RBSE Solutions Class 11 Chemistry Chapter 11 p-Block Elements

Question 1. Which of the following is a Lewis acid?
(a) PCl3
(b) AlCl3
(c) NCl3
(d) AsCl3
Answer: (b) AlCl3
In simple words: Aluminium chloride (AlCl3) is a Lewis acid because it has an empty orbital and can accept a pair of electrons.

🎯 Exam Tip: Remember that Lewis acids are electron-pair acceptors. Look for compounds with incomplete octets or available empty orbitals.

Question 3. The nature of aqueous solution of borax is :
(a) Acidic
(b) Neutral
(c) Amphoteric
(d) Basic
Answer: (d) Basic
In simple words: When borax is dissolved in water, the resulting solution is basic.

🎯 Exam Tip: Recall that borax hydrolyzes in water to form a strong base (NaOH) and a weak acid (boric acid), leading to an alkaline solution.

Question 4. Which of the following does not give Borax Bead test?
(a) Manganese salts
(b) Barium salts
(c) Nickel salts
(d) Cobalt salts
Answer: (b) Barium salts
In simple words: Barium salts do not produce a colored bead in the Borax Bead test, unlike salts of manganese, nickel, and cobalt.

🎯 Exam Tip: The Borax Bead test is used to identify certain metal ions by the characteristic colored beads they form. Metals like Barium do not form colored borate beads.

Question 5. What is the formula of dry ice from the following:
(a) CO2
(b) CO
(c) SiO2
(d) Al2O3
Answer: (a) CO2
In simple words: Dry ice is the solid form of carbon dioxide, so its chemical formula is `CO2`.

🎯 Exam Tip: Remember that dry ice sublimes directly from solid to gas, making it useful for refrigeration without liquid mess.

Rbse Class 11 Chemistry Chapter 11 Very Short Answer Type Questions

Question 7. Write the general electronic configuration of p-block elements.
Answer: The general electronic configuration of p-block elements is \( ns^2 np^{1-6} \).
In simple words: P-block elements have their outermost electrons arranged in a way that includes two 's' electrons and between one and six 'p' electrons.

🎯 Exam Tip: Knowing the general electronic configuration helps to understand the valency and chemical properties of elements in a block.

Question 8. Which compound of boron is used to make bullet proof jackets?
Answer: Boron nitride (BN) is used to make bullet proof jacket.
In simple words: Boron nitride, which is abbreviated as BN, is the compound of boron that is used for making bullet-proof jackets.

🎯 Exam Tip: Boron nitride exists in various forms, some of which are very hard and have high thermal stability, making them suitable for protective materials.

Question 9. Write the names and formula of two ores of aluminium.
Answer: Bauxite: \( Al_2O_3 \cdot 2H_2O \)
Cryolite: \( Na_3AlF_6 \)
In simple words: Two important ores from which aluminium is extracted are Bauxite, which is hydrated aluminium oxide, and Cryolite, which is a sodium aluminium fluoride.

🎯 Exam Tip: For ores, remember both the common name and the chemical formula as they are frequently asked in exams.

Question 10. Draw the tetrahedral dimer structure of AlCl3?
Answer: Aluminium chloride (AlCl3) forms a dimer, \( Al_2Cl_6 \), in which two chlorine atoms bridge the two aluminium atoms, forming a tetrahedral arrangement around each aluminium.
In simple words: Aluminium chloride molecules link up in pairs, forming a dimer. In this dimer, two chlorine atoms act as bridges between the two aluminium atoms, creating a structure where each aluminium has a tetrahedral arrangement around it.

🎯 Exam Tip: For dimer structures like \( Al_2Cl_6 \), remember to show the bridging atoms and the correct geometry around the central atoms.

