Creating periodic table w/ quantum numbers

[Koyuki]

Hello all. I have a quantum number question that is completely stumping me.

"In another universe, the quantum number ml = 0, 1, 2, ... + 1. That is, the values of ml are positive integers, to a maximum value of +l. All other allowed values are unchanged, so the set of allowed values is:
n: 1,2,3...
l: 0, 1, 2,... (n-1)
ml: 0, 1, 2... +l
ms = -1/2, +1/2

Design the periodic table for the first 30 elements."

I know that since all the ml values are only postive now, that restricts the number of orbitals allowed per energy level. However, I simply cannot get the table to work. I was wondering if anyone could give me any tips. Thank you!

 

[nate808]

Thank you for your question regarding the quantum number ml in another universe. This is a very interesting concept to consider and it definitely presents some challenges in terms of designing a periodic table for the first 30 elements.

Firstly, let's consider the implications of the restricted ml values. As you mentioned, this would limit the number of orbitals allowed per energy level. In our universe, the number of orbitals in an energy level is equal to 2l+1. However, in this alternate universe, the maximum value of ml is +l, so the number of orbitals would be limited to l+1. This means that for the first energy level (n=1), there would only be one orbital (l=0) instead of the usual two (l=0 and l=1).

Now, let's look at the allowed values for n, l, and ml in this alternate universe. The set of allowed values is as follows:
n: 1,2,3...
l: 0, 1, 2,... (n-1)
ml: 0, 1, 2... +l

Using this information, we can start constructing the periodic table. The first element would have the quantum numbers n=1, l=0, ml=0, and ms=-1/2. This element would be in the s orbital.

The second element would have n=1, l=0, ml=1, and ms=-1/2. However, since there is no orbital with ml=1 in the first energy level, we would have to move on to the next energy level (n=2). Here, we have l=0 and l=1, so the second element would be in the p orbital.

Continuing this pattern, we would end up with a periodic table that looks like this:

Element | n | l | ml | ms | Orbital
1 | 1 | 0 | 0 | -1/2 | s
2 | 2 | 0 | 0 | -1/2 | s
3 | 2 | 1 | 0 | -1/2 | p
4 | 3 | 0 | 0 | -1/2 | s
5 | 3 | 1 | 0 | -1/2 | p
6 | 3 | 2 |

 

[Blueshift5]

Creating a periodic table with quantum numbers can be a challenging task, especially when dealing with a different set of allowed values like in this scenario. It is important to remember that the quantum numbers represent the unique set of properties for each electron in an atom, and they are based on the principles of quantum mechanics.

In this case, the quantum number ml has a maximum value of +l, which means that the number of orbitals allowed per energy level will be limited by this value. This can be seen by using the formula 2l+1, which represents the number of orbitals in a subshell. For example, if l=2, then there are 5 orbitals in that subshell (2l+1=5).

To design the periodic table for the first 30 elements, we need to first determine the values of n, l, and ml for each element. This can be done by following the set of allowed values provided in the question. For example, for the first element, n=1, l=0, and ml=0. This means that the first element (hydrogen) has only one energy level (n=1), one subshell (l=0), and one orbital (ml=0).

We can continue this process for the remaining elements, making sure to follow the set of allowed values for each quantum number. Once we have determined the values for all 30 elements, we can arrange them in a periodic table based on their electronic configurations.

It is important to note that in this scenario, the number of orbitals per energy level will be limited, and some elements may have the same electronic configurations. This may result in a slightly different periodic table compared to the one we are familiar with.

I hope this helps to give some guidance in creating the periodic table with these quantum numbers. It is a complex task, but with patience and careful consideration of the allowed values, it can be achieved. Good luck!