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1. Where are the p-block elements located in the modern periodic table?
ⓐ. Groups 1 and 2
ⓑ. Groups 3 to 12
ⓒ. Groups 13 to 18
ⓓ. Lanthanides and Actinides
Correct Answer: Groups 13 to 18
Explanation: The p-block elements are those in which the last differentiating electron enters any of the three p-orbitals of their outermost (valence) shell. These elements are placed on the right side of the periodic table, encompassing Groups 13, 14, 15, 16, 17, and 18. Class 11 chemistry specifically focuses on Groups 13 and 14.
2. What is the general valence shell electronic configuration for the p-block elements?
ⓐ. $ns^2 np^{1-6}$
ⓑ. $(n-1)d^{1-10} ns^{1-2}$
ⓒ. $ns^1$ to $ns^2$
ⓓ. $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$
Correct Answer: $ns^2 np^{1-6}$
Explanation: For all p-block elements, the outermost s-orbital is completely filled with two electrons. The differentiating (last) electron enters the p-orbital of the same outermost shell. The p-orbital can accommodate a maximum of six electrons. Therefore, the general valence shell electronic configuration for these elements is $ns^2 np^{1-6}$, where ‘n’ is the principal quantum number.
3. Which of the following elements belongs to Group 13 (Boron family)?
ⓐ. Silicon (Si)
ⓑ. Nitrogen (N)
ⓒ. Gallium (Ga)
ⓓ. Lead (Pb)
Correct Answer: Gallium (Ga)
Explanation: Group 13 of the periodic table is known as the Boron family. The elements in this group are Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), and Thallium (Tl). Silicon (Si) and Lead (Pb) belong to Group 14 (Carbon family), and Nitrogen (N) belongs to Group 15.
4. The non-metallic character in the p-block elements:
ⓐ. Decreases down the group and increases from left to right across a period.
ⓑ. Increases down the group and decreases from left to right across a period.
ⓒ. Remains the same down the group.
ⓓ. Remains the same across a period.
Correct Answer: Decreases down the group and increases from left to right across a period.
Explanation: Non-metallic character is related to electronegativity and ionization enthalpy. As we move down a group, atomic size increases and ionization enthalpy decreases, making the element more metallic. Thus, non-metallic character decreases. As we move from left to right across a period, atomic size decreases and electronegativity increases, making the element more non-metallic.
5. What is the “inert pair effect”?
ⓐ. The inability of p-electrons to participate in bonding.
ⓑ. The reluctance of the outermost $ns^2$ electrons to participate in bonding.
ⓒ. The increase in shielding effect by d- and f-orbitals.
ⓓ. The pairing of electrons in the d-orbital.
Correct Answer: The reluctance of the outermost $ns^2$ electrons to participate in bonding.
Explanation: The inert pair effect is a phenomenon observed in the heavier elements of the p-block (especially Groups 13, 14, and 15). It describes the tendency of the two electrons in the outermost s-orbital (the $ns^2$ pair) to remain unshared or “inert” in compounds. This effect is due to the poor shielding of the nuclear charge by the intervening d- and f-orbitals, which pulls the $ns^2$ electrons closer to the nucleus, making them less available for bonding.
6. Due to the inert pair effect, which oxidation state becomes more stable for the heavier elements of Group 14?
ⓐ. +4
ⓑ. +2
ⓒ. +3
ⓓ. +1
Correct Answer: +2
Explanation: The group oxidation state for Group 14 (Carbon family) is +4 (by losing $ns^2$ and $np^2$ electrons). However, due to the inert pair effect, the heavier elements like Tin (Sn) and Lead (Pb) show a stable +2 oxidation state (by losing only the $np^2$ electrons). For Lead, the +2 state is significantly more stable than the +4 state.
7. Which of the following sets of elements constitutes the Group 14 (Carbon family)?
ⓐ. B, Al, Ga, In, Tl
ⓑ. C, Si, Ge, Sn, Pb
ⓒ. N, P, As, Sb, Bi
ⓓ. O, S, Se, Te, Po
Correct Answer: C, Si, Ge, Sn, Pb
Explanation: Group 14 of the periodic table is known as the Carbon family. The elements in this group are Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb). Option A represents Group 13, Option C represents Group 15, and Option D represents Group 16.
8. The first element of each p-block group shows anomalous behavior. This is primarily due to:
ⓐ. Its large size and high metallic character.
ⓑ. The presence of d-orbitals.
ⓒ. Its small size, high ionization enthalpy, high electronegativity, and absence of d-orbitals.
ⓓ. Its low electronegativity and low ionization enthalpy.
Correct Answer: Its small size, high ionization enthalpy, high electronegativity, and absence of d-orbitals.
Explanation: The first element of a p-block group (e.g., Boron in Group 13, Carbon in Group 14) differs significantly from the other members of its group. This anomalous behavior is a combined result of its (i) exceptionally small atomic size, (ii) very high ionization enthalpy, (iii) high electronegativity, and (iv) the complete absence of d-orbitals in its valence shell, which limits its maximum covalency (e.g., Boron to 4, Carbon to 4).
9. Which element in Group 13 is a non-metal, while the others are metals?
ⓐ. Aluminium (Al)
ⓑ. Gallium (Ga)
ⓒ. Boron (B)
ⓓ. Thallium (Tl)
Correct Answer: Boron (B)
Explanation: Boron (B) is the first element of Group 13 and is a typical non-metal (or sometimes classified as a metalloid). It has a very small size and high ionization enthalpy, leading to covalent bonding. All other elements in Group 13 (Aluminium, Gallium, Indium, and Thallium) are distinctly metallic in nature, with metallic character increasing down the group.
10. The p-block elements are unique because their members include:
ⓐ. Only metals
ⓑ. Only non-metals
ⓒ. Only metalloids
ⓓ. Metals, non-metals, and metalloids
Correct Answer: Metals, non-metals, and metalloids
Explanation: The p-block is the only block in the periodic table that contains all three types of elements. For example, in the groups studied in Class 11, Boron (Group 13) is a non-metal/metalloid, while Aluminium (Group 13) is a metal. In Group 14, Carbon is a non-metal, Silicon and Germanium are metalloids, and Tin and Lead are metals. This diversity is a key feature of the p-block.
11. What is the specific general electronic configuration for Group 13 elements?
ⓐ. $ns^1 np^1$
ⓑ. $ns^2 np^1$
ⓒ. $ns^2 np^2$
ⓓ. $ns^2$
Correct Answer: $ns^2 np^1$
Explanation: The Group 13 elements are the first group in the p-block. Their valence shell contains three electrons. Two of these electrons are in the s-orbital, and one electron enters the p-orbital of the same shell. Therefore, their general valence shell electronic configuration is $ns^2 np^1$, where ‘n’ ranges from 2 (for Boron) to 6 (for Thallium).
12. How many valence electrons are present in the elements of the Boron family?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 13
Correct Answer: 3
Explanation: The valence electrons are the electrons in the outermost shell, which for Group 13 is the $n^{th}$ shell. The electronic configuration is $ns^2 np^1$. By adding the electrons in the outermost s and p orbitals ($2 + 1$), we find that all elements in the Boron family (Group 13) have exactly 3 valence electrons, which they can use for chemical bonding.
13. What is the complete electronic configuration of Aluminium (Al, Z=13)?
ⓐ. $1s^2 2s^2 2p^6 3s^2 3p^2$
ⓑ. $1s^2 2s^2 2p^6 3s^2$
ⓒ. $1s^2 2s^2 2p^6 3s^1 3p^2$
ⓓ. $1s^2 2s^2 2p^6 3s^2 3p^1$
Correct Answer: $1s^2 2s^2 2p^6 3s^2 3p^1$
Explanation: Aluminium (Al) has an atomic number (Z) of 13, meaning it has 13 electrons. Following the Aufbau principle, electrons fill the lowest energy orbitals first. The configuration is filled as: $1s^2$ (2 electrons), $2s^2$ (2 electrons), $2p^6$ (6 electrons), $3s^2$ (2 electrons), and the final electron goes into the 3p orbital, $3p^1$. The valence shell is $n=3$, with the configuration $3s^2 3p^1$.
14. The electronic configuration of Gallium (Ga, Z=31) is $[Ar] 3d^{10} 4s^2 4p^1$. Why is the $3d$ orbital completely filled?
ⓐ. Gallium is a d-block element.
ⓑ. The $3d$ orbital has lower energy than the $4p$ orbital but higher energy than the $4s$ orbital.
ⓒ. The $3d$ orbital is part of the valence shell for Gallium.
ⓓ. This configuration is an exception to the Aufbau principle.
Correct Answer: The $3d$ orbital has lower energy than the $4p$ orbital but higher energy than the $4s$ orbital.
Explanation: Gallium is in the 4th period. According to the Aufbau principle (or the $(n+l)$ rule), the energy order for filling orbitals is $…4s < 3d < 4p$. After filling Argon's core ($1s^2...3p^6$), the next electrons fill the $4s$ orbital ($4s^2$), then the $3d$ orbital ($3d^{10}$), and finally the $4p$ orbital ($4p^1$). Because Gallium (Z=31) is the first p-block element after the first transition series (d-block), its configuration includes a completely filled $3d$ subshell.
15. Which element in Group 13 has the valence shell electronic configuration $2s^2 2p^1$?
ⓐ. Boron (B)
ⓑ. Aluminium (Al)
ⓒ. Gallium (Ga)
ⓓ. Thallium (Tl)
Correct Answer: Boron (B)
Explanation: The principal quantum number $n$ for the valence shell corresponds to the period number. The configuration $2s^2 2p^1$ indicates the valence shell is $n=2$, which corresponds to the second period. Boron (B, Z=5) is the Group 13 element in the second period. Its full configuration is $1s^2 2s^2 2p^1$. Aluminium is $n=3$ ($3s^2 3p^1$), Gallium is $n=4$ ($4s^2 4p^1$), and Thallium is $n=6$ ($6s^2 6p^1$).
16. The electronic configuration of Thallium (Tl, Z=81) includes a filled $4f$ subshell. What is its valence shell configuration?
ⓐ. $[Xe] 4f^{14} 5d^{10} 6s^2 6p^1$
ⓑ. $[Rn] 5f^{14} 6d^{10} 7s^2 7p^1$
ⓒ. $[Xe] 4f^{14} 5d^{10} 6p^3$
ⓓ. $[Xe] 5d^{10} 6s^2 6p^1$
Correct Answer: $[Xe] 4f^{14} 5d^{10} 6s^2 6p^1$
Explanation: Thallium (Tl) is in the 6th period and Group 13. Its atomic number is 81. The noble gas core before it is Xenon (Xe, Z=54). After Xe, we fill the orbitals: $6s^2$ (2 e-), then the $4f$ subshell ($4f^{14}$, 14 e-), then the $5d$ subshell ($5d^{10}$, 10 e-), and finally the $6p$ subshell ($6p^1$, 1 e-). Total electrons = 54 + 2 + 14 + 10 + 1 = 81. The valence shell is the outermost shell ($n=6$), so its configuration is $6s^2 6p^1$.
