Tuesday, April 16, 2019

physical chemistry - How does one find the ground-state term symbol for a configuration that is exactly half-filled?



For instance, the ground-state configuration of N atom is a p3 configuration of all parallel spins and one electron in each 2p orbital, which has:



  • Total spin angular momentum, S=312=32

  • Spin multiplicity, 2S+1=2(32)+1=4

  • Total orbital angular momentum, L=(1)+(0)+(1)=0


and so, J=|LS|,|LS+1|,...,|L+S1|,|L+S|=12,32. Thus, I've found that the two candidates are 4S1/2 and 4S3/2.


But my textbook (Pilar, from the '60s!) says that the ground state is 4S3/2. How could I figure something like this out?


Hund's Rules from my text (as far as I can tell---this text is hard for me to read) don't specify how we treat exactly half-filled subshells, only less-than-half-filled, or more-than-half-filled, assuming equal Lmax and Smax already.


(As a side note, my usual McQuarrie textbook is not with me this week, so I cannot simply refer to McQuarrie.)




Answer



The full calculation is laid out below. Start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number s=1/2 and magnetic quantum number ms=±1/2. Orbitals have angular momentum of s=0,p=1,d=2,f=3 etc.


The total spin angular momentum is the series of values S=|s1+s2||s1s2| where s1,s2,s3=+1/2 are the quantum numbers of each electron. The total angular momentum is L=|l1+l2||l1l2| and total angular momentum J=|L+S||LS| Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by 1.


The term symbol has the form 2S+1LJ. The super-prefix is the spin multiplicity, for spin angular momentum S this is 2S+1 or in general 2X+1 for angular momentum X.


When there are three 3 electrons the the spin and orbital angular momentum terms have to be added in two parts. First as a pair, with the equations above then again with each of the values in the series with the last electron or orbital.


For the spin the S values are
S=|s1+s2||s1s2|=|1/2+1/2||1/21/2|=1, 0 . We call these values S1 and S0 To calculate the total with the third electron gives Ss1=|S1+1/2||S11/2|=3/2, 1/2 and for the S0 there is one value S0=1/2. Thus for the spin the values are 3/2 and 1/2. This means that the multiplicity produces quartet and doublet states, thus so far we have 4LJ and 2LJ.


The same method is followed for the orbital angular momentum. The first two p orbitals give L=|l1+l2||l1l2|=|1+1||11|=2, 1, 0 then for each of these values called L1,L2,L3 we combine with the last orbital angular momentum, for example LL1=|L1+l1||L1l2|=|2+1||21|=3, 2, 1 and so forth for the other values. The L values are 3,2,1,0


Putting all these values into a table gives LSJterm symbol33/29/2,7/2,5/2,3/24F9/2,7/2,5/2,3/223/27/2,5/2,3/2,1/24D7/2,5/2,3/2,1/213/25/2,3/2,1/24P5/2,3/2,1/203/23/24S3/231/27/2,5/22F7/2,5/221/25/2,3/22D5/2,3/211/23/2,1/22P3/2,1/201/21/22S1/2 Finally to find the lowest energy term use Hund’s rules, i.e. of the terms given by equivalent electrons, i.e. having the same L values, the one with the greatest multiplicity is lowest in energy and of these the lowest is that one with the greatest L value. The only terms remaining after considering Pauli exclusion are 4S which is lowest in energy and next 2D then 2P


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