Imagine a universe in which the four quantum numbers can have the same possible values as in our universe except that the angular-momentum quantum number l can have integral values of 0, 1, 2...n + 1 (instead of 0, 1, 2..., n - 1). (c) Draw an orbital-filling diagram for the element with atomic number 12.

Question

Accepted Answer

Hello everyone in this video. We want to consider a hypothetical world in which the angular momentum quantum number L can have integral values of L equal in 2012 and so on. Instead of L equal to 012 and n minus one. The difference here is just the N and then n minus one. The other three quantum numbers have the same possible values as they do in our world provide the orbital diagram for the element with an atomic number of 18. Alright, so first let's kind of break down what these quantum numbers is. So of course we're starting off with what we have known to be the principal quantum number and this is denoted as N. So for our end value the possible values are going to be 12, infinity. The next one is going to be the angular momentum quantum number and the possible values here are going to be zero. I forget what the symbol is for the angular momentum quantum number and this is L. So the possible value for L is going to be from zero to N. Next we have the magnetic quantum number. So this is two noted as M sub L. The possible values here is going to be ranging from negative L two, positive L. And our last quantum number sat here is going to be the spin quantum number were still known as M so S. And this deals with the spin of the electron. So the possible values for R M. S is either negative one half which means spin down or positive one half, which means spin up. So the atomic number is going to be 18. So this just means that we have 18 electrons. So we have N equals to one. R. L value is equal to zero which is us. And then the L value equal into one is R. P. And then we have the M. L. To be zero negative 10 and positive one. In total here, this is eight electrons are ready now for N equaling 22. We have the L equaling to zero, which is just the S. And then the L. A. Going to one which is the P orbital for M. L number. This is zero negative one 01, negative two negative 101 and positive two. So then this gives us a total now of 18 electrons. So let's write out our sort of orbital's here. So starting off with the one, one S N one P. So one s this and then we have our p orbital's this one, P. Now moving on to the two S two p.m. To deem So we have one line being to us. Same thing for P. We have three. This two P for D. We have 1234 and five. And this is the two D. Set here. So let's fill out the 18 electrons. So we have 123456789, 10, 11, 12, 13, 14, 15 and 16. So the other two are the remaining two is just going to go to two D. So that's these two here. So this is going to be my final answer for this problem. Thank you all so much for watching.

Verified Solution

Key Concepts

Quantum Numbers

Orbital Filling Order

Electron Configuration of Element 12