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101. Which form of naturally occurring carbon is the hardest known natural substance and possesses a giant tetrahedral structure?
ⓐ. Graphite
ⓑ. Fullerenes
ⓒ. Carbon black
ⓓ. Diamond
Correct Answer: Diamond
Explanation: Diamond is an allotrope of carbon where each carbon atom is $sp^3$ hybridized and covalently bonded to four other carbon atoms in a repeating, rigid, three-dimensional tetrahedral lattice. This strong, extended network is responsible for its extreme hardness and high density.
102. In which of the following natural states does carbon primarily exist in the atmosphere?
ⓐ. Carbonyl sulfide (${COS}$)
ⓑ. Carbon dioxide (${CO}_2$)
ⓒ. Elemental soot
ⓓ. Graphite dust
Correct Answer: Carbon dioxide (${CO}_2$)
Explanation: Carbon dioxide (${CO}_2$) is the most significant form of carbon in the atmosphere, representing the primary combined state of carbon. It is essential for photosynthesis and is a major greenhouse gas.
103. Which of the following is NOT a form in which carbon occurs in the free (elemental) state?
ⓐ. Graphite
ⓑ. Coal
ⓒ. Limestone (${CaCO}_3$)
ⓓ. Diamond
Correct Answer: Limestone (${CaCO}_3$)
Explanation: Limestone (${CaCO}_3$) is a form of combined carbon, primarily found in sedimentary rocks. Graphite, coal (though impure), and diamond are all forms of carbon existing in the free (elemental) state.
104. Which mineral is the most abundant compound of Silicon on the Earth’s crust?
ⓐ. Mica (${KAl}_2({AlSi}_3{O}_{10})({OH})_2$)
ⓑ. Feldspar (${KAlSi}_3{O}_8$)
ⓒ. Talc (${Mg}_3{Si}_4{O}_{10}({OH})_2$)
ⓓ. Silica (${SiO}_2$) in its various forms (e.g., Quartz)
Correct Answer: Silica (${SiO}_2$) in its various forms (e.g., Quartz)
Explanation: Silicon is the second most abundant element on the Earth’s crust (after Oxygen). Its most common compound is Silicon dioxide, or Silica (${SiO}_2$), which occurs in crystalline forms like quartz, tridymite, and cristobalite, making up about 12% of the Earth’s crust by weight.
105. What is the fundamental building unit of all naturally occurring silicates?
ⓐ. Trigonal planar ${SiO}_3^{2-}$ ion
ⓑ. Tetrahedral ${SiO}_4^{4-}$ ion
ⓒ. Planar ${Si}_2{O}_7^{6-}$ ion
ⓓ. Linear ${SiO}_2$ molecule
Correct Answer: Tetrahedral ${SiO}_4^{4-}$ ion
Explanation: The basic structural unit of all silicates is the ${SiO}_4^{4-}$ anion, where a small Silicon atom is at the center of a tetrahedron, covalently bonded to four larger Oxygen atoms at the corners. This unit then links with other ${SiO}_4^{4-}$ units in various ways (sharing corners, edges, or faces) to form the diverse silicate structures.
106. In the structure of Quartz, a form of silica (${SiO}_2$), how many oxygen atoms are shared between adjacent ${SiO}_4$ tetrahedra?
ⓐ. One
ⓑ. Two
ⓒ. Three
ⓓ. Four
Correct Answer: Four
Explanation: Crystalline silica (like Quartz) is a giant three-dimensional network solid. In this structure, every Oxygen atom is shared between two adjacent ${SiO}_4$ tetrahedra. Consequently, every Silicon atom is linked to four Oxygen atoms, and every Oxygen atom is linked to two Silicon atoms, giving the empirical formula ${SiO}_{4/2}$ or ${SiO}_2$.
107. Feldspars and Zeolites are examples of which major class of silicates, characterized by a three-dimensional interlocking structure?
ⓐ. Framework silicates (Tectosilicates)
ⓑ. Chain silicates (Pyroxenes)
ⓒ. Ring silicates (Cyclosilicates)
ⓓ. Sheet silicates (Phyllosilicates)
Correct Answer: Framework silicates (Tectosilicates)
Explanation: Framework silicates (Tectosilicates) are characterized by the sharing of all four oxygen atoms of every ${SiO}_4^{4-}$ tetrahedron, creating a complex 3D network. Feldspars and Quartz are classic examples. Zeolites are framework silicates where some ${Si}$ atoms are replaced by ${Al}$, giving them a porous structure.
108. Which naturally occurring form of carbon is generally used as a lubricant and as an electrode material due to its layer structure?
ⓐ. Diamond
ⓑ. Graphite
ⓒ. Coal
ⓓ. Carbon nanotubes
Correct Answer: Graphite
Explanation: Graphite has a unique layered structure where carbon atoms are $sp^2$ hybridized, forming hexagonal rings in a plane. These layers are held together only by weak van der Waals forces, allowing them to slide over each other easily. This property makes graphite an excellent lubricant and enables it to conduct electricity, making it ideal for electrodes.
109. 109: What is the combined state of carbon that is the main constituent of marble, chalk, and limestone?
ⓐ. Calcium carbonate (${CaCO}_3$)
ⓑ. Carbon disulfide (${CS}_2$)
ⓒ. Carbon monoxide (${CO}$)
ⓓ. Methane (${CH}_4$)
Correct Answer: Calcium carbonate (${CaCO}_3$)
Explanation: Calcium carbonate (${CaCO}_3$) is the most abundant combined state of carbon in the Earth’s crust, forming massive deposits as the main component of limestone, marble, and chalk.
110. The vast diversity of silicate minerals is a result of the ability of the ${SiO}_4^{4-}$ units to link up by sharing various numbers of which atoms?
ⓐ. Silicon atoms
ⓑ. Metal cations (${Na}^+, {Mg}^{2+}$, etc.)
ⓒ. Hydrogen atoms
ⓓ. Oxygen atoms
Correct Answer: Oxygen atoms
Explanation: The structural classification of silicates (e.g., single chain, double chain, ring, sheet, or framework) is determined entirely by the number of Oxygen atoms shared between adjacent ${SiO}_4^{4-}$ tetrahedral units. Sharing zero, one, two, three, or four Oxygen atoms leads to distinct mineral structures.
111. Moving down Group 14 (C to Pb), how does the change in atomic radius typically proceed?
ⓐ. It increases steadily from C to Pb.
ⓑ. It decreases steadily due to increasing nuclear charge.
ⓒ. It increases from C to Si, but remains nearly constant thereafter.
ⓓ. It increases, but the increase from Sn to Pb is less than expected.
Correct Answer: It increases, but the increase from Sn to Pb is less than expected.
Explanation: The atomic radius generally increases down the group due to the addition of new electron shells. However, the increase from Tin (${Sn}$) to Lead (${Pb}$) is very slight. This is due to the Lanthanide Contraction, which results from the poor shielding of the intervening $4f$ and $5d$ orbitals. This causes a greater effective nuclear charge, pulling the valence shell slightly inward.
112. The first ionization enthalpy (${IE}_1$) generally decreases down Group 14. Which element disrupts this smooth trend by exhibiting a higher ${IE}_1$ than the element immediately above it?
ⓐ. Silicon (${Si}$)
ⓑ. Germanium (${Ge}$)
ⓒ. Tin (${Sn}$)
ⓓ. Lead (${Pb}$)
Correct Answer: Lead (${Pb}$)
Explanation: While ${IE}_1$ generally decreases down the group, ${IE}_1$ of Lead (${Pb}$) is slightly higher than that of Tin (${Sn}$). This inversion is due to the Lanthanide Contraction (poor shielding by $4f$ electrons), which causes the valence electrons in ${Pb}$ to be held more tightly than expected, making them harder to remove.
113. What is the most characteristic and common oxidation state shown by all elements in Group 14?
ⓐ. $+2$
ⓑ. $+3$
ⓒ. $+4$
ⓓ. $\pm 4$
Correct Answer: $+4$
Explanation: The general valence configuration of Group 14 is ${ns}^2{np}^2$. By sharing or losing all four valence electrons, these elements achieve the $\mathbf{+4}$ oxidation state. This is the maximum positive oxidation state and is exhibited by all elements, typically in covalent compounds.
114. The stability of the $+2$ oxidation state among Group 14 elements:
ⓐ. Decreases steadily from Carbon to Lead.
ⓑ. Increases steadily from Carbon to Lead.
ⓒ. Is stable only for Carbon and Silicon.
ⓓ. Is equal for all elements in the group.
Correct Answer: Increases steadily from Carbon to Lead.
