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

Option B (${\mathrm{O}_2}$) is correct because $\mathrm{Pb}{\mathrm{O}_2}$ on reaction with $\mathrm{HN}{\mathrm{O}_3}$ gives oxygen gas.

Option 14 is correct because elements in group 14 can form two chlorides with a difference of 2 in the oxidation state (e.g., SnCl2 and SnCl4).

The correct answer is B (Some of the $\mathrm{SiO}_4^{4 - }$ units are replaced by $\mathrm{AlO}_4^{5 - }$ and $\mathrm{AlO}_6^{9 - }$ ions in zeolite) because the correct replacement is $\mathrm{AlO}_4^{4-}$ not $\mathrm{AlO}_4^{5 - }$ and $\mathrm{AlO}_6^{9 - }$.

The correct answer is A (Diamond > Borazon > Carborundum > Corundum) because it represents the correct order of decreasing hardness based on the materials' crystal structures and bonding.

The correct answer is A (Diamond) because diamond has a giant covalent network structure with strong covalent bonds throughout the lattice, requiring a large amount of energy to break, hence the highest boiling point.

The correct answer is D ((c) and (d) only) because ZSM-5 is indeed used to convert alcohols to gasoline, and CO is a colorless and odorless gas. CO2 is used as a refrigerant, but the description of C60's structure is incorrect.

Option C is correct because carbon suboxide (C3O2) is a linear molecule, not bent, and each carbon atom is sp-hybridized.

The correct answer is D (Has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds) because the layered structure of graphite, with strong covalent bonds within the layers and weak van der Waals forces between them, allows it to act as a lubricant and makes it difficult to melt.

The correct answer is B (Solids) because silicon dioxide and other silicon oxides exist as solid materials at room temperature.

Option D (All of these) is correct because hydrolysis of $Be_2C$, $Al_4C_3$, and $Mn_3C$ all produce methane ($CH_4$).

Option A (Carbon monoxide is oxidized to carbon dioxide) is correct because the water-gas shift reaction involves the oxidation of CO to CO2 by steam.

Option A (Passing steam through a red hot coke bed) is correct because this process produces a mixture of carbon monoxide and hydrogen, which is water gas.

Option C is correct because in $SiO_4^{4-}$, silicon is tetrahedrally surrounded by oxygen atoms, resulting in $sp^3$ hybridization.

Option B is correct because graphite's structure consists of strongly bonded carbon layers with weak interlayer forces, making it soft and difficult to melt.

Option D (Carborundum) is correct because heating H4SiO4 at 1000°C gives silica (SiO2), which reacts with carbon at high temperatures to form carborundum (SiC).

Option B (Flint glass) is correct because flint glass contains lead oxide, which increases its refractive index, making it suitable for optical lenses.

Option D (Animal charcoal) is correct because animal charcoal contains the highest percentage of impurities compared to the other forms of carbon listed.

Option A ($\mathrm{SiC}$) is correct because carborundum is the common name for silicon carbide (SiC), a very hard ceramic compound used as an abrasive.

The correct answer is A ($\mathrm{C}{\mathrm{H}_4} > Si{\mathrm{H}_4} > Ge{\mathrm{H}_4} > Sn{\mathrm{H}_4} > Pb{\mathrm{H}_4}$) because the stability of hydrides decreases down the group due to decreasing bond energy and increasing atomic size.

Option A (Coefficient of expansion) is correct because the difference in the coefficient of expansion causes stress and cracking when the metal and glass are fused.

Option D (Stannic sulphate) is correct because tin reacts with concentrated sulfuric acid to produce stannic sulfate ($Sn(SO_4)_2$) and sulfur dioxide.

Option D is correct because both the assertion and the reason are false. PbI4 is not a stable compound, and iodide does not stabilize higher oxidation states of lead.

Option A (4740) is correct because it is calculated using the half-life formula for radioactive decay, determining the time elapsed for the given ratio of C-14 to C-12.

Option C (Destructive distillation) is correct because coke is a residue obtained from the destructive distillation of coal.

Option C (A is true but R is false) is correct because ${Pb}^{4+}$ can be reduced easily to ${Pb}^{2+}$ due to the inert pair effect, but ${Pb}^{2+}$ is diamagnetic, not paramagnetic.

Option A (CO) is correct because the molar mass of CO (28 g/mol) is closest to the average molar mass of air (approximately 29 g/mol), making its vapor density near to air.

The correct answer is A (Potassium alumino silicate having sheet structure) because mica has this chemical composition and structure.

The correct answer is D (+2, +4) because lead exhibits +2 and +4 oxidation states due to the inert pair effect.

Option C (Si) is correct because silicon, due to its size and electronic configuration, typically exhibits a coordination number of four in its compounds.

Option C is the correct answer because zeolites are aluminosilicates that do exist naturally.