Sodium nitroprusside is formulated as : | 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 nitroprusside is formulated as :

Options:

Answer:

Solution:

Sodium nitroprusside, Na₂[Fe(CN)₅NO], contains iron in the +2 oxidation state and the nitrosyl group as a neutral NO ligand (or +1 depending on the bonding model, but traditionally +2 for Fe). In this complex, iron is Fe²⁺, which has a $3 d^{6}$ configuration. Because CN⁻ and NO⁺ are strong field ligands, they cause all six electrons to pair up in the $t_{2 g}$ orbitals, making the complex diamagnetic. Therefore, the correct formulation is an NO⁺ complex of Fe²⁺, as indicated by Option B.

Chemistry - Coordination Chemistry Question with Solution | TestHub

Options:

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