Explanation: The stability of the $+2$ oxidation state increases down the group (${Pb} > {Sn} > {Ge} > {Si} > {C}$). This phenomenon is known as the Inert Pair Effect, where the ${ns}^2$ electrons of the heavier elements become increasingly reluctant to participate in bonding, favoring the loss or sharing of only the two ${np}^2$ electrons.
115. Which of the following compounds is the strongest reducing agent?
ⓐ. ${SnCl}_4$
ⓑ. ${SnCl}_2$
ⓒ. ${PbCl}_4$
ⓓ. ${PbCl}_2$
Correct Answer: ${SnCl}_2$
Explanation: The stability of the $+4$ state decreases down the group, while the stability of the $+2$ state increases. ${Sn}({II})$ is less stable than ${Sn}({IV})$, meaning ${SnCl}_2$ has a strong tendency to convert to the more stable ${SnCl}_4$ (i.e., lose electrons). Thus, ${SnCl}_2$ is a strong reducing agent. Conversely, ${PbCl}_4$ is a strong oxidizing agent because ${Pb}({IV})$ is unstable and prefers to convert to ${Pb}({II})$.
116. Carbon, the first element of Group 14, is known to form extended chains and rings through self-linking. What is this property called?
ⓐ. Catenation
ⓑ. Tautomerism
ⓒ. Hybridization
ⓓ. Allotropy
Correct Answer: Catenation
Explanation: Catenation is the ability of an element to form bonds with atoms of the same element, leading to long chains or large rings. Carbon exhibits this property to the maximum extent due to the small size of the carbon atom and the strong ${C}-{C}$ bond energy. This tendency decreases significantly down the group (${C} \gg {Si} > {Ge} \approx {Sn}$).
117. Due to the availability of vacant ${d}$-orbitals, Silicon readily shows an expanded octet in its compounds, such as in ${[SiF}_6]^{2-}$. What is the maximum covalency exhibited by Silicon in this ion?
ⓐ. 3
ⓑ. 4
ⓒ. 5
ⓓ. 6
Correct Answer: 6
Explanation: In the complex ion ${[SiF}_6]^{2-}$, the Silicon atom is covalently bonded to six Fluorine atoms (${Si} \leftarrow 6{F}$). Since Silicon has vacant ${3d}$ orbitals, it can expand its octet and achieve a maximum covalency of 6. Carbon, lacking the ${d}$-orbitals, cannot exceed a covalency of 4.
118. Which statement correctly describes the nature of the halides (${MX}_4$) formed by Group 14 elements?
ⓐ. They are generally covalent and hydrolyse easily.
ⓑ. They are all ionic and non-hydrolysable.
ⓒ. They are ionic for ${C}$ and ${Si}$ but covalent for ${Sn}$ and ${Pb}$.
ⓓ. They are covalent but do not react with water.
Correct Answer: They are generally covalent and hydrolyse easily.
Explanation: Due to the relatively high electronegativity of Group 14 elements (except ${Pb}$) and their tendency to share electrons, the ${MX}_4$ halides are predominantly covalent and molecular. They hydrolyze easily in water because the central atom (${Si}, {Ge}, {Sn}, {Pb}$) has vacant ${d}$-orbitals to accept lone pairs from water molecules, initiating the reaction.
119. Carbon exhibits a maximum covalency of 4, while the rest of the Group 14 elements can show a maximum covalency of 6. What is the fundamental reason for this difference?
ⓐ. Carbon has a higher electronegativity.
ⓑ. Carbon lacks vacant ${d}$-orbitals in its valence shell.
ⓒ. The ${C}-{C}$ bond is much stronger than the ${Si}-{Si}$ bond.
ⓓ. Carbon is typically found in the $-4$ oxidation state.
Correct Answer: Carbon lacks vacant ${d}$-orbitals in its valence shell.
Explanation: Carbon is in the second period ($n=2$) and only has $2s$ and $2p$ orbitals available for bonding. It lacks the ${2d}$ orbitals necessary to accommodate more than 8 electrons (four shared pairs) in its valence shell, thus limiting its maximum covalency to 4. All other Group 14 elements have available ${d}$-orbitals (${3d}, {4d}, {5d}$), allowing them to expand their octet.
120. When comparing the stability of the tetrahalides (${MX}_4$) and dihalides (${MX}_2$) of Lead (${Pb}$), which conclusion is correct?
ⓐ. ${PbX}_4$ is more stable than ${PbX}_2$.
ⓑ. Both are equally stable due to the ${ns}^2{np}^2$ configuration.
ⓒ. ${PbX}_4$ is a strong reducing agent.
ⓓ. ${PbX}_2$ is more stable than ${PbX}_4$.
Correct Answer: ${PbX}_2$ is more stable than ${PbX}_4$.
Explanation: Lead (${Pb}$), being the heaviest element, shows the most pronounced Inert Pair Effect. This means the $+2$ oxidation state (${PbX}_2$) is significantly more stable than the $+4$ oxidation state (${PbX}_4$). ${PbX}_4$ compounds are unstable and act as powerful oxidizing agents, converting readily to the more stable ${PbX}_2$.
121. The ability of an element to form bonds with atoms of the same element, leading to long chains or rings, is specifically termed:
ⓐ. Allotropy
ⓑ. Polymerization
ⓒ. Isomerism
ⓓ. Catenation
Correct Answer: Catenation
Explanation: Catenation is the unique property of self-linking, which is exhibited to the maximum extent by Carbon due to its small size and strong ${C}-{C}$ bond energy.
122. Among the elements of Group 14 (${C}, {Si}, {Ge}, {Sn}, {Pb}$), which element exhibits the maximum tendency for catenation?
ⓐ. Silicon (${Si}$)
ⓑ. Germanium (${Ge}$)
ⓒ. Carbon (${C}$)
ⓓ. Tin (${Sn}$)
Correct Answer: Carbon (${C}$)
Explanation: Carbon has the highest catenation property in Group 14 due to its small atomic size, which allows for very effective orbital overlap and the formation of very strong ${C}-{C}$ single bonds.
123. Which factor is the primary determinant of the extent of catenation for an element?
ⓐ. Ionization enthalpy
ⓑ. Electronegativity difference with Hydrogen
ⓒ. Bond dissociation enthalpy of the element-element single bond (${E}-{E}$)
ⓓ. Atomic radius of the element
Correct Answer: Bond dissociation enthalpy of the element-element single bond (${E}-{E}$)
Explanation: Catenation requires the resulting chain or ring to be stable. This stability is directly related to the energy required to break the single bond between two atoms of the element (${E}-{E}$), with a higher bond enthalpy favoring greater catenation.
124. When comparing the elements of Group 14, the ${E}-{E}$ bond energy (and thus the catenation tendency) decreases most significantly after which element?
ⓐ. Silicon (${Si}$)
ⓑ. Tin (${Sn}$)
ⓒ. Germanium (${Ge}$)
ⓓ. Lead (${Pb}$)
Correct Answer: Silicon (${Si}$)
Explanation: Catenation decreases in the order ${C} \gg {Si} > {Ge} \approx {Sn}$. The sharpest drop occurs after Carbon, but Silicon is the last element to show any substantial catenation; the tendency is negligible for Germanium and Tin, and virtually non-existent for Lead.
125. Which comparison of bond dissociation enthalpies for Group 14 single bonds is correct?
ⓐ. ${Si}-{Si}$ bond energy is much greater than ${C}-{C}$ bond energy.
ⓑ. ${Sn}-{Sn}$ bond energy is greater than ${Ge}-{Ge}$ bond energy.
ⓒ. ${C}-{C}$ bond energy ($348 \\, {kJ} \\, {mol}^{-1}$) is significantly greater than ${Si}-{Si}$ bond energy ($297 \\, {kJ} \\, {mol}^{-1}$).
ⓓ. ${Pb}-{Pb}$ bond energy is the highest in the group.
Correct Answer: ${C}-{C}$ bond energy ($348 \\, {kJ} \\, {mol}^{-1}$) is significantly greater than ${Si}-{Si}$ bond energy ($297 \\, {kJ} \\, {mol}^{-1}$).
Explanation: The strong ${C}-{C}$ bond enthalpy is the fundamental chemical reason for carbon’s extensive catenation, and it is significantly higher than the ${Si}-{Si}$ bond enthalpy, which results in weaker catenated structures for Silicon.
126. Silicon’s catenation tendency is low compared to Carbon because the ${Si}-{Si}$ bond is easily attacked and cleaved by which common reagent?