Question 11. Explain the structure of diborane.
Answer: Diborane (\( B_2H_6 \)) has a unique structure involving two 3-center-2-electron (banana) bonds. In this structure, two boron atoms are bridged by two hydrogen atoms, while each boron atom is also bonded to two terminal hydrogen atoms.
In simple words: Diborane has a unique "banana bond" structure where two boron atoms are connected by two hydrogen atoms that act as bridges. Each boron atom also has two other hydrogen atoms attached directly to it.

🎯 Exam Tip: For diborane, always highlight the two types of hydrogen atoms (terminal and bridging) and the 3-center-2-electron bonds.

Question 12. Write two uses of borane and aluminium.
Answer:
Uses of Borane
• It is used as a reducing agent in organic reactions.
• It is used as a fuel for supersonic rockets.
Uses of Aluminium
• It is used in packing of materials.
• It is used in making utensils.
In simple words: Borane compounds are useful for making organic reactions happen and can power supersonic rockets. Aluminium is commonly used to package food and make kitchen utensils.

🎯 Exam Tip: When listing uses, provide specific applications rather than general properties to score full marks.

Question 13. Write the number of six membered and five membered rings in Fullerenes.
Answer: In Fullerenes, the number of six membered rings is 20 and five membered rings is 12.
In simple words: Fullerenes, like the Buckyball, have 20 rings made of six carbon atoms each, and 12 rings made of five carbon atoms each.

🎯 Exam Tip: The C60 fullerene (Buckminsterfullerene) is the most common example; remember its specific number of 5-membered and 6-membered rings.

Question 15. Explain the hybridisation of carbon present in diamond and graphite.
Answer:
Hybridisation of Carbon in Diamond: In diamond, each carbon atom undergoes \( sp^3 \) hybridisation and is linked to four other carbon atoms using hybridised orbitals in a tetrahedral fashion. The C-C bond length is 154 pm.
Hybridisation of Carbon in Graphite: In graphite, each layer is composed of planar hexagonal rings of carbon atoms. Each carbon atom in a hexagonal ring undergoes \( sp^2 \) hybridisation and makes three sigma bonds with three neighbouring carbon atoms. The fourth electron forms a \( \pi \)-bond. C-C bond length is 14.15 pm.
In simple words: In diamond, carbon atoms use \( sp^3 \) hybridisation to form a very strong 3D network, making it hard. In graphite, carbon atoms use \( sp^2 \) hybridisation to form flat hexagonal layers held together by weak forces, making it soft and slippery.

🎯 Exam Tip: Focus on the different hybridisation states (\( sp^3 \) for diamond, \( sp^2 \) for graphite) and how they explain the distinct physical properties.

Question 16. The atomic radius of Ga is less than atomic radius of Al. Why?
Answer: The atomic radius of Ga is less than atomic radius of Al, because in between Ga and Al, there are ten elements of the first transition series which have electrons in inner d-orbital. The inner d-orbital do not shield the nucleus effectively due to their shape and poor penetration power. As a result, the effective nuclear charge in Ga becomes more than in Al and atomic radius of Ga is less than Al.
In simple words: Gallium's (Ga) atomic radius is smaller than Aluminium's (Al) because of the d-block elements found between them. These d-electrons poorly shield the nucleus, increasing the effective nuclear pull on Ga's outer electrons, making its atom smaller.

🎯 Exam Tip: This is an important exception to the general trend of increasing atomic radius down a group. Clearly explain the role of poor shielding by d-electrons.

Question 18. Write the stability of +1 oxidation state of group 13 elements in increasing order.
Answer: The increasing order of stability of \( +1 \) oxidation state of group 13 elements is given below:
Al < Ga < In < Tl (due to inert pair effect).
In simple words: The \( +1 \) oxidation state becomes more stable for Group 13 elements as you go down the group from aluminium to thallium. This is because of the 'inert pair effect', where the 's' electrons are less likely to be involved in bonding.

🎯 Exam Tip: The inert pair effect explains why lower oxidation states become more stable for heavier elements in p-block groups.

