Relevant chemistry in Group III analysis

In qualitative analysis, both $Fe^{3+}$ and $Cr^{3+}$ are precipitated as hydroxides in Group III: $Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3 \downarrow$ $Cr^{3+} + 3OH^- \rightarrow Cr(OH)_3 \downarrow$

So, under ordinary conditions, both give hydroxide precipitates and are not easily distinguished directly.

Key difference between the hydroxides

• $Fe(OH)_3$ is less soluble and not amphoteric in the same useful way.
• $Cr(OH)_3$ is more soluble compared to $Fe(OH)_3$ and its precipitation is more sensitive to the concentration of $OH^-$.

Therefore, if the effective $OH^-$ concentration is lowered, $Cr(OH)_3$ tends to dissolve or may fail to precipitate, while $Fe(OH)_3$ still precipitates.

Effect of increasing $NH_4^+$ concentration

In Group III, $NH_4OH$ is used in the presence of $NH_4Cl$. The ionization equilibrium is: $NH_4OH \rightleftharpoons NH_4^+ + OH^-$

Increasing $NH_4^+$ shifts this equilibrium to the left due to the common ion effect, thereby decreasing $[OH^-]$.

So, increasing $NH_4^+$ ion concentration reduces the hydroxide ion concentration.

Effect of decreasing $OH^-$ concentration directly

If $[OH^-]$ is decreased, then $Cr^{3+}$ can be kept in solution more readily than $Fe^{3+}$. Hence this also helps in differentiating between them.

Evaluate options

A: by taking excess of $NH_4OH$ solution
Increasing $NH_4OH$ generally increases $OH^-$, so both hydroxides precipitate; this does not help differentiation. Hence, A is incorrect.

B: by increasing $NH_4^+$ ion concentration
This lowers $OH^-$ via common ion effect and helps separate/differentiate $Fe^{3+}$ and $Cr^{3+}$. Hence, B is correct.

C: by decreasing $OH^-$ ion concentration
This is the direct chemical reason for differentiation. Hence, C is also correct.

D: both (b) and (c)
Since both B and C are correct, D is correct.

Conclusion

The best answer is: $\boxed{D \text{ : both (b) and (c)}}$

Comparison with stored correct answer

Stored correct answer: B
Derived answer: D

I disagree with the stored answer because increasing $NH_4^+$ concentration works by decreasing $OH^-$ concentration, so both statements are chemically valid.