ⓐ. Dilute ${HNO}_3$
ⓑ. Dry ${HCl}$ gas
ⓒ. Concentrated ${H}_2{SO}_4$
ⓓ. Oxidizing agents (like ${O}_2$) or Water/Acids/Bases
Correct Answer: Oxidizing agents (like ${O}_2$) or Water/Acids/Bases
Explanation: Silicon chains (${Si}_n{H}_{2n+2}$, called silanes) are thermally unstable and highly reactive. They are easily oxidized by air (${O}_2$) and are readily hydrolysed by trace amounts of water, acids, or bases, leading to cleavage of the ${Si}-{Si}$ bonds and termination of the chain.
127. While Carbon forms long stable chains with itself, Silicon catenation is often observed in chains where the Silicon atoms are separated by which bridging atom?
ⓐ. Nitrogen (${N}$)
ⓑ. Oxygen (${O}$)
ⓒ. Sulphur (${S}$)
ⓓ. Hydrogen (${H}$)
Correct Answer: Oxygen (${O}$)
Explanation: The ${Si}-{O}$ bond ($452 \\, {kJ} \\, {mol}^{-1}$) is much stronger than the ${Si}-{Si}$ bond ($297 \\, {kJ} \\, {mol}^{-1}$). This means Silicon forms far more stable chains and networks when atoms are linked through Oxygen bridges, as seen in the extremely stable silicone polymers (${Si}-{O}-{Si}$ chains) and natural silicates.
128. For the catenated compounds of Germanium (germananes, ${Ge}_n{H}_{2n+2}$), the degree of catenation typically does not exceed:
ⓐ. ${n}=6$
ⓑ. ${n}=2$
ⓒ. ${n}=10$
ⓓ. ${n}=100$
Correct Answer: ${n}=6$
Explanation: The tendency for catenation is very weak for Germanium. Stable germananes have been prepared, but the maximum number of Germanium atoms in a stable chain is typically low, rarely exceeding six or seven (${n}=6$ or ${n}=7$).
129. Why does Lead (${Pb}$) show virtually no catenation?
ⓐ. Its high electronegativity prevents self-bonding.
ⓑ. It is limited to the $+2$ oxidation state.
ⓒ. The ${Pb}-{Pb}$ bond is very weak due to large atomic size.
ⓓ. The Inert Pair Effect stabilizes the ${sp}^3$ hybridization.
Correct Answer: The ${Pb}-{Pb}$ bond is very weak due to large atomic size.
Explanation: Lead has a very large atomic size, which leads to poor orbital overlap between two ${Pb}$ atoms. This results in a very low ${Pb}-{Pb}$ bond dissociation enthalpy (the lowest in the group), making any potential ${Pb}$ chain highly unstable and prone to breaking.
130. What is the fundamental difference in bonding that allows Carbon to form stable ${C}={C}$ and ${C} \equiv {C}$ multiple bonds, while the other elements in Group 14 primarily form single bonds, hindering their catenation potential?
ⓐ. The availability of ${d}$-orbitals.
ⓑ. Effective ${p}\pi-{p}\pi$ orbital overlap due to small atomic size.
ⓒ. The ability to use $sp^3$ hybridization.
ⓓ. The formation of coordinate covalent bonds.
Correct Answer: Effective ${p}\pi-{p}\pi$ orbital overlap due to small atomic size.
Explanation: Carbon’s small size allows its ${2p}$ orbitals to overlap laterally very effectively with the ${2p}$ orbitals of adjacent atoms, forming stable $\pi$ bonds. The larger atomic size of ${Si}, {Ge}, {Sn}$, and ${Pb}$ makes their valence ${p}$-orbitals too diffuse for effective ${p}\pi-{p}\pi$ overlap, preventing the formation of stable multiple bonds.
131. Which of the following statements correctly describes the bonding ability that is unique to Carbon within Group 14?
ⓐ. Ability to form strong single bonds ($\sigma$ bonds).
ⓑ. Ability to achieve a covalency of six.
ⓒ. Ability to form stable ${p}\pi – {p}\pi$ multiple bonds.
ⓓ. Ability to form linear chain compounds.
Correct Answer: Ability to form stable ${p}\pi – {p}\pi$ multiple bonds.
Explanation: Carbon is the only element in Group 14 that can form stable double (${C}={C}, {C}={O}$) and triple (${C} \equiv {C}, {C} \equiv {N}$) bonds by effective lateral overlap of $2{p}$ orbitals (${p}\pi – {p}\pi$ bonding). The larger atomic size of ${Si}$ and the heavier elements makes their ${p}$ orbitals too diffuse for effective ${p}\pi – {p}\pi$ overlap.
132. The stability of ${C}={C}$ double bonds is primarily a result of Carbon’s:
ⓐ. High atomic mass.
ⓑ. Small atomic size (and high electronegativity).
ⓒ. Availability of $3{d}$ orbitals.
ⓓ. Low ionization enthalpy.
Correct Answer: Small atomic size (and high electronegativity).
Explanation: Carbon’s small size allows its ${2p}$ orbitals to approach closely, leading to strong, effective lateral overlap and the formation of stable $\pi$ bonds. This short bond length is essential for strong ${p}\pi – {p}\pi$ bonding.
133. Which of the following compounds is gaseous at room temperature due to Carbon’s ability to form multiple bonds?
ⓐ. ${SiCl}_4$
ⓑ. ${PbCl}_2$
ⓒ. ${CO}_2$
ⓓ. ${SiO}_2$
Correct Answer: ${CO}_2$
Explanation: Carbon forms stable ${C}={O}$ double bonds, resulting in discrete, neutral, non-polar ${O}={C}={O}$ molecules. The weak intermolecular forces (London dispersion forces) between these molecules lead to a low boiling point, making ${CO}_2$ a gas at room temperature.
134. Silicon forms ${SiO}_2$ with a giant three-dimensional network structure. Why does Silicon not form a simple ${Si}={O}$ double-bonded molecule analogous to ${CO}_2$?
ⓐ. Silicon has a much lower electronegativity than Carbon.
ⓑ. The ${Si}-{O}$ bond is too weak compared to the ${C}-{O}$ bond.
ⓒ. The ${p}\pi$ orbitals of Silicon are too large for effective lateral overlap with Oxygen’s ${p}\pi$ orbitals.
ⓓ. Silicon only exists in the $+2$ oxidation state.
Correct Answer: The ${p}\pi$ orbitals of Silicon are too large for effective lateral overlap with Oxygen’s ${p}\pi$ orbitals.
Explanation: Silicon’s valence shell is $n=3$. The ${3p}$ orbitals are larger and more diffuse than Carbon’s ${2p}$ orbitals. This leads to poor ${3p}\pi – {2p}\pi$ overlap with oxygen, making the hypothetical ${Si}={O}$ double bond unstable. Silicon maximizes stability by forming four strong ${Si}-{O}$ single bonds in a stable network.
135. The stability of ${C}={C}$ double bonds over ${C}-{C}$ single bonds is quantified by the difference in their bond enthalpies. For ${C}$, the average bond enthalpy of ${C}={C}$ is approximately:
ⓐ. Less than twice the ${C}-{C}$ single bond enthalpy.
ⓑ. Greater than twice the ${C}-{C}$ single bond enthalpy.
ⓒ. Exactly twice the ${C}-{C}$ single bond enthalpy.
ⓓ. Equal to the ${C} \equiv {C}$ bond enthalpy.
Correct Answer: Less than twice the ${C}-{C}$ single bond enthalpy.
Explanation: The bond enthalpy of a ${C}-{C}$ single bond is $\approx 348 \\, {kJ} \\, {mol}^{-1}$, and the ${C}={C}$ double bond is $\approx 614 \\, {kJ} \\, {mol}^{-1}$. While a double bond has both a $\sigma$ and a $\pi$ component, the $\pi$ bond is generally weaker than the $\sigma$ bond. Therefore, the double bond energy is less than $2 \times 348 = 696 \\, {kJ} \\, {mol}^{-1}$.
136. Which of the following compounds of Silicon is unstable and highly reactive, contrasting with the stable double-bonded organic compounds of Carbon?
ⓐ. ${SiCl}_4$
ⓑ. ${Si}_n{H}_{2n+2}$ (Silanes)
ⓒ. ${SiC}$ (Silicon Carbide)
ⓓ. ${SiH}_2$ (Silylene)
Correct Answer: ${SiH}_2$ (Silylene)
Explanation: Simple Silicon compounds that attempt to mimic Carbon’s unsaturated state, like ${SiH}_2$ (silylene), are often highly unstable and reactive intermediates because Silicon cannot form stable ${Si}={Si}$ or ${Si}={H}$ multiple bonds due to poor ${p}\pi – {p}\pi$ overlap.
137. Which element, due to its size difference from Oxygen, primarily forms stable ${p}\pi – {d}\pi$ bonds (if any), rather than ${p}\pi – {p}\pi$ bonds?