Question 19. Write the reaction for amphoteric nature of aluminium.
Answer: Aluminium dissolves in mineral acids and aqueous alkalies, showing its amphoteric nature.
(i) Aluminium dissolves in dilute acid and liberates dihydrogen, showing its basic nature:
\( 2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}_{(aq)} + 6Cl^-_{(aq)} + 3H_2(g) \)
(ii) Aluminium reacts with aqueous alkali and liberates dihydrogen, showing its acidic nature:
\( 2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na^+ [Al(OH)_4]^-(aq) + 3H_2(g) \)
In simple words: Aluminium is amphoteric, meaning it can act as both an acid and a base. It reacts with acids to produce hydrogen gas, showing its basic side. It also reacts with strong bases to produce hydrogen gas, showing its acidic side.

🎯 Exam Tip: To show amphoteric nature, it's crucial to provide balanced chemical equations for reactions with both an acid and a base.

Question 20. Why boron can form \( \mathbf{BF}_{4}^{-} \) but not \( \mathbf{B F}_{6}^{-} \)?
Answer: Boron can form \( \mathbf{BF}_{4}^{-} \) but not \( \mathbf{B F}_{6}^{-} \) because boron does not have d-orbitals, so its maximum covalency is four. Boron can form a maximum of four bonds.
In simple words: Boron can form \( BF_4^- \) but not \( BF_6^- \) because it does not have d-orbitals available. This means boron can only form up to four bonds, limiting its maximum covalency to four.

🎯 Exam Tip: The absence of d-orbitals in period 2 elements is a key reason for their maximum covalency being limited to four.

Question 21. Boric acid is considered as a weak acid. Why?
Answer: Boric acid is considered as a weak acid because it is not able to release \( H^+ \) ions on its own. It receives \( OH^- \) from water, behaving as a Lewis acid.
In simple words: Boric acid is weak because it doesn't give off \( H^+ \) ions by itself. Instead, it acts as a Lewis acid by accepting \( OH^- \) ions from water.

🎯 Exam Tip: Understand that boric acid is a weak monobasic acid and acts as a Lewis acid, not a Brønsted acid, by accepting hydroxide ions.

Rbse Class 11 Chemistry Chapter 11 Short Answer Type Questions

Question 22. Explain the various oxidation states found in group 14 elements.
Answer: The group 14 elements have four electrons in their outermost shell, with an electronic configuration of \( ns^2 np^2 \). They can form \( M^{4+} \) or \( M^{4-} \) ions to show ionic character or exhibit tetravalent covalent nature by sharing four electrons to achieve stability. Forming \( M^{4+} \) or \( M^{4-} \) ions requires a huge amount of energy, which is usually not possible. So, these elements primarily form covalent bonds in both oxidation states. For heavier elements like lead, compounds in the \( +2 \) state are more stable, and those in the \( +4 \) state are strong oxidizing agents.
In simple words: Group 14 elements can have \( +4 \) or \( -4 \) oxidation states. They mostly form covalent bonds because making ions needs a lot of energy. For heavier elements like lead, the \( +2 \) oxidation state is more stable, and \( +4 \) compounds tend to be strong oxidizing agents.

🎯 Exam Tip: Remember the inert pair effect significantly impacts the stability of oxidation states for heavier elements in Group 14, favoring the lower oxidation state.

Question 23. Pbl2 is formed but not Pbl4. Why?
Answer: \( PbI_2 \) is formed but not \( PbI_4 \) because the iodide ion (\( I^- \)) is a reducing agent. It reduces \( Pb(IV) \) to \( Pb(II) \) and gets oxidized itself to \( I_2 \). Therefore, only \( PbI_2 \) is formed.
In simple words: Lead forms \( PbI_2 \) but not \( PbI_4 \) because the iodide ion is a strong reducing agent. It converts \( Pb(IV) \) to the more stable \( Pb(II) \), preventing \( PbI_4 \) from forming.

🎯 Exam Tip: The reducing power of the halide ion plays a crucial role in determining the stability of higher oxidation states of lead.

