Problem 1D 23

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Saachi_Kotia_4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

Problem 1D 23

Postby Saachi_Kotia_4E » Wed Oct 17, 2018 11:39 pm

How do you do 1D 23? Can someone explain the concepts behind it as well as explain the answers?

2c_britneyly
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

Re: Problem 1D 23

Postby 2c_britneyly » Thu Oct 18, 2018 2:54 am

To do this we need to know quantum numbers, which is another way to describe orbitals.
The variable "n" represents principal energy levels and can take the values of 1,2,3,4,etc. "l" represents subshells (s,p,d,f) and takes the values of 0,1,2,...,n-1. "m" represents orientation, which takes the values -l through +l.
For example, if I were to ask you to express the orbital 3px as quantum numbers, then you would say n=3, l=1, m=-1.

Aside from understanding quantum numbers, we also need to understand the number of orbitals in a subshell increases by 2 as you move up in subshell. The s subshell has only 1 orbital, the p has 3, the d has 5, the f has 7, and so on.

The way you want to approach these questions is by figuring out if they are giving you a subshell or orbital. If they give you all three values, then you know the answer will be 1 because they have specified the orbital, and each orbital is unique to that subshell. Therefore, the answers to part (b) and (d) will be 1. However, (a) and (c) introduce ambiguity because we are not given 3 quantum numbers.

(a) 2p; You know the p subshell has 3 orbitals, and because they haven't specified "m," we don't know if it is 2px, 2py, or 2pz. Therefore, there are 3 possible orbitals that can have the following quantum numbers.

(c) 2; you are only given the principal energy level. You know "l" can only take values up to n-1. Given n=2, 2-1=1, so "l" can be 0 or 1, which corresponds to the s subshell and p subshell, respectively. As we said before, the s subshell has 1 orbital, and the p subshell has 3. That's a total of 4 possible orbitals.

Sorry this answer was kind of long, but I hope this helps.

Theodore_Herring_1A
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

Re: Problem 1D 23

Postby Theodore_Herring_1A » Thu Oct 18, 2018 3:14 pm

The number of possible orbitals can be found by either n, l or ml. Always look to ml first if it's given, and if not, then to l, then to n.

If ml is given, this describes the orientation of the orbital, and thus only the one orbital is allowed. So in part b, since ml= -2, this tells us the orientation of the orbital and only one orbital can exist in that coordinate (4,2-2). So number of orbitals is 1 (this is the case for part d as well).

If only l is given, then there can be more than one possible orbital for this subshell, as there could be more than one orientation. For example, in part a, l=1, representing the p subshell. The p orbital can have 3 different orbitals (found by the equation 2l+1, so 2*1+1=3). So number of orbitals is 3.

If only n is given, then more than one subshell can be allowed. For example, in part c, when n=2, l can equal 0 or 1. When l=0, there is one possible orbital (2*0 + 1 = 1), and when l=1, there are 3 possible orbitals (2*1 + 1 = 3). Adding them up, 3 + 1 equals 4, so there are four possible orbitals when n=2.


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