ⓐ. Carbon (${C}$)
ⓑ. Silicon (${Si}$)
ⓒ. Oxygen (${O}$)
ⓓ. Nitrogen (${N}$)
Correct Answer: Silicon (${Si}$)
Explanation: While ${p}\pi – {p}\pi$ bonding is virtually absent for Silicon, it can potentially participate in ${p}\pi – {d}\pi$ bonding with ${p}$-block elements that have lone pairs (like ${O}$ or ${N}$) by accepting the electron pair into its empty ${3d}$ orbitals. This mechanism is debated but is an important theoretical contrast to Carbon’s ${p}\pi – {p}\pi$ bonding.
138. Which of the following carbon-containing ions derives its unique stability and planar structure from the presence of delocalized ${p}\pi$ bonds?
ⓐ. ${CH}_3^+$ (Methyl carbocation)
ⓑ. ${CH}_3^-$ (Methyl carbanion)
ⓒ. ${CO}_3^{2-}$ (Carbonate ion)
ⓓ. ${C}_2{H}_6$ (Ethane)
Correct Answer: ${CO}_3^{2-}$ (Carbonate ion)
Explanation: The carbonate ion (${CO}_3^{2-}$) has a central ${C}$ atom that is ${sp}^2$ hybridized, resulting in a planar structure. The remaining unhybridized ${p}$ orbital on ${C}$ overlaps with the ${p}$ orbitals of all three ${O}$ atoms, leading to a single, delocalized $\pi$ bond that stabilizes the ion through resonance.
139. In the ${C} \equiv {C}$ triple bond, how many $\pi$ bonds are present?
ⓐ. Zero
ⓑ. Four
ⓒ. Three
ⓓ. Two
Correct Answer: Two
Explanation: A triple bond is composed of one strong sigma ($\sigma$) bond formed by axial overlap of $sp$ orbitals and two $\pi$ bonds formed by lateral overlap of two pairs of unhybridized ${p}$ orbitals.
140. Due to its inability to form stable ${p}\pi – {p}\pi$ bonds, Silicon is prone to reacting with nucleophiles like water. This reaction is called:
ⓐ. Polymerization
ⓑ. Catenation
ⓒ. Hydrolysis
ⓓ. Oligomerization
Correct Answer: Hydrolysis
Explanation: Since Silicon compounds often lack the stabilizing effect of multiple bonds, they are highly susceptible to attack by polar reagents like water. For example, ${SiCl}_4$ hydrolyzes rapidly because the ${Si}$ atom can accept a lone pair from the water molecule into its empty ${3d}$ orbital, leading to the cleavage of the ${Si}-{Cl}$ bonds and subsequent hydrolysis.
141. The vast majority of organic compounds are based on Carbon chains and rings. What is the geometry of the Carbon atoms in a simple saturated linear chain, such as an $n$-alkane?
ⓐ. Linear ($sp$ hybridization)
ⓑ. Trigonal planar ($sp^2$ hybridization)
ⓒ. Tetrahedral ($sp^3$ hybridization)
ⓓ. Trigonal bipyramidal ($sp^3d$ hybridization)
Correct Answer: Tetrahedral ($sp^3$ hybridization)
Explanation: In saturated carbon chains (alkanes), each carbon atom is bonded to four other atoms (either ${C}$ or ${H}$) by single bonds. This requires $sp^3$ hybridization, resulting in a tetrahedral geometry around each carbon center with bond angles near $109.5^\circ$.
142. The simple catenated compounds of Silicon, analogous to alkanes, are highly reactive and known as:
ⓐ. Germanes
ⓑ. Carboranes
ⓒ. Silanes
ⓓ. Silicones
Correct Answer: Silanes
Explanation: Silanes have the general formula ${Si}_n{H}_{2n+2}$ and are the direct ${Si}$ analogues of alkanes (${C}_n{H}_{2n+2}$). Disilane (${Si}_2{H}_6$) and trisilane (${Si}_3{H}_8$) are examples of catenated silanes, though they are much less stable than their carbon counterparts.
143. Which factor causes small cycloalkanes, such as cyclopropane (${C}_3{H}_6$), to be highly reactive compared to larger rings like cyclohexane?
ⓐ. Steric hindrance between hydrogen atoms.
ⓑ. Angle strain (Baeyer strain) due to deviation from the ideal tetrahedral angle.
ⓒ. Instability of the ${C}-{C}$ single bond.
ⓓ. High degree of $sp^2$ character in the bonds.
Correct Answer: Angle strain (Baeyer strain) due to deviation from the ideal tetrahedral angle.
Explanation: The ideal $sp^3$ tetrahedral angle is $109.5^\circ$. In cyclopropane, the bond angles are forced to be $60^\circ$ to close the ring, leading to significant angle strain (or ring strain). This strain increases the potential energy and makes the ring highly susceptible to ring-opening reactions.
144. The formation of stable ring and chain structures in Silicon chemistry is mainly achieved when the Silicon atoms are separated by which bridging element?
ⓐ. Hydrogen
ⓑ. Nitrogen
ⓒ. Carbon
ⓓ. Oxygen
Correct Answer: Oxygen
Explanation: Silicon forms much more stable and durable polymers when the ${Si}$ atoms are linked by Oxygen atoms (${Si}-{O}-{Si}$) rather than being directly linked (${Si}-{Si}$). These ${Si}-{O}-{Si}$ chains and rings form the backbone of silicones and silicates.
145. What is the general name given to the class of organosilicon polymers that form both chains and rings with the ${Si}-{O}-{Si}$ backbone?
ⓐ. Silicones or Polysiloxanes
ⓑ. Silanes
ⓒ. Organogermanes
ⓓ. Borazines
Correct Answer: Silicones or Polysiloxanes
Explanation: Silicones (formally polysiloxanes) are polymers characterized by a flexible chain or ring structure with alternating silicon and oxygen atoms (${R}_2{Si}-{O}-$ backbone, where ${R}$ is an organic group). These polymers are highly stable and chemically inert.
146. What is the fundamental repeating unit that links together to form long chain silicone polymers?
ⓐ. ${Si}({OH})_4$
ⓑ. ${SiCl}_4$
ⓒ. ${R}_3{SiOH}$
ⓓ. ${R}_2{SiO}$ (Dialkylsilanone unit)
Correct Answer: ${R}_2{SiO}$ (Dialkylsilanone unit)
Explanation: Silicone polymers are formed by the controlled hydrolysis and condensation of dialkyldichlorosilanes (${R}_2{SiCl}_2$). This process yields a linear chain with the repeating unit $\mathbf{{R}_2{Si}-{O}-}$, often referred to as a dialkylsilanone unit.
147. When Germanium forms chains (germanes, ${Ge}_n{H}_{2n+2}$), the maximum chain length observed is typically much shorter than that for alkanes. This is primarily because:
ⓐ. Germanium is highly electropositive.
ⓑ. The ${Ge}$ atom is too small to form a long chain.
ⓒ. The ${Ge}-{Ge}$ bond is thermodynamically weaker than the ${C}-{C}$ bond.
ⓓ. Germanium compounds are limited to the $+2$ oxidation state.
Correct Answer: The ${Ge}-{Ge}$ bond is thermodynamically weaker than the ${C}-{C}$ bond.
Explanation: As atomic size increases down the group, the ${E}-{E}$ bond length increases and the orbital overlap decreases, resulting in a lower bond dissociation energy. The weak ${Ge}-{Ge}$ bond means catenated germanes are thermodynamically unstable and easily break down.
148. Which structural feature characterizes the difference between a linear silicone polymer and a cross-linked (3D) silicone polymer?
ⓐ. Presence of ${Si}-{C}$ bonds in cross-linked structures.
ⓑ. The use of ${RSiCl}_3$ (tri-functional) monomers during synthesis for cross-linking.
ⓒ. The degree of hydrolysis of ${SiCl}_4$.
ⓓ. The chain length (${n}$) of the polymer.
Correct Answer: The use of ${RSiCl}_3$ (tri-functional) monomers during synthesis for cross-linking.
Explanation: Linear silicones are made from ${R}_2{SiCl}_2$ (di-functional), which can only form two bonds per ${Si}$ to extend the chain. Cross-linked silicones are formed when a tri-functional monomer, ${RSiCl}_3$, is introduced. This monomer can form three ${Si}-{O}$ bonds, creating side branches and linking chains together into a rigid 3D network.
149. The smallest stable cyclic alkane that exhibits planar geometry is:
ⓐ. Cyclopropane
ⓑ. Cyclohexane
ⓒ. Cyclobutane
ⓓ. Cyclooctane
Correct Answer: Cyclopropane
Explanation: Cyclopropane (${C}_3{H}_6$) is the smallest ring and the only cyclic alkane that is perfectly planar. While it is less stable than larger rings due to significant angle strain, it is a stable molecule that exists readily.
