Question

Calculate $\left| {C.F.S.E} \right|$ (mod value) is term of $Dq$ For complex ion ${\left| {Mn{F_6}} \right|^{3 - }}$.

Answer 1

A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals can lead to stabilization for some electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration.

Complete step by step answer:
C.F.S.E value = $- \left( {3 \times 0.4{\Delta _0}} \right) + \left( {1 \times 0.6{\Delta _0}} \right)$ = $- 0.6{\Delta _0}$​= $- 6Dq$
$\Rightarrow \left| {C.F.S.E} \right|$ = $6Dq$
Hence the CFSE value of ${\left| {Mn{F_6}} \right|^{3 - }}$ as calculated from the above procedure is $6Dq$

Additional Information :
- Describe the Hybridization of $Mn$ in ${\left[ {Mn{F_6}} \right]^{3 - }}$ .
Ans: Hybridization of Mn in ${\left[ {Mn{F_6}} \right]^{3 - }}$ is ${d^2}s{p^3}$.
How about we investigate an external electronic setup of unbiased $Mn$
$4s3{d^5}4{p^0}$
However, $Mn$ in ${\left[ {Mn{F_6}} \right]^{3 - }}$ is +3. So now $Mn$ has
$4s3{d^5}4{p^0}$
$F$ will take 5 void orbitals $\left( {1s,1d,3p} \right)$. At that point two electrons in the d orbital will combine and clear one orbital for the 6th $F$ .
- A particle of ${\left[ {Mn{F_6}} \right]^{3 - }}$ has octahedral shape
- Hybridization is ${d^2}s{p^3}$ in light of the fact that f is a frail ligand and matching of electrons is absurd. Or on the other hand, $Mn$ oxidation state is +3. Along these lines, the electronic setup is $4{s^0}3{d^4}$ so an empty orbital will shape hybridization. It is a manganese particle +3 and 6 fluoride particles that are complex on their external shells to fill the manganese octet that is the reason it is known as a complex.

Note: The principle use of manganese tetrafluoride is in the cleansing of natural fluorine. Fluorine gas is delivered by electrolysis of anhydrous hydrogen fluoride (with a modest quantity of potassium fluoride added as a help electrolyte) in a Moissan cell.