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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: unsu on August 14, 2014, 01:09:47 PM

Title: Term symbol for Cr
Post by: unsu on August 14, 2014, 01:09:47 PM
Does anyone know how to derive a term symbol for a Cr atom?
Because its configuration is 4s(1)3d(5)
To get a term symbol we must consider a sub-shell with a specific value of l.
In Cr atom there are two unoccupied orbitals with l=0 and l=1
So, any idea how to derive a term symbol for Cr?
Title: Re: Term symbol for Cr
Post by: Corribus on August 14, 2014, 03:33:39 PM
The same way you do any other electron configuration. Are you familiar with how to determine term symbols - generally?
Title: Re: Term symbol for Cr
Post by: unsu on August 14, 2014, 05:52:10 PM
oookk... then:

4s(1)3d5
ML = 0 + 2 + 1 + 0 + (-1) + (-2) = 0
L = S
max Ms = 6*(1/2) = 3
S = 3
Ground state configuration term symbol = 7S

Am I correct? :)
Title: Re: Term symbol for Cr
Post by: Corribus on August 14, 2014, 06:16:48 PM
This is what I got as well. Now can you determine the rest of them? :)
Title: Re: Term symbol for Cr
Post by: unsu on August 14, 2014, 06:58:51 PM
if we allow the electrons to pair up, even on 4s orbital,
then there are 12 positions for 6 electrons, there must be 924 micro states  >:D

if we can do whatever we want with d electrons but leave one electron on 4s orbital, then there are 2x256=504 micro states (2 because 4s :spinup: and 4s  :spindown:)

no pairing = 26 = 64 microstates

plus different J values  :o  :o  :o
Title: Re: Term symbol for Cr
Post by: Corribus on August 14, 2014, 09:20:00 PM
if we can do whatever we want with d electrons but leave one electron on 4s orbital, then there are 2x256=504 micro states (2 because 4s :spinup: and 4s  :spindown:)
Well if you don't adhere to this restriction, then you're in another electron configuration, so...

I went through it quickly and (excluding J states) I found the following: 7S, 3I, 1I, 3H, 1H, 5G, 5F, 5D, 5P, 3x3G, 3x3F, 3x3D, 2x3P, 2x1G, 4x1D, 1P.

I wonder if I'm right. Maybe we'll never know. :)