150. In the natural world, the most common form of ring structure containing Group 14 elements is found in silicates, specifically those where ${SiO}_4^{4-}$ tetrahedra link to form closed rings. What is the structural classification for silicates containing the ${Si}_3{O}_9^{6-}$ ion (like in Beryl)?
ⓐ. Sheet Silicate (Phyllosilicate)
ⓑ. Chain Silicate (Pyroxene)
ⓒ. Ring Silicate (Cyclosilicate)
ⓓ. Orthosilicate
Correct Answer: Ring Silicate (Cyclosilicate)
Explanation: Cyclosilicates (or Ring Silicates) are formed when ${SiO}_4^{4-}$ tetrahedra share two oxygen atoms each, forming a closed ring structure. The ${Si}_3{O}_9^{6-}$ ion is a six-membered ring containing three silicon atoms bridged by oxygen.
151. The most significant reason for the differences in the chemical properties of Carbon compared to Silicon is:
ⓐ. Carbon has a half-filled valence shell (${ns}^2{np}^2$).
ⓑ. Carbon lacks available ${d}$-orbitals in its valence shell.
ⓒ. Silicon is a semiconductor while Carbon is not.
ⓓ. Silicon has a much higher electronegativity.
Correct Answer: Carbon lacks available ${d}$-orbitals in its valence shell.
Explanation: Carbon is in the second period ($n=2$) and does not have access to ${2d}$ orbitals, restricting its maximum covalency to 4 and making it unable to act as an electron acceptor. Silicon, being in the third period ($n=3$), has vacant ${3d}$ orbitals, which allows it to expand its covalency to 6 and readily accept electron pairs.
152. What are the maximum covalent bonding capacities (covalencies) typically exhibited by Carbon and Silicon, respectively?
ⓐ. Carbon: 6; Silicon: 4
ⓑ. Carbon: 3; Silicon: 5
ⓒ. Carbon: 4; Silicon: 6
ⓓ. Carbon: 5; Silicon: 5
Correct Answer: Carbon: 4; Silicon: 6
Explanation: Carbon’s maximum covalency is 4 because it can only hybridize its four valence orbitals ($2s$ and $2p$) to form a maximum of four bonds. Silicon, having vacant $3d$ orbitals, can expand its octet and achieve a maximum covalency of 6, as seen in compounds like ${[SiF}_6]^{2-}$.
153. How does the stability of ${C}={C}$ multiple bonds compare to that of hypothetical ${Si}={Si}$ multiple bonds?
ⓐ. Both are highly stable due to ${p}\pi – {p}\pi$ overlap.
ⓑ. ${Si}={Si}$ bonds are more stable due to the larger size of the ${Si}$ atom.
ⓒ. ${C}={C}$ bonds are highly stable; ${Si}={Si}$ bonds are unstable and rarely form.
ⓓ. ${C}={C}$ bonds are less stable than ${C}-{C}$ single bonds.
Correct Answer: ${C}={C}$ bonds are highly stable; ${Si}={Si}$ bonds are unstable and rarely form.
Explanation: Carbon’s small size ensures strong ${2p}\pi – {2p}\pi$ overlap, making ${C}={C}$ and ${C} \equiv {C}$ stable. Silicon’s larger ${3p}$ orbitals lead to poor lateral overlap, making ${Si}={Si}$ double bonds unstable. Silicon prefers to form $\sigma$-bonds in extended networks.
154. Which statement accurately compares the hydrolysis of the tetrachlorides of Carbon and Silicon?
ⓐ. ${SiCl}_4$ hydrolyzes easily, but ${CCl}_4$ does not.
ⓑ. ${CCl}_4$ hydrolyzes easily, but ${SiCl}_4$ does not.
ⓒ. Both ${CCl}_4$ and ${SiCl}_4$ hydrolyze easily due to high polarity.
ⓓ. Neither compound is prone to hydrolysis.
Correct Answer: ${SiCl}_4$ hydrolyzes easily, but ${CCl}_4$ does not.
Explanation: ${SiCl}_4$ readily hydrolyzes because the ${Si}$ atom can accept a lone pair from the nucleophilic water molecule into its empty ${3d}$ orbitals, initiating the reaction. ${CCl}_4$ does not hydrolyze because the Carbon atom lacks available ${d}$-orbitals to accommodate the incoming water molecule, thus preventing the necessary intermediate complex formation.
155. Compare the bond dissociation enthalpies (${BDE}$) of the single bonds formed with Oxygen for Carbon and Silicon (${E}-{O}$ bond).
ⓐ. ${C}-{O}$ BDE is much higher than ${Si}-{O}$ BDE.
ⓑ. ${C}-{O}$ BDE and ${Si}-{O}$ BDE are nearly equal.
ⓒ. Both ${C}-{O}$ and ${Si}-{O}$ BDEs are the lowest in Group 14.
ⓓ. ${Si}-{O}$ BDE is significantly higher than ${C}-{O}$ BDE.
Correct Answer: ${Si}-{O}$ BDE is significantly higher than ${C}-{O}$ BDE.
Explanation: The ${Si}-{O}$ bond is one of the strongest single bonds known ($452 \\, {kJ} \\, {mol}^{-1}$), substantially higher than the ${C}-{O}$ bond ($358 \\, {kJ} \\, {mol}^{-1}$). This strength explains why ${SiO}_2$ forms a stable network and why silicones (${Si}-{O}-{Si}$) are so inert.
156. Carbon dioxide (${CO}_2$) is a gas, whereas silicon dioxide (${SiO}_2$) is a high melting point solid. This difference is due to:
ⓐ. Carbon having a higher atomic mass than Silicon.
ⓑ. ${CO}_2$ being ionic and ${SiO}_2$ being covalent.
ⓒ. ${CO}_2$ being a discrete molecular gas and ${SiO}_2$ being a giant network solid.
ⓓ. ${SiO}_2$ forming stable ${Si}={O}$ double bonds.
Correct Answer: ${CO}_2$ being a discrete molecular gas and ${SiO}_2$ being a giant network solid.
Explanation: ${CO}_2$ forms stable ${C}={O}$ double bonds resulting in weak intermolecular forces. ${SiO}_2$, due to the instability of ${Si}={O}$ double bonds, maximizes stability by forming a continuous tetrahedral structure where strong ${Si}-{O}$ single bonds extend throughout the crystal, leading to its high melting point.
157. Which element is capable of forming a stable hexacoordinate complex anion, where its ${E}$ atom is bonded to six ${F}$ atoms?
ⓐ. Carbon, ${[CF}_6]^{2-}$
ⓑ. Silicon, ${[SiF}_6]^{2-}$
ⓒ. Carbon and Silicon, both
ⓓ. Neither element
Correct Answer: Silicon, ${[SiF}_6]^{2-}$
Explanation: Silicon can form the stable hexacoordinate complex ion, ${[SiF}_6]^{2-}$, by utilizing its vacant ${3d}$ orbitals to expand its octet and achieve a covalency of 6 (octahedral geometry). Carbon cannot do this due to the absence of available ${d}$-orbitals.
158. In terms of chemical reactivity, how do the catenated hydrides of Carbon (${C}_n{H}_{2n+2}$) compare to the catenated hydrides of Silicon (${Si}_n{H}_{2n+2}$)?
ⓐ. Alkanes are chemically inert; Silanes are highly reactive and spontaneously flammable in air.
ⓑ. Silanes are more stable and less reactive towards air.
ⓒ. Both are equally stable and inert due to the protective hydrogen layer.
ⓓ. Alkanes are reactive towards water, while Silanes are not.
Correct Answer: Alkanes are chemically inert; Silanes are highly reactive and spontaneously flammable in air.
Explanation: Alkanes are stabilized by strong ${C}-{C}$ and ${C}-{H}$ bonds. Silanes are much less stable. The ${Si}-{H}$ bonds are less robust than ${C}-{H}$ bonds and the ${Si}$ atoms have low-lying empty ${3d}$ orbitals, which makes Silanes highly susceptible to nucleophilic attack (like ${O}_2$ or ${H}_2{O}$), leading to spontaneous combustion or hydrolysis.
159. Carbon’s maximum oxidation state of +4 is strongly preferred. While Silicon also prefers +4, it rarely shows a stable +2 state. This difference suggests that the Inert Pair Effect is:
ⓐ. Stronger for Carbon than for Silicon.
ⓑ. Negligible or non-existent for both Carbon and Silicon.
ⓒ. Responsible for the high stability of ${Si}({II})$ compounds.
ⓓ. Only found in compounds where the element is bonded to Oxygen.
