Which one of the following statements is correct?
(a) Manganese salts give a violet borax bead test

Q: Why does AgCl dissolve in ammonia but AgI does not?

AgI has a much lower solubility product ($K_{sp}$) than AgCl. The concentration of $Ag^+$ ions provided by AgI in solution is too low to shift the equilibrium and form the diamminesilver(I) complex with ammonia.

Q: What is the formula for Prussian blue?

Prussian blue is iron(III) ferrocyanide, with the chemical formula $Fe_4[Fe(CN)_6]_3$. It is formed when $Fe^{3+}$ reacts with $[Fe(CN)_6]^{4-}$.

Q: What happens to Mn in the reducing flame of the borax bead test?

In the reducing flame, the manganese ion is reduced from $Mn^{3+}$ to $Mn^{2+}$. The $Mn^{2+}$ state in the borax glass forms a colourless bead.

Q: How is temporary hardness of water removed by boiling?

Boiling causes the soluble calcium bicarbonate to thermally decompose into insoluble calcium carbonate, water, and carbon dioxide gas. The precipitate can then be filtered out.

Q: Does AgBr dissolve in ammonia?

Yes, AgBr dissolves in concentrated ammonia, but it is less soluble than AgCl. AgI, however, is completely insoluble even in concentrated ammonia.

Question

Which one of the following statements is correct?

Options:

(A) Manganese salts give a violet borax bead test in the reducing flame

(B) From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl

(C) Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution

(D) On boiling a solution having $K^+$, $Ca^{2+}$ and $HCO_3^-$ ions, we get a precipitate of $K_2Ca(CO_3)_2$

Sigma Percentile · Accepted Answer

Qualitative analysis is like being a chemical detective. You are given a set of clues—colors, precipitates, and reactions—and you have to deduce the truth. Let's walk through the four statements provided in this question and uncover the chemistry behind each one.

### Analyzing Option A: The Borax Bead Test

The borax bead test is a classic preliminary test for identifying transition metals. When a manganese salt is heated in a borax bead, it exhibits distinct colors depending on the nature of the flame. In an **oxidizing flame**, manganese forms $Mn^{3+}$, which gives a beautiful **pink or amethyst** color. However, in a **reducing flame**, it is reduced to $Mn^{2+}$, which makes the bead **colourless**. The statement claims it gives a violet bead in a reducing flame, which is factually incorrect.

### Analyzing Option B: The Solubility of Silver Halides

This is a fundamental concept in the qualitative analysis of halides. When you have a mixed precipitate of silver chloride ($AgCl$) and silver iodide ($AgI$), their behavior towards aqueous ammonia ($NH_3$) is very different.

Silver chloride reacts with ammonia to form a highly stable, soluble complex called diamminesilver(I) chloride:
$$AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]Cl$$

On the other hand, silver iodide has a much lower solubility product ($K_{sp}$). It is so insoluble that the concentration of $Ag^+$ ions it provides is insufficient to form the complex with ammonia. Therefore, **$AgI$ remains undissolved**. This makes statement (b) absolutely correct!

### Analyzing Option C: The Prussian Blue Test

When ferric ions ($Fe^{3+}$) react with potassium ferrocyanide ($K_4[Fe(CN)_6]$), a highly characteristic reaction occurs. The product is iron(III) ferrocyanide, a complex with the formula $Fe_4[Fe(CN)_6]_3$. This compound is famously known as **Prussian blue**. It forms a deep blue precipitate, not a deep green one. Hence, statement (c) is incorrect.

### Analyzing Option D: Temporary Hardness of Water

A solution containing $Ca^{2+}$ and $HCO_3^-$ ions is essentially hard water (temporary hardness). When you boil this solution, the soluble calcium bicarbonate decomposes into insoluble calcium carbonate, water, and carbon dioxide gas:
$$Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \downarrow + H_2O + CO_2 \uparrow$$

The precipitate formed is pure **calcium carbonate ($CaCO_3$)**, a white solid. It does not form a complex double salt like $K_2Ca(CO_3)_2$. Thus, statement (d) is also incorrect.

### Conclusion

After carefully evaluating all the options, it is clear that only statement (b) holds true. The differential solubility of silver halides in ammonia is a powerful tool for separating and identifying them in the laboratory.

Animated Solution for Chemistry - d and f-Block Elements: Which one of the following statements is correct?

Select Answer:

Visualized Solution

$Evaluating the Statements$

$Option (a): Borax Bead Test$

$Option (b): Solubility of Silver Halides$

$Option (c): Test for Ferric Ions$

$Option (d): Boiling Bicarbonates$

$Conclusion$

The Sigma Insight: d-block Elements

Analyzing Option A

Analyzing Option B

Analyzing Option C

Analyzing Option D

Conclusion

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