17. Why do Gallium (Ga) and Indium (In) show only a +3 oxidation state, while Thallium (Tl) also shows a stable +1 oxidation state?
ⓐ. Thallium is a non-metal.
ⓑ. Due to the high shielding effect of $f$-electrons in Thallium.
ⓒ. Due to the inert pair effect, which is prominent in Thallium.
ⓓ. Thallium has only one valence electron.
Correct Answer: Due to the inert pair effect, which is prominent in Thallium.
Explanation: The inert pair effect is the reluctance of the outermost $ns^2$ electrons to participate in bonding. This effect is very significant in heavier p-block elements like Thallium ($n=6$). This is because the $6s^2$ electrons are poorly shielded from the nucleus by the intervening $4f$ and $5d$ electrons. This poor shielding increases the effective nuclear charge on the $6s^2$ electrons, pulling them closer to the nucleus and making them “inert”. Thus, Thallium prefers to lose only its single $6p^1$ electron, leading to a stable $Tl^{+1}$ state.
18. What is the primary reason for the anomalous properties of Boron (B) compared to other Group 13 elements?
ⓐ. Its large size and presence of d-orbitals.
ⓑ. Its high metallic character.
ⓒ. Its small size, high ionization enthalpy, and absence of d-orbitals.
ⓓ. Its $1s^2 2s^2 2p^1$ configuration, which is unique.
Correct Answer: Its small size, high ionization enthalpy, and absence of d-orbitals.
Explanation: Boron, the first member, shows anomalous behavior. Its electronic configuration ($1s^2 2s^2 2p^1$) means it is in the $n=2$ shell. This shell has no d-orbitals, limiting Boron’s maximum covalency to 4 (using its $2s$ and three $2p$ orbitals). Furthermore, its atomic size is exceptionally small, and its ionization enthalpy and electronegativity are very high compared to Al, Ga, In, and Tl. This makes Boron a non-metal, whereas the others are metals.
19. Which of the following ions is not likely to exist based on the $ns^2 np^1$ configuration?
ⓐ. $Al^{3+}$
ⓑ. $Ga^{3+}$
ⓒ. $Tl^{1+}$
ⓓ. $B^{3+}$
Correct Answer: $B^{3+}$
Explanation: The formation of a $B^{3+}$ ion would require the removal of 3 valence electrons ($2s^2 2p^1$). The sum of the first three ionization enthalpies ($IE_1 + IE_2 + IE_3$) for Boron is extremely high due to its very small size and high effective nuclear charge. The energy required to form a free $B^{3+}$ cation is prohibitively large. Therefore, Boron does not form ionic compounds as $B^{3+}$; it forms covalent compounds by sharing its three valence electrons. $Al^{3+}$, $Ga^{3+}$, and $Tl^{1+}$ (due to inert pair effect) are common ions.
20. The presence of a completely filled $d$-subshell in Gallium ($[Ar] 3d^{10} 4s^2 4p^1$) leads to:
ⓐ. A very large atomic size compared to Aluminium.
ⓑ. A very low ionization enthalpy compared to Aluminium.
ⓒ. Poor shielding of the valence electrons by the $3d$ electrons.
ⓓ. Gallium being a transition metal.
Correct Answer: Poor shielding of the valence electrons by the $3d$ electrons.
Explanation: The $3d$ electrons in Gallium offer very poor shielding for the outer $4s$ and $4p$ electrons from the nucleus. This poor shielding increases the effective nuclear charge ($Z_{eff}$) experienced by the valence electrons. This effect, known as d-block contraction, causes Gallium’s atomic radius to be smaller than expected (in fact, slightly smaller than Aluminium’s) and its ionization enthalpy to be higher than Aluminium’s, which is a deviation from the general trend.
21. What is the chemical formula for Bauxite, which is the principal ore of aluminium?
ⓐ. $Al_2O_3$
ⓑ. $Al_2O_3 \cdot nH_2O$ (where $n=1$ or $2$)
ⓒ. $Na_3AlF_6$
ⓓ. $Na_2B_4O_7 \cdot 10H_2O$
Correct Answer: $Al_2O_3 \cdot nH_2O$ (where $n=1$ or $2$)
Explanation: Bauxite is not a single compound but a hydrated aluminium oxide. Its general formula is written as $Al_2O_3 \cdot nH_2O$ (or $AlO_x(OH)_{3-2x}$). It is a mixture of minerals like Gibbsite ($Al(OH)_3$ or $Al_2O_3 \cdot 3H_2O$) and Diaspore/Boehmite ($AlO(OH)$ or $Al_2O_3 \cdot H_2O$). Option A is anhydrous Alumina. Option C is Cryolite. Option D is Borax.
22. What is the correct chemical formula for Borax (also known as tincal)?
ⓐ. $Na_2B_4O_7 \cdot 4H_2O$
ⓑ. $Na_2B_4O_7$
ⓒ. $Na_2B_4O_7 \cdot 10H_2O$
ⓓ. $Ca_2B_6O_{11} \cdot 5H_2O$
Correct Answer: $Na_2B_4O_7 \cdot 10H_2O$
Explanation: Borax is a major ore of boron and is chemically known as sodium tetraborate decahydrate. Its formula is $Na_2B_4O_7 \cdot 10H_2O$. It is found in deposits from dried-up lakes. Option A is Kernite (or Rasorite), another boron ore. Option B is anhydrous borax (borax glass). Option D is Colemanite, a calcium borate ore.
23. By weight, which element is the most abundant metal in the Earth’s crust?
ⓐ. Iron (Fe)
ⓑ. Silicon (Si)
ⓒ. Aluminium (Al)
ⓓ. Boron (B)
Correct Answer: Aluminium (Al)
Explanation: Aluminium (Al) is the most abundant metal, making up about 8.3% of the Earth’s crust by weight. It is the third most abundant element overall, after oxygen (45.5%) and silicon (27.7%). Iron (Fe) is the second most abundant metal. Silicon (Si) is a metalloid, not a metal. Boron (B) is a very rare element in comparison.
24. “Alumina” is the common name for which chemical compound?
ⓐ. Anhydrous Aluminium Oxide ($Al_2O_3$)
ⓑ. Hydrated Aluminium Oxide ($Al_2O_3 \cdot nH_2O$)
ⓒ. Sodium Aluminate ($NaAlO_2$)
ⓓ. Aluminium Hydroxide ($Al(OH)_3$)
Correct Answer: Anhydrous Aluminium Oxide ($Al_2O_3$)
Explanation: Alumina is the common name for aluminium oxide, $Al_2O_3$. It is the compound that is extracted and purified from bauxite ore (which is the hydrated form, Option B) through processes like Bayer’s process. This purified alumina is then dissolved in molten cryolite and electrolyzed (Hall-Héroult process) to produce pure aluminium metal.
25. Kernite (or Rasorite) is the second most important ore of boron. What is its chemical formula?
ⓐ. $Na_2B_4O_7 \cdot 10H_2O$
ⓑ. $Ca_2B_6O_{11} \cdot 5H_2O$
ⓒ. $Na_2B_4O_7 \cdot 4H_2O$
ⓓ. $H_3BO_3$
Correct Answer: $Na_2B_4O_7 \cdot 4H_2O$
Explanation: Kernite is sodium tetraborate tetrahydrate, $Na_2B_4O_7 \cdot 4H_2O$. It is a significant ore of boron, differing from Borax (Option A) only in the number of water molecules of crystallization. Option B is Colemanite. Option D is orthoboric acid, which is a compound derived from boron ores, not a primary ore itself.
26. Cryolite is a mineral essential for the electrolytic extraction of aluminium. Its chemical formula is:
ⓐ. $Na_3AlF_6$
ⓑ. $AlF_3$
ⓒ. $CaF_2$
ⓓ. $Al_2(SiO_3)_3$
Correct Answer: $Na_3AlF_6$
Explanation: Cryolite is sodium hexafluoroaluminate, $Na_3AlF_6$. In the Hall-Héroult process, it is used as a solvent for alumina ($Al_2O_3$). Alumina has a very high melting point (over 2000°C), but dissolving it in molten cryolite (mixed with $AlF_3$ and $CaF_2$) reduces the melting point of the electrolyte bath to around 950-1000°C, making electrolysis feasible and more economical.
27. How does Boron (B) primarily occur in nature?
ⓐ. As a free element.
ⓑ. As compounds with metals called borides.
ⓒ. As boron nitride ($BN$).
ⓓ. As oxygen-containing compounds called borates.
Correct Answer: As oxygen-containing compounds called borates.
Explanation: Boron is a rare element and is too reactive to be found as a free element (Option A) in nature. It is found in concentrated deposits in combined states. Its most common and important ores are oxygen-containing salts known as borates, such as Borax ($Na_2B_4O_7 \cdot 10H_2O$), Kernite ($Na_2B_4O_7 \cdot 4H_2O$), and Colemanite ($Ca_2B_6O_{11} \cdot 5H_2O$).
28. The anhydrous crystalline form of Alumina ($Al_2O_3$) is known as Corundum. What are the gem-quality varieties of Corundum?
ⓐ. Diamond and Graphite
ⓑ. Beryl and Topaz
ⓒ. Quartz and Feldspar
ⓓ. Ruby and Sapphire
Correct Answer: Ruby and Sapphire
Explanation: Corundum is the extremely hard, anhydrous crystalline form of $Al_2O_3$. While plain corundum is used as an abrasive, its gemstone varieties are highly valued. Ruby is corundum that is colored red by trace impurities of chromium. Sapphire is corundum of any other color (most famously blue, caused by iron and titanium impurities), including pink, yellow, and green.
29. Which of the following is the chemical formula for the boron ore Colemanite?
ⓐ. $Ca_2B_6O_{11} \cdot 5H_2O$
ⓑ. $Na_2B_4O_7 \cdot 10H_2O$
ⓒ. $CaCO_3 \cdot MgCO_3$
ⓓ. $Al_2O_3 \cdot 2H_2O$
Correct Answer: $Ca_2B_6O_{11} \cdot 5H_2O$
Explanation: Colemanite is an important borate mineral, specifically a calcium borate hydrate. Its chemical formula is $Ca_2B_6O_{11} \cdot 5H_2O$. It is a major source for producing boric acid and other boron compounds. Option B is Borax, Option C is Dolomite (a magnesium/calcium carbonate), and Option D is Bauxite.
30. Which statement accurately compares the natural occurrence of Boron (B) and Aluminium (Al)?
ⓐ. Both are very abundant and found as free metals.
