When H<sub>2</sub>S is passed through Hg<sub>2</sub>S, we get:

255) When H₂S is passed through Hg₂S, we get:

Answer is:

Hg₂S + Hg

Inorganic Chemistry MCQ with Answers

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256) How do we differentiate between Fe³⁺ and Cr³⁺ in group III?

Answer is:

Both (b) and (c)

257) In the precipitation of the iron group in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to:

Answer is:

Decrease concentration of OH⁻ ions

Explanation:

As NH₄Cl is a strong electrolyte, it supresses the ionization of NH₄OH. So, the concentration of OH⁻ ions in the solution is decreased, but it is sufficient to precipitate the group III basic radicals because the solubility product of group III hydroxides is lower than group IV, V and VI hydroxides.

258) Ferric iron forms a Prussian blue coloured ppt. due to:

Answer is:

Fe₄[Fe(CN)₆]₃

259) A black sulphide is formed by the action of H₂S on:

Answer is:

Cupric chloride

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255) When H2S is passed through Hg2S, we get:

Hg2S + Hg

Inorganic Chemistry MCQ with Answers

Related Inorganic Chemistry MCQ with Answers

256) How do we differentiate between Fe3+ and Cr3+ in group III?

Both (b) and (c)

257) In the precipitation of the iron group in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to:

Decrease concentration of OH− ions

258) Ferric iron forms a Prussian blue coloured ppt. due to:

Fe4[Fe(CN)6]3

259) A black sulphide is formed by the action of H2S on:

Cupric chloride

Inorganic Chemistry MCQ with Answers

255) When H₂S is passed through Hg₂S, we get:

Hg₂S + Hg

256) How do we differentiate between Fe³⁺ and Cr³⁺ in group III?

Decrease concentration of OH⁻ ions

Fe₄[Fe(CN)₆]₃

259) A black sulphide is formed by the action of H₂S on: