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

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

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

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

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 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.

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

Option C is correct because Statement-1 is true (the C-O bond in CO+ is longer than in CO), and Statement-2 is false. When an electron is removed from CO to form CO+, it is removed from a bonding molecular orbital, which weakens the bond and increases the bond length.

Option C (Both are correct) is correct because six large chloride ions cannot be accommodated around the small $Si^{4+}$ ion, and the interaction between the lone pair of $Cl^-$ and $Si^{4+}$ is weak.

Option D ($\mathrm{P}{\mathrm{b}^{2 + }}$ salt) is correct because lead(II) chloride precipitates with HCl, dissolves on heating, and lead(II) sulfide precipitates with H2S in acidic solution.

Option C (Catenation tendency is higher in nitrogen when compared with other elements of the same group) is the wrong statement because nitrogen has a lower catenation tendency compared to other group 15 elements.

Option A is correct because the C-F bond is the strongest due to the small size and high electronegativity of fluorine, making $CF_4$ thermally more stable.

Option B (Wood’s metal) is correct because Wood's metal is a low-melting alloy that melts at around 70°C, which is below the boiling point of water.

Option C (For lead +2 , for tin +4) is correct because lead prefers the +2 oxidation state due to the inert pair effect, while tin prefers +4.

Option C (If Assertion is true but Reason is false) is correct because graphite is a good conductor, but it's due to delocalized electrons, not firmly held electrons.