GSEB Class 11 Chemistry Solutions Chapter 11 The p-Block Elements

 

Question 1. Discuss the pattern of variation of oxidation states in
(i) B to Tl
(ii) C to Pb.
Answer:
(i) The electronic configuration of all the elements listed is \( ns^2np^1 \). As we move down group 13, the two electrons in the s-orbital tend to remain paired. This means they do not easily participate in forming chemical bonds. This phenomenon is termed the 'inert-pair effect'. Due to this effect, as we go down the group, the +1 oxidation state becomes increasingly common and stable.

(ii) For carbon to lead, the inert-pair effect also plays a role. As we move down group 14, the \( ns^2 \) electrons prefer to stay paired and do not readily take part in bond formation. Therefore, the heavier elements in this group show an increasingly stable +2 oxidation state alongside the +4 oxidation state.

In simple words: As you go down a group in the periodic table, the electrons in the s-orbital sometimes prefer to stay together and not get involved in bonding. This 'inert pair' makes the lower oxidation states more stable for the heavier elements in that group.

Exam Tip: Remember to link the increasing stability of lower oxidation states in heavier p-block elements directly to the inert pair effect. Provide examples from both groups 13 and 14 to show your understanding.

 

Question 2. How can you explain higher stability of BCl3 as compared to TlCl3?
Answer: BCl3 is more stable than TlCl3 because the inert pair effect is not present in BCl3. Due to the inert pair effect, thallium will prefer to form TlCl rather than TlCl3. It tends to show an oxidation state of +1 because its \( ns^2 \) electrons are reluctant to participate in bonding.
In simple words: BCl3 is more stable than TlCl3 because thallium prefers to be in a +1 oxidation state (like in TlCl) due to the inert pair effect, which means its s-electrons do not readily form bonds. Boron does not have this effect.

Exam Tip: When comparing stability of halides within a group, always consider the inert pair effect for heavier elements and its impact on preferred oxidation states.

 

Question 3. Why does boron triflouride behave as a Lewis acid?
Answer: The Lewis structure of BF3 shows that boron has only six electrons around it, meaning its octet is incomplete. This makes BF3 an electron-deficient compound. All such compounds that are electron-deficient tend to accept a pair of electrons to achieve a stable electronic configuration, and because of this, they behave as Lewis acids. Thus, BF3 is an electron-deficient compound and acts as a Lewis acid.
In simple words: Boron trifluoride is a Lewis acid because boron in BF3 has only six electrons around it, not eight, so it wants to accept a pair of electrons to become stable.

Exam Tip: For Lewis acid/base questions, always focus on the availability of electron pairs (for bases) or empty orbitals/incomplete octets (for acids) to explain the behavior.

 

Question 4. Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.
Answer: BCl3 is completely hydrolyzed by water to give boric acid and hydrochloric acid. This reaction causes BCl3 to fume in air due to the production of HCl gas.
\( BCl_3 + 3H_2O \rightarrow H_3BO_3 + 3HCl \)
CCl4, on the other hand, cannot undergo hydrolysis. This is because carbon cannot expand its coordination number beyond four, as it lacks accessible d-orbitals. Consequently, CCl4 is insoluble in water.
In simple words: BCl3 reacts with water to form boric acid and hydrochloric acid, which you can see as fumes. CCl4 does not react with water because carbon cannot make more than four bonds and has no d-orbitals to help it react.

Exam Tip: When discussing hydrolysis of chlorides, consider the availability of vacant d-orbitals on the central atom. Elements with available d-orbitals (like boron in some contexts, or heavier elements) can expand their octet and undergo hydrolysis, whereas those without cannot.

 

Question 5. Is boric acid a protic acid? Explain.
Answer: Boric acid is not a protonic acid. Instead, it acts as a Lewis acid by readily accepting electrons from a hydroxyl ion. It does not donate a proton directly but rather accepts a hydroxide ion, causing water to release a proton.
\( B(OH)_3 + 2HOH \rightarrow [B(OH)_4]^- + H_3O^+ \)
In simple words: Boric acid is not a regular proton-donating acid. It works as a Lewis acid by taking electrons from water's hydroxyl ions, making the water molecule release a proton.

Exam Tip: Be precise with definitions. A protic acid donates a proton directly, while a Lewis acid accepts an electron pair. Boric acid's behavior is a classic example of a Lewis acid in aqueous solution.

 

Question 6. Explain what happens when boric acid is heated?
Answer: When boric acid is heated, it gradually loses water in three distinct stages at different temperatures, eventually yielding boron trioxide. First, at 370 K, it forms metaboric acid. Then, upon further heating, it converts to tetraboric acid. Finally, strong heating leads to boron trioxide.
\( H_3BO_3 \xrightarrow{370 K} HBO_2 + H_2O \) (Metaboric acid)
\( H_2B_4O_7 \xrightarrow{Red Heat} 2B_2O_3 + H_2O \) (Boron trioxide)
In simple words: When you heat boric acid, it loses water in steps. First, it turns into metaboric acid, then tetraboric acid, and finally, after much heat, it becomes boron trioxide.

Exam Tip: Remember the sequence of dehydration products when heating boric acid: from boric acid to metaboric acid, then tetraboric acid, and finally boron trioxide. It's important to know the specific formulas for each intermediate.

 

Question 7. Describe the shapes of BF3 and BH¯4. Assign the hybridisation of boron in these species.
Answer: In BF3, boron undergoes \( sp^2 \) hybridisation. This results in a trigonal planar structure, where the B-F bond angle is \( 120^\circ \). The electronic configuration of boron is \( 1s^2, 2s^2 2p^1 \). In the excited state, one 2s electron moves to a 2p orbital, leading to one 2s and two 2p atomic orbitals mixing to form three \( sp^2 \) hybrid orbitals. These orbitals then overlap with fluorine orbitals to form the bonds.
For BH¯4, boron undergoes \( sp^3 \) hybridisation. This hybridization creates a tetrahedral shape for the molecule, with bond angles of approximately \( 109.5^\circ \). In this case, one 2s and three 2p atomic orbitals of boron combine to form four \( sp^3 \) hybrid orbitals. These hybrid orbitals then form bonds with hydrogen atoms.
In simple words: For BF3, boron mixes its orbitals to make a flat, triangular shape called trigonal planar. For BH¯4, boron mixes its orbitals in a different way to make a 3D shape like a pyramid with a triangular base, called tetrahedral.

