Question

(a) Figure out the electron configurations (in the notation of Equation

5.33) for the first two rows of the Periodic Table (up to neon), and check your

results against Table 5.1.

${\left(1s\right)}^2{\left(2s\right)}^2{\left(2p\right)}^2$(5.33).

(b) Figure out the corresponding total angular momenta, in the notation of

Equation 5.34, for the first four elements. List all the possibilities for boron,

carbon, and nitrogen.

${}^{2S+1}{L}_J$ (5.34).

Short answer

(a)

$$\begin{gathered} \mathit{H}\mathit{y}\mathit{d}\mathit{r}\mathit{o}\mathit{g}\mathit{e}\mathit{n}\mathbf{:}\mathbf{}\left(1s\right)\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{h}\mathit{e}\mathit{l}\mathit{i}\mathit{u}\mathit{m}\mathbf{:}\mathbf{}{\left(1s\right)}^{\mathbf{2}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{l}\mathit{i}\mathit{t}\mathit{h}\mathit{i}\mathit{u}\mathit{m}\mathbf{:}\mathbf{}{\left(1s\right)}^{\mathbf{2}}\left(2s\right)\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{b}\mathit{e}\mathit{r}\mathit{y}\mathit{l}\mathit{l}\mathit{i}\mathit{u}\mathit{m}\mathbf{:}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}\mathbf{;} \\ \mathit{b}\mathit{o}\mathit{r}\mathit{o}\mathit{n}\mathbf{:}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}\left(2p\right)\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathit{c}\mathit{a}\mathit{r}\mathit{b}\mathit{o}\mathit{n}\mathbf{:}\mathbf{}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}{\left(2p\right)}^{\mathbf{2}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathit{n}\mathit{i}\mathit{t}\mathit{r}\mathit{o}\mathit{g}\mathit{e}\mathit{n}\mathbf{:}\mathbf{}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}{\left(2p\right)}^{\mathbf{3}}\mathbf{;} \\ \mathit{o}\mathit{x}\mathit{y}\mathit{g}\mathit{e}\mathit{n}\mathbf{:}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}{\left(2p\right)}^{\mathbf{4}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\mathit{f}\mathit{l}\mathit{u}\mathit{o}\mathit{r}\mathit{i}\mathit{n}\mathit{e}\mathbf{:}\mathbf{}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}{\left(2p\right)}^{\mathbf{5}}\mathbf{;}\mathbf{}\mathbf{}\mathit{n}\mathit{e}\mathit{o}\mathit{n}\mathbf{:}\mathbf{}\mathbf{}\mathbf{}{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}{\left(2p\right)}^{\mathbf{6}}\mathbf{;}\mathbf{} \end{gathered}$$

(b)$$\begin{gathered} {}^{\mathbf{2}}{\mathbf{S}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{S}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{2}}{\mathbf{P}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{2}}{\mathbf{P}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{P}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{P}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{P}}_{\mathbf{5}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{2}}{\mathbf{D}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{2}}{\mathbf{D}}_{\mathbf{5}\mathbf{/}\mathbf{2}}\mathbf{.} \\ {}^{\mathbf{4}}{\mathbf{D}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{D}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{D}}_{\mathbf{5}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{D}}_{\mathbf{7}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{2}}{\mathbf{F}}_{\mathbf{5}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{2}}{\mathbf{F}}_{\mathbf{7}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{F}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{F}}_{\mathbf{5}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}{\mathbf{F}}_{\mathbf{7}\mathbf{/}\mathbf{2}}\mathbf{,}{}^{\mathbf{4}}\mathit{F}_{\mathbf{9}\mathbf{/}\mathbf{2}}\mathbf{.} \end{gathered}$$

Step-by-step solution

(a) Figuring out the electron configuration

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

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

(b) Figuring out the corresponding total angular momenta 

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

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:

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