Answer
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Hint: Remember that nitric oxide is a non-innocent ligand i.e. a ligand in a metal complex where its oxidation state is unclear. This concept results in the assumption that redox reactions in metal complexes are either localized to their metals or ligands, which is a useful simplification of the overarching concept.
Complete step by step answer:
Let us now analyse Sodium Nitroprusside as a compound to help us answer this question.
Sodium Nitroprusside is an inorganic compound with the formula, $N{{a}_{2}}[Fe{{(CN)}_{5}}NO]$ which is usually found in its dihydrate form, $N{{a}_{2}}[Fe{{(CN)}_{5}}NO]\cdot 2{{H}_{2}}O$. It is a red-coloured sodium salt which dissolves in either water or ethanol to give solutions which contain the complex dianion ${{[Fe{{(CN)}_{5}}NO]}^{2-}}$.
We observe that Nitroprusside is a complex anion which possesses an octahedral iron(III) centre (since Iron’s oxidation state in this compound is +3) which is surrounded by five tightly bound cyanide ligands and one linear nitric oxide ligand with its bond angle Fe-N-O angle being ${{176.2}^{\circ }}$ according to its structure which VSEPR theory.
Now, even though nitric oxide is a non-innocent ligand, the linear Fe-N-O angle results in the relatively short N-O distance of 113 pm and the subsequently high stretching frequency of $1947c{{m}^{1-}}$ , the complex is thus found to be containing an $N{{O}^{+}}$ ligand. Consequently, iron is assigned an oxidation state of +2; as a result of which the iron centre has a diamagnetic low-spin d6 electron configuration but a paramagnetic long-lived metastable state has been observed by EPR spectroscopy.
Thus, we can safely conclude that the answer to this question is (c).
Note: This compound decomposes to form sodium ferrous ferrocyanide, sodium ferrocyanide, nitric oxide, and cyanogen at about 450 °C. It decomposes in aqueous acid to liberate hydrocyanic acid (HCN) and if shielded from light, the concentrated solution is stable for more than two years at room temperature.
Complete step by step answer:
Let us now analyse Sodium Nitroprusside as a compound to help us answer this question.
Sodium Nitroprusside is an inorganic compound with the formula, $N{{a}_{2}}[Fe{{(CN)}_{5}}NO]$ which is usually found in its dihydrate form, $N{{a}_{2}}[Fe{{(CN)}_{5}}NO]\cdot 2{{H}_{2}}O$. It is a red-coloured sodium salt which dissolves in either water or ethanol to give solutions which contain the complex dianion ${{[Fe{{(CN)}_{5}}NO]}^{2-}}$.
We observe that Nitroprusside is a complex anion which possesses an octahedral iron(III) centre (since Iron’s oxidation state in this compound is +3) which is surrounded by five tightly bound cyanide ligands and one linear nitric oxide ligand with its bond angle Fe-N-O angle being ${{176.2}^{\circ }}$ according to its structure which VSEPR theory.
Now, even though nitric oxide is a non-innocent ligand, the linear Fe-N-O angle results in the relatively short N-O distance of 113 pm and the subsequently high stretching frequency of $1947c{{m}^{1-}}$ , the complex is thus found to be containing an $N{{O}^{+}}$ ligand. Consequently, iron is assigned an oxidation state of +2; as a result of which the iron centre has a diamagnetic low-spin d6 electron configuration but a paramagnetic long-lived metastable state has been observed by EPR spectroscopy.
Thus, we can safely conclude that the answer to this question is (c).
Note: This compound decomposes to form sodium ferrous ferrocyanide, sodium ferrocyanide, nitric oxide, and cyanogen at about 450 °C. It decomposes in aqueous acid to liberate hydrocyanic acid (HCN) and if shielded from light, the concentrated solution is stable for more than two years at room temperature.
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