). You have the answer written out in your provided configuration. The only other number that remains (if neglecting spin), is the magnetic quantum number(m_l). $\endgroup$ – LordStryker Apr 29 '14 at 18:00
  • $\begingroup$ Note: We're getting closer. Try using this site as a reference. chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html $\endgroup$ – LordStryker Apr 29 '14 at 18:02
  • $\begingroup$ the website was really helpful! :) $\endgroup$ – krazkat Apr 30 '14 at 4:01
3
$\begingroup$

1s2 2s2 2p6 3s2 3p6 4s1 is the electronic configuration of the 19th electron of chromium.

since the last shell is 4, therefore n=4,

l=0(as there is only one orbital in the s sub shell) ,

m=0 (since we're still in the s sub shell only)

and s=+1/2 or s=-1/2(arbitrarily assigned values)

note:+1/2 or -1/2 are used to distinguish between electrons as no two electrons can have the same electronic configuration, but here the question isn't that specific, so we need not bother.

$\endgroup$
  • $\begingroup$ Electron spin is not necessarily arbitrary in the absence of another electron. Interestingly, the nucleus is a point of reference for spin to be defined and there is a very real, physical consequence with respect to which spin the electron has. This comment is included just for fun. $\endgroup$ – LordStryker Apr 30 '14 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.