ⓑ. Aluminium is the most abundant metal in the crust; Boron is a rare element.
ⓒ. Boron is the most abundant element in the crust; Aluminium is rare.
ⓓ. Both are rare elements and occur as borates and aluminates.
Correct Answer: Aluminium is the most abundant metal in the crust; Boron is a rare element.
Explanation: There is a significant difference in abundance. Aluminium is the most abundant metal (8.3% by weight) and third most abundant element in the Earth’s crust, found widely in aluminosilicate rocks and bauxite ore. Boron, in contrast, is a rare element (less than 0.001% by weight) and is only found in concentrated deposits of its borate ores (like borax and kernite). Neither is found as a free element due to their reactivity.
31. What is the correct order of atomic radii for Group 13 elements?
ⓐ. $B < Al < Ga < In < Tl$
ⓑ. $B < Ga < Al < In < Tl$
ⓒ. $B < Al \approx Ga < In < Tl$
ⓓ. $B < Ga < Al < Tl < In$
Correct Answer: $B < Ga < Al < In < Tl$
Explanation: The general trend is that atomic radius increases down the group. However, there is a significant anomaly. Gallium (Ga) has a smaller atomic radius (135 pm) than Aluminium (Al) (143 pm). This is due to the “d-block contraction” or “scandide contraction.” In Gallium, the valence electrons are preceded by 10 d-electrons (in the 3d subshell), which offer very poor shielding from the nucleus. This poor shielding increases the effective nuclear charge ($Z_{eff}$) on Gallium’s valence electrons, pulling them closer and resulting in a smaller-than-expected radius.
32. The first ionization enthalpy ($IE_1$) of Group 13 elements shows a complex trend. What is the correct order?
ⓐ. $B > Al > Ga > In > Tl$
ⓑ. $B > Tl > Ga > Al > In$
ⓒ. $B > Ga > Al > Tl > In$
ⓓ. $B > Tl > Ga > In > Al$
Correct Answer: $B > Tl > Ga > Al > In$
Explanation: The $IE_1$ trend is highly irregular. Boron (B) has the highest $IE_1$ due to its very small size. $IE_1$ drops significantly from B to Al as expected. However, $IE_1$ of Gallium (Ga) is slightly higher than Al due to the poor shielding by 3d electrons (d-block contraction). $IE_1$ drops from Ga to Indium (In). $IE_1$ of Thallium (Tl) is higher than both Al and In. This is due to the very poor shielding by the 14 $4f$-electrons (lanthanoid contraction) and 10 $5d$-electrons, which strongly increases $Z_{eff}$ on the $6p$ electron.
33. Why is the atomic radius of Gallium (Ga) surprisingly smaller than that of Aluminium (Al)?
ⓐ. Poor shielding of the nucleus by the intervening $3d^{10}$ electrons in Ga.
ⓑ. Gallium has a higher nuclear charge than Aluminium.
ⓒ. The inert pair effect is stronger in Gallium than in Aluminium.
ⓓ. Gallium is a non-metal, while Aluminium is a metal.
Correct Answer: Poor shielding of the nucleus by the intervening $3d^{10}$ electrons in Ga.
Explanation: Aluminium (Period 3) has the configuration $[Ne] 3s^2 3p^1$. Gallium (Period 4) has the configuration $[Ar] 3d^{10} 4s^2 4p^1$. The 10 electrons in the $3d$ orbital of Gallium do not shield the outer $4s$ and $4p$ electrons effectively from the pull of the nucleus. This increased effective nuclear charge ($Z_{eff}$) pulls the valence shell closer, resulting in a smaller atomic radius (135 pm) for Ga compared to Al (143 pm).
34. Boron (B) has an exceptionally high first ionization enthalpy ($IE_1 = 801 \, kJ/mol$) compared to the rest of Group 13. This is primarily because:
ⓐ. Boron is a metalloid.
ⓑ. Boron only has $p$-electrons in its valence shell.
ⓒ. Boron has a very small atomic size and high effective nuclear charge.
ⓓ. Boron exists as a large covalent network.
Correct Answer: Boron has a very small atomic size and high effective nuclear charge.
Explanation: Ionization enthalpy is the energy required to remove the outermost electron. Boron is the first member of the group and is in the second period ($n=2$). It has an exceptionally small atomic radius. This small size means its three valence electrons ($2s^2 2p^1$) are held very tightly by the nucleus (high $Z_{eff}$). A large amount of energy is required to overcome this strong attraction and remove an electron, resulting in a very high $IE_1$.
35. What is the correct trend for electronegativity (EN) down Group 13?
ⓐ. Decreases regularly from B to Tl.
ⓑ. Decreases from B to Al, and then increases slightly from Al to Tl.
ⓒ. Increases regularly from B to Tl.
ⓓ. Remains almost constant for all elements.
Correct Answer: Decreases from B to Al, and then increases slightly from Al to Tl.
Explanation: Electronegativity is expected to decrease down a group as atomic size increases. This holds true from Boron (2.0) to Aluminium (1.5). However, after Aluminium, the electronegativity increases slightly or remains almost constant: Ga (1.6), In (1.7), Tl (1.8). This anomaly is again due to the poor shielding by the intervening $d$-electrons (in Ga) and $d$- and $f$-electrons (in In and Tl), which increases the effective nuclear charge and the atom’s ability to attract a shared electron pair.
36. What is the most common and stable oxidation state for Boron (B) and Aluminium (Al)?
ⓐ. +3
ⓑ. +1
ⓒ. +2
ⓓ. -1
Correct Answer: +3
Explanation: The valence shell configuration for Group 13 is $ns^2 np^1$. By losing or sharing all three valence electrons, the elements can achieve a stable +3 oxidation state. For the lighter elements, Boron and Aluminium, the energy required to involve all three electrons in bonding is easily compensated. The sum of their first three ionization enthalpies is relatively low (except for $B^{3+}$ cation formation), and the +3 state is their only significant oxidation state.
37. The stability of the +1 oxidation state increases as we move down Group 13 (Al < Ga < In < Tl). This phenomenon is known as:
ⓐ. Lanthanoid Contraction
ⓑ. d-Block Contraction
ⓒ. Diagonal Relationship
ⓓ. Inert Pair Effect
Correct Answer: Inert Pair Effect
Explanation: The inert pair effect is the reluctance of the outermost $ns^2$ electrons to participate in bonding. In Group 13, this means the $ns^2$ electrons (e.g., $6s^2$ in Thallium) are held tightly by the nucleus. This effect becomes prominent for heavier elements (In, Tl) due to poor shielding by $d$- and $f$-electrons. Consequently, these elements prefer to lose only the single $np^1$ electron, leading to a +1 oxidation state. For Thallium (Tl), the +1 oxidation state is more stable than its +3 state.
38. Which element in Group 13 almost exclusively shows a +3 oxidation state in its compounds?
ⓐ. Gallium (Ga)
ⓑ. Boron (B)
ⓒ. Indium (In)
ⓓ. Thallium (Tl)
Correct Answer: Boron (B)
Explanation: Boron, being the first member, is very small and has a high ionization enthalpy, preventing the formation of a $B^{3+}$ ion. It forms only covalent compounds by sharing all three valence electrons ($2s^2 2p^1$), leading to an exclusive +3 oxidation state (e.g., in $BF_3$, $BCl_3$, $B_2H_6$). The heavier elements (Ga, In, Tl) can show both +3 and +1 oxidation states, even if +3 is more common for Ga and In.
39. Among the Group 13 elements, which one forms the most stable unipositive ion ($M^+$)?
ⓐ. Aluminium (Al)
ⓑ. Gallium (Ga)
ⓒ. Thallium (Tl)
ⓓ. Indium (In)
Correct Answer: Thallium (Tl)
Explanation: The stability of the +1 oxidation state increases down the group due to the inert pair effect. This effect is strongest in Thallium (Tl), the heaviest element in the group. The $6s^2$ electrons in Tl are strongly attracted to the nucleus due to poor shielding by $4f$ and $5d$ electrons, making them “inert”. Thus, Tl readily loses its single $6p^1$ electron to form the stable $Tl^+$ ion. $TlCl$ is a stable ionic compound, whereas $AlCl$ is not.
40. How does the first ionization enthalpy ($IE_1$) of Aluminium compare to that of Gallium?
ⓐ. $IE_1$ of Al is lower than $IE_1$ of Ga.
ⓑ. $IE_1$ of Al is higher than $IE_1$ of Ga.
ⓒ. $IE_1$ of Al is identical to $IE_1$ of Ga.
ⓓ. $IE_1$ of Al is twice the $IE_1$ of Ga.
Correct Answer: $IE_1$ of Al is lower than $IE_1$ of Ga.
Explanation: We would normally expect $IE_1$ to decrease down the group (so $Al > Ga$). However, dueem to the poor shielding by the ten $3d$ electrons in Gallium, its effective nuclear charge ($Z_{eff}$) is higher than expected. This increased $Z_{eff}$ holds the $4p$ valence electron more tightly. As a result, more energy is needed to remove the first electron from Gallium (579 kJ/mol) compared to Aluminium (577 kJ/mol). Therefore, $IE_1$ of Al is (slightly) lower than that of Ga.
41. Which of the following properties is not a reason for the anomalous behavior of Boron?
ⓐ. Very small atomic size.
ⓑ. High ionization enthalpy and electronegativity.
ⓒ. Presence of a completely filled $d$-orbital.
ⓓ. Absence of $d$-orbitals in its valence shell.
Correct Answer: Presence of a completely filled $d$-orbital.
Explanation: Boron’s anomalous behavior stems from its properties as a second-period element. These include its exceptionally small size, very high ionization enthalpy, and high electronegativity. Crucially, its valence shell ($n=2$) has only $s$ and $p$ orbitals and completely lacks $d$-orbitals. This limits its covalency. The presence of $d$-orbitals is a feature of elements from the third period onwards (like Aluminium) or fourth period (like Gallium), which actually contributes to their properties (e.g., d-block contraction in Ga).
42. What is the maximum covalency that Boron (B) can exhibit?
ⓐ. 6
ⓑ. 4
ⓒ. 3
ⓓ. 2
Correct Answer: 4
Explanation: Boron’s valence shell is $n=2$, which contains one $2s$ orbital and three $2p$ orbitals, for a total of four available valence orbitals. Although it has only 3 valence electrons ($2s^2 2p^1$), it can use all four orbitals to form four covalent bonds, as seen in the tetraborate ion, $[BH_4]^-$, or the tetrafluoroborate ion, $[BF_4]^-$. It cannot expand its covalency to 6 (like Aluminium in $[AlF_6]^{3-}$) because it has no $d$-orbitals available in its valence shell.
43. Boron trihalides (like $BF_3$) act as Lewis acids. This is primarily because:
ⓐ. Boron can expand its octet.
