This question involves analyzing a series of aqueous reactions and determining which statements about them are true. The image shows two reactions: first, zinc sulfate reacts with potassium ferrocyanide, and second, the filtrate from the first reaction reacts with iodine solution. Let's break down each reaction step by step.

Step 1: Analyze the First Reaction

The first reaction is: $2{\mathrm{ZnSO}}_4+K_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]\to {\mathrm{Zn}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]+2K_2\mathrm{SO}{4}_{}$

This is a double displacement reaction, not a redox reaction, because no change in oxidation states occurs. Zinc remains +2, iron remains +2, carbon remains +2, nitrogen remains -3. The product is zinc ferrocyanide, which is a white precipitate. It is often written as Zn₂[Fe(CN)₆], but the stoichiometry for a complete precipitate with Zn²⁺ and [Fe(CN)₆]⁴⁻ is actually K₂Zn₃[Fe(CN)₆]₂ or simply Zn₂[Fe(CN)₆] under certain conditions. However, the exact formula isn't critical; it's known that zinc ferrocyanide is white and insoluble.

Step 2: Analyze the Filtrate

After filtering out the white precipitate, the filtrate contains K₂SO₄ and possibly excess reactant. The filtrate is then reacted with iodine solution. Iodine (I₂) is known to react with reducing agents. Ferrocyanide ion, [Fe(CN)₆]⁴⁻, is a reducing agent and can reduce I₂ to I⁻ if present. But in the first reaction, all [Fe(CN)₆]⁴⁻ is precipitated out as zinc ferrocyanide. So, the filtrate should not contain any [Fe(CN)₆]⁴⁻; it mainly has K₂SO₄ and perhaps unreacted ZnSO₄ or K₄[Fe(CN)₆] if not in stoichiometric ratio. However, since the reaction is written with coefficients, it's likely balanced, and the filtrate has only K₂SO₄, which does not reduce iodine. Therefore, addition of filtrate to starch-I₂ solution should not cause decolorization or blue color formation because no reduction occurs. Starch gives blue color with I₂, but if I₂ is reduced, the blue color disappears. Here, since no reduction happens, the blue color persists if I₂ is added, but the statement says "addition of filtrate to starch solution" – it might imply starch solution already containing I₂ or not. Typically, it means adding filtrate to starch-I₂ mixture to test for reducing agents. Since filtrate has no reducing agent, it won't decolorize I₂, so no blue color change occurs. But the option says "gives blue colour", which would only happen if I₂ is present and not reduced. Actually, starch alone is colorless; it turns blue with I₂. So, if filtrate is added to starch solution (without I₂), no blue color. If added to starch-I₂, blue remains. The option is ambiguous, but likely it means the filtrate does not cause blue color formation because it lacks reducing agent to produce I⁻ for starch-I⁻ complex? Wait, starch gives blue with I₂, not I⁻. So, if filtrate has no effect on I₂, blue color remains if I₂ is present. But the statement says "gives blue colour", which might be misinterpreted. Actually, the correct interpretation is that since filtrate has no reducing agent, it won't reduce I₂, so when added to starch-I₂, the blue color is still seen. But the option says "gives blue colour", which is true only if I₂ is present. However, in standard tests, addition of a reducing agent to starch-I₂ decolorizes it. Here, no reducing agent, so blue color persists. So, technically, it does "give" blue color in the sense that the mixture is blue. But this is tricky.

Upon reevaluation, the filtrate from the first reaction is primarily K₂SO₄, which is not reducing. When this filtrate is added to a solution of iodine and starch, it should not affect the blue color. So, the blue color is observed, meaning the statement "Addition of filtrate to starch solution gives blue colour" is true, assuming the starch solution contains iodine. Often, "starch solution" in context means starch indicator used with iodine.

Step 3: Check the Precipitate Identity and Properties

The white precipitate is zinc ferrocyanide. The formula is often given as Zn₂[Fe(CN)₆], but sometimes as K₂Zn₃[Fe(CN)₆]₂. It is not Zn₃[Fe(CN)₆]₂; that would be for ferricyanide or different. For Zn²⁺ and [Fe(CN)₆]⁴⁻, the precipitate is Zn₂[Fe(CN)₆]. It is white and is soluble in NaOH because zinc compounds are amphoteric. Zn₂[Fe(CN)₆] dissolves in NaOH to form [Zn(OH)₄]²⁻ complex.

Step 4: Evaluate Each Statement

• Addition of filtrate to starch solution gives blue colour: True, as filtrate (K₂SO₄) does not reduce I₂, so when added to starch-I₂, blue color is seen.
• White precipitate is Zn₃[Fe(CN)₆]₂: False, the correct formula is Zn₂[Fe(CN)₆] for zinc ferrocyanide.
• The first reaction is a redox reaction: False, it is double displacement with no change in oxidation numbers.
• White precipitate is soluble in NaOH solution: True, as zinc ferrocyanide dissolves due to formation of soluble zincate ion.

Final Answer

The true statements are: "Addition of filtrate to starch solution gives blue colour" and "White precipitate is soluble in NaOH solution."

Related Topics

This involves qualitative analysis of ions, precipitation reactions, redox chemistry, and properties of coordination compounds. Key concepts include:

• Double displacement reactions and precipitation
• Redox reactions and reducing agents
• Starch-iodine test for reducing agents
• Amphoteric behavior of zinc compounds

Important Formulae

Precipitation reaction: $2{\mathrm{Zn}}^{2+}+{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\mathrm{Zn}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]\left(s\right)$

Solubility in NaOH: ${\mathrm{Zn}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]+4{\mathrm{OH}}^{-}\to 2{\left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_4\right]}^{2-}+{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$