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

Option C is correct because group 14 elements have a valence shell electronic configuration of $ns^2np^2$, and $[Ne]3s^23p^2$ fits this configuration.

Option B ($CO$) is correct because carbon monoxide is a colorless gas that burns with a blue flame.

Option D (4) is correct because P4O10 has four P=O double bonds, which are shorter than the P-O single bonds.

Option A (Diamond) is correct because both quartz (SiO2) and diamond (C) form giant network covalent structures.

Option C (Si) is correct because silicon (Si) has the highest bond energy for catenation after carbon (C) among the given options.

Both the assertion and the reason are true, but the reason doesn't explain the assertion. The difference in physical state is due to the difference in bonding and structure, not just the linear pi-bonded structure of CO2.

Option D ($\mathrm{Pb}{\mathrm{I}_4}$) is correct because lead(IV) iodide is unstable due to the strong oxidizing power of $Pb^{4+}$ and the reducing power of $I^-$, leading to redox reaction and formation of $PbI_2$ and $I_2$.

Option A (Diamond is the hardest and lamp black is the softest) is correct because diamond is known for its hardness, and lamp black is a very soft form of carbon.

Option C (Vander waal’s forces between layers) is correct because the weak forces allow the layers to slide easily, making graphite a good lubricant.

Option B ($\mathrm{P}{\mathrm{b}^{2 + }}$) is correct because lead chloride is insoluble in cold water but soluble in hot water.

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

Option D (Three dimensional network silicate) is correct because quartz has a three-dimensional network structure of ${\left( {\mathrm{Si}{\mathrm{O}_2}} \right)_n}$ silicate.

Option B (Fullerene) is correct because fullerenes, such as C60, are a recently discovered allotrope of carbon.

Option A (Calcium carbonate) is correct because the furring in kettles is primarily due to the precipitation of calcium carbonate from hard water upon heating.

Option C (Refrigerant) is correct because solid CO2, also known as dry ice, sublimes at low temperatures, absorbing heat and providing a cooling effect, thus acting as a refrigerant.

Option D (CO2) is correct because CO2 is the anhydride of carbonic acid (H2CO3).

Option C (CO) is correct because CO is a neutral oxide, while the others are amphoteric or acidic.

Option C (Coal gas) is correct because coal gas typically has a composition of around 34% CH₄, 48% H₂, 15% O₂, and 3% CO.

Option D (Carborundum) is correct because carborundum (SiC) is silicon carbide, a compound, and not an allotrope of carbon.

Option C ($C > Si > Ge \approx Sn$) is correct because catenation ability decreases down the group due to decreasing bond strength.

Option C ($sp$) is correct because carbon monoxide (CO) has one sigma and two pi bonds, requiring $sp$ hybridization for the carbon atom.

Option A (dry ice) is correct because rapid expansion of liquefied CO2 results in the formation of dry ice, which is solid CO2.

The correct answer is C (hot KOH) because silicon dioxide (sand) reacts with hot potassium hydroxide to form potassium silicate and water.

Option C (Carbon family, $[Rn]5{f^{14}}6{d^{10}}7{s^2}7{p^2}]$) is correct because element Z=114 belongs to the carbon family (group 14), and its electronic configuration reflects this placement with two electrons in the 7p orbitals.

Option B (28.1 amu) is correct because it is the weighted average of the isotopic masses based on their natural abundances.

Option C (-4 to +4) is correct because the oxidation number of carbon changes from -4 in CH4 to +4 in CO2.

Option C (allotropes) is correct because diamond and graphite are different structural forms of the same element, carbon.

Option D (Pb) is correct because lead forms only one stable hydride, $PbH_4$, due to its size and inert pair effect.

Option D ($SiC$) is correct because carborundum is the commercial name for silicon carbide, a very hard material used as an abrasive.

Option D is correct because Boric acid ($H_3BO_3$) acts as a Lewis acid by accepting $OH^-$ ions, making it a weak acid.