ⓑ. Boron is a non-metal.
ⓒ. The $B-F$ bond is ionic.
ⓓ. Boron has an incomplete octet (electron deficient).
Correct Answer: Boron has an incomplete octet (electron deficient).
Explanation: In $BF_3$, the central Boron atom forms three single covalent bonds with three fluorine atoms. This gives Boron a total of only 6 electrons in its valence shell (3 from B, 3 from F). Its octet is incomplete, making the $BF_3$ molecule electron-deficient. To complete its octet, the Boron atom has a strong tendency to accept a pair of electrons from a donor molecule (a Lewis base), thus behaving as a strong Lewis acid. For example, it reacts with ammonia ($:NH_3$) to form an adduct $F_3B \leftarrow NH_3$.
44. While Boron (B) is a non-metal (or metalloid) and an insulator, other elements of Group 13 (Al, Ga, In, Tl) are:
ⓐ. All non-metals
ⓑ. All metalloids
ⓒ. Soft, metallic conductors
ⓓ. Noble gases
Correct Answer: Soft, metallic conductors
Explanation: This is a key difference. Due to its small size and high ionization energy, Boron prefers to form strong covalent network solids, making it a non-metal (or metalloid) with a very high melting point and low electrical conductivity. In contrast, Aluminium, Gallium, Indium, and Thallium are all typical metals. They have lower ionization enthalpies, readily form metallic lattices, and are good conductors of electricity and heat.
45. The nature of Boron’s oxide ($B_2O_3$) is _______, while Aluminium’s oxide ($Al_2O_3$) is _______.
ⓐ. Acidic, Amphoteric
ⓑ. Basic, Acidic
ⓒ. Amphoteric, Acidic
ⓓ. Basic, Amphoteric
Correct Answer: Acidic, Amphoteric
Explanation: As a non-metal, Boron forms an acidic oxide. Boron trioxide ($B_2O_3$) reacts with water to form orthoboric acid ($H_3BO_3$) and reacts with basic oxides to form borate salts. As we move down the group, metallic character increases. Aluminium is a metalloid/metal on the borderline, and its oxide ($Al_2O_3$) is amphoteric, meaning it can react with both strong acids (acting as a base) and strong bases (acting as an acid). The oxides of Ga, In, and Tl become progressively more basic.
46. Boron is unable to form $B^{3+}$ ions, whereas Aluminium can form $Al^{3+}$ ions. Why?
ⓐ. Boron is larger than Aluminium.
ⓑ. The sum of the first three ionization enthalpies ($IE_1+IE_2+IE_3$) for Boron is extremely high.
ⓒ. Boron has a stronger inert pair effect than Aluminium.
ⓓ. Aluminium has $d$-orbitals, while Boron does not.
Correct Answer: The sum of the first three ionization enthalpies ($IE_1+IE_2+IE_3$) for Boron is extremely high.
Explanation: Due to Boron’s very small atomic size, its three valence electrons ($2s^2 2p^1$) are held very tightly by the nucleus. The energy required to remove all three electrons (the sum of the first three ionization enthalpies) is prohibitively large. This energy cannot be compensated by the lattice energy or hydration enthalpy that would be released. Therefore, Boron forms covalent compounds. Aluminium is larger, its valence electrons are further from the nucleus, and the sum of its $IE_1+IE_2+IE_3$ is much lower, allowing it to form $Al^{3+}$ ions in ionic compounds.
47. Unlike other Group 13 elements, Boron forms stable hydrides called boranes (e.g., Diborane, $B_2H_6$). What is a key feature of these compounds?
ⓐ. They are ionic compounds containing $H^-$ ions.
ⓑ. They contain $p\pi-p\pi$ back-bonding.
ⓒ. They are electron-precise, with complete octets.
ⓓ. They are electron-deficient and feature multicenter (banana) bonds.
Correct Answer: They are electron-deficient and feature multicenter (banana) bonds.
Explanation: Boron’s hydrides (boranes) are a unique class of compounds. A simple molecule like $BH_3$ is unstable as it is electron-deficient (6 valence electrons). To achieve stability, it dimerizes to form diborane ($B_2H_6$). This molecule does not have enough valence electrons to form 8 conventional 2-center-2-electron (2c-2e) bonds. Instead, it features two “three-center-two-electron” (3c-2e) bonds, also known as “banana bonds,” where two electrons hold three atoms ($B-H-B$) together. This is a hallmark of Boron’s chemistry, not seen in Al, Ga, In, or Tl hydrides.
48. The $B-F$ bond length in $BF_3$ (130 pm) is shorter than expected for a single bond. This is attributed to:
ⓐ. The $sp^2$ hybridization of Boron.
ⓑ. The ionic character of the $B-F$ bond.
ⓒ. $p\pi-p\pi$ back-bonding between Boron and Fluorine.
ⓓ. The small size of the Boron atom.
Correct Answer: $p\pi-p\pi$ back-bonding between Boron and Fluorine.
Explanation: In $BF_3$, the Boron atom is $sp^2$ hybridized and has an empty $2p$ orbital. Each Fluorine atom has filled $2p$ orbitals. A lone pair from a Fluorine $2p$ orbital can be partially donated back to the empty $2p$ orbital of Boron, forming a partial pi ($p\pi-p\pi$) bond. This “back-bonding” introduces double bond character into the $B-F$ bonds. This shortens the bond length and also strengthens the bond. This effect is not possible in $AlCl_3$ because the $3p$ orbital of Al is too large to overlap effectively with the $3p$ of Cl (it dimerizes instead).
49. How does the allotropic form of Boron differ from that of Aluminium?
ⓐ. Boron exists as soft, metallic crystals; Aluminium is a hard non-metal.
ⓑ. Boron exists as extremely hard, high-melting crystalline solids; Aluminium is a soft metal.
ⓒ. Both Boron and Aluminium exist as diatomic gases.
ⓓ. Boron exists as a simple molecular solid ($B_8$); Aluminium forms a covalent network.
Correct Answer: Boron exists as extremely hard, high-melting crystalline solids; Aluminium is a soft metal.
Explanation: This is a major anomalous property. Boron’s small size and high ionization energy lead it to form complex, stable, covalent network structures. Its crystalline allotropes (e.g., $\alpha$-rhombohedral, $\beta$-rhombohedral) are based on $B_{12}$ icosahedra units. These strong covalent networks make elemental Boron extremely hard (second only to diamond) and give it a very high melting point (over 2000°C). Aluminium, being a metal, forms a typical close-packed metallic lattice, making it a relatively soft, ductile, and lower-melting (660°C) solid.
50. Boron is the only element in Group 13 that:
ⓐ. Forms a stable +1 oxidation state.
ⓑ. Is a good conductor of electricity.
ⓒ. Is distinctly non-metallic.
ⓓ. Forms an amphoteric oxide.
Correct Answer: Is distinctly non-metallic.
Explanation: As the first member of the group, Boron’s small size and high electronegativity make its properties non-metallic. It forms covalent bonds, its oxide is acidic, and it is a semiconductor (not a good conductor). The other elements (Al, Ga, In, Tl) are all metals (though Al is borderline and its oxide is amphoteric). The stable +1 oxidation state is a feature of the heaviest element, Thallium (Tl), not Boron.
51. Borax can be commercially prepared from the mineral Colemanite (${Ca}_2{B}_6{O}_{11} \cdot 5{H}_2{O}$) by treating it with:
ⓐ. Sulphuric acid (${H}_2{SO}_4$)
ⓑ. Sodium Carbonate (${Na}_2{CO}_3$) solution
ⓒ. Hydrochloric acid (${HCl}$)
ⓓ. Boric acid (${H}_3{BO}_3$)
Correct Answer: Sodium Carbonate (${Na}_2{CO}_3$) solution
Explanation: Colemanite is a calcium borate ore. Borax is prepared by boiling the finely powdered Colemanite with a hot, saturated solution of sodium carbonate. The reaction precipitates calcium carbonate, leaving sodium borate (Borax) in solution, which is then crystallized.
$${Ca}_2{B}_6{O}_{11} + 2{Na}_2{CO}_3 \rightarrow 2{CaCO}_3 \downarrow + {Na}_2{B}_4{O}_7 + 2{NaBO}_2$$
52. What is the nature of the aqueous solution of Borax?
ⓐ. Neutral
ⓑ. Highly acidic
ⓒ. Amphoteric
ⓓ. Basic
Correct Answer: Basic
Explanation: When dissolved in water, Borax (${Na}_2{B}_4{O}_7$) undergoes hydrolysis. It is a salt formed from a strong base (${NaOH}$) and a weak acid (${H}_3{BO}_3$). The strong base component hydrolyzes water, releasing hydroxide ions (${OH}^-$), which makes the solution basic (alkaline).
$${Na}_2{B}_4{O}_7 + 7{H}_2{O} \rightleftharpoons 2{NaOH} + 4{H}_3{BO}_3$$
53. The correct structural representation of the tetraborate anion present in crystalline Borax (${Na}_2{B}_4{O}_7 \cdot 10{H}_2{O}$) is:
ⓐ. $[{B}({OH})_4]^-$
ⓑ. $[{B}_4{O}_7]^{2-}$
ⓒ. $[{B}_4{O}_5({OH})_4]^{2-}$
ⓓ. $[{B}_2{O}_3]^2-$
Correct Answer: $[{B}_4{O}_5({OH})_4]^{2-}$
Explanation: Although the simple formula is ${Na}_2{B}_4{O}_7 \cdot 10{H}_2{O}$, the actual structural unit in solid Borax is the complex ion $[{B}_4{O}_5({OH})_4]^{2-}$. This ion is a tetranuclear unit containing two tetrahedral and two triangular boron atoms bridged by oxygen atoms. The structure shows that three water molecules are coordinated to the sodium ions, and the remaining seven are hydrogen-bonded or present as water of crystallization.
