Q.Show by a chemical reaction with water that $Na_2O$ is a basic oxide and $Cl_2O_7$ is an acidic oxide.

The idea of a diagonal relationship is one of the most elegant patterns in the s- and p-blocks. Elements in the second period often resemble the element one step down and to the right in the third period, rather than their own group neighbours. This happens because the increase in nuclear charge down a group is offset by the decrease in size across a period — so Be (group 2, period 2) ends up with properties closer to Al (group 13, period 3) than to Mg. Similarly, Li (group 1, period 2) resembles Mg (group 2, period 3) more than it does Na.

The question asks about the oxides of X and Y, where X is the element diagonally related to Be, and Y is the element diagonally related to Li.

1. Identify X and Y.

   Be is in group 2, period 2. Its diagonal partner is Al (group 13, period 3). So X = Al.

   Li is in group 1, period 2. Its diagonal partner is Mg (group 2, period 3). So Y = Mg.

2. Determine the nature of Al₂O₃ (oxide of X).

   Aluminium oxide reacts with both acids and bases. With HCl it gives AlCl₃ and water; with NaOH it gives sodium aluminate and water. This amphoteric behaviour is a direct consequence of aluminium’s intermediate electronegativity and its ability to form both cations and oxoanions. So the oxide of X is amphoteric.

3. Determine the nature of MgO (oxide of Y). …

The question tests two separate properties of water and hydrogen sulfide: physical state and acid-base nature. Both arise from the same underlying difference — the strength of intermolecular forces and the polarity of the X–H bond.

Why water is a liquid but H₂S is a gas at room temperature

Water molecules form extensive hydrogen bonds with each other. Each H₂O molecule can participate in up to four hydrogen bonds (two through its H atoms, two through its lone pairs on oxygen). These bonds are strong enough to keep water molecules clustered together as a liquid at 25°C. Hydrogen sulfide, H₂S, also has a bent shape and polar S–H bonds, but sulfur is much less electronegative than oxygen (2.58 vs 3.44 on the Pauling scale). The S–H bond is far less polar, and the hydrogen bonds that do form are very weak. Consequently, H₂S molecules are held together only by weak van der Waals forces, and it boils at –60°C — a gas at room temperature. Statement I is correct.

Why water is neutral but H₂S is acidic

Water undergoes autoionisation to a tiny extent:

with $K_w = 1.0 \times 10^{-14}$ at 25°C. The concentration of H⁺ equals that of OH⁻, so pure water is neutral (pH 7). The O–H bond is strong and not easily broken to release a proton.

Hydrogen sulfide, on the other hand, is a weak diprotic acid. The S–H bond is weaker than the O–H bond because sulfur is larger and less electronegative, so the bond is longer and more easily polarised. In water, H₂S donates a proton:

with $K_{a1} \approx 9.5 \times 10^{-8}$. This makes the solution acidic. Statement II is also correct. …

Concept & Intuition

Chlorine gas ($\mathrm{Cl_2}$) has an oxidation state of 0. When it reacts with a base like NaOH, it disproportionates — one chlorine atom is reduced (to -1, as chloride) and the other is oxidized (to a positive state). The temperature and concentration of the base determine how far the oxidation goes: cold/dilute conditions stop at the hypochlorite stage (Cl = +1), while hot/concentrated conditions push it further to chlorate (Cl = +5). This is a classic example of how reaction conditions control the product in redox chemistry.

Step-by-step reasoning

1. Identify the known product

   In both reactions, $\mathrm{NaCl}$ is formed. Chlorine in $\mathrm{NaCl}$ has oxidation state $-1$. This is the reduced product. The other chlorine-containing product (X or Y) must therefore have a positive oxidation state.

2. Analyze the cold/dilute reaction

   Cold, dilute NaOH with $\mathrm{Cl_2}$ gives sodium hypochlorite ($\mathrm{NaClO}$) as the other product.

   • In $\mathrm{ClO^-}$, oxygen is $-2$, so chlorine must be $+1$ (since $-2 + x = -1$ → $x = +1$).
   • So Y is $\mathrm{NaClO}$ and its chlorine oxidation state is $+1$.

3. Analyze the hot/concentrated reaction …

Concept & Intuition

This question tests your ability to assign oxidation states and identify which elements are oxidized (lose electrons, oxidation number increases) and which are reduced (gain electrons, oxidation number decreases). The reaction is a decomposition where a single compound breaks into simpler substances. The key is to track the oxidation numbers of each element before and after the reaction — not to guess based on common ions alone.

Step-by-step reasoning

1. Assign oxidation numbers in the reactant $\text{KClO}_3$

   • Potassium (K) is always $+1$ in its compounds.
   • Oxygen (O) is usually $-2$ (except in peroxides, etc.). Here it’s $-2$.
   • Let the oxidation number of chlorine (Cl) be $x$.
   • The compound is neutral: $+1 + x + 3(-2) = 0 \Rightarrow 1 + x - 6 = 0 \Rightarrow x = +5$. So in $\text{KClO}_3$, Cl is $+5$, O is $-2$, K is $+1$.

2. Assign oxidation numbers in the products

   • In $\text{KCl}$: K is $+1$, so Cl must be $-1$ (since $+1 + (-1) = 0$).
   • In $\text{O}_2$: elemental oxygen has oxidation number $0$ (by definition).

