How many electrons have #n = 3#, #m_l = pm1#, and #m_s = +1/2#?
1 Answer
For
- In the
#n = 3# level, we have the values#color(green)(l = 0, 1)# , and#color(green)(2)# available, as#bb(l_max = n - 1)# and#n - 1 = 2# .
Therefore, we have the
#bb(3s)# ,#bb(3p)# , and#bb(3d)# orbitals available (#l = 0 harr s# ,#l = 1 harr p# ,#l = 2 harr d# ).
- For
#m_l = pm1# , we have exactly two specified orbitals (each orbital corresponds to one#m_l# ) for a given#bbl# . Now we examine each#l# and see what restriction#m_l = pm1# places on our set of orbitals.
For
#color(green)(l = 0)# ,#m_l = pm1# is not valid because it falls outside the range of#l# (we see that#pm1# is not in the "range" of#0# ). So, disregard this for the#3s# orbitals, because specifying#m_l = pm1# has disallowed the#3s# orbital, whose#m_l = bb(0)# .For
#color(green)(l = 1)# ,#m_l = pm1# is valid, since we would have allowed#m_l = {-1,0,+1}# . So we include two of the#3p# orbitals.For
#color(green)(l = 2)# ,#m_l = pm1# is also valid, since we would have allowed#m_l = {-2,-1,0,+1,+2}# . So we include two of the#3d# orbitals.
- For
#m_s = +"1/2"# , it just says that we specify a single electron with spin "up" of magnitude#"1/2"# . (Ordinarily#m_s# could have been#pm"1/2"# .)
As a result, our set of orbitals and electron(s) includes:
- Two
#3p# orbitals (one corresponds to#m_l = -1# , the other#m_l = +1# ) - Two
#3d# orbitals (one corresponds to#m_l = -1# , the other#m_l = +1# ) - One spin-up electron in each (
#m_s = +"1/2"# ).
Therefore, we have an (isolated, theoretical) configuration of