Question 24. Explain hydrolysis of SiCl4 .
Answer: Silicon tetrachloride (\( SiCl_4 \)) undergoes hydrolytic decomposition by water to form silica and hydrochloric acid.
\( SiCl_4 + 2H_2O \rightarrow SiO_2 + 4HCl \)
In simple words: When silicon tetrachloride reacts with water, it breaks down into silica and hydrochloric acid.

🎯 Exam Tip: For hydrolysis reactions, remember the products formed and provide a balanced chemical equation.

Question 25. Diamond is hard but graphite is soft and slippery. Explain why?
Answer: Diamond is hard because the carbon atoms in diamond are bonded in a stronger tetrahedron pattern, creating a rigid three-dimensional structure. Graphite is soft and slippery because its carbon atoms are bonded in layers with only weak van der Waals forces holding these layers together. These weak forces allow the layers to slide past each other easily, making graphite soft and slippery.
In simple words: Diamond is hard because its carbon atoms form a strong 3D network. Graphite is soft and slippery because its carbon atoms form flat layers that can slide over each other easily due to weak forces between them.

🎯 Exam Tip: Explain the structural differences (3D network vs. layered structure) and the types of forces involved (covalent bonds vs. van der Waals forces) to differentiate their properties.

Question 27. Explain the poisonous nature of carbon monoxide.
Answer: Carbon monoxide (CO) is highly poisonous. Even a small amount in the air can be fatal if inhaled for 10 to 15 minutes. It combines with hemoglobin in the blood to form carboxyhemoglobin, which is more stable than oxyhemoglobin. This reduces the blood's ability to carry oxygen, stopping its normal circulation to various body parts and potentially causing death by suffocation.
In simple words: Carbon monoxide is dangerous because it strongly binds to the hemoglobin in our blood, preventing oxygen from being carried to our body. This lack of oxygen can lead to suffocation and death.

🎯 Exam Tip: Highlight carboxyhemoglobin formation and its effect on oxygen transport as key points when explaining carbon monoxide's toxicity.

Question 28. Hydrolysis of SiCl4 can take place but not of CCl4 . Explain with reason.
Answer: In \( SiCl_4 \), vacant 3d orbitals are present, allowing it to undergo hydrolysis. Water molecules can attack silicon by donating a lone pair of electrons to these vacant orbitals. In contrast, carbon in \( CCl_4 \) lacks vacant d-orbitals and cannot accept a lone pair of electrons from water. Therefore, \( CCl_4 \) does not undergo hydrolysis.
In simple words: \( SiCl_4 \) can react with water because silicon has empty 'd' orbitals to accept electrons. \( CCl_4 \) cannot react with water because carbon does not have any empty 'd' orbitals to accept electrons.

🎯 Exam Tip: The availability of vacant d-orbitals is a critical factor for hydrolysis in compounds of elements from the third period onwards.

Question 29. Write the formula of unit of Silicones.
Answer: Silicones are a group of organosilicon polymers which have \( (R_2SiO) \) as a repeating unit.
In simple words: The basic repeating part of a silicone polymer molecule is represented by the formula \( (R_2SiO) \), where R stands for an organic group.

🎯 Exam Tip: Remember the general repeating unit for silicones; it helps in understanding their polymeric structure.

Question 30. In which of the metal the property of catenation is maximum?
Answer: Maximum tendency of catenation is found in carbon because carbon is small in size, and due to this, the bond between C-C is strong and p-p overlapping is also strong.
In simple words: Carbon shows the strongest ability to form long chains with itself, a property called catenation. This is because carbon atoms are small and form strong bonds with each other.

🎯 Exam Tip: Catenation is highest for carbon due to its small size and strong C-C single bond energy, which are key factors to mention.