54. When Borax is heated strongly, it loses water of crystallization and swells up. On further heating, it turns into a transparent, glassy mass known as:
ⓐ. Boric acid
ⓑ. Orthoboric anhydride
ⓒ. Borax glass or Borax bead
ⓓ. Sodium metaborate
Correct Answer: Borax glass or Borax bead
Explanation: Heating Borax in two stages: First, mild heating causes it to lose its ten water molecules and swell up. Second, strong heating converts the remaining anhydrous sodium tetraborate into a transparent liquid that solidifies into a glassy bead. This bead is a mixture of ${NaBO}_2$ (sodium metaborate) and ${B}_2{O}_3$ (boric anhydride or boron trioxide).
$${Na}_2{B}_4{O}_7 \cdot 10{H}_2{O} \xrightarrow{\Delta} {Na}_2{B}_4{O}_7 \xrightarrow{{Strong } \Delta} 2{NaBO}_2 + {B}_2{O}_3 ({Borax Glass})$$
55. The Borax Bead Test is used to identify colored metal ions. When a Borax bead is heated with a cobalt salt in the oxidizing flame, the resulting bead is:
ⓐ. Green
ⓑ. Blue
ⓒ. Yellow
ⓓ. Brown
Correct Answer: Blue
Explanation: The Borax Bead Test involves fusing metal salts with the borax glass (${NaBO}_2 + {B}_2{O}_3$). The metal oxide reacts with boric anhydride (${B}_2{O}_3$) to form a colored metal metaborate (${M}({BO}_2)_n$). For cobalt (${Co}^{2+}$), the reaction forms cobalt metaborate (${Co}({BO}_2)_2$), which is blue in color and characteristic of cobalt.
$${CoO} + {B}_2{O}_3 \rightarrow {Co}({BO}_2)_2$$
56. When a solution of Borax is treated with a calculated amount of an acid (like ${HCl}$ or ${H}_2{SO}_4$), the main product formed is:
ⓐ. Boron trioxide (${B}_2{O}_3$)
ⓑ. Sodium metaborate (${NaBO}_2$)
ⓒ. Orthoboric acid (${H}_3{BO}_3$)
ⓓ. Diborane (${B}_2{H}_6$)
Correct Answer: Orthoboric acid (${H}_3{BO}_3$)
Explanation: Borax reacts with strong inorganic acids to produce orthoboric acid (also known simply as boric acid), which is sparingly soluble and precipitates out. This reaction is one of the primary methods for preparing boric acid from borax.
$${Na}_2{B}_4{O}_7 + {H}_2{SO}_4 + 5{H}_2{O} \rightarrow {Na}_2{SO}_4 + 4{H}_3{BO}_3$$
57. The chemical name of Borax (${Na}_2{B}_4{O}_7 \cdot 10{H}_2{O}$) is:
ⓐ. Sodium tetraborate decahydrate
ⓑ. Sodium triborate decahydrate
ⓒ. Sodium metaborate decahydrate
ⓓ. Sodium orthoborate decahydrate
Correct Answer: Sodium tetraborate decahydrate
Explanation: The ${B}_4{O}_7^{2-}$ unit in the simplest formula is the tetraborate ion. The chemical name is derived from this simple formula and the 10 molecules of water of crystallization. Thus, Borax is chemically called sodium tetraborate decahydrate.
58. In the structural tetraborate anion $[{B}_4{O}_5({OH})_4]^{2-}$ of crystalline Borax, the number of Boron atoms that are $sp^2$ hybridized and $sp^3$ hybridized are:
ⓐ. $sp^2 = 4, sp^3 = 0$
ⓑ. $sp^2 = 2, sp^3 = 2$
ⓒ. $sp^2 = 0, sp^3 = 4$
ⓓ. $sp^2 = 3, sp^3 = 1$
Correct Answer: $sp^2 = 2, sp^3 = 2$
Explanation: The complex tetraborate anion contains a fused ring structure. Two Boron atoms are bonded to three oxygen atoms, forming a triangle and are thus $sp^2$ hybridized. The other two Boron atoms are bonded to four groups (three ${O}$ and one ${OH}$), forming a tetrahedron and are thus $sp^3$ hybridized.
59. Borax is widely used as a flux in metallurgy (e.g., soldering or welding). The function of Borax as a flux is to:
ⓐ. Increase the melting point of the metal.
ⓑ. Provide a cooling effect.
ⓒ. Dissolve the metal oxides/impurities from the surface.
ⓓ. Act as an oxidizing agent.
Correct Answer: Dissolve the metal oxides/impurities from the surface.
Explanation: A flux is a chemical cleaning agent. In metallurgical processes, Borax acts as a flux by melting and reacting with the unwanted metal oxides (the ‘slag’ or ‘impurities’) present on the surface of the metal being worked. The formation of fusible glass-like metal borates (e.g., the colored metaborates) helps to dissolve these oxides, leaving a clean metal surface for a stronger, clearer joint.
60. Which of the following is a primary industrial use of Borax, apart from its use in the Borax Bead Test and as a flux?
ⓐ. As a strong bleaching agent for clothes.
ⓑ. As a component in the manufacture of heat-resistant glass (Pyrex).
ⓒ. For the production of liquid fuels.
ⓓ. As a fertilizer to supply potassium to plants.
Correct Answer: As a component in the manufacture of heat-resistant glass (Pyrex).
Explanation: Borax is an essential component in the manufacture of borosilicate glass (like Pyrex and Kimax), fiberglass, and ceramic glazes. The addition of Boron compounds increases the glass’s resistance to thermal shock (sudden temperature changes) and chemical corrosion, making it suitable for laboratory and cooking equipment.
61. Orthoboric acid (${H}_3{BO}_3$) can be prepared by treating Borax (${Na}_2{B}_4{O}_7 \cdot 10{H}_2{O}$) with:
ⓐ. Sodium hydroxide (${NaOH}$)
ⓑ. Calcium carbonate (${CaCO}_3$)
ⓒ. Hydrochloric acid (${HCl}$)
ⓓ. Ethanol (${C}_2{H}_5{OH}$)
Correct Answer: Hydrochloric acid (${HCl}$)
Explanation: Orthoboric acid is prepared by acidifying a hot, concentrated solution of Borax with a mineral acid like hydrochloric acid (${HCl}$) or sulfuric acid (${H}_2{SO}_4$). The reaction is:
$${Na}_2{B}_4{O}_7 + 2{HCl} + 5{H}_2{O} \rightarrow 2{NaCl} + 4{H}_3{BO}_3$$
Since boric acid is sparingly soluble in cold water, it separates out as white crystals upon cooling.
62. The structure of solid orthoboric acid (${H}_3{BO}_3$) consists of:
ⓐ. Discrete tetrahedral ${B}({OH})_4^-$ ions.
ⓑ. Infinite three-dimensional network of ${B}_4{O}_7^{2-}$ units.
ⓒ. Planar ${BO}_3$ units joined by strong ionic bonds.
ⓓ. Planar ${BO}_3$ units linked together by weak hydrogen bonds.
Correct Answer: Planar ${BO}_3$ units linked together by weak hydrogen bonds.
Explanation: Orthoboric acid is a layered structure. Each Boron atom is $sp^2$ hybridized and surrounded by three hydroxyl (${OH}$) groups, resulting in a planar trigonal structure. These two-dimensional sheets of ${BO}_3$ units are held together by extensive, weak hydrogen bonds (${O}-{H} \cdot \cdot \cdot {O}$), which explains why boric acid is a soft, soapy solid.
63. When ${H}_3{BO}_3$ is heated strongly, it first forms metaboric acid, ${HBO}_2$, and finally yields:
ⓐ. Boron nitride (${BN}$)
ⓑ. Diborane (${B}_2{H}_6$)
ⓒ. Boron trioxide (${B}_2{O}_3$)
ⓓ. Borax (${Na}_2{B}_4{O}_7$)
Correct Answer: Boron trioxide (${B}_2{O}_3$)
Explanation: On heating at $373 \, {K}$, orthoboric acid loses a molecule of water to form metaboric acid (${HBO}_2$). On further heating above $413 \, {K}$, ${HBO}_2$ loses another water molecule to form tetraboric acid (${H}_2{B}_4{O}_7$). Finally, at red heat, both forms decompose to give the glassy substance Boron trioxide (Boric anhydride), ${B}_2{O}_3$.
$$2{H}_3{BO}_3 \xrightarrow{{Red heat}} {B}_2{O}_3 + 3{H}_2{O}$$
64. Orthoboric acid acts as a weak acid in water, but it is unique because it is not a proton donor. Instead, it acts as a Lewis acid by:
ⓐ. Forming an ionic bond with the ${H}^+$ ion.
ⓑ. Accepting a pair of electrons from the ${OH}^-$ ion of water.
ⓒ. Donating a pair of electrons to the ${H}^+$ ion of water.
ⓓ. Dissociating to form ${H}^+$ and ${BO}_3^{3-}$ ions.
Correct Answer: Accepting a pair of electrons from the ${OH}^-$ ion of water.
Explanation: Orthoboric acid (${H}_3{BO}_3$) is a monobasic Lewis acid. Due to the presence of an empty $2p$ orbital on the Boron atom, it can accept a lone pair of electrons from the ${OH}^-$ ion released from the water molecule, and then releases a proton from the ${H}_2{O}$ molecule it reacted with.
$${B}({OH})_3 + {H}_2{O} \rightleftharpoons [{B}({OH})_4]^- + {H}^+$$
65. What is the hybridization of the Boron atom in the anion formed when orthoboric acid reacts with water?
ⓐ. $sp^3$
ⓑ. $sp^2$
ⓒ. $sp$
ⓓ. $dsp^2$
Correct Answer: $sp^3$
Explanation: In orthoboric acid (${B}({OH})_3$), Boron is $sp^2$ hybridized (trigonal planar). When it accepts a lone pair from the ${OH}^-$ ion of water, it forms the tetrahydroxoborate ion, $[{B}({OH})_4]^-$. In this ion, the Boron atom is bonded to four ${OH}$ groups, making its coordination number four, and thus the hybridization changes to $sp^3$ (tetrahedral geometry).
66. The ${p}K_a$ value of ${H}_3{BO}_3$ is $9.25$. This indicates that it is a:
ⓐ. Strong acid
ⓑ. Strong base
ⓒ. Very weak monobasic acid
ⓓ. Weak tribasic acid
Correct Answer: Very weak monobasic acid
Explanation: ${p}K_a$ is the negative logarithm of the acid dissociation constant, ${K}_a$. A high ${p}K_a$ value (like $9.25$) corresponds to a very small ${K}_a$ value, meaning the acid dissociates very poorly in water. Boric acid is therefore a very weak acid. Furthermore, since it only releases one ${H}^+$ ion (indirectly, by reacting with water), it is a monobasic acid.
67. Orthoboric acid is often used in the form of a dilute aqueous solution for:
ⓐ. As a strong oxidizing agent.
ⓑ. As an antiseptic and eyewash.
ⓒ. For manufacturing strong hydrochloric acid.
ⓓ. As a general laboratory reagent for titration.
Correct Answer: As an antiseptic and eyewash.
Explanation: Because orthoboric acid is a very weak acid and non-toxic (at low concentrations), its dilute solution is mild enough to be used as a weak antiseptic (like ‘Boracic lotion’ or ‘Daktarin’) for minor cuts and bruises and, traditionally, as a soothing eyewash, taking advantage of its mild, bacteriostatic properties.