Correct Answer: Negligible or non-existent for both Carbon and Silicon.
Explanation: The Inert Pair Effect (the reluctance of ${ns}^2$ electrons to participate in bonding) becomes significant only in the heavier elements of the group (${Sn}$ and ${Pb}$). For both Carbon ($n=2$) and Silicon ($n=3$), the effect is minimal or absent, meaning the ${ns}^2$ electrons are easily promoted or shared, making the +4 oxidation state overwhelmingly dominant for both elements.
160. Which statement is true regarding the acidity/hydrolysis of the oxides of Carbon and Silicon in the presence of strong base?
ⓐ. ${CO}_2$ reacts as an amphoteric oxide; ${SiO}_2$ reacts as a basic oxide.
ⓑ. ${CO}_2$ is a neutral oxide, ${SiO}_2$ is an acidic oxide.
ⓒ. ${CO}_2$ is a basic oxide, ${SiO}_2$ is a neutral oxide.
ⓓ. Both ${CO}_2$ and ${SiO}_2$ react as acidic oxides.
Correct Answer: Both ${CO}_2$ and ${SiO}_2$ react as acidic oxides.
Explanation: Both Carbon and Silicon are non-metals, and their higher oxides are acidic.
${CO}_2$ dissolves in water to form carbonic acid (${H}_2{CO}_3$) and reacts with ${NaOH}$ to form ${Na}_2{CO}_3$.
${SiO}_2$ is insoluble in water but reacts with strong bases (like fused ${NaOH}$ or ${Na}_2{CO}_3$) to form silicates (${Na}_2{SiO}_3$), demonstrating its acidic character.
161. What is the hybridization of the carbon atoms in the diamond crystal structure?
ⓐ. $sp$
ⓑ. $sp^2$
ⓒ. $sp^3$
ⓓ. $dsp^2$
Correct Answer: $sp^3$
Explanation: In diamond, each carbon atom is covalently bonded to four other carbon atoms in a symmetrical, three-dimensional lattice. To accommodate these four single bonds, the carbon atom must utilize $sp^3$ hybridization.
162. Diamond is classified as which type of solid?
ⓐ. Molecular solid
ⓑ. Ionic solid
ⓒ. Metallic solid
ⓓ. Covalent network solid
Correct Answer: Covalent network solid
Explanation: Diamond is a covalent network solid (or giant molecule) where the entire crystal is composed of atoms held together by strong, continuous covalent bonds (${C}-{C}$). This structure extends throughout the whole crystal, leading to its extreme properties.
163. What is the specific geometry surrounding each carbon atom in the diamond lattice?
ⓐ. Trigonal planar
ⓑ. Tetrahedral
ⓒ. Octahedral
ⓓ. Linear
Correct Answer: Tetrahedral
Explanation: Since each carbon atom is $sp^3$ hybridized, the four surrounding carbon atoms are arranged in a tetrahedral configuration, with the bond angles being precisely $109.5^\circ$.
164. Diamond is the hardest known natural substance. This property is directly attributable to:
ⓐ. The presence of van der Waals forces.
ⓑ. High degree of ionic character in its bonds.
ⓒ. The strong, three-dimensional network of covalent bonds.
ⓓ. Its exceptionally high density.
Correct Answer: The strong, three-dimensional network of covalent bonds.
Explanation: The extreme hardness of diamond results from the fact that to scratch or break it, a huge amount of energy is required to break the numerous strong covalent bonds that make up the continuous 3D network structure.
165. Diamond is a non-conductor of electricity. What structural feature explains this property?
ⓐ. Its high density.
ⓑ. All valence electrons are localized in ${C}-{C}$ covalent bonds.
ⓒ. The repulsion between the tetrahedral units.
ⓓ. Its ability to absorb light.
Correct Answer: All valence electrons are localized in ${C}-{C}$ covalent bonds.
Explanation: For a substance to conduct electricity, it must have mobile electrons or ions. In diamond, all four valence electrons of each carbon atom are tied up in four strong $\sigma$ (${C}-{C}$) bonds, meaning there are no free or delocalized electrons available to move and carry current.
166. Diamond possesses an exceptionally high melting point (subliming at around $3550 \, ^{\circ}{C}$). This is due to the need to overcome which forces/bonds?
ⓐ. Strong covalent $\sigma$ bonds in the giant molecule.
ⓑ. Weak London dispersion forces.
ⓒ. Ionic forces between ${C}^+$ and ${C}^-$ ions.
ⓓ. Metallic bonding forces.
Correct Answer: Strong covalent $\sigma$ bonds in the giant molecule.
Explanation: To melt or sublime diamond, one must break the strong covalent bonds linking every atom in the lattice. This requires a massive input of thermal energy, leading to its extremely high sublimation temperature.
167. When heated strongly in the presence of air or pure oxygen, diamond will react to form:
ⓐ. Graphite
ⓑ. Carbon monoxide (${CO}$)
ⓒ. ${C}_2{O}_3$
ⓓ. Carbon dioxide (${CO}_2$)
Correct Answer: Carbon dioxide (${CO}_2$)
Explanation: Like all forms of carbon, when diamond is heated to high temperatures in the presence of oxygen, it undergoes complete combustion, reacting with oxygen to form carbon dioxide.
$${C}({diamond}) + {O}_2({g}) \xrightarrow{{heat}} {CO}_2({g})$$
168. How does the density of diamond compare to the density of graphite?
ⓐ. Diamond’s density is slightly lower.
ⓑ. Diamond and graphite have approximately equal densities.
ⓒ. Diamond’s density is significantly higher.
ⓓ. Density comparison varies depending on temperature.
Correct Answer: Diamond’s density is significantly higher.
Explanation: Diamond ($\approx 3.51 \, {g/cm}^3$) has a much higher density than graphite ($\approx 2.25 \, {g/cm}^3$). This is because the ${sp}^3$ tetrahedral structure of diamond is closely packed, while the layered ${sp}^2$ structure of graphite is much more open and bulky.
169. What gives diamond its exceptional brilliance and sparkle (high lustre)?
ⓐ. Its high electrical conductivity.
ⓑ. Its very high refractive index.
ⓒ. The reflection of light off its layered structure.
ⓓ. Its low thermal conductivity.
Correct Answer: Its very high refractive index.
Explanation: Diamond has one of the highest refractive indices of any natural material. This property causes incident light to be greatly slowed down and refracted (bent), leading to total internal reflection within a properly cut stone, which produces its characteristic brilliance and fire.
170. Which statement is true regarding the thermal conductivity of diamond?
ⓐ. Diamond is an excellent thermal conductor, superior to most metals.
ⓑ. Diamond is a poor thermal conductor, similar to glass.
ⓒ. Thermal conductivity increases sharply when diamond is subjected to a magnetic field.
ⓓ. Diamond is only a thermal conductor in the presence of an electric current.
Correct Answer: Diamond is an excellent thermal conductor, superior to most metals.
Explanation: While diamond is an electrical insulator, it is an exceptionally efficient conductor of heat. Heat is transferred efficiently through the rapid vibration of atoms within its rigid, strong covalent lattice, making it the best natural conductor of heat known.
171. What is the hybridization of the carbon atoms within the layers of the graphite structure?
ⓐ. $sp$
ⓑ. $sp^2$
ⓒ. $sp^3$
ⓓ. $dsp^2$
Correct Answer: $sp^2$
Explanation: In graphite, each carbon atom is bonded to three other carbon atoms within the same plane. This requires $sp^2$ hybridization, resulting in a planar hexagonal arrangement.
172. The structure of graphite is composed of flat, two-dimensional layers. The bonding within these layers is:
ⓐ. Ionic, leading to brittle layers.
ⓑ. Strong covalent ($\sigma$ bonds), forming hexagonal rings.
ⓒ. Metallic, allowing free movement of atoms.
ⓓ. Weak ${C}-{C}$ $\pi$ bonds only.
Correct Answer: Strong covalent ($\sigma$ bonds), forming hexagonal rings.
Explanation: Within each layer, the carbon atoms form strong covalent bonds with three neighbors, creating extended sheets of hexagonal rings. This strong bonding within the layer is similar to a giant molecule.
173. What type of intermolecular forces holds the parallel hexagonal layers of graphite together?
ⓐ. Strong ionic forces
ⓑ. Strong covalent bonds
ⓒ. Weak van der Waals forces (London dispersion forces)
ⓓ. Hydrogen bonds
Correct Answer: Weak van der Waals forces (London dispersion forces)
Explanation: The layers are held together by very weak van der Waals forces (specifically London dispersion forces). These weak forces allow the layers to easily slide past each other, explaining graphite’s softness and lubricity.
