S-block elements practice · Mixed difficulty

Option A (Density of lithium is more than sodium) is correct because lithium has a lower density than sodium due to its smaller size and weaker metallic bonding.

Option A ($\mathrm{NaFe}{\mathrm{O}_2}$) is correct because the reaction between sodium carbonate and ferric oxide at high temperatures produces sodium metaferrite and carbon dioxide.

Option A is correct because $\mathrm{KO}_2$ reacts with carbon dioxide to produce oxygen, making it suitable for use in space and submarines to regenerate breathable air.

Option A (Absorbs $\mathrm{C}{\mathrm{O}_2}$ and increase ${\mathrm{O}_2}$ content) is correct because potassium superoxide reacts with carbon dioxide to produce oxygen, thereby purifying the air.

Option C is correct because sodium thiosulphate reacts with silver bromide to form a soluble complex, sodium dithiosulphatoargentate(I).

The correct answer is Barium sulphate and zinc sulphide because Lithopone is a mixture of these two compounds.

Option A ($1: 2$) is correct because lithium nitrate decomposes to lithium oxide, nitrogen dioxide, and oxygen, while sodium nitrate decomposes to sodium nitrite and oxygen. The stoichiometry results in a 1:2 ratio of oxygen produced.

Option B ($\mathrm{MgS}{\mathrm{O}_4} \cdot 7{\mathrm{H}_2}\mathrm{O}$) is correct because Magnesium sulfate (Epsom salt) is commonly used as a purgative.

Option A (LiCl) is correct because, contrary to the general trend, LiCl has the highest melting point due to its more covalent character and smaller size, leading to stronger interactions.

Option B is correct because in $CaC_2$, each carbon atom is bonded to the calcium atom and to the other carbon atom via one sigma and two pi bonds, resulting in a triple bond between the two carbon atoms.

Option A (Golden yellow colour) is correct because when sodium is heated in a flame, it emits light with a golden yellow color.

Option A ($\mathrm{AlC}{\mathrm{l}_3} + \mathrm{NaOH} \to$) is correct because the reaction of aluminum chloride with sodium hydroxide initially forms a precipitate of aluminum hydroxide, which then dissolves in excess NaOH to form sodium aluminate, without liberating a gaseous product.

Option A is correct because the reaction of potassium chloride, concentrated sulfuric acid, and potassium dichromate produces chromyl chloride ($\mathrm{CrO_2Cl_2}$), which is a reddish-brown volatile liquid.

Option B ($\mathrm{Na}\left( {\mathrm{N}{\mathrm{H}_4}} \right)\mathrm{HP}{\mathrm{O}_4}$) is correct because it is the chemical formula for microcosmic salt.

Strontium (Option D) produces a characteristic scarlet red flame in the Bunsen flame test.

Option A is correct because sodium readily loses its single valence electron to achieve a stable octet configuration, making the formation of $\mathrm{Na}^{2+}$ energetically unfavorable.

Option C ($\mathrm{Ca}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2} \cdot \mathrm{CaO}$) is the correct formula for Norwegian saltpetre, which is a naturally occurring form of calcium nitrate.

Option D is correct because synthetic resins can be designed to exchange both cations and anions, while zeolites primarily exchange cations.

Option A (Li) is correct because Lithium has the highest oxidation potential among the given options due to its small size and high hydration energy.

Option A is correct because $KNO_3$ on heating decomposes to $KNO_2$ and $O_2$, not $NO_2$. The other nitrates decompose to give $NO_2$.

Option A (2 moles of CO2) is correct because MgCO3 decomposes to MgO and CO2, while Li2CO3 decomposes to Li2O and CO2. Na2CO3 does not decompose under normal heating.

Option D is correct because the solubility of alkaline earth metal sulfates decreases down the group due to decreasing hydration energy. Therefore, $BeSO_4$ is the most soluble.

Option C is correct because lithium hydroxide ($\mathrm{LiOH}$) has the lowest solubility among the given hydroxides due to its high lattice energy and strong interionic forces.