68. When orthoboric acid (${H}_3{BO}_3$) is treated with excess ethanol (${C}_2{H}_5{OH}$) in the presence of concentrated ${H}_2{SO}_4$, the product formed is:
ⓐ. Sodium borate
ⓑ. Boron trihydride
ⓒ. Metaboric acid
ⓓ. Triethyl borate
Correct Answer: Triethyl borate
Explanation: This is the Borate Ester Test. Orthoboric acid reacts with alcohols (like ethanol) to form volatile borate esters (Triethyl borate, ${B}({OC}_2{H}_5)_3$) in the presence of concentrated ${H}_2{SO}_4$ (as a dehydrating agent). The triethyl borate ester burns with a characteristic bright green-edged flame, confirming the presence of Boron.
$${H}_3{BO}_3 + 3{C}_2{H}_5{OH} \xrightarrow{{Conc. } {H}_2{SO}_4} {B}({OC}_2{H}_5)_3 + 3{H}_2{O}$$
69. What chemical reagent is added during the titration of a Boric acid solution to make it strong enough to be titrated against a strong base (${NaOH}$)?
ⓐ. Sodium chloride (${NaCl}$)
ⓑ. Methyl alcohol
ⓒ. Glycerol or Mannitol
ⓓ. Acetic acid (${CH}_3{COOH}$)
Correct Answer: Glycerol or Mannitol
Explanation: Boric acid is too weak to be titrated directly with ${NaOH}$ using common indicators. To strengthen it, polyhydroxy compounds like Glycerol, Mannitol, or Sorbitol are added. These compounds form stable, chelated complexes (esters) with the ${H}_3{BO}_3$ molecule, which are much stronger acids than ${H}_3{BO}_3$ itself. This stronger acid can then be accurately titrated.
70. The empirical formula of Orthoboric acid is ${H}_3{BO}_3$. What is the corresponding general formula for the acid derived from Boron trioxide (${B}_2{O}_3$)?
ⓐ. Boron(III) hydroxide
ⓑ. ${B}({OH})_3$
ⓒ. Both A and B
ⓓ. ${B}_2{O}_3$
Correct Answer: Both A and B
Explanation: The empirical formula ${H}_3{BO}_3$ can be written as ${B}({OH})_3$, which chemically corresponds to Boron(III) hydroxide. However, because of its unique Lewis acid nature (it accepts ${OH}^-$ rather than donating ${H}^+$), it is classified as ${H}_3{BO}_3$ and not a base. Both ${B}({OH})_3$ and Boron(III) hydroxide are alternative ways to represent or name the compound, which is derived by dissolving ${B}_2{O}_3$ in water (${B}_2{O}_3 + 3{H}_2{O} \rightarrow 2{H}_3{BO}_3$).
71. Diborane (${B}_2{H}_6$) can be prepared in the laboratory by the reaction of iodine (${I}_2$) with which reducing agent?
ⓐ. Sodium tetrahydroxoborate (${Na}[{B}({OH})_4]$)
ⓑ. Sodium borohydride (${NaBH}_4$)
ⓒ. Lithium aluminium hydride (${LiAlH}_4$)
ⓓ. Borax (${Na}_2{B}_4{O}_7 \cdot 10{H}_2{O}$)
Correct Answer: Sodium borohydride (${NaBH}_4$)
Explanation: The convenient and common laboratory method for the synthesis of diborane involves the reaction of ${NaBH}_4$ (a good source of ${H}^-$ ions) with iodine (${I}_2$) in diglyme (diethylene glycol dimethyl ether) as a solvent. The balanced reaction is:
$$2{NaBH}_4 + {I}_2 \rightarrow {B}_2{H}_6 + 2{NaI} + {H}_2$$
72. Which set of reactants is used for the large-scale industrial preparation of Diborane?
ⓐ. Sodium borohydride (${NaBH}_4$) with iodine (${I}_2$)
ⓑ. Boron trichloride (${BCl}_3$) with sodium hydride (${NaH}$)
ⓒ. Boron trioxide (${B}_2{O}_3$) with water
ⓓ. Orthoboric acid (${H}_3{BO}_3$) with lithium aluminium hydride (${LiAlH}_4$)
Correct Answer: Boron trichloride (${BCl}_3$) with sodium hydride (${NaH}$)
Explanation: For large-scale industrial production, cheaper and more readily available reagents are used. Diborane is prepared by the reduction of boron halides (${BCl}_3$) using inexpensive metal hydrides, such as sodium hydride (${NaH}$), at high temperatures ($\sim 450 \, {K}$).
$$2{BCl}_3 + 6{NaH} \xrightarrow{450 \, {K}} {B}_2{H}_6 + 6{NaCl}$$
73. Diborane (${B}_2{H}_6$) is an electron-deficient compound. What is its common classification in terms of chemical bonding?
ⓐ. Electron-precise hydride
ⓑ. Covalent network solid
ⓒ. Electron-rich hydride
ⓓ. Electron-deficient hydride
Correct Answer: Electron-deficient hydride
Explanation: Diborane is termed electron-deficient because it has fewer valence electrons (12 electrons total: 3 from each ${B}$ and 1 from each ${H}$) than required to form simple, conventional $2c-2e$ (two-center, two-electron) bonds between all adjacent atoms. If it formed 6 normal ${B}-{H}$ bonds and one ${B}-{B}$ bond, it would require 14 electrons. This deficiency forces it to adopt the unique bridge-bonded structure.
74. In the structure of diborane (${B}_2{H}_6$), how many $2c-2e$ (two-center, two-electron) terminal ${B}-{H}$ bonds are present?
ⓐ. 2
ⓑ. 4
ⓒ. 6
ⓓ. 0
Correct Answer: 4
Explanation: The structure of diborane contains two distinct types of ${B}-{H}$ bonds. The molecule has four terminal ${H}$ atoms (two attached to each ${B}$ atom) that are bonded via normal single covalent bonds (${B}-{H}$). These are $2c-2e$ bonds. The remaining two ${H}$ atoms form the bridge bonds, which are $3c-2e$ bonds.
75. The unique bonding in diborane (${B}_2{H}_6$) involves the formation of $3c-2e$ bonds. What are these bonds commonly called?
ⓐ. Double bonds
ⓑ. Metallic bonds
ⓒ. Hydrogen bonds
ⓓ. Bridge bonds or Banana bonds
Correct Answer: Bridge bonds or Banana bonds
Explanation: Diborane’s structure is stabilized by the formation of two $3c-2e$ bonds (three-center, two-electron bonds). In this type of bonding, two electrons are shared among three atomic nuclei (${B}-{H}-{B}$). This geometry, which resembles the shape of a banana, gives rise to the popular name “banana bonds” or more formally, “bridge bonds.”
76. What is the hybridization of the Boron atom in diborane (${B}_2{H}_6$)?
ⓐ. $sp$
ⓑ. $sp^2$
ⓒ. $sp^3$
ⓓ. $dsp^2$
Correct Answer: $sp^3$
Explanation: In ${B}_2{H}_6$, each Boron atom is attached to four neighboring atoms: two terminal hydrogen atoms and two bridge hydrogen atoms. The arrangement around each boron is approximately tetrahedral. To accommodate these four surrounding atoms, the Boron atom must undergo ${sp}^3$ hybridization, utilizing all three $2p$ orbitals and the $2s$ orbital.
77. Which of the following describes the geometry of the terminal ${B}-{H}$ bonds in diborane, ${B}_2{H}_6$?
ⓐ. They lie in a plane perpendicular to the bridge ${B}-{H}-{B}$ bond plane.
ⓑ. They are oriented linearly with the ${B}-{B}$ axis.
ⓒ. They form a plane that is the same as the bridge ${B}-{H}-{B}$ bond plane.
ⓓ. They are ionic bonds with no directional character.
Correct Answer: They lie in a plane perpendicular to the bridge ${B}-{H}-{B}$ bond plane.
Explanation: The diborane molecule has two distinct geometrical planes. The two Boron atoms and the four terminal hydrogen atoms lie in one plane. The two bridging hydrogen atoms lie in a separate plane that is perpendicular (${90}^\circ$) to the plane containing the terminal atoms.
78. Which statement correctly compares the bond lengths in diborane (${B}_2{H}_6$)?
ⓐ. Terminal ${B}-{H}$ bond length is equal to bridge ${B}-{H}$ bond length.
ⓑ. Terminal ${B}-{H}$ bond length is shorter than bridge ${B}-{H}$ bond length.
ⓒ. Terminal ${B}-{H}$ bond length is longer than bridge ${B}-{H}$ bond length.
ⓓ. The ${B}-{B}$ bond length is the shortest bond in the molecule.
Correct Answer: Terminal ${B}-{H}$ bond length is shorter than bridge ${B}-{H}$ bond length.
Explanation: The terminal ${B}-{H}$ bonds are conventional $2c-2e$ covalent bonds, with a length of $119 \, {pm}$. The bridge ${B}-{H}$ bonds are $3c-2e$ bonds, where the electron pair is shared among three nuclei, leading to a weaker and longer bond ($134 \, {pm}$). Therefore, the terminal bonds are significantly shorter and stronger.
79. The ${B}-{H}$ bridge bond angle ($\angle{H}_{{bridge}}-{B}-{H}_{{bridge}}$) in diborane is approximately:
ⓐ. ${120}^\circ$
ⓑ. ${180}^\circ$
ⓒ. ${97}^\circ$
ⓓ. ${109.5}^\circ$
Correct Answer: ${97}^\circ$
Explanation: The bond angles around the Boron atom are not perfectly tetrahedral (${109.5}^\circ$) due to the constraints of the bridge structure. The terminal ${H}-{B}-{H}$ angle is wider (approx. ${120}^\circ$), while the internal bridge ${H}_{{bridge}}-{B}-{H}_{{bridge}}$ bond angle is compressed to about ${97}^\circ$. This compression is characteristic of bridge bonding.
80. Borane (${BH}_3$), the monomer of diborane, is highly unstable at room temperature because:
ⓐ. It is a highly polar molecule.
ⓑ. Borane has an expanded octet.
ⓒ. It possesses $p\pi-p\pi$ back-bonding, stabilizing the monomer.
ⓓ. It is an electron-deficient species and dimerizes immediately.
Correct Answer: It is an electron-deficient species and dimerizes immediately.
Explanation: Borane (${BH}_3$) has only six valence electrons, making it extremely electron-deficient and a strong Lewis acid. To minimize this deficiency and utilize the Boron atom’s available ${sp}^3$ orbitals, it instantly dimerizes at room temperature to form the more stable, less electron-deficient diborane (${B}_2{H}_6$), which uses the unique $3c-2e$ bridge bonds.
81. When hydrated aluminium chloride, ${AlCl}_3 \cdot 6{H}_2{O}$, is heated strongly in air, what is the major stable solid product formed?