Question 31. Explain the oxidation state of carbon in Be2 C and Al4 C3.
Answer: The oxidation state of carbon in \( Be_2C = -4 \).
Let the oxidation state of carbon in \( Al_4C_3 = x \).
For \( Al_4C_3 \): \( 4(+3) + 3x = 0 \)
\( 12 + 3x = 0 \)
\( 3x = -12 \)
\( x = -4 \)
Therefore, the oxidation state of carbon in \( Al_4C_3 = -4 \).
In simple words: In both beryllium carbide (\( Be_2C \)) and aluminium carbide (\( Al_4C_3 \)), the carbon atoms have an oxidation state of \( -4 \). This means each carbon atom in these compounds has gained four electrons.

🎯 Exam Tip: When calculating oxidation states, always know the standard oxidation states of other elements (like +2 for Be and +3 for Al) and ensure the total charge of the compound is zero.

Question 32. Tl shows (+1) and (+3) oxidation states. Why?
Answer: Thallium (Tl) shows \( +1 \) and \( +3 \) oxidation states. On moving down group 13, the \( +1 \) state becomes more stable compared to \( +3 \) due to the inert pair effect. The two electrons present in the s-shell (\( ns^2np^1 \)) are strongly attracted by the nucleus and do not participate in bonding.
In simple words: Thallium (Tl) can exist in both \( +1 \) and \( +3 \) oxidation states. The \( +1 \) state becomes more stable for thallium because of the 'inert pair effect', where the 's' electrons are held tightly and are less likely to join in chemical bonding.

🎯 Exam Tip: The inert pair effect is significant for heavier elements in p-block, causing the \( ns^2 \) electrons to remain unshared and favoring lower oxidation states.

Question 33. Why Boron forms electron deficient compounds?
Answer: Boron forms electron deficient compounds because of its small size and high charge density. Boron has a very small atomic radius and only three valence electrons. When it forms covalent compounds, it can only form three bonds, resulting in an incomplete octet (six electrons around boron instead of eight). Due to this incomplete octet, boron forms electron deficient compounds.
In simple words: Boron forms electron-deficient compounds because it is a small atom with only three valence electrons. This means it can only form three bonds, leaving it with less than eight electrons, so it wants more electrons.

🎯 Exam Tip: Remember that electron deficiency in boron compounds makes them act as Lewis acids, readily accepting electron pairs.

Question 34. The halides of boron do not form dimer but AlCl3 form dimer (Al2Cl6). Why?
Answer: \( BCl_3 \) and \( AlCl_3 \) are both electron-deficient compounds, as there are only six electrons in the valence shell of the central atom after molecule formation. However, the electron deficiency of boron in \( BCl_3 \) is compensated by the formation of a coordinate bond (pn - pn backbonding) between the lone pair of electrons of the chlorine atom and the empty unhybridized p-orbital of the boron atom. Aluminium is larger than boron and lacks effective backbonding, so \( AlCl_3 \) overcomes its electron deficiency by forming a dimer, \( Al_2Cl_6 \), where chlorine atoms bridge the two aluminium atoms.
In simple words: Boron halides don't form dimers because boron can use its available orbitals to pull electrons from chlorine, filling its electron gap. But aluminium chloride forms a dimer (\( Al_2Cl_6 \)) because aluminium is larger and completes its electron needs by sharing chlorine atoms between two \( AlCl_3 \) units.

🎯 Exam Tip: The key difference lies in the ability of boron to form effective backbonding, which is less significant for larger elements like aluminium, leading to dimer formation.

Question 35. Why property of catenation decreases down the group in group 14?
Answer: The property of catenation decreases down the group in group 14 due to an increase in atomic size and a decrease in electronegativity. As atoms get larger, the bond strength between similar atoms decreases, making longer chains less stable. The order of catenation is given as below:
C >> Si > Ge \( \approx \) Sn
Lead does not show catenation.
In simple words: As you go down Group 14, the ability of elements to form long chains with themselves (catenation) decreases. This happens because the atoms become larger, and the bonds they form with each other get weaker.

🎯 Exam Tip: For catenation, remember that bond strength between identical atoms is key; it decreases with increasing atomic size due to weaker orbital overlap.

