When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is solubl

Q: Why doesn't Zinc fit the description?

While Zinc is amphoteric and its hydroxide dissolves in excess NaOH, the resulting Zinc Oxide ($ZnO$) is not typically used as a primary adsorbent in chromatography like Alumina is.

Q: What is the formula of the soluble complex formed?

The soluble complex is Sodium metaaluminate, which can be written as $NaAlO_2$ or as the hydrated complex ion $Na[Al(OH)_4]$.

Q: Why does the precipitate dissolve in excess NaOH?

Because $Al(OH)_3$ is amphoteric. It reacts with the strong base ($OH^-$ ions) to form a soluble complex ion, $[Al(OH)_4]^-$, effectively dissolving the solid.

Q: What happens if we use $NH_4OH$ instead of $NaOH$?

$Al(OH)_3$ will precipitate out, but it will NOT dissolve in excess $NH_4OH$ because ammonium hydroxide is a weak base and cannot force the amphoteric hydroxide to act as an acid.

Q: What is the color of the iron precipitate?

Fe(II) gives a green precipitate of $Fe(OH)_2$, and Fe(III) gives a brown precipitate of $Fe(OH)_3$. Neither of these dissolves in excess NaOH.

Question

When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is

Options:

(A) Zn

(B) Ca

(C) Al

(D) Fe

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are a chemical detective, and you've just been handed a mystery metal, $M$. Your first test is to treat it with sodium hydroxide ($NaOH$). Immediately, a white, gelatinous precipitate, $X$, forms in your test tube.

But the investigation doesn't stop there. When you add an excess of $NaOH$, the precipitate completely dissolves, leaving a clear solution. Finally, when you take that precipitate $X$ and heat it strongly, it transforms into an oxide that is famously used as an adsorbent in chromatography.

### The Amphoteric Clue
Let's decode the first major clue: the precipitate dissolves in excess base. This is the classic signature of an **amphoteric** metal. Amphoteric substances can act as both acids and bases. When the initial hydroxide precipitate reacts with more strong base, it forms a soluble complex ion.

Metals like aluminum ($Al$), zinc ($Zn$), lead ($Pb$), and tin ($Sn$) are well-known for this behavior. However, the second clue is the clincher. The oxide formed upon heating is used in chromatography. **Alumina** ($Al_2O_3$) and **Silica** ($SiO_2$) are the undisputed kings of chromatography adsorbents. Since our starting material is a metal, it must be aluminum!

### The Chemical Confirmation
Let's verify our hypothesis by writing out the chemical reactions. When aluminum ions react with hydroxide ions, they form aluminum hydroxide:
$$Al^{3+} + 3NaOH \rightarrow Al(OH)_3 \downarrow + 3Na^+$$
This $Al(OH)_3$ is exactly the white, gelatinous precipitate $X$ mentioned in the problem.

Now, what happens when we add excess sodium hydroxide? The amphoteric aluminum hydroxide reacts further to form sodium metaaluminate, which is highly soluble in water:
$$Al(OH)_3 + NaOH \rightarrow NaAlO_2 + 2H_2O$$

Finally, let's heat the precipitate $X$. Strong heating of aluminum hydroxide drives off water, leaving behind aluminum oxide:
$$2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$$
This confirms that the oxide is indeed alumina, perfectly matching all the clues.

### Why Not the Others?
Before we wrap up, let's quickly see why the other options fail. Zinc forms zinc oxide ($ZnO$), which isn't a primary chromatography adsorbent. Iron forms a green or brown precipitate ($Fe(OH)_2$ or $Fe(OH)_3$) that doesn't dissolve in excess base. And calcium forms a white precipitate ($Ca(OH)_2$) that also remains insoluble in excess base.

By carefully following the chemical breadcrumbs, we've successfully identified the mystery metal as aluminum!

Animated Solution for Chemistry - s and p-Block Elements: When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is

Select Answer:

Visualized Solution

$M N a O H X N a O H Y$

$X + N a O H \to Y$

$A l^{3 +} + 3 O H^{-} \to A l (O H)_{3}$

$A l (O H)_{3} + N a O H \to N a A l O_{2} + 2 H_{2} O$

$2 A l (O H)_{3} Δ A l_{2} O_{3} + 3 H_{2} O$

$M = A l$

$Z n, F e, C a$

The Sigma Insight: Group 13 Elements

Analyzing the Setup Imagine you are a chemical detective, and you've just been handed a mystery metal, $M$

The Amphoteric Clue Let's decode the first major clue: the precipitate dissolves in excess base

The Chemical Confirmation Let's verify our hypothesis by writing out the chemical reactions

Why Not the Others? Before we wrap up, let's quickly see why the other options fail

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Animated Solution for Chemistry - s and p-Block Elements: When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MNaOH​XNaOH​Y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X+NaOH→Y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Al3++3OH−→Al(OH)3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Al(OH)3​+NaOH→NaAlO2​+2H2​O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2Al(OH)3​Δ​Al2​O3​+3H2​O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=Al

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Zn,Fe,Ca