Question

reluza la configuracion electronica indicando niveles subniveles orbitales y espines para los atomos de los siguientes elementos tunsgteno(W) platino (Pt) y telurio(Te)

Answer 1

$Hola~~~\mathbb{KARINA} \\ \\ El~tunsgtenio~(W)~tiene~n\acute{u}mero~at\acute{o}mico~(Z)=74 \\ \\ Haciendo~la~configuraci\acute{o}n~electr\acute{o}nico~tenemos: \\ \\ _{74} W= 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{2} 4f^{14} 5d^{4} \\ Nivel ---\ \textgreater \ 6 \\ Subnivel--\ \textgreater \ 6s \\ \\ Orbitales ~y~espines$

$n=1~2~3~4~\boxed{5}~6~7 \\ n=5 \\ l=s~p~\boxed{d}~f \\~~~~~~~~~~~~| \\ ~~~~~~0~1~\boxed{2}~3~4 \\ \\ l=2~--\ \textgreater \ quiere~decir~que~\boxed{l}~va~desde~-2 ~a~+2~vea: \\ \\ ~~~~~~~~~~~~-2~-1~~~0~+1~+2 \\~~~~~~~~~~~~~~ \boxed{\uparrow~|~~\uparrow~|\uparrow~|\uparrow~|~~~~} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{+1}=m \\ s=- \frac{1}{2}$

$Entonces~que: \\ \\ n=5 \\ l=2 \\ m=1 \\ s=- \frac{1}{2} \\ \\ \mathbb{qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq}\\ Ahora~con~el~platino~(Pt) \\ Pt~~tiene~de~n\acute{u}mero~at\acute{o}mico~(Z)=78 \\ \\ Haciendo~su~configuraci\acute{o}n~electr\acute{o}nica~tenemos: \\ _{78}Pt=\ \textgreater \ 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{6} 6s^{1} 4f^{14} 5d^{9} \\ \\ Nivel---\ \textgreater \ 6 \\ Subnivel--\ \textgreater \ 6s$

$Sus~orbitales~y~espines~son: \\ \\ n=1~2~3~4~\boxed{5}~6~7 \\ n=5 \\ \\ l=s~~p~~d~~f \\ ~~~~~~0~~1~\boxed{2}~3 \\ l=2~~--\ \textgreater \ quiere~decir~que~\boxed{l}~va~de~-2 ~a~+2 \\ \\ tenemos~el~\acute{u}ltimo~subnivel~energ\acute{e}tico~de~9~electrones~tenemos: \\ \\ ~~~~~~~~~~~~-2~~-1~~~0~~+1~~+2 \\ ~~~~~~~~~~~~~\boxed{\uparrow\downarrow~~|\uparrow \downarrow~|\uparrow \downarrow~|\uparrow \downarrow~|~\uparrow}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{+1}=m \\ \\ s=+ \frac{1}{2}$

$Entonces~tenemos: \\ \\ n=5 \\ l=2 \\ m=1 \\ s=+ \frac{1}{2}$

$\mathbb{qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq} \\ El~telurio~(Te)~tiene~n\acute{u}mero~at\acute{o}mico~(Z)=52 \\ \\ _{52}Te=\ \textgreater \ 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{4} \\ \\ niveles--\ \textgreater \ 5 \\ subnivel--\ \textgreater \ 5s~5p \\ \\ Orbital~y~espines \\ \\ n=1~2~3~4~\boxed{5}~6~7 \\ n=5$

$l=s~~~p~~~d~~~f~~~\\ ~~~~~~0~~\boxed{1}~~~2~~~3 \\ \\ l=1~~--\ \textgreater \ l~=-1~~0+1 \\ Tiene~4~electrones~en~el~\acute{u}ltimo~subnivel~entonces: \\ \\~~~~~~~~ -1~~~~~0~~~+1 \\~~~~~~~~~~\boxed{\uparrow \downarrow~|~~\uparrow~|~~\uparrow ~}\\~~~~~~~~~~\boxed{-1}=m \\ \\ m=-1 \\ s=+ \frac{1}{2}$

$entonces~tenemos: \\ \\ n=5 \\ l=1 \\ m=-1 \\ s=+ \frac{1}{2} \\ \\ \mathbb{qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq} \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Espero~te~sirva~saludos!!$

Answer 2

La configuración electrónica de los elementos son:

• tunsgteno (W) =4f14 5d4 6s2
• platino (Pt) = 4f14 5d9 6s1
• telurio (Te) = 4d10 5s2 5p4

La configuración electrónica nos indica la forma en como se distribuyen los electrones, como se estructuran, se comunican y organizan entre sí, con el modelo de capas electrónicas.

Para realizar la configuración electrónica se utiliza el diagrama de Moeller, en el cual se sigue el camino hasta completar la cantidad total de electrones.

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