$N a_{2} S + N a_{2} [F e (C N)_{5} N O] ⟶ X (p u r p l e c o l o r)$
Find the value of magnetic moment of X.

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Solution

here X is,
$N a_{4} [F e (C N)_{5} (N O S)]$
+2
so $F e^{+ 2} = d^{6}$ , due to strong ligand pairing happens and low spin complex forms.
So it have zero unpaired electron and zero magnetic moment.
$F e = d^{2} s p^{3}$ hybridization so two d-orbitals involved in hybridization.
which have 6 ligands.
and two ambidentate ligands.

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