Question #a3028

1 Answer
Dec 23, 2016

Answer:

Here's what I got.

Explanation:

As you know, we can use four quantum numbers to describe the location and spin of an electron in an atom.

figures.boundless.com

These four quantum numbers describe

  • #n-># the energy shell in which the electron is located
  • #l-># the subshell in which the electron resides
  • #m_l -># the exact orbital that holds the electron
  • #m_s -># the spin of the electron

As you can see, the exact orbital that holds the electron is given by the magnetic quantum number, #m_l#. In other words, you can only know the identity of the orbital, i.e. its orientation, if you know the value of #m_l#.

One exception to this rule is actually your first example

#n=1, l=0#

For the first energy shell, which is described by #n=1#, you can only have one value for #l#. Consequently, you can only have one value for #m_l#, since

#m_l = {-l, ..., -1, 0, 1, ..., l}#

Therefore, you can say for sure that an electron that has #n=1# and #l=0# will reside in the #1s# orbital, since that's the only value that can exist for #m_l#.

For your second set

#n=3, l=1#

you must keep in mind that you have

  • #l=0 -># the s subshell
  • #l=1 -># the p subshell
  • #l=2 -># the d subshell
  • #l=3 -># the f subshell
  • #vdots#

and so on. This means that this set describes an electron located in the third energy shell, since #n=3#, and also in the #p# subshell, since #l=1#.

However, you have #3# possible values for #m_l# here

#m_l = {-1, 0, 1}#

This means that you can't say for sure which exact orbital holds this electron. The same can be said for the third set

#n=4, l=2#

since the #d# subshell can have

#m_l = {-2, -1, 0, 1, 2}#

On the other hand, the fourth set is not valid because it doesn't follow the rule

#l = {0, 1, ..., (n-1)}#

For #n=1# you can only have #l=0#, which is why the set is said to be invalid.