Exam Tip: To determine hybridisation and shape, first draw the Lewis structure, then apply VSEPR theory. For BF3, count 3 bonding pairs and 0 lone pairs on boron. For BH4-, count 4 bonding pairs and 0 lone pairs on boron (considering the extra electron for the negative charge). This will correctly lead to \( sp^2 \) (trigonal planar) and \( sp^3 \) (tetrahedral), respectively.

 

Question 8. Write reactions to justify amphoteric nature of aluminium.
Answer: Aluminium dissolves in both mineral acids and aqueous alkalies, which demonstrates its amphoteric character. This means it can react as both a base and an acid.
1. Aluminium reacts with dilute HCl, liberating dihydrogen gas:
\( 2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^-(aq) + 3H_2(g) \)
2. Aluminium also reacts with aqueous alkali, also liberating dihydrogen gas:
\( 2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na^+[Al(OH)_4]^-(aq) + 3H_2(g) \)
(Sodium tetrahydroxoaluminate (III) ion)
In simple words: Aluminium is amphoteric because it can react with strong acids like HCl and also with strong bases like NaOH, producing hydrogen gas in both cases.

Exam Tip: When asked to justify amphoteric behavior, always provide balanced chemical equations showing the element reacting with both an acid and a base, demonstrating the release of hydrogen gas.

 

Question 9. What are electron-deficient compounds? Are BCl3 and SiCl4 electron-deficient species? Explain.
Answer: Electron-deficient compounds are those in which the central atom has fewer than eight electrons in its outermost shell after forming covalent bonds. Such molecules tend to accept an electron pair to attain a stable octet configuration and therefore behave as Lewis acids.
In BCl3, boron is in its trivalent state and has only six electrons around the central boron atom (three from itself and three from the three chlorine atoms). Thus, BCl3 is an electron-deficient compound and acts as a Lewis acid.
On the other hand, SiCl4 has a tetrahedral shape due to \( sp^3 \) hybridisation of silicon. In SiCl4, the central silicon atom has completed its octet (four bonds with chlorine, meaning eight electrons around it). Therefore, SiCl4 is not an electron-deficient compound. Silicon's electronic structure allows it to achieve a stable octet.
In simple words: Electron-deficient compounds have less than eight electrons around their central atom. BCl3 is electron-deficient because boron has only six electrons. SiCl4 is not electron-deficient because silicon has a full eight electrons around it.

Exam Tip: To identify an electron-deficient compound, count the total valence electrons around the central atom after bonding. If it's less than eight, it's electron-deficient. Always specify the hybridisation and geometry as part of your explanation.

 

Question 10. Write the resonance structures of \( CO_3^{2-} \) and \( HCO_3^- \).
Answer:
Resonance structures of \( CO_3^{2-} \):
\[ :O: \\ \ || \\ \ C = O: \quad \longleftrightarrow \quad :O: \\ \ \ || \\ \ C \\ \ / \ \ \ \\ \ :O: \quad :O: \\ \ \quad \quad \quad \quad \ \longleftrightarrow \quad :O: \\ \ \ / \ \ \ \\ \ C = O: \quad :O: \\ \ \ || \] Resonance structures of \( HCO_3^- \):
\[ :O: \\ \ || \\ \ C = O: \\ \ / \ \\ \ :O:H \quad \longleftrightarrow \quad :O: \\ \ || \\ \ C \\ \ / \ \ \ \\ \ :O: \quad :OH \\ \ \longleftrightarrow \quad :O: \\ \ / \ \ \ \\ \ C = O: \\ \ \quad \quad \quad \quad \ \quad \quad \quad \ \ \quad \quad \quad \ \quad \quad \quad \quad :O:H \]In simple words: Resonance structures show that the actual bonding in these ions is a mix of several possible ways to draw the double and single bonds, making the electrons spread out (delocalized) over the whole ion.

Exam Tip: When drawing resonance structures, remember that only the positions of electrons (especially pi electrons and lone pairs) can move, not the atoms themselves. Ensure all formal charges are correctly assigned and that all valid resonance forms are included.

 

Question 11. What is the state of hybridisation of carbon in
(a) \( CO_3^{2-} \)
(b) diamond
(c) graphite ?
Answer:
(a) Carbon shows \( sp^2 \) hybridisation in \( CO_3^{2-} \) ion.
(b) Carbon shows \( sp^3 \) hybridisation in diamond.
(c) Carbon shows \( sp^2 \) hybridisation in graphite.
In simple words: In carbonate, diamond, and graphite, carbon atoms mix their orbitals in different ways. Carbon in carbonate and graphite uses sp2 hybridisation, while in diamond, it uses sp3 hybridisation.

Exam Tip: To determine hybridisation, count the number of sigma bonds and lone pairs around the central atom. For 3 regions of electron density, it's \( sp^2 \). For 4 regions, it's \( sp^3 \).

 

Question 12. Explain the difference in the properties of diamond and graphite on basis of their structures.
Answer: Diamond and graphite, both allotropes of carbon, have significantly different properties due to their distinct atomic structures.
In diamond, each carbon atom is \( sp^3 \) hybridised and covalently linked to four other carbon atoms in a tetrahedral arrangement. These strong C-C \( sp^3-sp^3 \) sigma bonds create a highly rigid, three-dimensional network structure that extends throughout the crystal. This extensive network makes diamond extremely hard and gives it a high melting point. The C-C bond length in diamond is 154 pm.
In contrast, in graphite, each carbon atom is \( sp^2 \) hybridised, forming a sheet-like (layered) structure. Each carbon atom is bonded to three other carbon atoms in a hexagonal planar arrangement, similar to benzene rings fused together. The C-C bond length in graphite is 142 pm, which is shorter than in diamond due to partial double bond character. The various sheets or layers in graphite are held together by weak van der Waals forces, allowing them to slip over each other easily. The distance between layers is 340 pm.
Based on these structural differences, the properties can be summarized as follows:
Diamond:
- Purity: It is the purest form of carbon.
- Bond length: Due to \( sp^3 \) hybridisation, the C-C bond length is 154 pm.
- Hardness: It is the hardest known substance with high density and a very high melting point.
- Conductivity: Since there are no free electrons, diamond is a poor conductor of electricity.
- Transparency: Because of its high refractive index (2.5), it can reflect and refract light well, making it a transparent substance.
Graphite:
- Purity: Like diamond, graphite is also a very pure form of carbon.
- Bond lengths: C-C bonds in graphite, due to \( sp^2 \) hybridisation, are 142 pm.
- Softness and Lubrication: Since its successive layers can easily slip over one another, graphite is soft and acts as a good lubricant.
- Conductivity: The fourth valence electron of each carbon atom in graphite is free and delocalized, making graphite a good conductor of electricity and heat.
- Opaqueness: Unlike diamond, graphite is a black substance and possesses a metallic luster.
In simple words: Diamond is very hard and transparent because its carbon atoms are tightly linked in a 3D network. Graphite is soft, dark, and conducts electricity because its carbon atoms form flat layers that can slide past each other, and it has free electrons.

