Question

In its ground state, nitrogen's 2p electrons interact to produce ${\mathbf{j}}_{\mathbf{T}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}$. Given Hund's rule, how might the orbit at angular momenta of these three electrons combine?

Short answer

The total orbital quantum number l_T are${\mathrm{l}}_{\mathrm{T}}=0,1,2,3$.

Step-by-step solution

total spin

The total spin S_Tis given by

${\mathbf{S}}_{\mathbf{T}}\mathbf{=}\sqrt{{\mathbf{s}}_{\mathbf{T}}\left(s_T+1\right)}\sout{\mathbf{h}}$

total quantum number

The total quantum number j_T is ${\mathrm{j}}_{\mathrm{T}}=\frac{3}{2}$.

total spin quantum number

The spin quantum number of each electron is ${\mathrm{s}}_1={\mathrm{s}}_2={\mathrm{s}}_3=\frac{1}{2}$.

By using Hund's rule, we find the total spin quantum number s_T as

$\begin{array}{rcl}{\mathrm{s}}_{\mathrm{T}}& =& {\mathrm{s}}_1+{\mathrm{s}}_2+{\mathrm{s}}_3\\ & =& \frac{1}{2}+\frac{1}{2}+\frac{1}{2}\\ {\mathrm{s}}_{\mathrm{T}}& =& \frac{3}{2}\end{array}$

the total orbital quantum number lT

Now, the 'candidates' for the total orbital quantum number l_T are

${\mathrm{l}}_{\mathrm{t}}=0,1,2,3$

observation

As we see from the values of j_T and s_T, all of the 'candidates' for l_T are possible, since the difference between them and the total quantum number j_T is not greater than$\frac{3}{2}$ (the rules of addition of the angular momenta are satisfied).