00:01
In this example, we're being asked to think about what a periodic table would look like if it was from another universe with different physical laws, but similar quantum numbers, although they're p, q, r, and s with the following rules.
00:21
P equals integers 1, 2, 3, 4, 5, etc.
00:25
Q is positive odd integers, and q is less than or equal to p.
00:28
R takes on all even integer values from negative q to positive q and s is one half or negative one half.
00:38
And the first thing we're being asked to do is to sketch what that periodic table would look like.
00:43
Based upon these rules, we would have a periodic table for the first, we'll just do the first few elements here, but we would have something that looked a little.
01:02
Little bit more like this.
01:23
There we go.
01:27
So this would be one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, fifteen, sixteen, sixteen, eighteen, nineteen, nineteen, and twenty.
01:51
And it's because of that the odd integers and the, the changes between our q and our r that are going to make our table look a little bit different from the original.
02:04
The next question asks, which ones, which atomic numbers of the first four elements would be expected to be the least reactive.
02:16
And just like with our periodic table, we would expect those to be in the area of our noble gases.
02:22
So it would be the ones to the furthest of the right.
02:25
So 2, 4, 12, and 20 would be the least reactive of the elements.
02:33
The next part is asking us to come up with, using these elements, come up with some examples of some different ionic compounds that could be created using the formulas x, oops, put it to my pen, there we go, using x, y, x y squared x squared y and x y to the third or cubed and x to y3 or cubed and x to y 3.
03:18
So there's a variety of different examples that you could come up with.
03:22
So x y for example could be one with 11, which would be similar to what we see with an ionic compound.
03:33
And for x, y, 2, we could have 6 with 11.
03:44
So this would have 2 and that would have 7, so we would need two of those.
03:50
For x2 .y, you could have 1 and 10.
03:56
For x and y 3, it could be 7 and 11 because 7 would have 3 valence electrons.
04:05
And then finally for x2y 3, we could do 7 and 10 because this would have 3 and this would have 6.
04:16
So you would use the least common multiple be 2 and 3.
04:23
Okay, so for the next part, it's asking us to figure out how many electrons could be held in different orbits...