Exam Tip: When explaining the differences between diamond and graphite, always relate their distinct properties (hardness, conductivity, transparency) directly back to their unique atomic structures and bonding types ( \( sp^3 \) tetrahedral network vs. \( sp^2 \) layered structure with delocalized electrons).

 

Question 13. Rationalise the given statements and give chemical reactions:
1. Lead (II) chloride reacts with Cl2 to give PbCl4.
2. Lead (IV) chloride is highly unstable towards heat.
3. Lead is known not to form an iodide PbI4.
Answer:
1. Lead (II) chloride, PbCl2, does not react with Cl2 to form PbCl4. Instead, PbCl4 is a strong oxidising agent and actually decomposes when heated to yield PbCl2 and Cl2.
\( PbCl_4 \xrightarrow{\Delta} PbCl_2 + Cl_2 \)
This statement is incorrect because Pb(IV) is less stable than Pb(II) due to the inert pair effect, making PbCl4 more prone to reduction rather than formation from PbCl2 and Cl2.
2. Lead (IV) chloride, PbCl4, is indeed highly unstable towards heat. This is again explained by the inert pair effect, where the +2 oxidation state becomes more stable for heavier group 14 elements like lead. Therefore, PbCl4 easily decomposes into PbCl2 and Cl2 when heated, as shown in the reaction above.
3. PbI4 does not exist. This is because the Pb-I bond that would form initially during the reaction of lead with iodine does not release sufficient energy to unpair the 6s² electrons and excite one of them to a higher orbital. This unpairing and excitation would be necessary to achieve four unpaired electrons for the +4 oxidation state. The large size of the iodide ion also makes it a weaker oxidising agent compared to chloride, meaning it cannot readily oxidise Pb(II) to Pb(IV) nor stabilise the higher oxidation state due to insufficient lattice energy or bond strength.
In simple words: 1. Lead (II) chloride doesn't react with chlorine to make lead (IV) chloride; instead, lead (IV) chloride breaks down into lead (II) chloride and chlorine when heated. 2. Lead (IV) chloride is not stable when hot. 3. Lead (IV) iodide does not form because lead's electrons are too stable to unpair and bond easily with iodine.

Exam Tip: For lead compounds, the inert pair effect is a key concept. It explains why the +2 oxidation state is more stable than +4 for lead, influencing reactivity, decomposition, and compound existence. Always consider the energy required for electron unpairing and the nature of the bond formed (bond energy/lattice energy).

 

Question 14. Suggest reasons why the B-F bond lengths in BF3 (130 pm) and \( BF_4^- \) (143 pm) differ?
Answer: In BF3, boron undergoes \( sp^2 \) hybridisation, which results in a trigonal planar structure. The B-F bond length in BF3 is shorter (130 pm) due to a partial double bond character arising from p\( \pi \)-p\( \pi \) backbonding between the filled p-orbital of fluorine and the vacant p-orbital of boron.
In \( BF_4^- \), boron undergoes \( sp^3 \) hybridisation, leading to a tetrahedral structure. In this ion, there is no vacant p-orbital on boron available for backbonding. Therefore, the B-F bonds are purely single bonds, which are longer (143 pm) compared to the bonds in BF3 that have some double bond character.
In simple words: The B-F bond in BF3 is shorter because boron forms a partial double bond with fluorine due to electron sharing. In \( BF_4^- \), boron forms only single bonds, so the B-F bond is longer.

Exam Tip: When explaining bond length differences, always consider factors like hybridisation, bond order (single, double, partial double), and the presence of backbonding. Backbonding usually shortens bond lengths due to increased electron density between atoms.

 

Question 15. If B-Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
Answer: BCl3 is a trigonal planar molecule with a symmetrical shape, a direct consequence of boron undergoing \( sp^2 \) hybridisation. Although each individual B-Cl bond possesses a dipole moment (because chlorine is more electronegative than boron), the overall BCl3 molecule has a zero dipole moment. This occurs because the three individual bond dipoles are oriented symmetrically at \( 120^\circ \) to each other in a plane, causing them to cancel out perfectly. The vector sum of these three equal and symmetrically arranged dipoles is zero.
In simple words: Even though each B-Cl bond has a tiny magnetic pull, the BCl3 molecule as a whole has no pull. This is because its flat, triangular shape means all the pulls cancel each other out perfectly.

Exam Tip: For molecules with polar bonds but zero net dipole moment, always explain how the symmetrical molecular geometry causes the individual bond dipoles to cancel each other out. Common examples include CCl4 (tetrahedral), CO2 (linear), and BCl3 (trigonal planar).

 

Question 16. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
Answer: AlF3 is insoluble in anhydrous HF because of its highly ionic nature and strong lattice energy, making it difficult for HF to solvate the ions. However, AlF3 dissolves when NaF is added. This is because NaF provides fluoride ions, which react with AlF3 to form a soluble complex ion, sodium hexafluoroaluminate (III), \( [AlF_6]^{3-} \).
\( AlF_3 + 3NaF \rightarrow Na_3[AlF_6] \)
When gaseous BF3 is bubbled through the solution of \( Na_3[AlF_6] \), aluminium trifluoride (AlF3) precipitates out. This happens because BF3 is a stronger Lewis acid than AlF3. BF3 readily accepts fluoride ions from the \( [AlF_6]^{3-} \) complex to form \( [BF_4]^- \), which is a very stable complex. This removal of fluoride ions causes the equilibrium for the formation of \( [AlF_6]^{3-} \) to shift backward, leading to the precipitation of insoluble AlF3.
\( Na_3[AlF_6] + BF_3 \rightarrow AlF_3 \downarrow + Na_3BF_6 \)
In simple words: AlF3 doesn't dissolve in plain HF but does in NaF because it forms a complex. But if you bubble BF3 through that solution, AlF3 comes out again because BF3 is a stronger acid and takes the fluoride ions away.

Exam Tip: This question tests your understanding of Lewis acid-base interactions and complex ion formation. Remember that a stronger Lewis acid can displace a weaker Lewis acid from its complex, leading to precipitation or dissolution changes.

