Consider the carbon atom with an electron configuration of [He]2s 2 2p 2 : Attac
ID: 529099 • Letter: C
Question
Consider the carbon atom with an electron configuration of [He]2s22p2:
Attach term symbols to the states:
Start with the highest value of ML and, within that value, the highest value of MS.
Within that group, start with the highest value of MJ.
Assign all of the microstates belonging to that term symbol;
(a 1P1 state, for example, will contain three microstates with MJ=0, 1, and -1, while a 1P0 state will only contain a state with MJ=0.)
Which of the states described above will have the lowest energy? Explain your reasoning.
Explanation / Answer
Electronic configuration of C = 1s2 2s2 2p2 (p2configuration)
For two equivalent electrons l1= l2=1 & s1=s2=1/2
L= 2,1,0 & S =1,0
Terms corresponding to these values are calculated by spin multiplicity (2S+1)
when S = 1 ; spin multiplicity = 3 (triplet)
when S = 0 ; spin multiplicity = 1 (singlet)
Six states = 3D,3P,3S & 1D,1P,1S
J value for these states are ; 3D: J=3,2,1 ; 3P: J=2,1,0 ; 3S: J=1; 1D :J=2 ; 1P: J=1 ; 1S: J= 1
Total term symbols are ; 3D33D23D1 ; 3P2,3P1,3P0 ; 3S1 ; 1D2, 1P1, 1S0
First choose the max. Ms value & max. ML associated with it.
Ms=1 & ML=1 & corresponds to a group of terms,L=1 & S=1
L= 1 ; P state & S= 1 multiplicity (2S+1) = 3 So 3P state
If L = 1 then ML= +1,0,-1 & if S= 1 then MS= +1,0,-1.
There are 9 combination of these two terms:
ML=+1 Ms=+1,0,-1
ML=0 Ms=+1,0,-1
ML=-1 Ms=+1,0,-1
when Ms= 0 & ML= 2
S = 0 & L= 2 this is D state since S= 0 so multiplicity (2,0+1)= 1 hence singlet 1D state
If L = 2 then ML = +2,+1,0,-1,-2 & since S= 0, MS=0
this gives 5 combination of ML & MS : 3P &1D = 9+ 5 = 14
If ML= 0; MS= 0 last
L =0 & S = 0 this gives singlet S state 1S.
thus 9+5+1= 15 electronic arengements by 1D,3P,1S.
Permisible microstate of a configuration is specified by giving the vale of ml & ms of the electron of the configuration.If we write the values of ml for each elctron & denotes ms= +1/2 & -1/2 by + & -, the 15 possible microstates of a p2 configuration are :
1-,0+
1-,-1-
The microstates 1+,1- & 1-,1+ are identical & are not listed separatly because individual electrons can't be distinguished. Microstates such as 1+,1+are excluded by Pauli's exclusion principle for they would correspond to two electrons with the same spin in the same orbitals. Interelectronic repulsions results in some microstates which have the same energy & some having different energies.
3P state have lowest energy. The terms are placed in order depending on their multiplicities i.e. S value.The most stable state has the largest S value & stability decreases as S decreases.The ground state therefor has the most unpaired spins (electrons). For the given value of S, the state with highest L is the most stable. Greater is the stability lesser is the energy.
1+,1- 1+,0+ 1+,0-
1-,0+
1-,0- 0+,0- 1+,-1+ 1+,-1- 1-,-1+1-,-1-
0+,-1+ 0+,-1- 0-,-1+ 0-,-1- -1+,-1-