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If n = 6, the correct sequence of filling of electrons will be
A
$ns \rightarrow np \rightarrow (n-1)d \rightarrow (n-2)f$
B
$ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np$
C
$ns \rightarrow (n-1)d \rightarrow (n-2)f \rightarrow np$
D
$ns \rightarrow (n-2)f \rightarrow np \rightarrow (n-1)d$
Detailed Solution
For n = 6 the orbitals concerned are 6s, 4f [(n − 2)f], 5d [(n − 1)d] and 6p.
By the (n + l) rule, orbitals are filled in increasing order of (n + l); for equal (n + l), the orbital with lower n is filled first.
6s: n + l = 6 + 0 = 6
4f: n + l = 4 + 3 = 7; 5d: n + l = 5 + 2 = 7; 6p: n + l = 6 + 1 = 7
6s has the lowest value and is filled first. Among 4f, 5d and 6p (all 7), the order follows increasing n: 4f, then 5d, then 6p.
Sequence: $6s \rightarrow 4f \rightarrow 5d \rightarrow 6p$, i.e., $ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np$
By the (n + l) rule, orbitals are filled in increasing order of (n + l); for equal (n + l), the orbital with lower n is filled first.
6s: n + l = 6 + 0 = 6
4f: n + l = 4 + 3 = 7; 5d: n + l = 5 + 2 = 7; 6p: n + l = 6 + 1 = 7
6s has the lowest value and is filled first. Among 4f, 5d and 6p (all 7), the order follows increasing n: 4f, then 5d, then 6p.
Sequence: $6s \rightarrow 4f \rightarrow 5d \rightarrow 6p$, i.e., $ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np$