Question

For principle quantum number `n = 4` the total number of orbitals having `l = 3`.
A. 3
B. 7
C. 5
D. 9

Answer 1

Correct Answer - B
(b) `n = 4 rarr 1 s^2 2 s^2 2 p^6, 3 s^2, 3p^6, 3 d^10, 4 s^2 ,4 p^6, 4 d^10, 4 f^14`
So `f = (n - 1) = 4 - 1 = 3` which is `h` orbit contains `7` orbital.