Question

Which orbital fills completely immediately before the 4f ?

• A. 6s
• B. 5p
• C. 5d
• D. 4d

Answer 1

The correct option is A 6s

The $n+l$ for the $4f$ orbital is $4+3=7$.
To be filled completely before $4f$ the orbital must have an $n+l=6$.
Consider the n+l for the options:-
A) $6s$ - $6+0=6$
B) $5p$ - $5+1=6$
C) $5d$ - $5+2=7$
D)$4d$ - $4+2=6$
Hence the highest $n$ will be filled first if the $n+l$ is same. Therefore, option A is the right answer.