
Aqueous solution of Nickel sulphate on treating with pyridine and then adding a solution of sodium nitrate gives dark blue crystals of:
A. \[[Ni{(py)_4}]S{O_4}\]
B. \[[Ni{(py)_2}{(N{O_2})_2}]\]
C. \[[Ni{(py)_4}{(N{O_2})_2}]\]
D. \[{[Ni{(py)_3}(N{O_2})]_2}S{O_4}\]
Answer
561.6k+ views
Hint: Nickel sulphate has the molecular formula \[NiS{O_4}\]. In an aqueous solution, it dissociates to give nickel ion and sulphate ion. Nickel is present under d-block elements since nickel has outermost electrons present in the d- subshell. The d-block elements exhibit variable valency and are mostly formed as stable complexes. The ionization of nickel sulphate in an aqueous solution can be written as,
\[NiS{O_4} \to N{i^{2 + }} + S{O_4}^{2 - }\].
Complete answer:
> When nickel sulphate is treated with pyridine, nickel (ii) pyridine complex is formed. Pyridine is a strong field ligand and hence, it replaces sulphate from nickel sulphate solution. The reaction can be written as follows,
\[NiS{O_4} + Py \to {[Ni{(Py)_4}]^{2 + }}S{O_4}^{2 - }\]
> The formed complex is further treated with sodium nitrate, it forms dark blue crystals. The reaction can be written as follows,
\[{[Ni{(Py)_4}]^{2 + }}S{O_4}^{2 - } + NaN{O_3} \to [Ni{(Py)_4}{(N{O_2})_2}] + N{a_2}S{O_4}\].
> The complex appeared as the dark blue crystal is Dinitritotetrapyridine-N-nickel (II).
> Thus, an aqueous solution of Nickel sulphate on treating with pyridine and then adding a solution of sodium nitrate gives dark blue crystals of \[[Ni{(Py)_4}{(N{O_2})_2}]\].
So, the correct answer is Option C.
Note:
\[NiS{O_4} \to N{i^{2 + }} + S{O_4}^{2 - }\].
Complete answer:
> When nickel sulphate is treated with pyridine, nickel (ii) pyridine complex is formed. Pyridine is a strong field ligand and hence, it replaces sulphate from nickel sulphate solution. The reaction can be written as follows,
\[NiS{O_4} + Py \to {[Ni{(Py)_4}]^{2 + }}S{O_4}^{2 - }\]
> The formed complex is further treated with sodium nitrate, it forms dark blue crystals. The reaction can be written as follows,
\[{[Ni{(Py)_4}]^{2 + }}S{O_4}^{2 - } + NaN{O_3} \to [Ni{(Py)_4}{(N{O_2})_2}] + N{a_2}S{O_4}\].
> The complex appeared as the dark blue crystal is Dinitritotetrapyridine-N-nickel (II).
> Thus, an aqueous solution of Nickel sulphate on treating with pyridine and then adding a solution of sodium nitrate gives dark blue crystals of \[[Ni{(Py)_4}{(N{O_2})_2}]\].
So, the correct answer is Option C.
Note:
> The molecular formula for pyridine is \[{C_6}{H_5}N\]. It is a neutral and strong magnetic field ligand. Generally, pyridine is used as a solvent. It is used in the preparation of vitamins, drugs, and insecticides.
> Nickel sulphate present in an aqueous solution can be obtained as hexahydrate and heptahydrate.
> Nickel sulphate hexahydrate is blue to emerald in colour.
> Nickel sulphate heptahydrate is green in colour.
> Nickel sulphate helps to synthesize other nickel compounds.
> It acts as mordants in dyes. It is also used in coatings and ceramics. Sodium nitrate is a white crystalline solid.
> It is used as a fertilizer.
> It is also used in explosives and propellants.
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