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

Option B ($SnC{l_4}.5{H_2}O$) is correct because this is the chemical formula for butter of tin.

Option B is correct because in two-dimensional sheet silicates, three oxygen atoms are shared, while in cyclic and single-strand chain silicates, two oxygen atoms are shared per ${SiO}_4^{4-}$ tetrahedron.

Option D (Both (A) and (R) are false) is correct because $PbI_4$ is unstable and iodide does not generally stabilize higher oxidation states.

Option B (Antiknock compound) is correct because tetraethyl lead (TEL) was historically used as an antiknock agent in gasoline to increase its octane rating and prevent engine knocking.

Option A (Quartz) is correct because Quartz is the purest form of silica.

Option D ($\mathrm{N}{\mathrm{a}_2}\mathrm{O}\mathrm{.CaO}.6\mathrm{Si}{\mathrm{O}_2}$) is correct because it represents the correct composition of common soda-lime glass.

Option A (Graphite) is correct because graphite is the most thermodynamically stable allotrope of carbon under standard conditions.

Option A (C) is correct because electronegativity generally decreases down a group in the periodic table, and carbon is the highest element in the given group.

Option D (Sheet silicates) is correct because clay minerals are composed of layers or sheets of silicate tetrahedra.

Option B (PbO2) is correct because lead(IV) oxide can act as both an acid and a base, depending on the reaction conditions.

The correct answer is B ($\left( {Si{O_3}} \right)_n^{2n - }$) because chain silicates have the general formula (SiO3)n2n-.

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

Option C (Silicones) is correct because silicones are used in cosmetic surgery due to their biocompatibility and flexibility.

Option A ($-\left({R}_2 {SiO}\right)-$) is correct because organosilicon polymers (silicones) have a repeating unit of $-\left({R}_2 {SiO}\right)-$.

Option A (4740) is correct because it is calculated using the half-life of C-14 and the given ratio of C-14 to C-12.

Option A is correct because the melting points of the Carbon family elements increase down the group due to stronger metallic bonding from an increased number of valence electrons.

Option D (CO) is correct because carbon monoxide (CO) reacts with hemoglobin in blood to form carboxyhemoglobin, which has a cherry-red color.

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 D (None of these) is correct because lead sulfate ($pb\mathrm{S}{\mathrm{O}_4}$) is generally considered insoluble in common reagents like concentrated nitric acid, potassium permanganate, or potassium dichromate.

Option D (Chlorosilanes) is correct because silicones are polymers formed by the hydrolysis and subsequent polymerization of chlorosilanes.

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 (20 and 12) is correct because C60 fullerene contains 20 six-membered rings and 12 five-membered rings.