Samacheer Kalvi Class 12 Chemistry Solutions Chapter 2 pBlock Elements I

I. Choose the Correct Answer

 

Question 1. An aqueous solution of borax is _______.
(a) neutral
(b) acidic
(c) basic
(d) amphoteric
Answer: (c) basic
In simple words: Borax, when dissolved in water, makes the solution basic. This means it has a pH greater than 7, acting like a weak base.

🎯 Exam Tip: Remember that borax is a salt of a strong base (NaOH) and a weak acid (boric acid), which is why its aqueous solution is basic.

 

Question 2. Boric acid is an acid because its molecule (NEET)
(a) contains replaceable H⁺ ion
(b) gives up a proton
(c) combines with proton to form water molecule
(d) accepts OH⁻ from water, releasing proton
Answer: (d) accepts OH⁻ from water, releasing proton
In simple words: Boric acid acts as an acid by taking in hydroxide ions (OH⁻) from water. When it takes OH⁻, it causes more H⁺ ions to be left in the water, making the solution acidic.

🎯 Exam Tip: Boric acid is a Lewis acid, not a Brønsted-Lowry acid, as it accepts a pair of electrons (from OH⁻) rather than donating a proton.

 

Question 3. Which among the following is not a borane?
(a) \( \text{B}_{2}\text{H}_{6} \)
(b) \( \text{B}_{3}\text{H}_{6} \)
(c) \( \text{B}_{4}\text{H}_{10} \)
(d) none of these
Answer: (b) \( \text{B}_{3}\text{H}_{6} \)
In simple words: Boranes are compounds of boron and hydrogen. Among the options, \( \text{B}_{3}\text{H}_{6} \) is not a known stable borane structure.

🎯 Exam Tip: Familiarize yourself with common borane formulas like diborane (\( \text{B}_{2}\text{H}_{6} \)) and tetraborane (\( \text{B}_{4}\text{H}_{10} \)).

 

Question 4. Which of the following metals has the largest abundance in the earth's crust?
(a) Aluminium
(b) Calcium
(b) Magnesium
(d) Sodium
Answer: (a) Aluminium
In simple words: Aluminium is the metal found most often in the Earth's outer layer. It is a very common element.

🎯 Exam Tip: Remember the order of elemental abundance in the Earth's crust: Oxygen, Silicon, Aluminium, Iron, Calcium, etc. Aluminium is the third most abundant element overall, and the most abundant metal.

 

Question 5. In diborane, the number of electrons that accounts for banana bonds is
(a) six
(b) two
(c) four
(d) three
Answer: (c) four
In simple words: In a diborane molecule, there are two special bonds, often called "banana bonds," which each use two electrons. So, a total of four electrons make up these unique bonds.

🎯 Exam Tip: Diborane (\( \text{B}_{2}\text{H}_{6} \)) has two 3-center-2-electron bonds (banana bonds), and each uses 2 electrons, making a total of 4 electrons for these bonds. It is important to know this structure.

 

Question 6. The element that does not show catenation among the following p-block elements is
(a) Carbon
(b) Silicon
(c) Lead
(d) germanium
Answer: (c) Lead
In simple words: Catenation is the ability of an element to form long chains with itself. Carbon shows this most strongly, but lead does not.

🎯 Exam Tip: Catenation decreases down a group due to decreasing bond strength and increasing atomic size. Carbon exhibits the highest catenation, while lead shows almost no catenation.

 

Question 7. Carbon atoms in fullerene with formula \( \text{C}_{60} \) have
(a) sp³ hybridised
(b) sp hybridised
(c) sp² hybridised
(d) partially sp² and partially sp³ hybridised
Answer: (c) sp² hybridised
In simple words: In a fullerene like \( \text{C}_{60} \), each carbon atom forms bonds by mixing one s-orbital and two p-orbitals. This creates three new hybrid orbitals that lie in a flat plane.

🎯 Exam Tip: Fullerenes are spherical molecules, and each carbon atom is bonded to three other carbon atoms, leading to sp² hybridization. This is similar to graphite, but fullerenes form closed cages.

 

Question 8. Oxidation state of carbon in its hydrides
(a) +4
(b) -4
(c) +3
(d) +2
Answer: (a) +4
In simple words: When carbon forms compounds with hydrogen, its typical charge is usually a positive four. This shows how it shares electrons in these specific compounds.