ⓐ. Anhydrous ${AlCl}_3$
ⓑ. Aluminium oxide, ${Al}_2{O}_3$
ⓒ. Aluminium hydroxide, ${Al}({OH})_3$
ⓓ. Aluminium metal, ${Al}$
Correct Answer: Aluminium oxide, ${Al}_2{O}_3$
Explanation: Heating the hydrated salt strongly releases water of crystallization, which then reacts with the ${AlCl}_3$ in a hydrolysis reaction due to the high charge-to-size ratio of ${Al}^{3+}$. The overall reaction leads to the formation of stable aluminium oxide:
$${AlCl}_3 \cdot 6{H}_2{O} \xrightarrow{{heat}} {Al}_2{O}_3 + {HCl}({g}) + {H}_2{O}({g})$$
82. Which set of reactants is preferred for the preparation of anhydrous aluminium chloride, ${AlCl}_3$?
ⓐ. Aqueous ${Al}({OH})_3$ with concentrated ${HCl}$
ⓑ. ${Al}_2{O}_3$ with ${H}_2$ at $500^{\circ}{C}$
ⓒ. Heated aluminium metal (${Al}$) with dry chlorine gas (${Cl}_2$)
ⓓ. Hydrated ${AlCl}_3$ heated with excess ${HCl}$ gas
Correct Answer: Heated aluminium metal (${Al}$) with dry chlorine gas (${Cl}_2$)
Explanation: Anhydrous ${AlCl}_3$ is prepared by direct synthesis from the elements using dry conditions to prevent the formation of the hydrate or subsequent hydrolysis. The reaction is:
$$2{Al}({s}) + 3{Cl}_2({g}) \xrightarrow{{heat}} 2{AlCl}_3({s})$$
83. How many valence electrons surround the central aluminium atom in a gaseous ${AlCl}_3$ monomer, and what does this imply about its properties?
ⓐ. Six; it is electron-deficient and acts as a Lewis acid.
ⓑ. Eight; it is electron-precise and chemically inert.
ⓒ. Four; it must form triple bonds to satisfy its valency.
ⓓ. Six; it is electron-deficient and acts as a Lewis base.
Correct Answer: Six; it is electron-deficient and acts as a Lewis acid.
Explanation: In the ${AlCl}_3$ monomer, the aluminium atom forms three covalent bonds ($3 \times 2 = 6$ valence electrons). Since it has less than eight electrons, it possesses an incomplete octet. This electron deficiency makes it highly prone to accepting an electron pair, which is the definition of a Lewis acid.
84. The dimeric structure of aluminium chloride, ${Al}_2{Cl}_6$, forms primarily to:
ⓐ. Increase the ionic character of the ${Al}-{Cl}$ bonds.
ⓑ. Minimize repulsion between the chlorine atoms.
ⓒ. Decrease the sublimation point of the compound.
ⓓ. Allow the aluminium atoms to achieve a stable octet.
Correct Answer: Allow the aluminium atoms to achieve a stable octet.
Explanation: The monomeric ${AlCl}_3$ is electron-deficient (6 valence electrons). By dimerizing to ${Al}_2{Cl}_6$, each ${Al}$ atom is coordinated to four ${Cl}$ atoms (two terminal and two bridging), giving it a full octet (8 valence electrons) and increasing its stability.
85. What type of bonding is responsible for holding the two ${AlCl}_3$ units together via the bridging chlorine atoms in the ${Al}_2{Cl}_6$ dimer?
ⓐ. Ionic bonds
ⓑ. Hydrogen bonds
ⓒ. Coordinate covalent (dative) bonds
ⓓ. Simple ${2c-2e}$ covalent bonds
Correct Answer: Coordinate covalent (dative) bonds
Explanation: The bridging bonds are formed when a lone pair of electrons from a chlorine atom (Lewis base site) of one ${AlCl}_3$ unit is donated to the vacant $p$-orbital on the electron-deficient aluminium atom (Lewis acid site) of the second ${AlCl}_3$ unit. This is a coordinate covalent bond.
86. When ${AlCl}_3$ reacts with a Lewis base such as ammonia (${NH}_3$), what is the term for the resulting product?
ⓐ. Salt
ⓑ. ${Al}_2{Cl}_6$
ⓒ. Adduct or Coordination Complex
ⓓ. Carbocation
Correct Answer: Adduct or Coordination Complex
Explanation: The product formed by the reaction of a Lewis acid (${AlCl}_3$) and a Lewis base (${NH}_3$) is called an adduct or coordination complex. In this specific case, the product is ${Cl}_3{Al} \leftarrow {NH}_3$, where the nitrogen atom donates its lone pair to the aluminium atom.
87. What are the hybridization and geometry of the aluminium atom in the ${AlCl}_4^-$ ion?
ⓐ. $sp^3$, Tetrahedral
ⓑ. $sp^2$, Trigonal pyramidal
ⓒ. $sp^2$, Trigonal planar
ⓓ. $sp^3d$, Trigonal bipyramidal
Correct Answer: $sp^3$, Tetrahedral
Explanation: When ${AlCl}_3$ accepts a chloride ion (${Cl}^-$) to form ${AlCl}_4^-$, the aluminium atom becomes bonded to four chlorine atoms, achieving a stable octet. This involves four $\sigma$ bonds and requires $sp^3$ hybridization, resulting in a tetrahedral geometry.
88. In the Friedel-Crafts alkylation reaction, which species is created when the Lewis acid ${AlCl}_3$ reacts with the alkyl halide (${R}-{Cl}$)?
ⓐ. The ${AlCl}_3$ dimer
ⓑ. A neutral organic compound, ${R}-{H}$
ⓒ. A highly reactive electrophile (${R}^+$ or a polarized complex)
ⓓ. A Lewis base, ${AlCl}_2$
Correct Answer: A highly reactive electrophile (${R}^+$ or a polarized complex)
Explanation: ${AlCl}_3$ accepts the lone pair from the halogen (${Cl}$) of the alkyl halide, polarizing the ${C}-{Cl}$ bond and either generating the free carbocation electrophile (${R}^+$) or a highly polarized ion-pair complex (${R}^+[{AlCl}_4^-]$). This electrophile then attacks the aromatic ring.
89. Why is ${AlCl}_3$ considered a stronger Lewis acid than ${BCl}_3$ despite both having an incomplete octet?
ⓐ. Aluminium is more electronegative than Boron.
ⓑ. ${BCl}_3$ is a linear molecule, reducing its reactivity.
ⓒ. ${AlCl}_3$ has a greater tendency to dimerize.
ⓓ. Less effective ${p}-{p}$ back-bonding in ${AlCl}_3$.
Correct Answer: Less effective ${p}-{p}$ back-bonding in ${AlCl}_3$.
Explanation: In ${BCl}_3$, the small size of the Boron atom allows effective ${p}-{p}$ overlap between the lone pairs of chlorine and the vacant $p$-orbital of boron (back-bonding). This partially compensates for boron’s electron deficiency. In ${AlCl}_3$, the larger size of the ${Al}$ atom leads to poorer orbital overlap, making back-bonding less effective. Consequently, the ${Al}$ atom remains more electron-deficient, resulting in ${AlCl}_3$ being the stronger Lewis acid.
90. When anhydrous ${AlCl}_3$ dissolves in water, the solution becomes acidic. Which chemical equilibrium explains this acidity?
ⓐ. ${AlCl}_3 \rightleftharpoons {Al}^{3+} + 3{Cl}^-$
ⓑ. $[{Al}({H}_2{O})_6]^{3+} + {H}_2{O} \rightleftharpoons [{Al}({H}_2{O})_5({OH})]^{2+} + {H}_3{O}^+$
ⓒ. ${Cl}^- + {H}_2{O} \rightleftharpoons {HCl} + {OH}^-$
ⓓ. ${Al}_2{Cl}_6 + 12{H}_2{O} \rightleftharpoons 2{Al}({OH})_3 + 6{HCl} + 6{H}_2{O}$
Correct Answer: $[{Al}({H}_2{O})_6]^{3+} + {H}_2{O} \rightleftharpoons [{Al}({H}_2{O})_5({OH})]^{2+} + {H}_3{O}^+$
Explanation: The small, highly charged ${Al}^{3+}$ ion strongly polarizes the electron density in the ${O}-{H}$ bonds of the coordinated water molecules in the hexa-aqua complex ion, $[{Al}({H}_2{O})_6]^{3+}$. This makes the coordinated water molecules acidic, allowing them to release a proton (${H}^+$) to the bulk solvent (${H}_2{O}$), increasing the concentration of ${H}_3{O}^+$ and making the solution acidic.
91. Which of the following elements has the ground state valence shell electronic configuration ${4s}^2{4p}^2$?
ⓐ. Carbon (${C}$)
ⓑ. Silicon (${Si}$)
ⓒ. Germanium (${Ge}$)
ⓓ. Tin (${Sn}$)
Correct Answer: Germanium (${Ge}$)
Explanation: The general electronic configuration for Group 14 elements is ${ns}^2{np}^2$. For Germanium (${Ge}$), the valence shell is the $n=4$ shell. Therefore, its valence configuration is ${4s}^2{4p}^2$, following the core configuration $[{Ar}]{3d}^{10}$.
92. In the excited state, an element with the general configuration ${ns}^2{np}^2$ promotes an electron to achieve a valency of 4. What is the resulting hybridization in a simple tetravalent compound like ${CCl}_4$?
ⓐ. $sp^3$
ⓑ. $sp^2$
ⓒ. $sp$
ⓓ. $dsp^2$
Correct Answer: $sp^3$
Explanation: In the ground state (${ns}^2{np}^2$), there are only two unpaired electrons, suggesting a valency of 2. To achieve a valency of 4 (as seen in ${CCl}_4$), one electron from the filled ${ns}$ orbital is promoted to the empty ${np}$ orbital (${ns}^1{np}^3$). These four singly occupied orbitals then hybridize to form four equivalent $sp^3$ hybrid orbitals, resulting in a tetrahedral geometry.
93. The ${ns}^2{np}^2$ configuration results in a common oxidation state of +4 for all Group 14 elements. Which of the following is also a prominent and stable oxidation state, particularly for the heavier elements (Sn, Pb)?
ⓐ. +1
ⓑ. +2
ⓒ. +3
ⓓ. +5
Correct Answer: +2
Explanation: Due to the inert pair effect, the paired ${ns}^2$ electrons in the heavier elements become increasingly reluctant to participate in bonding. This means only the two ${np}^2$ electrons are lost or shared, leading to a stable $\mathbf{+2}$ oxidation state for Tin (${Sn}$) and Lead (${Pb}$). This effect increases down the group.
94. Which of the following characteristics is directly caused by the ${ns}^2{np}^2$ configuration where the valence shell is exactly half-filled?
ⓐ. Exceptionally high reactivity with halogens.
ⓑ. A large tendency towards catenation (self-linking).
ⓒ. Strong metallic character.
ⓓ. Exclusive formation of ionic compounds.