174. Graphite is an excellent conductor of electricity. This property is due to the presence of:
ⓐ. Delocalized $\pi$ electrons within each layer.
ⓑ. Strong metallic bonds.
ⓒ. Mobile ${C}^+$ ions.
ⓓ. Vacant ${d}$-orbitals.
Correct Answer: Delocalized $\pi$ electrons within each layer.
Explanation: Since each carbon atom is $sp^2$ hybridized, only three of its four valence electrons are used for the $\sigma$ bonds. The remaining one valence electron per atom resides in a $p$-orbital perpendicular to the layer. These $p$-orbitals overlap laterally to form a continuous delocalized electron cloud (a $\pi$ system) above and below the plane, allowing electrons to move freely.
175. In which direction is the electrical conductivity of graphite most efficient?
ⓐ. Perpendicular to the layers (across the layers).
ⓑ. Parallel to the layers (within the planes).
ⓒ. Conductivity is isotropic (equal in all directions).
ⓓ. Conductivity depends on the length of the sheet.
Correct Answer: Parallel to the layers (within the planes).
Explanation: The delocalized $\pi$ electrons, which are responsible for conductivity, are confined to the plane of the layers. Therefore, electrical conduction is much easier and faster parallel to the layers than across the layers. This makes graphite an anisotropic conductor.
176. Graphite is commonly used as a lubricant and in pencil lead because of which physical property?
ⓐ. Its magnetic properties.
ⓑ. Its high melting point.
ⓒ. Its low density.
ⓓ. Its softness and greasy feel, allowing layers to slide.
Correct Answer: Its softness and greasy feel, allowing layers to slide.
Explanation: The extremely weak van der Waals forces between the carbon layers allow them to slip easily over one another when subjected to pressure or shear force. This sliding motion is what gives graphite its lubricity and makes it effective as pencil lead.
177. The ${C}-{C}$ bond length within a graphite layer is approximately $141.5 \, {pm}$. This value lies between the bond length of a ${C}-{C}$ single bond ($154 \, {pm}$) and a ${C}={C}$ double bond ($134 \, {pm}$). This is evidence for:
ⓐ. Strong van der Waals forces.
ⓑ. $sp^3$ hybridization.
ⓒ. Partial double bond character (resonance/delocalization).
ⓓ. Metallic bonding.
Correct Answer: Partial double bond character (resonance/delocalization).
Explanation: All ${C}-{C}$ bonds in the hexagonal rings of graphite are identical, with a length intermediate between a single and a double bond. This indicates that the electrons are delocalized over the entire layer, giving each bond partial double bond character (aromatic resonance).
178. How does the density of graphite compare to that of diamond, and what is the structural reason for the difference?
ⓐ. Graphite is denser because its layers are tightly packed.
ⓑ. Graphite is less dense because its layered structure is relatively open.
ⓒ. Graphite is less dense because its ${C}-{C}$ bonds are stronger.
ⓓ. Graphite and diamond have the same density as they are made of the same element.
Correct Answer: Graphite is less dense because its layered structure is relatively open.
Explanation: Graphite ($\approx 2.25 \, {g/cm}^3$) is significantly less dense than diamond ($\approx 3.51 \, {g/cm}^3$). The large distance between the layers ($\approx 335 \, {pm}$) makes the overall arrangement much more open and bulky, leading to a lower density.
179. Graphite is used as a moderator in nuclear reactors. This is due to its ability to:
ⓐ. Slow down (moderate) fast neutrons without capturing them.
ⓑ. Completely absorb high-energy neutrons.
ⓒ. Reflect neutrons back into the core.
ⓓ. Conduct the intense electricity generated by the reactor.
Correct Answer: Slow down (moderate) fast neutrons without capturing them.
Explanation: A nuclear moderator slows down fast neutrons, making them thermal neutrons that are capable of sustaining a fission chain reaction. Graphite is an excellent moderator because it is light enough to effectively transfer kinetic energy away from the neutrons in collisions, but it is chemically inert and has a low neutron absorption cross-section (it does not capture the neutrons).
180. Unlike diamond, graphite is thermodynamically more stable at standard conditions. This is reflected by the enthalpy change ($\Delta H$) of the conversion reaction: ${C}({graphite}) \to {C}({diamond})$. The value of $\Delta H$ for this reaction is:
ⓐ. Large negative (exothermic).
ⓑ. Zero.
ⓒ. Small positive (endothermic).
ⓓ. Large positive (highly endothermic).
Correct Answer: Small positive (endothermic).
Explanation: Since graphite is the reference state and the most stable form of carbon under standard conditions, the conversion of graphite to diamond is slightly endothermic ($\Delta H \approx +1.897 \, {kJ} \, {mol}^{-1}$). This confirms that diamond is thermodynamically less stable than graphite.
181. The discovery of Buckminsterfullerene (${C}_{60}$) in 1985 led to the awarding of the Nobel Prize in Chemistry in 1996 to which trio of scientists?
ⓐ. Linus Pauling, Robert Robinson, and Roger Adams
ⓑ. John Dalton, Jöns Jacob Berzelius, and Antoine Lavoisier
ⓒ. Robert F. Curl, Harold W. Kroto, and Richard E. Smalley
ⓓ. Peter W. Higgs, François Englert, and Satoshi Ōmura
Correct Answer: Robert F. Curl, Harold W. Kroto, and Richard E. Smalley
Explanation: The discovery of fullerenes, a new allotrope of carbon, was made by Curl, Kroto, and Smalley at Rice University and Sussex University, for which they were jointly awarded the Nobel Prize in Chemistry in 1996.
182. ${C}_{60}$ is the most stable and common fullerene molecule. Its structure closely resembles a soccer ball. What is the common name given to this specific ${C}_{60}$ allotrope?
ⓐ. Graphane
ⓑ. Diamondoid
ⓒ. Carbon nanotube
ⓓ. Buckminsterfullerene
Correct Answer: Buckminsterfullerene
Explanation: The molecule was named Buckminsterfullerene after the architect Buckminster Fuller, who designed geodesic domes with similar hexagonal and pentagonal patterns.
183. In the structure of ${C}_{60}$ (Buckminsterfullerene), what is the hybridization of every carbon atom?
ⓐ. $sp$
ⓑ. $sp^2$
ⓒ. $sp^3$
ⓓ. $dsp^2$
Correct Answer: $sp^2$
Explanation: Each carbon atom in ${C}_{60}$ is bonded to three other carbon atoms (forming three $\sigma$ bonds), which is characteristic of $sp^2$ hybridization. The remaining $p$-orbital electron is delocalized over the entire spherical surface, contributing to the stability and aromaticity of the molecule.
184. The structure of ${C}_{60}$ consists of a closed cage made up entirely of five-membered rings and six-membered rings. How many five-membered rings are present in the ${C}_{60}$ molecule?
ⓐ. 12
ⓑ. 13
ⓒ. 15
ⓓ. 20
Correct Answer: 12
Explanation: The structure of ${C}_{60}$ is defined by 12 pentagons (five-membered rings) and 20 hexagons (six-membered rings). The presence of exactly 12 pentagons is required to allow the sheet of carbon atoms to close into a sphere.
185. Following up on the previous question, how many six-membered rings (hexagons) are present in the ${C}_{60}$ molecule?
ⓐ. 12
ⓑ. 15
ⓒ. 20
ⓓ. 30
Correct Answer: 20
Explanation: Buckminsterfullerene has a truncated icosahedron structure which contains 12 pentagons and 20 hexagons. The carbon atoms sit at the corners where these rings meet.
186. 186: What is the total number of ${C}-{C}$ covalent bonds (single and double) present in the ${C}_{60}$ molecule?
ⓐ. 30
ⓑ. 60
ⓒ. 90
ⓓ. 120
Correct Answer: 30
Explanation: In the ${C}_{60}$ molecule (Buckminsterfullerene), there are two distinct types of bonds. The shorter bonds (which have double-bond character) are located where two hexagons meet. There are 30 of these bonds. The 60 longer bonds (which have single-bond character) are located where a hexagon and a pentagon meet.
187. Fullerenes are typically soluble in organic solvents, exhibiting a distinctive color. What is the color of the ${C}_{60}$ solution in solvents like toluene or benzene?
ⓐ. Colorless
ⓑ. Blue
ⓒ. Red-brown
ⓓ. Magenta or purple
Correct Answer: Magenta or purple
Explanation: Fullerenes, like ${C}_{60}$, form dark, crystalline solids. When dissolved in aromatic organic solvents such as toluene or benzene, the resulting solution has a characteristic bright magenta or deep purple color.
188. Fullerenes exhibit chemical properties that classify them as:
ⓐ. Strong Brønsted-Lowry acids.
