91Ó°ÊÓ

$$\begin{gathered} \mathrm{Hydrogen}:\left(1\mathrm{s}\right);\mathrm{helium}:{\left(1\mathrm{s}\right)}^2;\mathrm{lithium}:{\left(1\mathrm{s}\right)}^2\left(2\mathrm{s}\right);\mathrm{beryllium}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2; \\ \mathrm{boron}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2\left(2\mathrm{p}\right);\mathrm{carbon}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2{\left(2\mathrm{p}\right)}^2;\mathrm{nitrogen}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2{\left(2\mathrm{p}\right)}^3; \\ \mathrm{oxygen}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2{\left(2\mathrm{p}\right)}^4;\mathrm{fluorine}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2{\left(2\mathrm{p}\right)}^5;\mathrm{neon}:{\left(1\mathrm{s}\right)}^2{\left(2\mathrm{s}\right)}^2{\left(2\mathrm{p}\right)}^6; \end{gathered}$$

These values agree with those in Table 5.1-no surprises so far.

$\mathrm{Hydrogen}:{}^2{S}_{1/2};\mathrm{helium}:{}^1{S}_0;\mathrm{lithium}:{}^2{S}_{1/2};\mathrm{beryllium}:{}^1{S}_0;$

S0. (These four are unambiguous, because the orbital angular momentum is zero in all cases.) For boron, the spin (1/2) and orbital (1) angular momenta could add to give 3/2 or 1/2, so the possibilities are For carbon, the two p electrons could combine for orbital angular momentum 2, 1, or 0, and the spins could add to 1 or 0:

${}^1{S}_0,{}^3{S}_1,{}^1{P}_1,{}^3{P}_2,{}^3{P}_1,{}^3{P}_0,{}^1{D}_2,{}^3{D}_3,{}^3{D}_2,{}^3{D}_1,$

For nitrogen, the 3 p electrons can add to orbital angular momentum 3, 2, 1, or 0, and the spins to 3/2 or 1/2:

$$\begin{gathered} {}^2{S}_{1/2},{}^4{S}_{3/2},{}^2{P}_{1/2},{}^2{P}_{3/2},{}^4{P}_{1/2},{}^4{P}_{3/2},{}^4{P}_{5/2},{}^2{D}_{3/2},{}^2{D}_{5/2}. \\ {}^4{D}_{1/2},{}^4{D}_{3/2},{}^4{D}_{5/2},{}^4{D}_{7/2},{}^2{F}_{5/2},{}^2{F}_{7/2},{}^4{F}_{3/2},{}^4{F}_{5/2},{}^4{F}_{7/2},{}^{4}F_{9/2}. \end{gathered}$$