S-block elements practice · Medium

Option D is correct. Beryllium hydroxide is amphoteric, while barium hydroxide is strongly basic; thus, beryllium hydroxide is less basic than barium hydroxide.

Option A ($MgC{O_3}$) is correct because magnesium carbonate is less thermally stable than the carbonates of larger alkali metals and calcium.

Option D is correct. Calcium carbonate (CaCO3) decomposes to form calcium oxide (CaO), a basic oxide, and carbon dioxide (CO2), an acidic oxide.

Option A is correct because alkali metal carbonates are stable to heat except for lithium carbonate.

Option C ($\mathrm{N}{\mathrm{a}_2}\mathrm{Sn}{\mathrm{O}_3}$) is correct because tin reacts with boiling caustic soda to form sodium stannate, which is soluble.

Option D (It is the super oxide of sodium) is not correct because $\mathrm{Na}_2 \mathrm{O}_2$ is the peroxide of sodium, not the superoxide.

The correct answer is B. The blue color of the solution is due to solvated electrons, which absorb energy and impart the blue color.

Option B ($\mathrm{BaCO}_{3}$) is correct because barium ions ($Ba^{2+}$) are typically precipitated as barium carbonate ($BaCO_3$) in qualitative analysis.

Option A ($\mathrm{K}$) is correct because Potassium (K) is an alkali metal with a density less than that of water at $25^{\circ} \mathrm{C}$.

Option D (Both (1) and (3)) is correct because the solution contains both solvated sodium ions and solvated electrons, which contribute to its electrical conductivity.

The correct answer is A (NaHCO3 and NaOH) because NaOH will react with NaHCO3 to form Na2CO3 and H2O, thus they cannot coexist in solution.

Option A (Both A and R are true and R is the correct explanation of A) is correct because both ${BeCl}_2$ and ${AlCl}_3$ act as Lewis acids in Friedel-Crafts reactions by providing ${Cl}^{+}$.

Option D ($Na_2CO_3$ is partially converted into bi-carbonate) is the wrong statement because $NaHCO_3$ converts into $Na_2CO_3$ mainly, and the reverse does not happen.

Option D (Hydrated ${MgCl}_2$ gives anhydrous ${MgCl}_2$ on heating in dry air.) is incorrect because hydrated MgCl2 on heating gives MgO, not anhydrous MgCl2.

Option A (efflorescence) is correct because efflorescence is the process where a hydrated salt loses water molecules to the atmosphere.

Option A (Dry slaked lime) is correct because chlorine reacts with dry slaked lime [Ca(OH)₂] to produce bleaching powder [CaOCl₂].

Option A ($\mathrm{C}{\mathrm{O}_2}\mathrm{and N}{\mathrm{H}_3}$) is correct because carbon dioxide and ammonia are recycled in the Solvay process.

Option A ([3Ca3(PO4)2Ca(OH)2]) is correct because the formula for hydroxyapatite is [3Ca3(PO4)2Ca(OH)2], which is a major component of tooth enamel.

Option C (When heated to 1200 K, it decomposes to evolve ${O_2}$) is correct because calcium carbonate decomposes to produce calcium oxide and carbon dioxide, not oxygen.

Option C ($LiN{O_3}$) is correct because Lithium nitrate decomposes to give $NO_2$ gas upon heating, unlike other alkali metal nitrates.

Option D (Is zero) is correct because at the initial stage of the reaction, the concentration of products is negligible, making the reaction quotient close to zero.

Option D (All of the above) is correct because the solubility of sulphates down the Be group decreases due to a combination of factors, including higher coordination number.

Option D (Decrease, Increase, Increase) is correct because as we move down the group for alkali metal carbonates, covalent character decreases, solubility increases, and thermal stability increases.

Option A ($\mathrm{LiBr}$) is correct because Lithium salts are known to impart a bright red color to the flame during a flame test.

Option A ($\mathrm{KN}{\mathrm{O}_3}$ +Charcoal+S) is correct because gunpowder is a mixture of potassium nitrate, charcoal, and sulfur.

Option B ($Cs^+ > Rb^+ > K^+ > Na^+$) is correct because smaller hydrated radius leads to higher conductivity, and hydration decreases with increasing ionic size.

Option A (Reduction of beryllium halide with Mg) is correct because it is a highly efficient method for obtaining beryllium.

Option B (${Na}_2 {CO}_3$) is correct because the Solvay process is an industrial method for the production of sodium carbonate.

Option A (If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I) is correct because the hydrogen carbonates of alkali metals are soluble but less soluble than normal carbonates due to hydrogen bonding.

Option A is correct because alkali metals form ionic hydrides with H-, and their low electronegativity and conductive properties when fused support this.