ⓑ. Lewis acids, acting as electron acceptors.
ⓒ. Strong $\pi$-donating Lewis bases.
ⓓ. Inert gases due to their stable cage structure.
Correct Answer: Lewis acids, acting as electron acceptors.
Explanation: Fullerenes have a relatively low-lying ${LUMO}$ (Lowest Unoccupied Molecular Orbital) and are thus capable of accepting electrons into their $\pi$ system. This makes them relatively good Lewis acids (electron pair acceptors), capable of forming radical anions or reacting with nucleophiles.
189. A unique property of fullerenes is their ability to trap metal ions or noble gas atoms inside the carbon cage without chemical bonding. What is the term used to describe these encapsulated fullerene compounds?
ⓐ. Endofullerenes
ⓑ. Exofullerenes
ⓒ. Dendrimers
ⓓ. Nanotubes
Correct Answer: Endofullerenes
Explanation: Compounds where an atom or small molecule is completely enclosed within the fullerene cage are called Endohedral Fullerenes or simply Endofullerenes (e.g., ${La} @ {C}_{60}$). The $@$ symbol indicates the atom is inside the cage.
190. What is a key property that distinguishes fullerenes from diamond and graphite, which are purely covalent network solids?
ⓐ. High melting point
ⓑ. Electrical conductivity
ⓒ. Solubility in organic solvents
ⓓ. Ability to form single bonds
Correct Answer: Solubility in organic solvents
Explanation: Fullerenes are discrete, molecular species held together by weak van der Waals forces. This molecular nature allows them to dissolve in organic solvents (like benzene or toluene), forming true solutions. Diamond and graphite are extended network solids that are insoluble in all common solvents.
191. Graphene is structurally best described as:
ⓐ. A three-dimensional tetrahedral lattice.
ⓑ. A spherical cage-like molecule (${C}_{60}$).
ⓒ. A single, two-dimensional layer of ${sp}^2$-hybridized carbon atoms.
ⓓ. A rolled-up cylindrical sheet of ${sp}^3$-hybridized carbon.
Correct Answer: A single, two-dimensional layer of ${sp}^2$-hybridized carbon atoms.
Explanation: Graphene is essentially a single atomic layer of graphite. It is a one-atom-thick sheet of ${sp}^2$-hybridized carbon atoms arranged in a honeycomb (hexagonal) lattice. This 2D structure is the fundamental building block of graphite, fullerenes, and nanotubes.
192. What structural feature allows graphene to be an exceptionally good conductor of electricity?
ⓐ. Delocalized $\pi$ electrons moving within the hexagonal lattice.
ⓑ. The use of $sp^3$ orbitals.
ⓒ. Strong ${C}-{C}$ single bonds.
ⓓ. The presence of metallic ions embedded in the sheet.
Correct Answer: Delocalized $\pi$ electrons moving within the hexagonal lattice.
Explanation: In graphene, the unhybridized ${p}$ orbital on each ${sp}^2$-hybridized carbon atom overlaps with its neighbors, forming a continuous sheet of delocalized $\pi$ electrons above and below the plane. These highly mobile charge carriers give graphene its remarkable electrical conductivity.
193. Carbon Nanotubes (CNTs) are structurally derived from which other carbon allotrope?
ⓐ. Graphene
ⓑ. Fullerenes (${C}_{60}$)
ⓒ. Diamond
ⓓ. Amorphous carbon
Correct Answer: Graphene
Explanation: A Carbon Nanotube is conceptually formed by rolling up a single sheet of graphene into a seamless cylinder. The size and chirality (twist) of the tube depend on how the graphene sheet is rolled.
194. Carbon Nanotubes (CNTs) are famous for being the strongest and stiffest materials known. This is primarily due to:
ⓐ. The presence of ionic bonds in the cylinder walls.
ⓑ. The strong van der Waals forces between the tube ends.
ⓒ. The strength of the ${sp}^2$ covalent bonds connecting every atom in the lattice.
ⓓ. The hollow nature of the cylindrical structure.
Correct Answer: The strength of the ${sp}^2$ covalent bonds connecting every atom in the lattice.
Explanation: Like graphene, the walls of the carbon nanotube are made of highly stable, strong ${sp}^2$ covalent bonds ($\sigma$ and $\pi$ systems) that form a perfect hexagonal lattice. This flawless, extended covalent bonding gives CNTs exceptional tensile strength and mechanical rigidity.
195. The electrical properties of a single-walled carbon nanotube (SWCNT) can vary significantly depending on its “chirality.” A carbon nanotube can exhibit the electrical properties of which two material types?
ⓐ. Superconductor and insulator
ⓑ. Metal and ionic conductor
ⓒ. Semiconductor and insulator
ⓓ. Metal and semiconductor
Correct Answer: Metal and semiconductor
Explanation: Chirality describes the angle at which the graphene sheet is rolled. Depending on this chiral angle, the $\pi$ electron band structure can change, causing a nanotube to behave either as a metallic conductor (with high conductivity) or as a semiconductor (with a measurable band gap).
196. The thickness of a graphene sheet is defined by the diameter of a single atom. Approximately how thick is a sheet of graphene?
ⓐ. $10 \\, {nm}$
ⓑ. $1 \\, {nm}$
ⓒ. $0.34 \\, {nm}$
ⓓ. $100 \\, {nm}$
Correct Answer: $0.34 \\, {nm}$
Explanation: The thickness of a graphene sheet is defined by the interlayer spacing in graphite, which is $\approx 0.34 \\, {nm}$ ($3.4 \\, {\\AA}$), corresponding to the diameter of a single carbon atom. This extreme thinness contributes to its unique 2D properties.
197. Single-walled carbon nanotubes (SWCNTs) are often bundled together into ropes. The individual tubes within a rope are held together by:
ⓐ. Covalent bonds ($\sigma$ bonds)
ⓑ. Ionic bonds
ⓒ. Weak van der Waals forces
ⓓ. Metallic bonds
Correct Answer: Weak van der Waals forces
Explanation: Although the walls of the tubes are held together by strong covalent bonds, the individual tubes attract each other through relatively weak van der Waals forces (like the layers in graphite). This attraction causes them to spontaneously assemble into bundled ropes.
198. Why is graphene considered to be a much more promising material than graphite for transparent, flexible electronic devices?
ⓐ. Graphene is heavier and more durable.
ⓑ. Graphene is an insulator, while graphite is a conductor.
ⓒ. Graphite requires a vacuum to be stable, unlike graphene.
ⓓ. Graphene is optically transparent and extremely flexible.
Correct Answer: Graphene is optically transparent and extremely flexible.
Explanation: Graphene’s single-atom thickness makes it nearly perfectly transparent (absorbing only $2.3 \\, {\\%}$ of light). Coupled with its incredible strength and flexibility, it is an ideal candidate for applications like flexible touchscreens and transparent electrodes.
199. Multi-walled carbon nanotubes (MWCNTs) consist of several concentric tubes. The distance between these concentric walls is closest to the interlayer spacing found in:
ⓐ. Diamond
ⓑ. Graphite
ⓒ. Fullerenes
ⓓ. Coal
Correct Answer: Graphite
Explanation: The spacing between the concentric walls in an MWCNT is approximately $0.34 \\, {nm}$, which is essentially the same as the interlayer distance in bulk graphite. This indicates that the layers within an MWCNT interact via the same van der Waals forces as in bulk graphite.
200. Graphene exhibits a unique phenomenon where electrons behave as if they have no mass (known as massless Dirac fermions). This is responsible for which of its extreme properties?
ⓐ. Its magnetic levitation ability.
ⓑ. Its extreme hardness.
ⓒ. Its exceptionally high electron mobility/speed.
ⓓ. Its low density.
Correct Answer: Its exceptionally high electron mobility/speed.
Explanation: The “massless Dirac fermion” behavior of electrons in graphene is a quantum mechanical effect that gives them extremely high mobility (the speed at which they travel in an electric field). This high mobility is the basis for graphene’s potential use in ultrafast transistors.
You are on Class 11 Chemistry MCQs – Chapter 11: The p-Block Elements (Part 2). This section brings the second set of 100 solved MCQs covering deeper concepts of Group 14 (Carbon Family). These questions strengthen your understanding of CO, CO₂, silicones, silicates, and allotropes of carbon such as diamond, graphite, and fullerenes.
Practice each MCQ carefully and use the ❤️ Favourite icon to mark important ones for revision. Write short notes under the Workspace section below each question—your notes will be saved automatically. Click the Random button to shuffle questions and test your preparation like a mock test.
- 👉 Total in chapter: 277 MCQs (100 + 100 + 77)
- 👉 This page: Second set of 100 solved MCQs
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