Atomic theory in another planet - problem

[naren11]

In a planet, if the spin numbers were +1/2, 0 , and -1/2 and their magnetic quantum numbers take the values 0, 1, 2,...L

( Everything else including the ordering and naming of the orbits are same except the spin numbers and the magnetic quantum numbers as provided above.)

how would the perodic table and the energy level diagram may look like?

thanks!

 

[dextercioby]

There are 3 possible spin states (instead of 2) and "l+1" (instead of "2l+1") values for $m_{l}$...

Does that help in any way?

Daniel.

 

[naren11]

im not sure if that helps..
In our planet Earth , there are only 2 spin states which are +1/2 and
-1/2 and the magnetic quantum numbers are -L to +L (postive to negative)

Now, if there were an additional spin state which is 0 in another planet and the magnetic quantum numbers are only whole numbers (0,1,2...)

how would theperodic table has to be modified and the energy level diagram?

 

[dextercioby]

$$n=0\rightarrow \mbox{impossible}$$
$$n=1\rightarrow l=0\rightarrow m_{l}=0$$
$$n=2\rightarrow l=1\rightarrow m_{l}=0$$
$$n=2\rightarrow l=1\rightarrow m_{l}=1$$
$$n=2\rightarrow l=0\rightarrow m_{l}=0$$
$$n=3\rightarrow l=2\rightarrow m_{l}=2$$
$$n=3\rightarrow l=2\rightarrow m_{l}=1$$
$$n=3\rightarrow l=2\rightarrow m_{l}=0$$
$$n=3\rightarrow l=1\rightarrow m_{l}=1$$
$$n=3\rightarrow l=1\rightarrow m_{l}=0$$
$$n=3\rightarrow l=0\rightarrow m_{l}=0$$

Build the rest.Keep in mid that for every subshell,u can have max.3 electrons...

Daniel.

 

[Gokul43201]

What Dexter's trying to get you to do (if I may attempt to read his mind ) is get you to count how many allowed configurations exist for a particular subshell.

How many 2p configurations has he built (n=2, l=1, ml = 0,1 and ms = +, 0, -) ? In fact, he has even written out, in general how many configurations you will have for a particular valence subshell. Compare this with the number of configurations in a "normal" (meaning, on planet Earth) subshell [clearly, the number is 2*(2l+1) on Earth; what about the other planet?].

So, the obvious question to then ask is "what is the significance of the number of subshell configurations ?" Look at the different groups in your periodic table and see if you can find the relationship.

 

[naren11]

$$n=0\rightarrow \mbox{impossible}$$
$$n=1\rightarrow l=0\rightarrow m_{l}=0$$ 1s
$$n=2\rightarrow l=1\rightarrow m_{l}=0,1$$ 2p
$$n=2\rightarrow l=0\rightarrow m_{l}=0$$ 2s
$$n=3\rightarrow l=2\rightarrow m_{l}=0,1,2$$ 3d
$$n=3\rightarrow l=1\rightarrow m_{l}=0,1$$ 3p
$$n=3\rightarrow l=0\rightarrow m_{l}=0$$ 3s
$$n=4\rightarrow l=3\rightarrow m_{l}=0,1,2,3$$ 4f
$$n=4\rightarrow l=2\rightarrow m_{l}=0,1,2$$ 4d
$$n=4\rightarrow l=1\rightarrow m_{l}=0,1$$ 4p
$$n=4\rightarrow l=0\rightarrow m_{l}=0$$ 4s

ok, $$l=0\rightarrow s > orbital$$
$$l=1\rightarrow p > orbital$$
$$l=2\rightarrow d > orbital$$
$$l=3\rightarrow f > orbital$$

i guess iam right so far..

what is the formula we use to find the number of electrons in this case?
if u don't mind, could you show me how would it look like for sodium?

 

[dextercioby]

$$_{11}^{23}Na:1s^{3}2s^{3}2p^{5}$$

Daniel.

 

[naren11]

the p shell would have maximum of 7 electrons?
d > 10 electrons
f > 15 ?

 

[Gokul43201]

the p shell would have maximum of 7 electrons?
d > 10 electrons
f > 15 ?

No; how did you get these numbers ? You are making a mistake somewhere. The first two numbers are not even multiples of 3 (which they will have to be, since each m_l has 3 allowed values of m_s). I've written the formula which Dexter suggested, in my previous post. You just have to modify this with the new numbers for m_l, m_s.

 

[naren11]

ok, so this is how would the electron config. look like..?

$$1s^{3}2s^{3}2p^{6}3s^{3}3p^{6}4s^{3}3d^{9}4p^{6}4d^{9}4f^{12}$$

 

[Gokul43201]

Correct ! And don't forget 5s, 5p and 6s. If you're sticking with the Earth based ordering, these would figure before the 4f.

Of course, the ordering may change but that is surely beyond the scope of this problem.

 

[naren11]

im only doing upto fourth energy level. Just to make sure, since this is not an Earth based ordering, 5s, 5p, and 6s wouldn't come..right?

thanks alot

 

[Gokul43201]

im only doing upto fourth energy level.

You mean fourth shell. There is a difference (except in the Bohr Model)

Just to make sure, since this is not an Earth based ordering, 5s, 5p, and 6s wouldn't come..right?

thanks alot

This is hard to tell. If you are setting an energy cutoff, you will have to calculate the energies of the different states in the other planet. This is not a simple thing to do, and is surely beyond the scope of this problem.

I suggest you stick to Earth based ordering (even if it doesn't apply in the other planet) and put in 5s, 5p and 6s in the correct (earth-based) places.