Correct Answer: A large tendency towards catenation (self-linking).
Explanation: The configuration ${ns}^2{np}^2$ means these elements are tetravalent. They form four covalent bonds, which are strong and non-polar, especially for Carbon and Silicon. This strong, multi-directional covalent bonding capability is the fundamental reason for the extensive catenation seen in this group (Carbon $\gg$ Silicon $\gg$ Germanium $\gg$ Tin/Lead).
95. Moving down Group 14 from Carbon to Lead, how does the tendency of the ${ns}^2$ electrons to remain paired (the inert pair effect) change?
ⓐ. It decreases due to increasing atomic size.
ⓑ. It remains constant throughout the group.
ⓒ. It increases due to poor shielding of $d$ and $f$ orbitals.
ⓓ. It increases due to decreasing electronegativity.
Correct Answer: It increases due to poor shielding of $d$ and $f$ orbitals.
Explanation: The inert pair effect is caused by the poor shielding provided by the intervening ${d}$ and ${f}$ electrons in heavier atoms like ${Sn}$ and ${Pb}$. This poor shielding allows the effective nuclear charge on the ${ns}$ electrons to increase, holding them tighter and making them less available for bonding.
96. The ionization enthalpy ($IE$) generally decreases down a group. For Group 14, which element exhibits a slight increase in the first ionization enthalpy compared to the element immediately preceding it?
ⓐ. Carbon ($C$)
ⓑ. Silicon ($Si$)
ⓒ. Germanium ($Ge$)
ⓓ. Lead (${Pb}$)
Correct Answer: Lead (${Pb}$)
Explanation: The ionization energy typically decreases down the group. However, Lead (${Pb}$) has a slightly higher ionization enthalpy than Tin (${Sn}$) due to the presence of the intervening ${4f}$ orbitals (lanthanide contraction). The ${4f}$ orbitals provide very poor shielding, leading to a significant increase in the effective nuclear charge on the valence electrons of ${Pb}$.
97. Which of the following compounds is the most stable in the +2 oxidation state?
ⓐ. ${CCl}_2$
ⓑ. ${SiCl}_2$
ⓒ. ${SnCl}_2$
ⓓ. ${PbCl}_2$
Correct Answer: ${PbCl}_2$
Explanation: The stability of the +2 oxidation state increases down Group 14 because of the inert pair effect. Lead (${Pb}$) is the heaviest element and exhibits the most pronounced inert pair effect, making the ${ns}^2$ electrons very difficult to remove or share. Therefore, ${Pb}({II})$ compounds like ${PbCl}_2$ are the most stable in this oxidation state.
98. The structure of ${SiO}_2$ (silica) is a giant three-dimensional network, while ${CO}_2$ is a discrete monomeric molecule. This difference is mainly attributed to:
ⓐ. The difference in electronegativity between ${C}$ and ${Si}$.
ⓑ. ${SiO}_2$ being ionic while ${CO}_2$ is covalent.
ⓒ. The inability of Silicon to form stable $p\pi-p\pi$ multiple bonds.
ⓓ. The difference in their atomic mass.
Correct Answer: The inability of Silicon to form stable ${p}\pi-{p}\pi$ multiple bonds.
Explanation: Carbon forms stable ${p}\pi-{p}\pi$ double bonds with oxygen, resulting in a small, discrete ${O}={C}={O}$ molecule. However, Silicon’s larger size and dispersed orbitals make ${p}\pi-{p}\pi$ bonding weak and unstable. Instead, Silicon maximizes stability by forming strong $\sigma$ bonds in an extended network structure where each ${Si}$ is bonded tetrahedrally to four ${O}$ atoms.
99. Which of the following elements, due to its ${ns}^2{np}^2$ configuration and access to vacant $d$-orbitals, readily expands its covalency beyond 4?
ⓐ. Carbon (${C}$)
ⓑ. Silicon (${Si}$)
ⓒ. Lead (${Pb}$)
ⓓ. Boron (${B}$)
Correct Answer: Silicon (${Si}$)
Explanation: Silicon, Germanium, Tin, and Lead all possess vacant ${d}$-orbitals in their valence shell. Silicon, being lighter, is the first element in the group to show this characteristic strongly. It can use these vacant ${3d}$ orbitals to accept electron pairs from species like ${F}^-$, expanding its covalency to 5 or 6, as seen in complex ions like ${[SiF}_6]^{2-}$. Carbon lacks these vacant ${d}$-orbitals.
100. Carbon (${C}$) forms ${CO}$ (carbon monoxide), a stable compound in the +2 oxidation state. Which factor contributes to the stability of ${CO}$ despite Carbon’s strong preference for the +4 state?
ⓐ. The formation of a triple bond including a coordinate covalent bond.
ⓑ. The inert pair effect.
ⓒ. Resonance stabilization.
ⓓ. The high electronegativity of Carbon.
Correct Answer: The formation of a triple bond including a coordinate covalent bond.
Explanation: The stability of ${CO}$ is due to the strong triple bond between ${C}$ and ${O}$. This bond consists of two normal covalent bonds and one coordinate covalent bond where the oxygen atom donates a lone pair to the carbon atom (${C} \leftarrow {O}$), giving the structure $\mathbf{|C \equiv O|}$ and effectively completing the octet of both atoms.
Welcome to Class 11 Chemistry MCQs – Chapter 11: The p-Block Elements (Part of 3). This chapter is one of the most important and high-weightage topics in NCERT Class 11 Chemistry as well as in competitive exams like JEE Main, NEET, and state-level entrance tests. Here, you’ll explore the fascinating chemistry of Group 13 (Boron Family) and Group 14 (Carbon Family) through 100 concept-rich MCQs designed for exam success.
Navigation & parts: The full chapter contains 277 MCQs organized into 3 structured parts (100 + 100 + 77). This page presents the first 100 MCQs with detailed explanations and reasoning based on the latest NCERT syllabus and previous-year exam questions. Use the Part buttons above to continue with later sections.
What you will learn & practice (Chapter 11: The p-Block Elements)
- General trends of p-block elements — electronic configuration, atomic size, ionization energy, and oxidation states
- Group 13 elements (Boron family) — Boron, Aluminium, Gallium, Indium, Thallium; their compounds, reactions, and properties
- Boron compounds like borax, orthoboric acid, boric anhydride, diborane ($B_2H_6$), and their industrial significance
- Aluminium chemistry — amphoteric nature, reactions with acids & bases, and applications in metallurgy
- Group 14 elements (Carbon family) — Carbon, Silicon, Germanium, Tin, Lead and their compounds
- Allotropes of carbon — diamond, graphite, fullerenes, graphene; their structure, bonding, and uses
- Silicates, silicones, and oxides of carbon — their structure, formula, and practical applications
- Trends in covalency, catenation, and oxidation states across the p-block
- Important reactions, equations, and applications of boron trihalides, aluminium chloride, and silicon tetrachloride
- Real-world connection: Role of p-block elements in environment, glass industry, semiconductors, and daily life
How to use this site to master p-Block MCQs
- Start small: Attempt 10–15 MCQs at a time; analyze the explanations and note tricky ones.
- Mark important questions: Use the ❤️ Heart icon beside each MCQ to save it as Favourite. Toggle the Favourite Filter above the question list to revise only your selected MCQs.
- Take notes your way: Click the Workspace button below each MCQ to write short notes or formulas. Notes are auto-saved for future revision.
- Shuffle & test: Use the Random button to reshuffle questions and test your actual concept recall.
- Revise smartly: Before exams, open your Favourite list and review only key questions + Workspace notes for quick recall.
Why this chapter matters for Boards, JEE & NEET
- Boards: Commonly asked 1-mark questions on compounds of boron, aluminium, and carbon oxides.
- JEE Main/Advanced: Numerical and conceptual questions on bonding, trends, hybridization, and chemical reactions.
- NEET: Conceptual MCQs on periodic trends, borax bead test, and hybridization of boron compounds appear regularly.
- Applications: p-block elements form the foundation of environmental chemistry, material science, and inorganic industries.
Common mistakes students make
- Confusing oxidation states of boron and aluminium
- Ignoring diagonal relationship (B-Al, Si-Be)
- Misunderstanding catenation and inert pair effect
- Forgetting formulas of important compounds like $Na_2B_4O_7·10H_2O$, $B_2O_3$, $SiCl_4$
- Mixing up allotropes of carbon (diamond vs graphite vs fullerene)
This page is your complete guide to mastering p-block elements. Treat it as your digital notebook — mark favourites, write notes, revise periodically, and track your progress with each visit.
Explore more: Class 11 Chemistry – Chapter Index, Class 11 – Subject Index, All Classes – MCQ Library.
Quick FAQ
Is the p-block elements chapter tough?
It may seem vast, but most questions are concept-based. Focus on periodic trends and compound formulas, and revise through MCQs + Workspace notes.
Which group is most important for JEE/NEET?
Group 13 (Boron family) and Group 14 (Carbon family) are the most scoring. Expect 1–2 direct MCQs in almost every paper.
How to revise before the exam?
Use the Favourite Toggle to filter only your key MCQs and skim through Workspace notes — quick, effective, and stress-free.
- 👉 Total MCQs in this chapter: 277 (100 + 100 + 77)
- 👉 This page: First 100 MCQs with answers & explanations
- 👉 Best for: Boards • JEE/NEET • Periodic Trends • Concept Revision • Inorganic Chemistry Quizzes
- 👉 Next: Continue with the Part 2 button above
FAQs on The p-Block Elements ▼
▸ What are The p-Block Elements MCQs in Class 11 Chemistry?
These are multiple-choice questions from Chapter 11 of NCERT Class 11 Chemistry – The p-Block Elements. They assess your understanding of Group 13 (boron family) and Group 14 (carbon family) trends, important compounds, and their applications.
▸ How many MCQs are available in this chapter?
There are a total of 277 The p-Block Elements MCQs. They are divided into 3 structured parts – two sets of 100 questions each and one set of 77 questions.
▸ Are The p-Block Elements MCQs important for JEE and NEET?
Yes, this chapter is frequently tested in JEE and NEET. High-yield areas include electron deficiency and Lewis acidity (BF3, AlCl3), diborane (B2H6) bonding, borax and boric acid, catenation, allotropy of carbon (diamond, graphite), carbides, SiO2 and silicones.
▸ Do these MCQs include correct answers and explanations?
Yes, every MCQ is accompanied by the correct answer and clear explanations to help you understand the reasoning and avoid rote learning.
▸ Which subtopics are covered in these The p-Block Elements MCQs?
Subtopics include electronic configuration and periodic trends of Group 13 and 14, anomalous behaviour of boron and carbon, diagonal relationship, oxidation states, important compounds (B2H6, BF3, borax, boric acid, AlCl3, CO, CO2, SiO2), catenation, allotropy, and silicones.