 

Question 17. Suggest a reason as to why CO is poisonous.
Answer: Carbon monoxide (CO) is poisonous because it has an extremely strong ability to form a complex with haemoglobin in red blood cells. This complex, called carboxyhaemoglobin, is approximately 300 times more stable than the oxygen-haemoglobin complex (oxyhaemoglobin). When CO binds to haemoglobin, it essentially blocks oxygen from binding, preventing the red blood corpuscles from carrying oxygen to the body's tissues and organs. This lack of oxygen eventually leads to cell death and ultimately, death of the organism.
\( Haemoglobin + CO \rightarrow Carboxyhaemoglobin \)
In simple words: Carbon monoxide is dangerous because it sticks to the blood's haemoglobin much better than oxygen does. This stops oxygen from getting to your body, which is deadly.

Exam Tip: For questions about the toxicity of CO, always mention its high affinity for haemoglobin compared to oxygen, leading to the formation of stable carboxyhaemoglobin and subsequent oxygen deprivation in the body.

 

Question 18. How is excessive content of CO2 responsible for global warming?
Answer: Carbon dioxide gas itself is not poisonous. It is normally present in the atmosphere at about 0.03% by volume. Green plants remove atmospheric CO2 through photosynthesis, converting it into carbohydrates like glucose.
\( 6CO_2 + 12H_2O \xrightarrow{hv/Chlorophyll} C_6H_{12}O_6 + 6O_2 + 6H_2O \)
However, the increasing combustion of fossil fuels and the decomposition of limestone for cement production in recent years have significantly raised the CO2 content in the atmosphere. Carbon dioxide is a greenhouse gas, meaning it traps heat within the Earth's atmosphere. An excessive amount of CO2 strengthens this greenhouse effect, causing the atmosphere's temperature to rise. This global temperature increase, known as global warming, can have severe environmental consequences, including climate change, rising sea levels, and extreme weather events.
In simple words: Too much CO2 in the air, mainly from burning fossil fuels, traps more heat from the sun. This makes the Earth's temperature go up, which is called global warming, and it causes many problems.

Exam Tip: When explaining global warming due to CO2, connect the increase in atmospheric CO2 (from human activities) to its role as a greenhouse gas that traps infrared radiation, leading to an overall rise in Earth's temperature.

 

Question 19. Explain structures of diborane and boric acid.
Answer: Structure of diborane \( (B_2H_6) \): Diborane has a unique structure. The four terminal hydrogen atoms and the two boron atoms lie in the same plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B-H bonds are normal two-centre, two-electron bonds. However, the two bridge (B-H-B) bonds are different; they are three-centre, two-electron bonds, often called 'banana bonds'. These bonds involve three atoms (two boron and one hydrogen) sharing two electrons, making the molecule electron-deficient but stable.
Structure of boric acid \( (H_3BO_3) \): Orthoboric acid, or simply boric acid, has a layered structure. In this structure, individual \( BO_3 \) units (boron bonded to three hydroxyl groups) are linked together by hydrogen bonds. Each boron atom is \( sp^2 \) hybridized and is surrounded by three oxygen atoms in a trigonal planar fashion. These planar \( BO_3 \) units form extensive two-dimensional layers through a network of weak hydrogen bonds between the hydrogen of one \( -OH \) group and the oxygen of an adjacent \( BO_3 \) unit. These layers are stacked, resulting in a flaky structure.
In simple words: Diborane has a special "banana bond" structure where two hydrogen atoms bridge the two boron atoms. Boric acid forms flat layers where its main units are held together by hydrogen bonds, like stacked sheets.

Exam Tip: For diborane, always highlight the presence of 'banana bonds' (three-centre, two-electron bonds) and distinguish them from terminal B-H bonds. For boric acid, emphasize its layered structure formed by extensive hydrogen bonding between planar \( BO_3 \) units.

 

Question 20. What happens when
(a) Borax is heated strongly
(b) Boric acid is added to water
(c) Aluminium is treated with dilute NaOH
(d) BF3 is treated with ammonia?
Answer:
(a) On heating borax \( (Na_2B_4O_7 \cdot 10H_2O) \) strongly, it first loses its water molecules and swells up significantly. Upon further intense heating, it transforms into a transparent liquid. This liquid then solidifies into a glass-like material known as a borax bead, which is a mixture of sodium metaborate \( (NaBO_2) \) and boron trioxide \( (B_2O_3) \).
\( Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 \rightarrow 2NaBO_2 + B_2O_3 \)
(Sodium metaborate, Boric anhydride)
(b) Boric acid \( (H_3BO_3) \) is sparingly soluble in cold water but shows better solubility in hot water. It does not function as a proton donor (i.e., it's not a protic acid). Instead, it acts as a weak monobasic Lewis acid by accepting a pair of electrons from a hydroxyl \( (OH^-) \) ion, primarily from water molecules, generating \( [B(OH)_4]^- \) and \( H_3O^+ \).
\( B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+ \)
(c) When aluminium is treated with dilute NaOH solution, it reacts to liberate hydrogen gas. This reaction showcases the amphoteric nature of aluminium, as it reacts with a strong base.
\( 2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na^+[Al(OH)_4]^-(aq) + 3H_2(g) \)
(Sodium tetrahydroxoaluminate (III) ion)
(d) BF3, being a strong Lewis acid, reacts readily with ammonia \( (NH_3) \), which acts as a Lewis base (electron pair donor). They form an adduct or complex where ammonia donates its lone pair of electrons to the electron-deficient boron in BF3. This complex has a tetrahedral structure around the boron atom.
\[ F \quad H \\ \quad \backslash \ / \\ F - B \quad + \quad N - H \\ \quad / \quad \backslash \\ F \quad H \] This forms the adduct: \( F_3B \leftarrow NH_3 \)
In simple words: (a) Borax swells and turns into a clear, glassy bead when heated strongly. (b) Boric acid dissolves in water and acts as a weak acid by taking electrons from water's hydroxide. (c) Aluminium reacts with dilute NaOH to release hydrogen gas. (d) BF3 combines with ammonia to form a new complex because BF3 wants electrons and ammonia has them to share.

Exam Tip: For chemical reactions, ensure correct reactants and products, and balance the equations. Also, understand the underlying chemical principles (e.g., dehydration, Lewis acid-base reactions, amphoteric behavior) for a complete explanation.

