If n = 6, the correct sequence for filling of | vedclass

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(A) According to the $(n + l)$ rule,electrons fill orbitals in order of increasing $(n + l)$ values.For $n = 6$:$ns$ corresponds to $6s$ $(n+l = 6+0 =

Vedclass · Accepted Answer

$ns$ $\rightarrow (n - 2)f$ $\rightarrow (n - 1)d$ $\rightarrow np$

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(A) According to the $(n + l)$ rule,electrons fill orbitals in order of increasing $(n + l)$ values.For $n = 6$:$ns$ corresponds to $6s$ $(n+l = 6+0 = 6)$$(n-2)f$ corresponds to $4f$ $(n+l = 4+3 = 7)$$(n-1)d$ corresponds to $5d$ $(n+l = 5+2 = 7)$$np$ corresponds to $6p$ $(n+l = 6+1 = 7)$Comparing the $(n+l)$ values and the $n$ values for orbitals with the same $(n+l)$,the filling order is $6s$ $\rightarrow 4f$ $\rightarrow 5d$ $\rightarrow 6p$.

If $n = 6,$ the correct sequence for filling of electrons will be

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