Question 36. What are silicones? How are they prepared?
Answer: Silicones are a group of organosilicon polymers which have \( (R_2SiO) \) as a repeating unit. They are characterized by their high thermal stability, water repellency, and chemical inertness.
Preparation of Silicones: The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides, \( R_nSiCl_{4-n} \), where R is an alkyl or aryl group. When methyl chloride reacts with silicon in the presence of copper as a catalyst at a temperature of 570 K, various types of methyl substituted chlorosilanes are formed. Hydrolysis of dimethyl dichloro silane, i.e., \( (CH_3)_2SiCl_2 \), followed by condensation polymerisation, forms a straight chain polymer, which is silicone.
In simple words: Silicones are special polymers with a repeating \( (R_2SiO) \) unit, known for being waterproof and stable. They are made by reacting organic chlorides with silicon, then treating the resulting compounds with water and allowing them to link up into long chains.

🎯 Exam Tip: Remember the general formula \( (R_2SiO)_n \) and the two main steps in silicone preparation: formation of organosilicon halides, followed by hydrolysis and polymerization.

Question 37. Write the formula of water gas and producer gas.
Answer:
Water gas: \( CO + H_2 \)
Producer gas: \( CO + N_2 \)
In simple words: Water gas is a mix of carbon monoxide and hydrogen, while producer gas is a mix of carbon monoxide and nitrogen.

🎯 Exam Tip: These are important industrial gases; remember their specific compositions to avoid confusion.

Question 38. What happens when formic acid is heated with concentrated sulphuric acid. Give equation also.
Answer: When formic acid is heated with concentrated sulphuric acid, it forms carbon monoxide and water. Concentrated sulphuric acid acts as a dehydrating agent.
\( HCOOH \xrightarrow{conc. H_2SO_4, 373 K} CO + H_2O \)
In simple words: When formic acid is heated with concentrated sulphuric acid, the sulphuric acid removes water from it, leaving behind carbon monoxide gas.

🎯 Exam Tip: Concentrated sulphuric acid is a strong dehydrating agent; remember its role in removing water in such reactions.

Rbse Class 11 Chemistry Chapter 11 Long Answer Type Questions

Question 39. Write short notes on:
(i) Zeolites
(ii) Silicates
(iii) Silicones.
Answer:
(i) Zeolites: Zeolites are microporous, aluminosilicate minerals commonly used as commercial adsorbents and catalysts. They possess small pores which provide a general surface area. These molecules contain tetrahedral \( AlO_4^{5-} \) and \( SiO_4^{4-} \) molecules bound by oxygen atoms. Zeolites are widely used as catalysts in petrochemical industries for cracking hydrocarbons and isomerization, e.g., ZSM-5 is used to convert alcohols directly into gasoline. Hydrated zeolites are also used as ion exchangers in the softening of hard water.
(ii) Silicates: The basic structural unit of silicates is \( SiO_4^{4-} \) in which a silicon atom is bonded to four oxygen atoms in a tetrahedral fashion. In silicates, either the discrete unit is present, or a number of such units are joined via corners by sharing 1, 2, 3, or 4 oxygen atoms per silicate unit. When silicate units are linked together, they form chain, ring, sheet, or three-dimensional structures. Negative charges on the silicate structure are neutralized by positively charged metal ions. If all four corners are shared with other tetrahedral units, it forms a three-dimensional network. Examples include Feldspar, Zeolites, Mica, and Asbestos. Man-made silicates are glass and cement.
(iii) Silicones: Silicones are a group of organosilicon polymers, which have \( (R_2SiO) \) as a repeating unit. Silicones, being surrounded by non-polar alkyl groups, are water repellent. They have high thermal stability, high dielectric strength, and resistance to oxidation and chemicals. The starting materials for manufacturing silicones are alkyl or aryl substituted silicon chlorides, \( R_nSiCl_{4-n} \), where R is an alkyl or aryl group.
In simple words: (i) Zeolites are porous minerals used as catalysts and water softeners, structured from connected \( AlO_4 \) and \( SiO_4 \) units. (ii) Silicates are compounds with silicon and oxygen, forming structures from basic \( SiO_4 \) units that can link in various ways like chains, rings, or 3D networks. (iii) Silicones are special polymers with a repeating \( (R_2SiO) \) unit. They are waterproof, stable in heat, and resistant to chemicals, and are made from organic silicon chlorides.