 

Question 21. Explain the following reactions
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper;
(b) Silicon dioxide is treated with hydrogen fluoride;
(c) CO is heated with ZnO;
(d) Hydrated alumina is treated with aqueous NaOH solution.
Answer:
(a) When silicon is heated with methyl chloride \( (CH_3Cl) \) at a high temperature (around 570 K) in the presence of copper powder as a catalyst, various types of methyl-substituted chlorosilanes are formed. These include MeSiCl3, Me2SiCl2, and Me3SiCl, with Me2SiCl2 being the main product, along with smaller amounts of Me3Si. Further hydrolysis of dimethyl dichlorosilane \( ((CH_3)_2SiCl_2) \) yields straight-chain polymers, which are silicones.
\( 2CH_3Cl + Si \xrightarrow{Cu-Powder, 570K} (CH_3)_2SiCl_2 \)
Then, \( (CH_3)_2SiCl_2 + 2H_2O \rightarrow (CH_3)_2Si(OH)_2 + 2HCl \)
Finally, \( n(CH_3)_2Si(OH)_2 \rightarrow HO-[Si(CH_3)_2-O]_n-H + nH_2O \)
(b) When silicon dioxide \( (SiO_2) \) is treated with hydrogen fluoride (HF), silicon tetrafluoride \( (SiF_4) \) is formed along with water. This reaction is often used to etch glass, as glass is primarily silicon dioxide.
\( SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O \)
(c) When carbon monoxide (CO) is heated with zinc oxide (ZnO), CO acts as a reducing agent. It reduces zinc oxide to zinc metal, while carbon monoxide itself gets oxidised to carbon dioxide.
\( ZnO + CO \xrightarrow{\Delta} Zn + CO_2 \)
(d) When hydrated alumina \( (Al_2O_3 \cdot 2H_2O) \) is treated with an aqueous NaOH solution, a reaction occurs that forms sodium meta-aluminate \( (NaAlO_2) \) and water. This reaction demonstrates the amphoteric nature of alumina.
\( Al_2O_3 \cdot 2H_2O + 2NaOH \rightarrow 2NaAlO_2 + 3H_2O \)
(Sodium meta-aluminate)
In simple words: (a) Heating silicon and methyl chloride with copper makes different silanes, which can then form long chains called silicones. (b) Silicon dioxide reacts with hydrogen fluoride to make silicon tetrafluoride and water. (c) Carbon monoxide reacts with zinc oxide to turn it into zinc metal. (d) Hydrated alumina reacts with NaOH to form sodium meta-aluminate and water.

Exam Tip: For each reaction, identify the type of reaction (e.g., reduction, hydrolysis, complexation), the key reactants, and the major products. Pay close attention to conditions like temperature and catalysts, as they are crucial.

 

Question 22. Give reasons:
(i) Conc. HNO3 can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv) Diamond is used as an abrasive
(v) Aluminium alloys are used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.
Answer:
(i) Concentrated nitric acid (HNO3) can be safely transported in aluminium containers because HNO3 renders aluminium passive. It forms a thin, protective, and non-porous oxide layer (Al2O3) on the surface of the aluminium. This oxide layer prevents further reaction between the acid and the metal, effectively passivating the aluminium.
\( 2Al + 3O \rightarrow Al_2O_3 \) (passive protective layer)
(ii) A mixture of dilute NaOH and aluminium pieces is used to open blocked drains because the reaction between aluminium and NaOH liberates hydrogen gas. The rapid production of gas creates pressure and turbulence, which helps dislodge and push away blockages in the drain.
\( 2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na^+[Al(OH)_4]^-(aq) + 3H_2(g) \)
(iii) Graphite is used as a lubricant because of its unique layered structure. Its layers can easily slip over each other due to weak van der Waals forces between them. This makes graphite very soft and slippery. It is particularly useful as a dry lubricant in machinery operating at high temperatures where oil-based lubricants might degrade.
(iv) Diamond is used as an abrasive because it is the hardest known substance on Earth. This extreme hardness comes from its rigid, three-dimensional network of carbon atoms, each \( sp^3 \) hybridised and strongly covalently bonded to four others. This robust structure makes diamond highly effective for cutting, grinding, and polishing other materials.
(v) Aluminium alloys are used to make aircraft bodies because they offer a superior combination of properties. When alloyed with other metals, aluminium becomes significantly more tensile and stronger than pure aluminium. Crucially, aluminium and its alloys are lightweight, which is essential for aircraft to maximize fuel efficiency and payload capacity. They can also be formed into various shapes like pipes, tubes, rods, wires, plates, or foils.
(vi) Aluminium utensils should not be kept in water overnight because in the prolonged presence of water and oxygen, a thin layer of aluminium oxide (Al2O3) forms on their surface. This Al2O3 layer can react with certain food components, and more importantly, it is a toxic substance. Consuming food prepared or stored in such utensils might pose health risks.
(vii) Aluminium wire is used to make transmission cables because aluminium possesses high electrical conductivity. On a weight-for-weight basis, aluminium's electrical conductivity is approximately twice that of copper, making it a lighter and often more cost-effective choice for transmitting electricity over long distances despite copper having higher absolute conductivity per volume.
In simple words: (i) Concentrated nitric acid can be stored in aluminium containers because aluminium forms a protective layer. (ii) A mix of NaOH and aluminium helps clear drains by releasing gas. (iii) Graphite is a lubricant because its layers slide easily. (iv) Diamond is an abrasive because it's the hardest material. (v) Aluminium alloys are used in aircraft because they are strong and light. (vi) Aluminium utensils shouldn't be left in water overnight because a toxic layer can form. (vii) Aluminium wire is used for power lines because it conducts electricity well and is lighter than copper.

Exam Tip: When providing reasons for practical applications or behaviors of elements, always link the property directly to the underlying chemical structure, bonding, or reactivity. For example, hardness to strong covalent networks, conductivity to free electrons, or corrosion resistance to passivation layers.

 

Question 23. Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon.
Answer: The electronic configuration of Carbon (C) is \( 1s^2, 2s^2 2p^2 \), and for Silicon (Si) it is \( 1s^2, 2s^2 2p^6, 3s^2 3p^2 \). As we move from Carbon to Silicon in group 14, there is a sharp increase in the covalent radius. For example, the covalent radius jumps from 77 pm for Phosphorous (P) to 118 pm for silicon. This rise in the size of the atom leads to a significant drop in the ionisation enthalpy value when moving from carbon to silicon.
In simple words: As you go from carbon to silicon, the atoms get much bigger. This larger size makes it easier to remove an electron, so the energy needed for that (ionization enthalpy) drops a lot.

