`Fe^(3+)overset(SNC^(Θ)("Excess"))rarrunderset("Bloodre"d)(A)overset(F^(ΘExcess))rarrunderset("Colourless")(B)`
Identify `A` and `B`
(a) Write the `IUPAC` name of `A` and `B`
(b) Find out the spin only magnetic moment of `B` .

Explanation

`NaS_(2)O_(3)+2HCI rarr 2NaCI +H_(2)O+SO_(2) +S`
A is `[Fe(SCN)(H_(2)O)_(5)]^(2+)`
The `IUPAC` name is pentaaquathiocyanatoferrate(III) ion `B` is `[FeF_(6)]^(3-)`
The `IUPAC` mane is hexafluridoferrate(III)
`Fe^(3+) + SCNbar rarr [Fe(SCN)(H_(2)O)_(5)]^(2+)`
`Fe[SCN(H_(2)O)_(5)]^(2+)+6F rarr [FeF_(6)]^(3-) + SCN^(Θ) +5H_(2)O`
`B[FeF_(6)]^(3-)`
The oxidation number of ion is + 3 Atomic number = 26
`Fe = [Ar]3d^(6) 4s^(2)`
`Fe =[Ar]3d^(5)`
Which means five unpaired electrons
Spin magnetic moment ` sqrt(n(+2))`
`=sqrt(5(5+2)) =sqrt35BM`
.