Question #f307f

1 Answer
Jan 17, 2018

Answer:

#n=3,l=2,m_l = 1, m_s = -1/2#

Explanation:

figures.boundless.com

The first quantum number set given to you is not correct because the spin quantum number, #m_s#, can only take two possible values.

#m_s = {+1/2, - 1/2}#

So the quantum number set that has

#m_s = +1#

cannot be used to describe the location and the spin of an electron inside an atom.

The second quantum number set is also incorrect because the angular momentum quantum number, #l#, cannot take the same value as the principal quantum number, #n#.

Similarly, the third quantum number set is incorrect because the angular momentum quantum number cannot take a negative value.

More specifically, the angular momentum quantum number can take

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

So the quantum number set that has

#n=1, l =1#

is not valid because the value of the angular momentum quantum number is not possible given the value of the principal quantum number.

The set that has

#l = -3#

is not valid because of the negative value assigned to the angular momentum quantum number.

The fourth quantum number set is correct because all the four quantum numbers take accepted values.

#n = 3, l = 2, m_l = 1, m_s = -1/2#

This set describes an electron located in the third energy shell, in the #3d# subshell, let's say in the #3d_(xy)# orbital, that has spin-down.

The value of the magnetic quantum number, #m_l#, is valid here because this quantum number can take the possible values

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

So for

#l = 2#

you can have

#m_l= 1#

as a valid value for the magnetic quantum number.