14 th group elements (p-block) practice · Mixed difficulty

Option C (As moderator) is correct because graphite is used as a moderator in nuclear reactors to slow down neutrons.

The correct answer is C ($\mathrm{Si}{\mathrm{O}_4}$ tetrahedron) because this is the fundamental building block of silicate structures.

Option D (High electric arc furnace) is correct because graphite is produced by heating coke with silica at very high temperatures in an electric arc furnace.

Option A ($\mathrm{Pb}{\mathrm{F}_2} > \mathrm{PbC}{\mathrm{l}_2} > \mathrm{PbB}{\mathrm{r}_2} > \mathrm{Pb}{\mathrm{I}_2}$) is correct because ionic character decreases with increasing size of the halide ion due to increasing covalent character (Fajan's rule).

Option C ($C > Si > Ge \approx Sn$) is correct because the catenation tendency decreases as you move down Group 14, with carbon having the highest tendency and germanium and tin having similar tendencies.

Option C is correct because carboxyhemoglobin is significantly MORE stable than oxyhemoglobin, making CO a dangerous poison.

Option C is correct because alumina reacts with carbon in a nitrogen atmosphere to form aluminum nitride (AlN) and carbon monoxide (CO).

Option D is correct because boric acid acts as a Lewis acid and is considered a weak acid.

Option B (Plumbane) is the least stable hydride because the stability of hydrides decreases down the group due to increasing size and decreasing bond strength.

Option A ($\mathrm{C}{\mathrm{H}_4} + \mathrm{CO} + {\mathrm{H}_2}$) is correct because coal gas primarily consists of methane (CH4), carbon monoxide (CO), and hydrogen (H2).

Option C (-4 to +4) is correct because the oxidation number of carbon changes from -4 in methane ($\mathrm{CH}_4$) to +4 in carbon dioxide ($\mathrm{CO}_2$).

Option B (White turning to grey) is correct because $SnCl_2$ reduces $HgCl_2$ to $Hg_2Cl_2$ (white) and then to Hg (grey).

Option D is correct because all the statements are true. Cyanamide ion is isoelectronic and isostructural with CO2, Mg2C3 reacts with water to form propyne, and CaC2 has a NaCl type lattice.

Option A ($\mathrm{C}{\mathrm{O}_2}$) is correct because carbon dioxide (CO2) is commonly used to create the fizz in aerated water (soda water).

Option A ($\mathrm{Na}_2 \mathrm{SO}_3$ and $\mathrm{Na}_2 \mathrm{S}$) is correct because $\mathrm{SO}_2$ released from sulfite turns dichromate paper green, and $\mathrm{H}_2\mathrm{S}$ from sulfide turns lead acetate paper black.

Option C (Oxalate) is correct because oxalates react with hot H2SO4 to produce CO2 and also decolorize acidified KMnO4 due to the oxidation of oxalate ions to CO2.

Option B (Benzene hexacarboxylic acid) is correct because hot concentrated nitric acid oxidizes graphite to benzene hexacarboxylic acid (mellitic acid).

Option A ($\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2} \cdot 2\mathrm{PbC}{\mathrm{O}_3}$) is correct because white lead, or basic lead carbonate, has the chemical formula $\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2} \cdot 2\mathrm{PbC}{\mathrm{O}_3}$.

Option A ($CO\mathrm{ } + {\mathrm{N}_2}$) is correct because producer gas is primarily a mixture of carbon monoxide and nitrogen.

Option B (Conversion of white tin to grey tin) is correct because tin plague refers to the allotropic transformation of white tin (β-tin) to grey tin (α-tin) at low temperatures.

Option A is correct because the network structure of quartz ($SiO_2$) resembles the tetrahedral network arrangement found in diamond.

Option B is correct because the stability of the +2 oxidation state increases down the carbon family due to the inert pair effect.

Option A (b, c) is correct because fullerenes are indeed cage-like molecules, and graphite is the thermodynamically most stable allotrope of carbon.

Option A is correct because $Pb^{2+}$ is more stable due to the inert pair effect, which is more pronounced in heavier elements.

Option B (Neso silicate) is correct because in Neso silicates, each tetrahedral unit has 4 monovalent oxygens (each bonded to only one Si) and zero divalent oxygens (bridging oxygens). The question states the number of monovalent and divalent oxygens are equal. This is only possible if there are 4 monovalent oxygens and 0 divalent oxygens. This is incorrect. The question is flawed. However, if we consider the ratio of monovalent to divalent oxygen, then Neso silicates have a ratio of infinity, Soro silicates have a ratio of 1, Pyroxenes have a ratio of 0, and Sheet silicates have a ratio of 0. The closest answer would be Neso silicates. The question is flawed.

Option C (+2 & +4) is correct because lead exhibits both +2 and +4 oxidation states, with +2 being more stable due to the inert pair effect.

Option D is correct because both the assertion and the reason are false. CO2 is a gas and SiO2 is a solid at room temperature. Also, both CO2 and SiO2 contain double bonds.

Option C (Brittleness) is correct because arsenic is added to lead shot to increase its brittleness, allowing it to shatter upon impact.

Option C is correct because CO (carbon monoxide) is a neutral oxide and does not affect litmus paper.

Option B (Propanoic acid) is correct because the reaction of CO2 with ethyl magnesium bromide followed by protonation yields propanoic acid.