Exam Tip: Remember that ionization enthalpy generally decreases down a group due to increasing atomic size and greater shielding effect, making outer electrons easier to remove.

 

Question 24. How would you explain the lower atomic radius of Ga as compared to Al?
Answer: The electronic configuration for Aluminium (Al) is \( [Ne] 3s^2 3p^1 \), while for Gallium (Ga) it is \( [Ar] 3d^{10} 4s^2 4p^1 \). Gallium possesses 10 extra d-electrons in its inner shell or core. These d-electrons provide a relatively poor screening effect, meaning they do not effectively shield the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm), despite gallium being lower in the group.
In simple words: Gallium has extra d-electrons that do not block the nuclear charge very well. This means the outer electrons are pulled in closer, making gallium's atomic radius smaller than aluminium's.

Exam Tip: The poor shielding effect of d- and f-electrons often leads to unexpected trends in atomic radii and other properties in transition elements and post-transition metals.

 

Question 25. What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite.
or
What is the impact of structure on physical properties of two allotropes?
Answer: When an element exists in two or more different physical forms, while having similar chemical properties, it is said to be present in allotropes or allotropic modifications. Carbon exists in two common allotropic forms: diamond and graphite.

Structure of Diamond: In diamond, each carbon atom is \( sp^3 \) hybridized and is tetrahedrally bonded to four other carbon atoms through strong C-C \( sp^3-sp^3 \) covalent bonds. This creates a highly rigid and extensive three-dimensional network structure. The C-C bond length is 154 pm.

Structure of Graphite: In graphite, each carbon atom is \( sp^2 \) hybridized, forming a sheet-like (layered) structure. These layers consist of numerous benzene rings fused together. The layers are held by weak van der Waals forces, while within each layer, strong C-C bonds with a length of 142 pm exist. The distance between layers is 340 pm.

Impact of the two structures on physical properties:

• Diamond is the hardest known substance and can cut glass, while graphite is so soft that it marks paper.
• Diamond is a transparent substance, whereas graphite is opaque.
• Diamond cannot conduct electricity because it lacks free electrons, whereas graphite can conduct electricity due to the presence of free electrons.
• The specific gravity of diamond is 3.5, while for graphite it is 2.5.

In simple words: Allotropes are different forms of the same element that look different but react similarly. Diamond is super hard and clear because its carbon atoms are tightly locked in a 3D network, but it doesn't conduct electricity. Graphite is soft, dark, and slippery because its carbon atoms are in flat layers that can slide, and it does conduct electricity.

Exam Tip: When discussing allotropes, always define the term first. For specific allotropes like diamond and graphite, describe their bonding, hybridization, and how these structural differences lead to distinct physical properties.

 

Question 26. Classify following oxides as neutral, acidic, basic or amphoteric : CO, \( B_2O_3 \), \( SiO_2 \), \( CO_2 \), \( Al_2O_3 \), \( PbO_2 \), \( Tl_2O_3 \).
(b) Write suitable chemical equations to show their nature.
Answer:
(a) Classification of oxides:

• CO is neutral.
• \( B_2O_3 \), \( SiO_2 \), \( CO_2 \) are acidic oxides.
• \( PbO_2 \), \( Al_2O_3 \) are amphoteric oxides.
• \( Tl_2O_3 \) is basic oxide.

(b) Chemical equations showing their nature:

• \( B_2O_3 + CuO \rightarrow Cu(BO_2)_2 \)
• \( SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H_2O \)
• \( CO_2 + H_2O \rightarrow H_2CO_3 \)
• \( PbO_2 + 4HCl \rightarrow PbCl_4 + 2H_2O \)
• \( PbO_2 + 2NaOH \rightarrow Na_2PbO_3 + H_2O \)
• \( Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O \)
• \( Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O \)
• \( Tl_2O_3 + 3H_2SO_4 \rightarrow Tl_2(SO_4)_3 + 3H_2O \)

In simple words: Some oxides are neutral (like CO), some are acidic (like those of boron, silicon, carbon), some are basic (like thallium oxide), and some can act as both acid and base (amphoteric, like lead and aluminium oxides). The equations show how these oxides react with acids or bases to prove their nature.

Exam Tip: For classifying oxides, remember that non-metal oxides are typically acidic, metal oxides are generally basic, and metalloid oxides or oxides of certain metals in higher oxidation states can be amphoteric or neutral.

 

Question 27. In some of the reactions thallium resembles aluminium, whereas in others it resembles with group-I metals. Support this statement by giving some evidences.
Answer: Thallium exhibits dual characteristics, resembling both aluminium and group-I metals in various reactions.

Resemblance with Aluminium: Like aluminium compounds, thallium also shows an oxidation state of +3 in compounds such as \( Tl_2O_3 \). This similarity occurs because both Aluminium (Al) and Thallium (Tl) belong to group 13 and possess \( ns^2np^1 \) electronic configuration in their valence shells.

Resemblance with Group-I Metals: Similar to alkali metals (Group-I), thallium shows an oxidation state of +1 in its halides. For instance, the reaction \( 2Tl + Cl_2 \rightarrow 2TlCl \) illustrates this. The preference for a +1 oxidation state in heavier elements like thallium is due to the inert pair effect. This effect causes the \( ns^2 \) electrons to remain paired and not participate in bond formation, leading to a more stable +1 oxidation state, much like alkali metals.

In simple words: Thallium acts like aluminium by forming compounds with a +3 charge, as both are in the same group. But it also acts like group 1 metals (like sodium) by forming compounds with a +1 charge. This happens because the heavier thallium atom prefers to keep two of its outer electrons paired up, an effect called the "inert pair effect," making the +1 state very stable.

Exam Tip: Remember the inert pair effect becomes more pronounced for heavier elements in p-block groups, causing lower oxidation states to become more stable than expected.

 

Question 28. When metal X is treated with sodium hydroxide, a wsate precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.
Answer: Based on the reactions described, we can identify the substances as follows:

X is metal aluminium (Al).
Compound A is aluminium hydroxide, \( Al(OH)_3 \).
Soluble complex B is sodium tetrahydroxoaluminate(III) ion, \( Na^+[Al(OH)_4]^- \).
Compound C is aluminium chloride, \( AlCl_3 \).
Compound D is aluminium oxide, \( Al_2O_3 \).