🎯 Exam Tip: For short notes, define the substance, describe its key structural features, and list important applications or properties.

Question 40. Explain the following:
(i) Catenation
(ii) Inert Pair Effect
(iii) Allotropy.
Answer:
(i) Catenation: It is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This property is due to strong C-C bonds.
(ii) Inert Pair Effect: This refers to the reluctance of the s-electrons in the outermost shell to participate in chemical bonding. It is due to the poor shielding effect of d and f-block electrons in the inner shell, which increases the effective nuclear charge. The term 'inert pair' was first proposed by N. Sidgwick in 1927. This means the s-electrons are more tightly bound to the nucleus and therefore more difficult to ionize or use in bonding. For example, in p-block elements of the 4th, 5th, and 6th periods, the intervening d- and f-orbitals do not effectively shield the s-electrons of the valence shell. Due to this, the inert pair of \( ns \) electrons remains more tightly held by the nucleus and participates less in bonding.
(iii) Allotropy: Allotropy is the property of some chemical elements to exist in two or more different forms, in the same physical state, known as allotropes. Allotropes are different structural modifications of an element, where the atoms of the element are bonded together in a different manner. For example, the allotropes of carbon include diamond, graphite, and fullerenes.
In simple words: (i) Catenation is the ability of atoms of an element, especially carbon, to link together and form long chains. (ii) The Inert Pair Effect means that in heavier elements, the outermost 's' electrons are less likely to participate in bonding due to poor shielding by inner d and f electrons. (iii) Allotropy is when an element can exist in different physical forms, called allotropes, such as diamond and graphite for carbon, which have different structures.

🎯 Exam Tip: For each term, define it clearly and provide a suitable example. For the inert pair effect, explain the underlying reason (poor shielding).

Question 41. Complete the following reactions:
(ii) \( C + H_2O \rightarrow \)
(iii) \( B_2H_6 + O_2 \rightarrow \)
(iv) \( H_3BO_3 \rightarrow \)
(v) \( Al + NaOH \rightarrow \)
(vi) \( BF_3 + NH_3 \rightarrow \)
Answer:
(i) \( ZnO + CO \rightarrow Zn + CO_2 \)
(ii) \( C + H_2O \xrightarrow{473-1273 K} CO + H_2 \) (Water gas)
(iii) \( 2H_3BO_3 \rightarrow B_2O_3 + 3H_2O \)
(iv) \( H_3BO_3 \xrightarrow{370 K} HBO_2 \xrightarrow{} B_2O_3 \)
(v) \( 2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2 \)
(vi) \( BF_3 + NH_3 \rightarrow [B(NH_3)F_3] \)
In simple words: (i) Zinc oxide reacts with carbon monoxide to form zinc and carbon dioxide. (ii) Carbon reacts with steam at high temperatures to produce water gas (carbon monoxide and hydrogen). (iii) Boric acid decomposes to boron oxide and water. (iv) Boric acid, when heated, first forms metaboric acid, then further heating leads to boron oxide. (v) Aluminium reacts with sodium hydroxide and water to form sodium tetrahydroxoaluminate and hydrogen gas. (vi) Boron trifluoride reacts with ammonia to form an adduct, where ammonia donates electrons to boron.

🎯 Exam Tip: Always balance chemical equations and remember the conditions (temperature, catalyst) where applicable.

