Question

$\begin{array}{ccccc}N{a}_{2}S& +& X& ⟶& N{a}_2\left[Fe\right(CN{)}_5\left(NOS\right)]\\ & & & & purplecolor\end{array}$

Find the value of magnetic moment of X.

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is A 0

With sodium nitrprusside solution, ionic sulphide gives a purple colour in alkaline medium.

$N{a}_{2}S+N{a}_2\left[Fe\right(CN{)}_{5}NO]⟶N{a}_4[Fe\left(CN{)}_{5}NOS\right]$

$Purplecolour$

Here X is,
$N{a}_2\left[Fe\right(CN{)}_5\left(NO\right)]$
+2 +1
$\Rightarrow F{e}^{+2}=d^6$ , due to strong ligand pairing happens and low spin complex forms as shown in the figure.
So, it have zero unpaired electron and zero magnetic moment.
Hence, the correct option is A.