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configuración electrónica= 1s2 2s2 2p6 3s2 3p6 4s1
Números cuánticos= según capa de valencia
n (nivel de energía)= 4
l (subnivel)= 0 -----> según la tabla s = 0 p = 1 d = 2 f = 3
m (orbital) = 0 -----> s = |__ 1 electrón
0
Se (selenio) 34e-
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
n = 4
l = 1
m = -1 ---> p = |_| |__ |__ 4 electrones
-1 0 +1
Au (oro) 79e-
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s24d10 5p6 6s1 4f14 5d10
n = 5
l = 2
m = +2 -----> d = |__| |__| |__| |__| |__|
-2 -1 0 +1 +2
Cl (cloro) 17e-
1s2 2s2 2p6 3s2 3p5
n = 3
l = 1
m = 0
I (yodo)
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s24d10 5p5
n= 5
l = 1
m = 0
Cualquier error me avisan.