The addition of dilute NaOH to Cr³⁺ salt solution will give:

🧠 Cr³⁺ + Dilute NaOH → Cr(OH)₃ Precipitate

$\mathrm{Cr^{3+}}$ in solution + dilute $\mathrm{NaOH}$ → grey-green precipitate of $\mathrm{Cr(OH)_3}$.

In excess NaOH, this precipitate dissolves to give $[\mathrm{Cr(OH)_4}]^-$ (chromite). But with dilute NaOH (limited amount), precipitation stops at $\mathrm{Cr(OH)_3}$.

🗺️ Cr(OH)₃ Amphoteric Behaviour

| Reagent | Product | |---|---| | Dilute NaOH | $\mathrm{Cr(OH)_3}$ (precipitate, grey-green) | | Excess NaOH | $[\mathrm{Cr(OH)_4}]^-$ (soluble chromite) | | Acid | $\mathrm{Cr^{3+}} + 3\mathrm{H_2O}$ |

🗺️ Eliminate Other Options

• $[\mathrm{Cr(OH)_4}]^-$: requires excess NaOH; not from dilute.
• $[\mathrm{Cr(OH)_6}]^{3-}$: not a common species; chromium hexahydroxide is not formed in dilute base.
• $\mathrm{Cr_2O_3 \cdot xH_2O}$: this is a dehydrated form of Cr(OH)₃ (hydrated Cr₂O₃ is essentially the same thing), but standard nomenclature in JEE = $\mathrm{Cr(OH)_3}$.

⚡ Group III Hydroxides (Qualitative Analysis)

• $\mathrm{Cr^{3+}}, \mathrm{Al^{3+}}, \mathrm{Fe^{3+}}$ all form gelatinous hydroxides on adding $\mathrm{NH_4OH}$ or dilute NaOH.
• $\mathrm{Al(OH)_3}$ and $\mathrm{Cr(OH)_3}$ are amphoteric (dissolve in excess base).
• $\mathrm{Fe(OH)_3}$ is not amphoteric (doesn't dissolve in NaOH).

⚠️ Dilute vs Excess Matters

Always check if NaOH is dilute (limited) or excess. Amphoteric hydroxides redissolve in excess.

$\boxed{\text{Answer: (4) precipitate of } \mathrm{Cr(OH)_3}}$