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

Option C (CO2) is the correct answer because CO2 is a gaseous oxide formed by a group 14 element (carbon), it dissolves in water to form carbonic acid (decreasing pH), and the addition of group 2 hydroxides leads to the precipitation of carbonates.

Option C (Refrigerant) is correct because solid CO2, also known as dry ice, is commonly used as a refrigerant due to its low temperature and sublimation properties.

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 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.

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

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 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.

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

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 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 C (Fullerenes do not have dangling edge or surface bonds) because the closed cage structure of fullerenes eliminates dangling bonds, making them chemically stable and relatively pure.