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Q.
The orbital nearest to the nucleus is
KCETKCET 2018Classification of Elements and Periodicity in Properties
Solution:
In general orbitals with lower energy are present near the nucleus.
This can be calculated by $(n+l)$ rule.
(i) Lower the sum of $(n+l)$ value, nearer the is the orbital to the nucleus.
(ii) For the same sum, lower the $n$ value, lower is the energy and the closer it is to the nucleus. Among the given options:
(a) $4 f \rightarrow(l=3) \therefore n+l=7$
(b) $5 d \rightarrow(l=2) \therefore n+l=7$
(c) $4 s \rightarrow(l=0) \therefore n+l=4$
(d) $7 p \rightarrow(l=1) \therefore n+l=8$