s-Block Elements

The compound(s) formed upon the combustion of sodium metal in the excess air is sodium peroxide (Na₂O₂) and sodium oxide (Na₂O).

$ \begin{aligned} & 4 \mathrm{Na}+\mathrm{O}_2 \rightarrow 2 \mathrm{Na}_2 \mathrm{O} \\\\ & 2 \mathrm{Na}+\mathrm{O}_2 \rightarrow \mathrm{Na}_2 \mathrm{O}_2 \end{aligned} $

The product formed the reaction is used for chlorination of water bodies. The compound Na$_2$S$_2$O$_3$ gets oxidised by chlorine to form NaHSO$_3$. The balanced chemical reaction is,

Na$_2$S$_2$O$_3$ + 4Cl$_2$ + 5H$_2$O $\to$ 2NaHSO$_4$ + 8HCl

Statement-1 :

Alkali metals dissolve in liquid ammonia to give blue solutions.

✅ True

Explanation: When alkali metals (Li, Na, K, etc.) are dissolved in liquid ammonia, they form deep blue conducting solutions at low concentrations.

Reaction:

$ \mathrm{M (s) \xrightarrow{liq.\,NH_3} M^+_{(solv)} + e^-_{(solv)}} $

The blue color arises from solvated electrons ($ e^-_{(NH_3)} $), which absorb visible light.

Statement-2 :

Alkali metals in liquid ammonia give solvated species of the type $\left[\mathrm{M}\left(\mathrm{NH}_{3}\right)_n\right]^+ (\mathrm{M} = \text{alkali metals})$.

✅ True

During dissolution, the metal forms metal cations solvated by ammonia molecules, i.e., $\mathrm{M^+}$ surrounded by several $\mathrm{NH_3}$ molecules.

$ \mathrm{M^+ + e^- \rightarrow [M(NH_3)_n]^+ + e^-_{(solv)}} $

But:

The blue color (Statement-1) is not due to $[\mathrm{M(NH_3)_n}]^+$, but due to the solvated electrons ($e^-_{(NH_3)}$).

Therefore, although both statements are true, Statement-2 does not explain the blue color.

✅ Final Answer :

Option B — Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

The chemical reaction between $\mathrm{MgSO}_4$, $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{Na}_2 \mathrm{HPO}_4$ is given below.

$\underset{\substack{\text { Magnesium } \\ \text { sulfate }}}{\mathrm{MgSO}_4}+\underset{\substack{\text { Ammonium } \\ \text { hydroxide }}}{\mathrm{NH}_4 \mathrm{OH}}+\underset{\substack{\text { Sodium hydrogen } \\ \text { phosphate }}}{\mathrm{Na}_2 \mathrm{HPO}_4} \rightarrow$\underset{\substack{\text { Magnesium ammonium } \\ \text { white precipitate }}}{\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4 \downarrow}+\underset{\text { Sodium sulfate }}{\mathrm{Na}_2 \mathrm{SO}_4}+\mathrm{H}_2 \mathrm{O}$

The magnesium ammonium phosphate hexahydrate $\left[\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}\right]$ formed has white coloured precipitate.

The other products formed during the chemical reaction include sodium sulfate and water.

Ammonium chloride is used to prevent magnesium hydroxide and ammonia solutions from precipitating.

The correct answer to the question is Option B: CaO.

Let's break down the reasoning:

CaO (Calcium oxide), also known as quicklime, vigorously reacts with water to form calcium hydroxide (Ca(OH)₂) and releases a significant amount of heat. This reaction is known as slaking of lime and can be represented as:

$ \text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 $

Additionally, calcium oxide has a strong affinity for carbon dioxide (CO₂). It absorbs CO₂ from the air to form calcium carbonate (CaCO₃) through the following reaction:

$ \text{CaO} + \text{CO}_2 \rightarrow \text{CaCO}_3 $

Thus, CaO fulfills both conditions: it absorbs CO₂ and reacts violently with water. The other substances listed do not meet these criteria:

CaCO₃: Calcium carbonate reacts with acids but does not react violently with water and does not absorb CO₂.

H₂SO₄: Sulfuric acid is a liquid that reacts vigorously with water (specifically, dilution of concentrated sulfuric acid is exothermic), but it does not absorb CO₂.

ZnO: Zinc oxide is an amphoteric oxide that does not react violently with water and does not absorb CO₂.

Therefore, the most appropriate answer is:

Option B: CaO