Based on quantum number restrictions in atoms, what are...?

The main constraints you should remember about quantum numbers are:

• #n = 1, 2, 3, . . . #, so #n > 0# is required.
• #l = 0, 1, 2, . . . , l_max#, where #0 <= l <= n-1#.
• #m_l = {-l, -l+1, . . . , l-1, l}#, meaning that #-l <= m_l <= l#.
• #m_s = pm1/2#, meaning that it cannot be otherwise.

And so, you should have these constraints in your head and cycle through them to see which ones you should pull up.

#1)#

Keep in mind that #m_s# is independent of which orbital you are in---it does not relate to what #n#, #l#, or #m_l# happen to be.

For #l = 2# and #m_l = 0#, that gives you some sort of #d# orbital (namely, the #d_(z^2)#), AND it specifies which #d# orbital for a given energy level #n#... which actually isn't given.

So, you can choose any #n# you want, as long as #l <= n-1#. If you flip it around, #n >= l + 1#.

Since #l = 2#, the minimum #n# you can have is #n = 3#. Therefore, you can only have

#color(blue)(n = 3, 4, 5, . . . , N)#

That corresponds to #3d_(z^2)#, #4d_(z^2)#, . . . , #Nd_(z^2)# orbitals.

#2)#

Here, you have #n = 2# and #m_l = -1#, and we can freely ignore the value of #m_s# because it's a valid #m_s#. That corresponds to one of the orbitals on the second energy level.

Since #m_l# is constrained by #-l <= m_l <= l#, and it is given that #m_l = -1#, we know that #l# must be at least #1# (if #l = 0#, then #m_l < -l#, and that's not allowed). However, since #n = 2# and #l <= n-1#, #l# can be at most #1# (otherwise #l > n-1#, which is not allowed).

Therefore, #color(blue)(l = 1)#.

#3)#

Hopefully by now you've realized that #m_s# is independent of what any of the other quantum numbers are. It's a property of an electron.

And so, #m_s# can be #color(blue)(pm1/2)#, whichever you want, so long as your orbital started out empty.

#4)#

Here, we only have to focus on #l = 0#. If #l = 0#, that must be an #s# orbital. But we know that there are many #s# orbitals, namely, #1s, 2s, 3s, . . . #, all the way to the #7s# as the latest known #s# orbital.

As you now know, #m_l# is restricted in range by the value of #l#, so since #l = 0#, we know that #pml = 0# as well and thus #m_l = {0}#, i.e. #m_l# has only one option and that is ZERO.

And as we mentioned, #n = 1, 2, 3, . . . # is fair game, so in the end, we have an infinity of choices of this PAIR of quantum numbers:

#(color(blue)(n), l, color(blue)(m_l), m_s)#

#= (color(blue)(1), 0, color(blue)(0), -1/2), (color(blue)(2), 0, color(blue)(0), -1/2), (color(blue)(3), 0, color(blue)(0), -1/2), . . . #