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

Option D (Both (A) and (R) are false) is correct because tin in +2 oxidation state is a reducing agent, and a substance that undergoes oxidation is a reducing agent.

Option D (A is not correct but R is correct) is correct because $GeF_4$ and $SiF_4$ act as Lewis acids, not bases, and both Ge and Si have d-orbitals to accept electrons.

The correct answer is C (Ordinary water) because plumbo-solvency refers to the ability of water to dissolve lead, especially in the absence of protective mineral coatings.

Option D (If the pressure on the system is increased by reducing the volume to half, more water vapour would be formed) is correct because increasing pressure shifts the equilibrium to the side with fewer moles of gas, favoring the reverse reaction and increasing water vapor.

Option B (A little below the tip of the flame) is correct because the hottest part of the Bunsen burner flame is located just below the tip of the inner blue cone, where complete combustion occurs.

Option B (${R_3}\mathrm{Si} - \mathrm{O} - \mathrm{Si}{R_3}$) is correct because the initial product R3SiOH is unstable and undergoes condensation to form R3Si-O-SiR3.

The correct answer is D (A mixture of calcium and sodium silicates with silica) because ordinary glass is composed of these substances.

Option A (I, II and III) is correct because CO has one sigma and two pi bonds, acts as an electron donor, and has one lone pair on its carbon atom.

Option B is correct because both the assertion and the reason are true, but the reason does not correctly explain why carbon dioxide is a gas and silicon dioxide is a solid. The difference in physical state is due to the ability of silicon to form extended network structures.

Option D (Tin) is correct because solder is an alloy of lead (Pb) and tin (Sn).

Option D ($\mathrm{CO}$) is correct because carbon monoxide (CO) binds strongly to hemoglobin in blood, forming carboxyhemoglobin, which has a characteristic cherry-red color.

The correct answer is A (Coal gas) because coal gas generally has a higher calorific value than the other options listed.

Option A (Water gas is used in the manufacture of methyl alcohol) is correct because water gas ($CO + H_2$) is a feedstock for methanol synthesis.

The correct answer is C (Calamine) because calamine is an ore of zinc, not lead.

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the stronger C-C bonds in diamond compared to Si-O-Si bonds in silica make diamond harder.

Option D (Pb) is correct because lead (Pb) does not exhibit allotropy under normal conditions, unlike carbon, tin, and silicon.

Option D is false because in zeolites, some $SiO_4^{4-}$ units are replaced by $AlO_4^{5-}$ units, not $AlO_6^{9-}$.

Option D (All of the above) is correct because Mg2C3 is indeed called magnesium allylide, contains Mg2+ and C34- ions, and hydrolyzes to give propyne.

Option D (${\left[ {SiC{l_6}} \right]^{2 - }}$) is correct because silicon's small size prevents it from coordinating six chloride ions due to steric hindrance.

Option A (18) is correct because the enthalpy change for the conversion of graphite to diamond is the difference between their enthalpies of combustion, which equals 1.8 kJ/mol or 1800 J/mol, hence P = 18.

Option A ($\mathrm{B}{\left( {\mathrm{OMe}} \right)_3}$) is correct because boron compounds, when heated in a flame, produce a characteristic green color. Boron trimethyl ($\mathrm{B}{\left( {\mathrm{OMe}} \right)_3}$) is a volatile boron compound.

Option D (${Pb}^{+2}$ is less stable than ${Pb}^{+4}$) is the incorrect statement because due to the inert pair effect, ${Pb}^{+2}$ is more stable than ${Pb}^{+4}$.

Option B ($\mathrm{GeC}{\mathrm{l}_2}$ is more stable than $\mathrm{GeC}{\mathrm{l}_4}$) is correct because, due to the inert pair effect, the +2 oxidation state becomes more stable than the +4 oxidation state as we go down the group.

Option C is correct because $Pb_3O_4$ reacts with $HNO_3$ to form $PbO_2$, $Pb(NO_3)_2$ and $H_2O$.

Option A (Carbon tetrachloride) is correct because the chlorination of carbon disulfide in the presence of AlCl3 yields carbon tetrachloride and sulfur chlorides.

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

Option D (Diamond) is correct because diamond has a giant covalent network structure with strong covalent bonds, making it an excellent insulator.

Option D ($\mathrm{Pb}{\mathrm{I}_4}$) is correct because lead(IV) iodide is unstable due to the strong oxidizing power of Pb(IV) and reducing power of I⁻.

Option C (Carbon family, $[Rn]5{f^{14}}6{d^{10}}7{s^2}7{p^2}$) is correct because element with Z=114 belongs to the carbon family and has the given electronic configuration.

The correct answer is D (Graphite) because graphite has a well-defined crystalline structure and is a relatively pure form of carbon compared to the other options.