Question

The orbital nearest to the nucleus is

• A. 4f
• B. 5d
• C. 4s
• D. 7p

Answer 1

Answer: (C)

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) $4f\to (l=3)\therefore n+l=7$
(b) $5d\to (l=2)\therefore n+l=7$
(c) $4s\to (l=0)\therefore n+l=4$
(d) $7p\to (l=1)\therefore n+l=8$