Sodium nitropruside is :- | TestHub

Passage / Comprehension

There are many known monosubstitution products of the $[F e (CN)_{6}]^{4 -}$ ion, the most important being the red salt sodium nitrosopentacyanoferrate(II), $N a_{2} [F e (CN)_{5} NO] \cdot 2 H_{2} O$ , commonly called sodium nitroprusside.

This results from the action of nitric acid or sodium nitrite on the hexacyano complex:

$[F e (CN)_{6}]^{4 -} + 4 H^{+} + N O_{3}^{-} \to [F e (CN)_{5} NO]^{2 -} + C O_{2} + N H_{4}^{+}$

$[F e (CN)_{6}]^{4 -} + N O_{2}^{-} + H_{2} O \to [F e (CN)_{5} NO]^{2 -} + C N^{-} + 2 O H^{-}$

The anion of this compound, which is diamagnetic, contains coordinated NO with an interatomic distance of $1.13 \overset{˚}{A}$ and a stretching vibration at $1947 c m^{- 1}$ . With sulfide ion, it gives the purple anion $[F e (CN)_{5} NOS]^{4 -}$ , a reaction which forms the basis of a sensitive test for sulfide.

Sodium nitropruside is :-

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Answer:

A, B, D

Solution:

Based on the passage provided, here is the solution for the characteristics of sodium nitroprusside, $N a_{2} [F e (CN)_{5} NO]$ :

Diamagnetic complex (A): The passage explicitly states that the anion $[F e (CN)_{5} NO]^{2 -}$ is diamagnetic.

Heteroleptic complex (B): It contains two different types of ligands ( $C N^{-}$ and $N O^{+}$ ) coordinated to the central metal atom.

Fe–C linkage (D): There are five cyanide ( $C N^{-}$ ) ligands, each forming a coordination bond with Iron ( $F e$ ) through the Carbon atom.

Note: It is not paramagnetic (C is incorrect) because all electrons are paired, resulting in a diamagnetic state.

Chemistry - Coordination Chemistry Question with Solution | TestHub

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Doubts & Discussion