# 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?

Jul 14, 2017

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

In ${\text{FeCl}}_{2}$, I think you know the oxidation state of iron is $+ 2$, which means its atomic electron configuration is

$\left[A r\right] 3 {d}^{6} \textcolor{red}{\cancel{\textcolor{b l a c k}{4 {s}^{2}}}}$,

or

$\underline{\uparrow \downarrow} \text{ "ul(uarr color(white)(dar))" "ul(uarr color(white)(dar))" "ul(uarr color(white)(dar))" } \underline{\uparrow \textcolor{w h i t e}{\mathrm{da} r}}$
$\underbrace{\text{ "" "" "" "" "" "" "" "" "" "" "" "" "" "" }}$
$\text{ "" "" "" "" "" "" } 3 d$

For the orbital angular momentum of the last electron in ${\text{Fe}}^{2 +}$... it's unclear to me why you would be asked for that, which is just $\textcolor{b l u e}{l = 2}$ for a $3 d$ electron...

We're going to get the magnetic moment next.

$\boldsymbol{{\mu}_{S + L} = 2.0023 \sqrt{S \left(S + 1\right) + \frac{1}{4} L \left(L + 1\right)}}$,

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...

$S = | 4 \times + \frac{1}{2} | = 2$
$L = | 2 \cdot \left(- 2\right) + \left(- 1\right) + 0 + 1 + 2 | = 2$

So, the magnetic moment is:

$\textcolor{b l u e}{{\mu}_{S + L}} = 2.0023 \sqrt{2 \left(2 + 1\right) + \frac{1}{4} \left(2 \left(2 + 1\right)\right)}$

$= 2.0023 \sqrt{7.5} = \textcolor{b l u e}{\text{5.484 Bohr Magnetons}}$

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.
• $2 S + 1$ is the spin multiplicity of the atom.
• $L = 0 , 1 , 2 , 3 , . . . \leftrightarrow S , P , D , F , . . .$

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:

${\text{Li"(g) -> "Li}}^{3 +} \left(g\right) + 3 {e}^{-}$, DeltaH_(IE_3)("Li") = "19800 kJ/mol" $\text{ } \boldsymbol{\left(1\right)}$

${\text{Li"(g) -> "Li}}^{+} \left(g\right) + {e}^{-}$, DeltaH_(IE_1)("Li") = "520 kJ/mol" $\text{ "" } \boldsymbol{\left(2\right)}$

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

${\text{Li"^(+) (g) -> "Li}}^{2 +} + {e}^{-}$, DeltaH_(IE_1)("Li"^(+),i) = ??? $\text{ } \boldsymbol{\left(3\right)}$

${\text{Li"^(2+) (g) -> "Li}}^{3 +} + {e}^{-}$, DeltaH_(IE_2)("Li"^(+),i) = ??? $\text{ } \boldsymbol{\left(4\right)}$

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

Try splitting $\left(1\right)$ 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 $1 - 3$)

$\cancel{\text{Li"^(+) (g)) + cancel(e^(-)) -> cancel("Li} \left(g\right)}$, -DeltaH_(IE_1)("Li") = -"520 kJ/mol"

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

${\text{Li"^(+) (g) -> "Li}}^{3 +} + 2 {e}^{-}$,

we currently have:

$\Delta {H}_{I {E}_{3}} \left({\text{Li") - DeltaH_(IE_1)("Li") = DeltaH_(IE_1)("Li"^(+),i) + DeltaH_(IE_2)("Li}}^{+} , i\right)$

$= \Delta {H}_{I {E}_{2}} \left({\text{Li}}^{+}\right)$ (cumulative)

$= \left(19800 - 520\right) \text{kJ/mol" = "19280 kJ/mol}$

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