Question #48bcb

When the subshell is half-filled, #mu_(S+L) = mu_S#.

Besides that, when #1/4L(L+1)# is small compared to #S(S+1)#, #mu_(S+L) ~~ mu_S# is a good approximation. That is when

• We have a light atom (which occupies orbitals of low #l#, i.e. #s# or #p# orbitals).
• The atom is isolated (i.e. not interacting with other atoms).

TOTAL MAGNETIC MOMENT

From Miessler et al., pg. 360, the total magnetic moment in units of Bohr magnetons is:

#barul|stackrel(" ")(" "mu_(S+L) = g_esqrt(S(S+1) + 1/4L(L+1))" ")|#

where:

• #g_e = 2.0023193043617# (note that the book had a typo, and it's not #2.00023#) is the electron g-factor.
• #S = |sum_i m_s(i)|# is the total spin angular momentum for the atom, summing over the #m_s# values for the #i#th electron in each orbital.
• #L = |sum_i m_l(i)|# is the total orbital angular momentum for the atom, calculated in a similar manner, summing over the #m_l# values for the #i#th electron in each orbital.

Because two electrons in the same orbital must be opposite spins, the sums only pick up nonzero values for singly-occupied orbitals, i.e. it allows determination of how many unpaired electrons an atom has.

SPIN-ONLY MAGNETIC MOMENT

For finding the number of unpaired electrons, it is generally sufficient, for light atoms, to ignore the total orbital angular momentum and only calculate the spin-only magnetic moment:

#barul|stackrel(" ")(" "mu_S = g_esqrt(S(S+1))" ")|#

i.e. where #1/4L(L+1) ~~ 0# so that #mu_(S+L) ~~ mu_S#.

EXAMPLE: IRON(II) CATION

Let's test it out on #"Fe"^(2+)# atom, with electron configuration #[Ar] 3d^6 4s^0#.

#underbrace(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr)))_(3d)#

For this, we'll first need to calculate #L# and #S#. Remember that the #d# orbitals have #m_l = {-2,-1,0,+1,+2}#. Each square bracket indicates a separate orbital.

#L = |[2 cdot -2] + [-1] + [0] + [1] + [2]|#

#= 2#

#S = [1/2 + (-1/2)] + [1/2] + [1/2] + [1/2] + [1/2]#

#= 2#

Therefore, theoretically, the total magnetic moment is:

#color(blue)(mu_(S+L)(Fe^(2+))) = 2.0023sqrt(2(2+1) + 1/4 2(2+1))#

#=# #color(blue)ul"5.484 BM"#

Theoretically, the spin-only magnetic moment is:

#color(blue)(mu_S(Fe^(2+))) = 2.0023sqrt(2(2+1)) = color(blue)ul"4.905 BM"#

The experimental values lie in the range #5.1# to #5.5# #"BM"# (Miessler et al., pg. 361). So, in the case of #"Fe"^(2+)#, it is necessary to calculate #mu_(S+L)# for agreement with experiment.

FINAL REMARKS

If in a classroom setting, you were given simply #mu_(S+L) = 5.2#, for relatively light atoms, you could still use the equation for #mu_S# as an approximation to determine that the number of unpaired electrons is #4#, since it is easier to solve:

#5.2 = 2.0023sqrt(S(S+1)) -= mu_(S+L)#

#=> S^2 + S - 6.745 = 0#

Solving this quadratic leads to a physical solution of #S = 2.145#. From this, the number of unpaired electrons seems to be:

#color(blue)("Number of unpaired electrons") = S/|m_s|#

#= 2.145/(1//2) = 4.29 ~~ color(blue)(4)#

So it looks like treating #mu_(S+L) ~~ mu_S# for #"Fe"^(2+)# is OK to determine the number of unpaired electrons.

For #"Fe"^(3+)# it should be that #mu_(S+L) = mu_S# perfectly, since the #3d# is half-filled there.

In fact, the approximation works better as the number of unpaired electrons increases, because #S# gets large compared to #L#, and also, #L = 0# for half-filled subshells.