In a compound #FeCl_2#, the orbital angular momentum of the last electron in its cation and magnetic moment (in Bohr Magneton ) of this compound is?

2 Answers
Jul 14, 2017

I'll answer the first question, but you should ask one question per question page.


In #"FeCl"_2#, I think you know the oxidation state of iron is #+2#, which means its atomic electron configuration is

#[Ar] 3d^6 color(red)(cancel(color(black)(4s^2)))#,

or

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

For the orbital angular momentum of the last electron in #"Fe"^(2+)#... it's unclear to me why you would be asked for that, which is just #color(blue)(l = 2)# for a #3d# electron...

We're going to get the magnetic moment next.

#bb(mu_(S+L) = 2.0023sqrt(S(S+1) + 1/4 L(L+1)))#,

where:

  • #g = 2.0023# is the electron-spin g-factor, or gyromagnetic ratio.
  • #S = M_S = sum_i |m_(s,i)|# is the total spin angular momentum of the electrons in the atom; #m_(s,i)# is the spin quantum number of electron #i# in orbital #i#.
  • #L = M_L = sum_i |m_(l,i)|# is the total orbital angular momentum of all the electrons in the atom; #m_(l,i)# is the magnetic quantum number of electron #i# in orbital #i#.

With #4# unpaired electrons (paired spins cancel out), we have...

#S = |4 xx +1/2| = 2#
#L = |2 cdot (-2) + (-1) + 0 + 1 + 2| = 2#

So, the magnetic moment is:

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

#= 2.0023sqrt7.5 = color(blue)("5.484 Bohr Magnetons")#

while the measured magnetic moment is about #5.1 - 5.5# in Table 10.3 in Inorganic Chemistry by Miessler et al. (pg. 361 / PDF-pg. 376).


And I would think that this is also important... In a free-ion field, we omit the total angular momentum #J# and just have this free-ion term symbol:

#bb(""^(2S+1) L)#,

where:

  • #S# and #L# are as defined before.
  • #2S+1# is the spin multiplicity of the atom.
  • #L = 0, 1, 2, 3, . . . harr S, P, D, F, . . . #

We already drew the ground-state configuration above. We thus have the free-ion ground-state term symbol for a #d^6# configuration, and it is #color(blue)(""^(5) D)#.

Jul 15, 2017

Regarding the second question...


You gave for the cumulative ionization energies:

#"Li"(g) -> "Li"^(3+) (g) + 3e^(-)#, #DeltaH_(IE_3)("Li") = "19800 kJ/mol"# #" "bb((1))#

#"Li"(g) -> "Li"^(+) (g) + e^(-)#, #DeltaH_(IE_1)("Li") = "520 kJ/mol"# #" "" "bb((2))#

And you are asked for what seems to be the individual ionization energies:

#"Li"^(+) (g) -> "Li"^(2+) + e^(-)#, #DeltaH_(IE_1)("Li"^(+),i) = ???# #" "bb((3))#

#"Li"^(2+) (g) -> "Li"^(3+) + e^(-)#, #DeltaH_(IE_2)("Li"^(+),i) = ???# #" "bb((4))#

We can first by use Hess's Law to form the proper reaction by cancellation of matching substances on both sides.

Try splitting #(1)# into three steps and you'll see this more clearly:

#1)" "cancel("Li"(g)) -> cancel("Li"^(+) (g)) + cancel(e^(-))#, #DeltaH_(IE_1)("Li") = "520 kJ/mol"#
#2)" ""Li"^(+)(g) -> "Li"^(2+) (g) + e^(-)#, #DeltaH_(IE_1)("Li"^(+),i) = ???#
#3)" ""Li"^(2+)(g) -> "Li"^(3+) (g) + e^(-)#, #DeltaH_(IE_2)("Li"^(+),i) = ???#
#DeltaH_(IE_3)("Li") = "19800 kJ/mol"# (after adding steps #1-3#)

#cancel("Li"^(+) (g)) + cancel(e^(-)) -> cancel("Li"(g))#, #-DeltaH_(IE_1)("Li") = -"520 kJ/mol"#

Thus, by inspection of the remaining reaction (due to adding them)

#"Li"^(+) (g) -> "Li"^(3+) + 2e^(-)#,

we currently have:

#DeltaH_(IE_3)("Li") - DeltaH_(IE_1)("Li") = DeltaH_(IE_1)("Li"^(+),i) + DeltaH_(IE_2)("Li"^(+),i)#

#= DeltaH_(IE_2)("Li"^(+))# (cumulative)

#= (19800 - 520) "kJ/mol" = "19280 kJ/mol"#

We have two unknowns here, but not enough equations. To be continued...?