🎯 Exam Tip: In organic chemistry, carbon can have varying oxidation states depending on the atoms it's bonded to. When bonded to less electronegative atoms like hydrogen, carbon's oxidation state can be negative, for example, in methane (\( \text{CH}_{4} \)), carbon is -4. However, the question asks about carbon in *its* hydrides in a general sense, and often refers to the maximum positive oxidation state which is +4. The source answer of +4 could be considering a more generalized or theoretical highest state, or is mistaken as it implies the carbon is the more electronegative atom in some contexts. We will stick to the provided answer. In many inorganic hydrides or hypothetical highest states, +4 might be considered. It's often complex for carbon in organic compounds.

 

Question 9. The basic structural unit of silicates is (NEET) (PTA – 1)
(a) \( (\text{SiO}_{3})^{2-} \)
(b) \( (\text{SiO}_{4})^{2-} \)
(c) \( (\text{SiO})^{-} \)
(d) \( (\text{SiO}_{4})^{4-} \)
Answer: (d) \( (\text{SiO}_{4})^{4-} \)
In simple words: All silicates are built from a simple four-sided shape where one silicon atom is in the middle and four oxygen atoms surround it. This basic building block has a charge of minus four.

🎯 Exam Tip: The fundamental unit of all silicates is the tetrahedral \( [\text{SiO}_{4}]^{4-} \) unit, which can link in various ways to form different silicate structures.

 

Question 10. The repeating unit in silicone is
(a) \( \text{SiO}_{2} \)
(b) R-Si(R)-O-
(c) R-O-Si(R)-O
(d) -Si(R)(R)-O-
Answer: (d) -Si(R)(R)-O-
In simple words: Silicones are polymers made of repeating units where a silicon atom is bonded to two organic groups (R) and also to two oxygen atoms, forming a chain.

🎯 Exam Tip: Silicones are polysiloxanes characterized by a silicon-oxygen backbone with organic side groups (R), giving them unique properties like water repellency and thermal stability. The repeating unit is \( [\text{R}_{2}\text{SiO}] \).

 

Question 11. Which of these is not a monomer for a high molecular mass silicone polymer?
(a) \( \text{Me}_{3}\text{SiCl} \)
(b) \( \text{PhSiCl}_{3} \)
(c) \( \text{MeSiCl}_{3} \)
(d) \( \text{Me}_{2}\text{SiCl}_{2} \)
Answer: (a) \( \text{Me}_{3}\text{SiCl} \)
In simple words: To make a long chain silicone polymer, the starting chemical needs at least two places to connect. \( \text{Me}_{3}\text{SiCl} \) only has one such connecting point, so it stops the chain from growing.

🎯 Exam Tip: Monomers for linear silicone polymers require two reactive sites (like in \( \text{R}_{2}\text{SiCl}_{2} \)), while those for branched or cross-linked polymers require three or more (like in \( \text{RSiCl}_{3} \)). \( \text{R}_{3}\text{SiCl} \) acts as a chain terminator.

 

Question 12. Which of the following is not sp² hybridised?
(a) Graphite
(b) graphene
(c) Fullerene
(d) dry ice
Answer: (d) dry ice
In simple words: Dry ice is frozen carbon dioxide (\( \text{CO}_{2} \)). In carbon dioxide, the carbon atom is sp hybridized, not sp². Graphite, graphene, and fullerenes all have carbon atoms with sp² hybridization.

🎯 Exam Tip: Recall the hybridization of carbon in different allotropes and compounds. Graphite, graphene, and fullerenes typically have sp² hybridized carbons, while diamond has sp³ hybridized carbons. In \( \text{CO}_{2} \), carbon is sp hybridized.

 

Question 13. The geometry at which carbon atom in diamond are bonded to each other is
(a) Tetrahedral
(b) hexagonal
(c) Octahedral
(d) None of these
Answer: (a) Tetrahedral
In simple words: In a diamond, each carbon atom is connected to four other carbon atoms in a specific three-dimensional shape like a pyramid with a triangular base. This shape is called tetrahedral.

🎯 Exam Tip: Diamond is known for its strong, rigid, three-dimensional network structure where each carbon atom is sp³ hybridized and forms four single bonds, resulting in a tetrahedral geometry around each carbon.

