Atomic Structure Question 312
Question: For principle quantum number $ n=4 $ the total number of orbitals having $ l=3 $
[AIIMS 2004]
Options:
A) 3
B) 7
C) 5
D) 9
Show Answer
Answer:
Correct Answer: B
Solution:
$ n=4,\to 1s^{2},,2s^{2},,2p^{6},,3s^{2},,3p^{6},,3d^{10},,4s^{2},,4p^{6},,4d^{10},,4f^{14} $ So $ l=(n-1)=4-1=3 $ which is f orbit contain 7 orbital.
