# 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# isnotvalid because it falls outside the range of#l# (we see that#pm1# is not in the "range" of#0# ). So,disregard thisfor 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 includetwoof 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 includetwoof 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 **four electrons**.