14 th group elements (p-block) practice · Medium

Option D (${\mathrm{H}_2}\mathrm{S}$) is correct because hydrogen sulfide reacts with white lead paint to form lead sulfide, which is black.

Option A ($s{p^2}$) is correct because each carbon atom in fullerene is bonded to three other carbon atoms, resulting in $s{p^2}$ hybridization.

Option B (Acetic acid) is correct because acetic acid can corrode lead pipes due to the formation of lead acetate, which is soluble.

Option A (Metastannic acid) is correct because tin reacts with dilute nitric acid to form metastannic acid ($H_2SnO_3$) as the main product.

Option C is correct because at atmospheric pressure, solid CO2 undergoes sublimation, directly changing into a gas without forming a liquid phase, hence it doesn't wet the surface.

Option B is correct because both the assertion and the reason are true, but the reason is not the correct explanation of the assertion. Nickel(II) compounds are thermodynamically more stable than platinum(II), and platinum(IV) compounds are relatively more stable than nickel(IV) compounds, but the latter does not explain the former.

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.

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

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 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.

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 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 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 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.

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.

Option A (Nickel) is correct because carbon monoxide (CO) reacts with nickel (Ni) to form a volatile compound, nickel carbonyl ($Ni(CO)_4$).

Option D (It decomposes into $\mathrm{PbO}\;and\;{\mathrm{O}_2}$) is correct because red lead ($\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}$) decomposes upon heating to lead(II) oxide (PbO) and oxygen.

Option A is correct because controlled hydrolysis and condensation of $R_3SiCl$ yields $R_3Si-O-SiR_3$.

Option A (Sn + Pb + Bi) is correct because Rose metal is an alloy composed of tin (Sn), lead (Pb), and bismuth (Bi).

Option B (More than $\mathrm{Si}{\mathrm{F}_4}\;$) is correct because the C-F bond is stronger than the Si-F bond due to smaller size and higher electronegativity of carbon.

Option D (Diamond is $s{p^3}$ hybridized and graphite is $s{p^2}$ hybridized) is correct because it accurately describes the hybridization states of carbon atoms in diamond and graphite.