Problema 1C20

Distribuição eletrônica de cada átomo: $$Mg:\;[Ne]\,3s^{2}$$ $$Al:\;[Ne]\,3s^2\,3p^{1}$$ $$Bi:\;[Xe]\,6s^{2}\,4f^{14}\,5d^{10}\,6p^{3}$$ $$As:\;[Kr]\,4s^{2}\,3d^{10}\,4p^{3}$$ $$Cu:\;[Ar]\,4s^{1}\,3d^{10}$$ Cálculo dos números quânticos: $Mg: 3s^{2}$ $$\underset{0}{\color{red}\boxed{\uparrow\downarrow}}$$ $$n=3\;;\;l=0\;;\;m_{l}=0$$ $Al:3p^{1}$ $$\underset{-1}{\color{red}\boxed{\uparrow\;\;}}\underset{0}{\boxed{\phantom\uparrow\;\;}}\underset{1}{\boxed{\phantom\uparrow\;\; }}$$ $$n=3\;;\;l=1\;;\;m_{l}=-1$$ $Bi: 6p^{3}$ $$\underset{-1}{\boxed{\uparrow\;\;}}\underset{0}{\boxed{\uparrow\;\;}}\underset{1}{\color{red}\boxed{\uparrow\;\; }}$$ $$n=6\;;\;l=1\;;\;m_{l}=1$$ $As:4p^{3}$ $$\underset{-1}{\boxed{\uparrow\;\;}}\underset{0}{\boxed{\uparrow\;\;}}\underset{1}{\color{red}\boxed{\uparrow\;\; }}$$ $$n=4\;;\;l=1\;;\;m_{l}=1$$ $Cu:3d^{10}$ $$\underset{-2}{\boxed{\uparrow\downarrow}}\underset{-1}{\boxed{\uparrow\downarrow}}\underset{0}{\boxed{\uparrow\downarrow}}\underset{1}{\boxed{{\uparrow\downarrow}}}\underset{2}{\color{red}\boxed{{\uparrow{\downarrow}}}}$$ $$n=3\;;\;l=2\;;\;m_{l}=2$$