Asked by Eric
Which one of the following sets of quantum numbers can correctly represent a 3p orbital?
a) (n = 3 l = 1, ml = 2)
b)(n = 1, l = 3, ml = 3)
c) (n = 3, l = 2, ml = 1)
d) (n = 3, l = 1, ml = -1)
e) (n = 3, l = 0, ml = 1)
a) (n = 3 l = 1, ml = 2)
b)(n = 1, l = 3, ml = 3)
c) (n = 3, l = 2, ml = 1)
d) (n = 3, l = 1, ml = -1)
e) (n = 3, l = 0, ml = 1)
Answers
Answered by
DrBob222
3p so
n = 3
l = 1 because it's a p
ml can be -1 or 0 or +1
You take it from here.
You can immediately rule out a because of mL, b because of n, c because l, etc.
n = 3
l = 1 because it's a p
ml can be -1 or 0 or +1
You take it from here.
You can immediately rule out a because of mL, b because of n, c because l, etc.
Answered by
Hmmmm
The answer should be Option (d) (n = 3, l = 1, mₗ = -1).
This is because a 3p orbital must have n = 3.
*That eliminates Option (b), because it has n = 1*
A p orbital must have l = 1.
*That eliminates Options (c), and (e), because they have l = 2, and 0*
The mₗ is required to be an integer satisfying −l ≤ m ≤ l, so mₗ could be any of −1, 0, or +1.
*That eliminates Option (a), because they have mₗ = 2*
The only correct option is:
(d) n = 3, l = 1, mₗ = -1
This is because a 3p orbital must have n = 3.
*That eliminates Option (b), because it has n = 1*
A p orbital must have l = 1.
*That eliminates Options (c), and (e), because they have l = 2, and 0*
The mₗ is required to be an integer satisfying −l ≤ m ≤ l, so mₗ could be any of −1, 0, or +1.
*That eliminates Option (a), because they have mₗ = 2*
The only correct option is:
(d) n = 3, l = 1, mₗ = -1
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