Supporting Equations:

1. Reaction of X (Aluminium) with NaOH to form A (Aluminium Hydroxide precipitate):
\( X(Al) \xrightarrow{NaOH} Al(OH)_3 \downarrow \) (White precipitate A)

2. Dissolution of A (Aluminium Hydroxide) in excess NaOH to form B (Soluble Complex):
\( Al(OH)_3 + NaOH \rightarrow Na^+[Al(OH)_4]^- \) (Soluble complex B - Sodium tetrahydroxoaluminate (III) ion)

3. Dissolution of A (Aluminium Hydroxide) in dilute HCl to form C (Aluminium Chloride):
\( Al(OH)_3 + 3HCl(aq) \rightarrow AlCl_3 + 3H_2O \) (Compound C)

4. Strong heating of A (Aluminium Hydroxide) to form D (Aluminium Oxide):
\( 2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O \) (Compound D)

5. Overall reaction for metal X (Aluminium) with NaOH (liberation of \( H_2 \)):
\( 2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na^+[Al(OH)_4]^-(aq) + 3H_2(g) \)
or simply put: \( 2Al(s) + 2NaOH(aq) \rightarrow 2NaAlO_2 + H_2(g) \) (Sodium meta-aluminate)

In simple words: Metal X is aluminium. When it reacts with NaOH, it first forms a white solid (A), which is aluminium hydroxide. This solid then dissolves in more NaOH to make a clear liquid (B). If solid A reacts with HCl, it forms a different compound (C), aluminium chloride. Heating solid A very strongly turns it into (D), aluminium oxide, which is used to get the metal.

Exam Tip: Aluminium is an amphoteric metal, meaning its hydroxide can react with both acids and bases. This dual reactivity is a key property often tested in such identification questions.

 

Question 29. What do you understand by
(a) Inert pair effect
(b) allotropy
(c) catenation.
Answer:
(a) Inert pair effect: The heavier elements of groups 13, 14, and 15 often show an oxidation state that is two less than the number of valence electrons they possess. This happens because as we move down these groups, the two electrons in the \( ns^2 \) orbital prefer to remain paired and do not participate in chemical bond formation. Such a pair of electrons is called an inert pair, and this phenomenon is known as the inert pair effect. For example:

• Group 13: Electronic configuration \( ns^2np^1 \). Last element (Tl) prefers +1 oxidation state over +3.
• Group 14: Electronic configuration \( ns^2np^2 \). Last element (Pb) prefers +2 oxidation state over +4.
• Group 15: Electronic configuration \( ns^2np^3 \). Last element (Bi) prefers +3 oxidation state over +5.

(b) Allotropy: Allotropy refers to the property of an element existing in two or more different forms that have distinct physical properties but similar chemical properties. These different forms are called allotropes. For example, phosphorus appears in three allotropic forms: white, red, and black. Sulfur also has two different solid forms: rhombic sulfur and monoclinic sulfur.

(c) Catenation: Catenation is the property of self-linking of atoms of the same element through strong covalent bonds to form chains or rings. This property is most prominent in carbon, which forms long C-C chains and rings in hydrocarbons and many other organic compounds. The strong C-C covalent bonds enable carbon to exhibit extensive catenation.
In simple words: (a) The "inert pair effect" means that for heavy elements, the two s-orbital electrons don't always join in chemical bonds, making a lower charge more common. (b) "Allotropy" is when an element can exist in different physical forms, like diamond and graphite for carbon. (c) "Catenation" is when atoms of the same element, especially carbon, link together in long chains or rings.

Exam Tip: For definitions like these, always provide a clear explanation and relevant examples to illustrate your understanding. For inert pair effect, linking it to the s-electrons and heavier elements is crucial.

 

Question 30. A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When cone. \( H_2SO_4 \) is added to a hot solution of X, white crystal of an acid Z separates out
Write equations for all the above reactions and identify X, Y and Z.
Answer: The salt X is BORAX (\( Na_2B_4O_7 \)).

(i) Its aqueous solution in water is alkaline due to hydrolysis. Borax is the salt of a strong base (NaOH) and a weak acid (\( H_2B_4O_7 \)).
\( Na_2B_4O_7 + 2H_2O \rightarrow 2NaOH + H_2B_4O_7 \)
(Strong base) (Weak acid)
Such a solution turns litmus alkaline.

(ii) It swells up to a glassy material Y on strong heating.
\( Na_2B_4O_7 \cdot 10H_2O \xrightarrow{Heat} Na_2B_4O_7 \)
Borax loses its water of crystallisation on heating and swells up to form a puffy mass of anhydrous borax, \( Na_2B_4O_7 \).
Further heating changes it into a transparent glassy bead (Y), which is a mixture of sodium metaborate and boric anhydride:
\( Na_2B_4O_7 \xrightarrow{Heat} 2NaBO_2 + B_2O_3 \)
(Sodium metaborate) (Boric anhydride)

(iii) When concentrated \( H_2SO_4 \) is added to a hot solution of X, white crystals of an acid Z separate out.
\( Na_2B_4O_7 + H_2SO_4 \rightarrow Na_2SO_4 + H_2B_4O_7 \)
\( H_2B_4O_7 + 5H_2O \rightarrow 4H_3BO_3 \)
Therefore, Z is boric acid (\( H_3BO_3 \)), whose white crystals separate out when concentrated \( H_2SO_4 \) is added to the hot solution of X.

Summary of identifications:
X is borax (\( Na_2B_4O_7 \)).
Y is anhydrous borax and a glassy mixture of \( 2NaBO_2 + B_2O_3 \).
Z is boric acid (\( H_3BO_3 \)).
In simple words: The salt X is borax. When borax is dissolved in water, it makes the solution alkaline. If you heat borax strongly, it puffs up and then melts into a clear, glassy substance (Y). If you add strong sulfuric acid to hot borax solution, white crystals of boric acid (Z) appear.

Exam Tip: This question tests your knowledge of borax chemistry, including its hydrolysis, thermal decomposition, and reaction with strong acids. Knowing the common names and formulas is essential.

 

Question 31. Write balanced equations for :
(i) \( BF_3 + LiH \)
(ii) \( B_2H_6 + H_2O \)
(iii) \( NaH + B_2H_6 \)
(iv) \( H_3BO_3 \)
(v) \( Al + NaOH \)
(vi) \( B_2H_6 + NH_3 \)
Answer:
(i) \( 2BF_3 + 6LiH \rightarrow B_2H_6 + 6LiF \)
(ii) \( B_2H_6 + 6H_2O(l) \rightarrow 2H_3BO_3(aq) + 6H_2(g) \)
(iii) \( 2NaH + B_2H_6 \rightarrow 2Na^+[BH_4]^- \)
(iv) \( H_3BO_3 \xrightarrow{Heat\ (370K)} HBO_2 + H_2O \)
(v) \( 2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2 \)
(vi) \( 3B_2H_6 + 6NH_3 \xrightarrow{Lowtemp.\ (473K)} 3[BH_2(NH_3)_2]^+ [BH_4]^- \)

In simple words: These are balanced chemical equations for various reactions. They show how different chemicals combine or break apart under specific conditions, like when boron trifluoride reacts with lithium hydride, or diborane reacts with water or ammonia.

