$N a_{2} S + N a_{2} [F e (C N)_{5} N O] ⟶ X (p u r p l e c o l o r)$
Find the number of type of ambidentate ligands in X.

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Solution

The correct option is D 2
here X is,
$N a_{4} [F e (C N)_{5} (N O S)]$ which has 6 ligands and two ambidentate ligands $C N a n d N O S$ .

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Similar questions

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)

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)

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)

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)

\begin{matrix} N a_{2} S & + & X & ⟶ & N a_{2} [F e (C N)_{5} (N O S)] p u r p l e c o l o r \end{matrix}

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