Question 42. What are electron deficient compounds? Are BCl3 and SiCl4 electron deficient compounds? Explain.
Answer: Electron deficient compounds are those compounds in which the central atom has fewer than eight electrons in its valence shell after forming the molecule. These compounds typically have an incomplete octet. Electron deficient bonds are sometimes described as 3-center-2-electron bonds.
\( BCl_3 \) is an electron-deficient compound because boron has only 6 valence electrons (three bonds with chlorine, \( 3 \times 2 = 6 \) electrons) around it, so it is short of 2 electrons to complete its octet. Therefore, it can accept a pair of electrons from nucleophiles, acting as a Lewis acid.
\( SiCl_4 \) is not an electron-deficient compound because silicon has 8 valence electrons (four bonds with chlorine, \( 4 \times 2 = 8 \) electrons) around it, completing its octet.
In simple words: Electron-deficient compounds are molecules where the central atom has less than eight electrons around it. \( BCl_3 \) is electron deficient because boron only has six electrons. \( SiCl_4 \) is not electron deficient because silicon has eight electrons, a complete octet.

🎯 Exam Tip: When identifying electron-deficient compounds, count the total valence electrons around the central atom after bonding. An octet rule violation (less than 8 electrons) indicates electron deficiency.

Question 43. Explain the following with reason :
(ii) Lead tetrachloride (PbCl4) is highly unstable towards heat because the stability of +4 oxidation state decreases on moving down the group in the periodic table due to inert pair effect. So, +2 oxidation state of Pb is more stable. It decomposes on heating to give lead dichloride and chlorine gas. \( \text{PbCl}_4 \rightarrow \text{PbCl}_2 + \text{Cl}_2(\text{g}) \)
Answer: Lead tetrachloride (\( \text{PbCl}_4 \)) is very sensitive to heat. This is because as we go down Group 14 in the periodic table, the +4 oxidation state becomes less stable due to the 'inert pair effect'. The +2 oxidation state of lead (\( \text{Pb} \)) is more stable. Therefore, \( \text{PbCl}_4 \) breaks down when heated, forming lead dichloride (\( \text{PbCl}_2 \)) and chlorine gas (\( \text{Cl}_2 \)).
In simple words: Lead tetrachloride is not strong against heat. Lead prefers to be in the +2 state, not +4, due to a special effect. So, when heated, it splits into lead dichloride and chlorine gas.

🎯 Exam Tip: Remember that the inert pair effect explains why lower oxidation states become more stable for heavier elements in p-block groups.

Question 43. (iii) Concentrated nitric acid reacts with aluminium to form a very thin layer of aluminium oxide which acts as a protective layer and protects from further reaction of nitric acid with aluminium. Due to this protective layer aluminium cannot react further with cone. HNO3. Hence, concentrated nitric acid can be transported in aluminium container.
Answer: When concentrated nitric acid comes into contact with aluminium, it creates a very thin coating of aluminium oxide on the metal's surface. This oxide layer acts as a shield, preventing the acid from reacting any further with the aluminium underneath. Because of this strong protective barrier, concentrated nitric acid can be safely stored and moved in containers made of aluminium.
In simple words: Strong nitric acid forms a thin, protective layer of aluminium oxide on aluminium. This layer stops any more reaction. So, we can store strong nitric acid in aluminium containers.

🎯 Exam Tip: The formation of an unreactive oxide layer that prevents further corrosion is known as passivation, an important concept in metal chemistry.

Question 43. (iv) Graphite has layered structure in which different layers are held together by weak Vander Waals forces and so each layer can slip over one another and hence graphite is used as a lubricant.
Answer: Graphite is structured in layers, and these layers are held together by very weak forces, known as Van der Waals forces. Because these forces are not strong, the layers can easily slide past each other. This unique property of graphite, allowing its layers to slip smoothly, is why it is used as an effective lubricant to reduce friction.
In simple words: Graphite has layers held by weak forces. These layers can slide easily over each other, making graphite a good lubricant.

🎯 Exam Tip: Always link the physical properties of a substance (like slipperiness) to its underlying atomic structure and the types of forces between its components.