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For principle quantum number n = 4 the t...

For principle quantum number `n = 4` the total number of orbitals having `l = 3`.

A

3

B

7

C

5

D

9

Text Solution

Verified by Experts

The correct Answer is:
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.
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