Question

How many degenerate orbitals are in each of the following subshells? (Enter 0 if there are none.)

(a) n = 6,l = 5
(b) n = 4,l = 1
(c) n = 2, l = 0

Answer 1

For n = 6, l = 5, there are 7 degenerate orbitals in the f subshell. For n = 4, l = 1, there are 3 degenerate orbitals in the p subshell. For n = 2, l = 0, there is 1 degenerate orbital in the s subshell.

Step-by-step explanation:

(a) For the given quantum numbers n = 6, l = 5, we can determine the subshell by matching the values of l with the corresponding subshell letters. The subshell with l = 5 corresponds to the f subshell. In the f subshell, there are 7 orbitals. Therefore, there are 7 degenerate orbitals in this subshell.

(b) For n = 4, l = 1, the subshell with l = 1 corresponds to the p subshell. In the p subshell, there are 3 orbitals. Therefore, there are 3 degenerate orbitals in this subshell.

(c) For n = 2, l = 0, the subshell with l = 0 corresponds to the s subshell. In the s subshell, there is only 1 orbital. Therefore, there is 1 degenerate orbital in this subshell.

Answer 2

For each subshell, the number of degenerate orbitals can be calculated using the formula 2l + 1, where l is the angular momentum quantum number.

Step-by-step explanation:

In each subshell, the number of degenerate orbitals is equal to the 2l + 1, where l is the angular momentum quantum number. The angular momentum quantum number (l) can have values ranging from 0 to (n-1), where n is the principal quantum number. Let's apply this to the given subshells:

(a) n = 6, l = 5: Since l > n-1, there are no degenerate orbitals in this subshell. The answer is 0.

(b) n = 4, l = 1: The angular momentum quantum number l has the value of 1. Therefore, there are (2 x 1) + 1 = 3 degenerate orbitals in this subshell.

(c) n = 2, l = 0: The angular momentum quantum number l has the value of 0. Therefore, there is 1 degenerate orbital in this subshell.