Carbon:
z = 6
n 



L  2 

Ý  Ý  
K  1  Ýß  C 
1s^{2} 2s^{2} 2p^{2}
The total spin can be 0 or 1. The total orbital momentum can be 0, 1 or 2 since either electron possesses an orbital momentum of 1. l_{1}−l_{2} ≤ L ≤ l_{1} + l_{2}. Now let's try to explain possible configurations with l ≠ 0. Primarily, both pelectrons have common quantum numbers n and l; therefore only the quantum numbers (m_{l1}, m_{s1}) and (m_{l2}, m_{s2}) remain to be considered. m_{l1} and m_{l2} may be equal to 1, 0, +1 and m_{s1} and m_{s2} may be equal to ½, +½.
Table 1 contains the results for the total orbit momentum M_{L} and the total spin M_{S}. These values represent the sum of the orbit momenta m_{l} and of the spins m_{s} of the two electrons.
 
Table 1: Determination of resultant pairs of M_{L} and M_{S} 
The recieved combinations (M_{L};M_{S}) belong to atomic states ^{2S+1}L _{J}. In a second step, we determine the respective states. We start with the highest value for L, i.e. M_{L} = 2. From this M_{L}, we conclude that the two electrons produce an atomic state with L = 2 which is denoted with a capital D in the term symbol. Both extremes, M_{L} = 2 and M_{L} = 2 are combined with M_{S} = 0, i.e. an antiparallel orientation of the two electron spins. Therefore, we have state ^{1}D. The calculated pairs (2;0), (1;0), (0;0), (1;0) and (2;0) are special cases for the carbon atom in state ^{1}D (compare Fig. 1a). As next value, M_{L} = 1 has been obtained, for example in (M_{L};M_{S}) = (1;1). This combination is one out of nine cases which emerge from assuming an atomic state with L = 1 and S = 1. The respective state is denoted as ^{3}P state. (compare Figure 1b). So far, 5 + 9 = 14 of our set of 15 have been identified as cases of two states. The remaining pair (0;0) is the only case for a state with L = 0 and S = 0 (compare Fig 1c); in other words, an atomic state ^{1}S_{0}.
Fig. 1a J=2 S=0 ^{1}D_{2} 
Fig. 1b J=1 S=1 ^{3}P_{2,1,0} 
Fig. 1c J=0 S=0 ^{1}S_{0} 
Therefore, dependent on the distribution of electrons and their spins, the carbon atom exists in states denoted by the term symbols ^{1}S_{0}, ^{3}P_{0}, ^{3}P_{1}, ^{3}P_{2} and ^{1}D_{2}. These symbole have been introduced in table 2.


Table 2: The combinations are representations of three atomic states 
According to Hund's rule (1) the ground state is ^{3}P and according to rule (2) is the ^{3}P_{0} state.
C : K 2s^{2} 2p^{2} ≡^{3}P_{0} 
Other energy terms than ^{1}S, ^{3}P und ^{1}D, i.e. the excited the carbon atom, are represented on the next illustration. To the right one can see the
energy levels which are obtained by (2s) electron excitation.
Figure: Energy levels of Carbon atoms.
To the left: the (2p)electron excitation. To the right: the (2s)electron excitation. 
The ground state of Catom is the combination of two (np)electrons. The table below lists the terms for an atom with three (np)electrons, i.e. p^{3} which leads to the states ^{4}S, ^{2}P and ^{2}D.
Configuration  Term  Configuration  Term  
s
s^{2} p, p^{5} p^{2}, p^{4} p^{3} 
^{2}S
^{1}S ^{2}P ^{1}S, ^{1}D, ^{3}P ^{4}S, ^{2}P, ^{2}D 
d, d^{9}
d^{2}, d^{8} d^{3}, d^{7} d^{4}, d^{6} d^{5} 
^{2}D
^{1}S, ^{1}D, ^{1}G, ^{3}P, ^{3}F ^{2}D, ^{2}F, ^{2}G, ^{2}H, ^{4}P, ^{4}F ^{1}S, ^{1}D, ^{1}F,^{1}G,^{1}I,^{3}P,^{3}D,^{3}F,^{3}G,^{3}H,^{5}D ^{2}S, ^{2}P, ^{2}D,^{2}F,^{2}G,^{2}H,^{2}I,^{4}P,^{4}D,^{4}F,^{4}G,^{6}S 
Nitrogen:
z = 7
n 



L  2 

Ý  Ý  Ý 
K  1  Ýß  N 
(1s^{2}) ( 2s^{2}) ( 2p^{3})
According to Hunds rule spins should be directed as parallel to each other as
possible. So that the total spin will be S = ^{3}/_{2}.
According to the above table 3 terms ^{4}S, ^{2}P,
^{2}D
is the combination of three pelectrons. And the state is:
^{4}S_{3/2} 
Oxygen:
z = 8
n 



L  2 

Ýß  Ý  Ý 
K  1  Ýß  O 
(1s^{2}) ( 2s^{2}) ( 2p^{4})
The abovementioned considerations would be very difficult to apply to this atom having 4 pelectrons: we will consider another way of 4 pelectrons combination here:
We've known that the shell is fulfilled with 6 pelectrons (p^{6})
and also L = 0 and S = 0. We will represent our Oxygen atom as 2 electrons of (p^{6})shell
would be absent. Now these missing electrons or also holes are combined into
"2holes configuration" since holes have also spin ½. Therefore we have
as for Carbon atom the following possible states ^{1}S,
^{3}P,
^{1}D.
The highest spin (Hunds rule (1))
is S = 1 i.e. ^{3}P is the ground state and total kinetic moment
J is for the ground state (Hunds rule (2))
J = L + S = 2. And finally:
^{3}P_{2} 
Fluorine:
z = 9
n 



L  2 

Ýß  Ýß  Ý 
K  1  Ýß  F 
(1s^{2}) ( 2s^{2}) (2p^{5}) : ^{2}P_{3/2}
There is no one pelectron i.e. L = 1 and S = ½.
J = ½ and ^{3}/_{2} are possible.
According to Hunds rule (2) the higher Jvalue is
possible for the ground state:
^{2}P_{3/2} 
n 



L  2 

Ýß  Ýß  Ýß 
K  1  Ýß  Ne 
(1s^{2}) (2s^{2}) (2p^{6}) : ^{1}S_{0}
2pshell is full ! High ionization energy 21,6 eV.
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