Question

What is the correct orbital designation of an electron with the quantum number, n = 4, l = 3, m = – 2, s = 1/2 ?
(a) 3s
(b) 4 f
(c) 5p
(d) 6s

Answer 1

${\huge {\mathfrak {Answer :-}}}$

➡️ Option (A)

That's it..

Answer 2

Option b is the correct answer.

1. The given problem can be solved using the concepts of quantum numbers.

• In quantum numbers, the value of n implies the value of the principal electron shell. n can never be zero.
• The value of l is calculated using the formula l = n - 1, The value of l varies from 0 to n-1.
• The values of m for a particular shell vary from -l to +l.
• The value of s can be either +1/2 or -1/2 depending upon the last electron of the element.

2. In the given question,

=> n = 4 implies the shell considered is 4, the possible answers are 4s, 4p, 4d, and 4f.

=> l = 3 implies f orbital, hence n = 4 and l = 3 implies 4f.

Hence, the correct orbital designation is 4f. Option B is the correct answer.