Question

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?

Answer 1

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

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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).

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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)`.

Answer 2

Regarding the second question...

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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...?