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If $n=6$, the correct sequence for filling of electrons will be
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The correct answer is:
$n s \longrightarrow(n-2) f \longrightarrow(n-1) d \longrightarrow n p$
For $n-6$ olectron aro filled as $6 s \rightarrow 4 f \rightarrow 5 d \rightarrow 6 p$
This is because electron first enters in that orbitals for which $(n+l)$ is lower. In case, if $(n+I)$ is same, it goes first in that orbital for which $n$ is lower.
This is because electron first enters in that orbitals for which $(n+l)$ is lower. In case, if $(n+I)$ is same, it goes first in that orbital for which $n$ is lower.
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