Exam Tip: When balancing equations, always ensure that the number of atoms for each element is the same on both the reactant and product sides. Pay close attention to subscripts and coefficients.

 

Question 32. Give one method for industrial preparation and one for laboratory preparation of CO and \( CO_2 \) each.
Answer:
Carbon Monoxide (CO):

• Laboratory Preparation: On a small scale, pure CO is prepared by dehydrating formic acid (HCOOH) using concentrated \( H_2SO_4 \) at 373K.
  \( HCOOH \xrightarrow{Conc. H_2SO_4\ (373K)} H_2O + CO \)
• Industrial Preparation: On a commercial scale, CO is prepared by passing steam over hot coke.
  \( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)

Carbon Dioxide (\( CO_2 \)):

• Laboratory Preparation: In the laboratory, \( CO_2 \) is prepared by the action of dilute HCl acid on calcium carbonate (\( CaCO_3 \)).
  \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \)
• Industrial Preparation: On a commercial scale, it is prepared by heating limestone.
  \( CaCO_3(s) \xrightarrow{Heat\ (1173K)} CaO(s) + CO_2(g) \)

In simple words: Carbon monoxide can be made in the lab by taking water out of formic acid, or in factories by passing steam over hot carbon. Carbon dioxide can be made in the lab by mixing calcium carbonate with acid, or in factories by heating limestone.

Exam Tip: For preparation methods, always mention both the reactants and the specific conditions (e.g., temperature, catalyst) required for the reaction, along with the balanced chemical equation.

 

Question 33. An aqueous solution of ______.
(a) neutral
(c) basic
Answer: (c) It is basic.
In simple words: An aqueous solution means something dissolved in water. When this solution is tested, it shows basic properties.

Exam Tip: Pay attention to the context of the question (e.g., what specific compound is being discussed) to determine the pH of its aqueous solution.

 

Question 34. Boric acid is polymeric due to ______.
(a) Its acidic nature
(b) The presence of hydrogen bonds
(c) Its monobasic nature
(d) Its geometry.
Answer: (b) The presence of hydrogen bonds
In simple words: Boric acid forms long chains or networks because its molecules connect to each other using hydrogen bonds, making it a polymer.

Exam Tip: Hydrogen bonding is a significant intermolecular force that can lead to properties like higher boiling points, solubility, and polymeric structures in many compounds.

 

Question 35. The type of hybridisation of boron in diborane is ______.
(a) sp
(b) \( sp^2 \)
(c) \( sp^3 \)
(d) \( dsp^2 \).
Answer: (c) \( sp^3 \) hybridisation of B in diborane \( B_2H_6 \).
In simple words: In the molecule diborane, each boron atom uses \( sp^3 \) hybridization, which means it forms four bonds with a tetrahedral shape.

Exam Tip: For hybridization questions, determine the number of sigma bonds and lone pairs around the central atom, as this total will dictate the hybridization state.

 

Question 36. Thermodynamically the most stable form of carbon is ______.
(a) Diamond
(b) Graphite
(c) Fullerenes
(d) Coal.
Answer: (b) Graphite is thermodynamically most stable allotrope of carbon.
In simple words: Of all the different forms carbon can take, graphite is the most stable one when thinking about energy, meaning it needs the least energy to exist.

Exam Tip: While diamond is harder, graphite is the more thermodynamically stable allotrope of carbon at standard conditions, meaning it has a lower free energy.

 

Question 37. Elements of group 14 ______.
(a) exhibit oxidation state of + 4 only .
(b) exhibit oxidation state of + 2 and + 4
(c) form \( M^{2-} \) and \( M^{2+} \) ion
(d) form \( M^{2+} \) and \( M^{4+} \) ions.
Answer: (b) Elements of group 14 exhibit oxidation states of + 2 and + 4.
In simple words: Elements in group 14 can usually form chemical bonds where they have a charge of either +2 or +4.

Exam Tip: Remember that for p-block elements, both the group oxidation state (e.g., +4 for group 14) and an oxidation state two units lower (e.g., +2 for group 14, due to the inert pair effect) are common.

 

Question 38. If the starting material for the manufacture of silicones is \( RSiCl_3 \), write the structure of the product formed.
Answer: If the starting material for the manufacture of silicones is \( RSiCl_3 \), cross-linked polymers are obtained as described below:

Initially, \( RSiCl_3 \) undergoes hydrolysis. The chlorine atoms are replaced by hydroxyl (-OH) groups:
\( RSiCl_3 + 3H_2O \xrightarrow{Heat\ (1173K)} RSi(OH)_3 + 3HCl \)

The resulting trihydroxy compound, \( RSi(OH)_3 \), then undergoes condensation polymerization. This means the -OH groups on different \( RSi(OH)_3 \) units react with each other, losing water molecules to form Si-O-Si linkages. Since each silicon atom has three -OH groups (and therefore three potential sites for linkage), this leads to the formation of a complex three-dimensional, cross-linked polymeric structure.

This cross-linked silicone polymer can be represented as:

    O     O     O
    |     |     |
---Si---Si---Si---
    |     |     |
    R     R     R
    |     |     |
---Si---Si---Si---
    |     |     |
    O     O     O
   / \   / \   / \
  O   O O   O O   O
   \ /   \ /   \ /
---Si---Si---Si---
    |     |     |
    R     R     R

This shows a simplified view where R is an alkyl group (like \( CH_3 \)), and the silicon atoms are connected by oxygen bridges in a complex, branched network.

In simple words: When \( RSiCl_3 \) is used to make silicones, it first reacts with water to form \( RSi(OH)_3 \). Because this compound has three reactive -OH groups, it can link up with many other similar units, forming strong bonds and creating a complicated, three-dimensional, cross-linked polymer.

Exam Tip: Understanding the number of functional groups (like Cl or OH) on the starting silane determines the type of polymer formed: \( R_2SiCl_2 \) yields linear polymers, \( RSiCl_3 \) yields cross-linked polymers, and \( R_3SiCl \) acts as a chain terminator.