3. Identify changes for each element

   • Chlorine: goes from $+5$ (in $\text{KClO}_3$) to $-1$ (in $\text{KCl}$). A decrease from $+5$ to $-1$ means it gains electrons → reduction.
   • Oxygen: goes from $-2$ (in $\text{KClO}_3$) to $0$ (in $\text{O}_2$). An increase from $-2$ to $0$ means it loses electrons → oxidation.
   • Potassium: remains $+1$ throughout — no change, so neither oxidized nor reduced. …

The key concept here is the amphoteric nature of aluminium: it reacts with both acids and bases to liberate hydrogen gas. Many students mistakenly think that a metal reacting with an acid gives chlorine or oxygen, but aluminium’s behaviour is special — it displaces hydrogen from both acidic and basic solutions.

1. Reaction with dilute HCl Aluminium is more reactive than hydrogen in the reactivity series, so it displaces hydrogen from hydrochloric acid:

The gas ‘A’ is hydrogen (H₂).

2. Reaction with aqueous alkali Aluminium also reacts with a strong base like NaOH (or KOH) because it is amphoteric. It forms a soluble aluminate and hydrogen gas:

The gas ‘B’ is also hydrogen (H₂).

3. Comparing the options
   • (A) Cl₂, O₂ — incorrect; no chlorine or oxygen is produced.
   • (B) O₂, O₂ — incorrect; oxygen is not liberated in either reaction. …

A disproportionation reaction is one where a single element in a single oxidation state simultaneously gets oxidised (its oxidation number increases) and reduced (its oxidation number decreases). The key is to check each reaction for an element that appears in at least two products with different oxidation numbers, starting from the same reactant.

Let’s assign oxidation numbers step by step for each reaction.

1. Reaction A: $\mathrm{Cl_2 + 2\,NaOH \rightarrow NaCl + NaOCl + H_2O}$

   Chlorine in $\mathrm{Cl_2}$ has oxidation number 0. In $\mathrm{NaCl}$, chlorine is $-1$ (reduction). In $\mathrm{NaOCl}$, chlorine is $+1$ (oxidation). The same element (Cl) goes from 0 to both $-1$ and $+1$ — this is a classic disproportionation.

   Tip

   Chlorine disproportionation in cold alkali is a standard example: one Cl atom gains an electron, the other loses one.

2. Reaction B: $\mathrm{H_2O_2 \rightarrow 2H_2O + O_2}$

   In $\mathrm{H_2O_2}$, oxygen is $-1$ (since H is $+1$, total $+2$, so each O is $-1$). In $\mathrm{H_2O}$, oxygen is $-2$ (reduction). In $\mathrm{O_2}$, oxygen is $0$ (oxidation). Oxygen in the $-1$ state is both reduced to $-2$ and oxidised to $0$ — yes, this is disproportionation.

   Watch out

   A common mistake is to think this is just decomposition. But the key is that the same element (oxygen) changes in two directions, so it is disproportionation.

3. Reaction C: $\mathrm{2\,KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2}$ …

Concept & Intuition

Soil acidity means the soil has a high concentration of hydrogen ions (H⁺). To “reduce acidity” means to raise the pH — that is, to neutralize the excess acid. This requires adding a base (alkaline substance). But not just any base will do: you need one that is safe for plants, doesn’t wash away too quickly, and doesn’t cause sudden, harmful chemical changes. Farmers and gardeners commonly use agricultural lime — powdered limestone (calcium carbonate, CaCO₃). It reacts slowly with soil acids, releasing calcium ions (which improve soil structure) and neutralizing H⁺ to form water and carbon dioxide. Let’s examine each option.

1. Option (A): Sprinkling dil. NaOH solution

   NaOH (sodium hydroxide) is a strong base. It would indeed neutralize acid very quickly. But it is too strong and caustic — it can burn plant roots, kill beneficial soil microbes, and cause a sudden, drastic pH spike. Also, sodium ions can accumulate in soil, damaging its structure. This is not a practical or safe method for agriculture.

2. Option (B): Spraying CO₂ dissolved water

   CO₂ dissolved in water forms carbonic acid (H₂CO₃), a weak acid. Adding an acid to acidic soil would increase acidity, not reduce it. This would make the problem worse. So this is clearly wrong.

3. Option (C): Adding powdered Limestone

   Limestone is mainly CaCO₃. It reacts with soil acids like this:

The H⁺ ions are consumed, raising pH. The reaction is slow and gentle, so plants aren’t shocked. Calcium also benefits soil structure. This is the standard, recommended method for reducing soil acidity. …

The question gives you a set of chemical clues about a metal M. Let’s unpack each one and see what they reveal.

1. M readily gives MSO$_4$, which is soluble in water.

   All alkaline earth metals (Be, Mg, Ca, Sr, Ba) form soluble sulfates — but solubility decreases down the group. BeSO$_4$ and MgSO$_4$ are highly soluble; CaSO$_4$ is sparingly soluble; BaSO$_4$ is practically insoluble. So this clue alone doesn’t narrow it down completely, but it rules out Ba (and nearly rules out Ca).

2. It forms its oxide MO which is amphoteric.

   This is the decisive clue. Among the alkaline earth metals, only beryllium oxide (BeO) is amphoteric — it reacts with both acids and bases. MgO, CaO, and BaO are all basic oxides.

   Watch out

   A common mistake is to think MgO is amphoteric because it’s “less basic” — but it is not. MgO dissolves only in acids, not in strong bases. Amphoteric behaviour in group 2 is unique to Be.

3. It forms an insoluble hydroxide M(OH)$_2$, which is soluble in NaOH solution.

   Be(OH)$_2$ is insoluble in water but dissolves in excess NaOH to form the beryllate ion, $[\text{Be(OH)}_4]^{2-}$: …