 

Question 14. Which of the following statements is not correct?
(a) Beryl is a cylic silicate
(b) \( \text{Mg}_{2}\text{SiO}_{4} \) is an orthosilicate
(c) \( \text{SiO}_{4}^{4-} \) is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer: (d) Feldspar is not aluminosilicate
In simple words: All the first three statements are true. Feldspar is actually a type of aluminosilicate, which means it contains both aluminium and silicon oxygen structures. So, the last statement claiming it's not an aluminosilicate is wrong.

🎯 Exam Tip: Feldspars are important minerals that are indeed aluminosilicates, meaning they incorporate aluminium into their silicate framework. This changes the charge and properties of the basic silicate unit.

 

Question 15. Match items in Column-I with the items of Column-II and assign the correct code.

(a) 2 1 4 3
(b) 1 2 4 3
(c) 1 2 4 3
(d) None of these
Answer: (a) 2 1 4 3
In simple words: This question matches common chemical names with their correct formulas. Borazole is \( \text{B}_{3}\text{N}_{3}\text{H}_{6} \), Boric acid is \( \text{B}(\text{OH})_{3} \), Quartz is \( \text{SiO}_{2} \), and Borax is \( \text{Na}_{2}[\text{B}_{4}\text{O}_{5}(\text{OH})_{4}].8\text{H}_{2}\text{O} \).

🎯 Exam Tip: It is crucial to memorize the chemical formulas for common compounds, especially for borates and silicates, as they frequently appear in exams.

 

Question 16. Duralumin is an alloy of
(a) Cu, Mn
(b) Cu, AZ, Mg
(c) AZ, Mn
(d) Al, Cu, Mn, Mg
Answer: (d) Al, Cu, Mn, Mg
In simple words: Duralumin is a type of strong metal mix that is mainly aluminium. It also contains smaller amounts of copper, manganese, and magnesium to make it even stronger and more useful.

🎯 Exam Tip: Duralumin is a significant alloy due to its high strength-to-weight ratio, making it useful in aerospace. Always remember its primary component (Aluminium) and key alloying elements.

 

Question 17. The compound that is used in nuclear reactors as protective shields and control rods is
(a) Metal borides
(b) Metal oxides
(c) Metal carbonates
(d) Metal carbide
Answer: (a) Metal borides
In simple words: Metal borides are used in nuclear reactors because they are very good at soaking up extra neutrons. This helps to keep the nuclear reaction under control and also protects people from radiation.

🎯 Exam Tip: Boron's high neutron absorption cross-section makes borides excellent materials for nuclear applications, specifically for controlling the chain reaction by absorbing excess neutrons.

 

Question 18. The stability of +1 oxidation state increases in the sequence
(a) Al < Ga < In < Tl
(b) Tl < In < Ga < Al
(c) In < Ga < Al
(d) Ga < In < Tl
Answer: (a) Al < Ga < In < Tl
In simple words: For elements in the p-block, as you go down the group, it becomes easier for them to have a charge of +1. This is because the outer electrons are less tightly held, making the +1 state more stable for heavier elements.

🎯 Exam Tip: The "inert pair effect" explains this trend. For heavier p-block elements, the \( ns^{2} \) electrons become less willing to participate in bonding, making the +1 oxidation state more stable than the group oxidation state (e.g., +3 for group 13).

II. Answer the Following Questions

 

Question 1. Write a short note on anamolous properties of the first element of p-block.
Answer: The first element in each p-block group shows properties that are different from the other elements in its group. This unusual behavior happens because of these reasons:
1. The first member is very small in size.
2. It has high ionization energy and high electronegativity, meaning it holds its electrons very tightly.
3. It does not have d-orbitals in its outermost electron shell.
These unique characteristics allow the first element to form multiple bonds and have different chemical behavior compared to its heavier group members.
Here is a table showing some properties:

In simple words: The first element in each p-block group is special because it's small, holds electrons strongly, and has no d-orbitals. This makes it act differently from the other elements below it in the group.

🎯 Exam Tip: When explaining anomalous behavior, always mention small size, high electronegativity/ionization enthalpy, and absence of d-orbitals as key factors.

 

Question 2. Describe briefly allotropism in p-block elements with specific reference to carbon.
Answer: Allotropism is a phenomenon where some elements can exist in more than one crystalline or molecular form, all in the same physical state.
* This phenomenon is called allotropism.
* The different forms of an element are called allotropes.
* For example, carbon exists in various allotropic forms such as diamond, graphite, graphene, and carbon nanotubes. These forms have different structures and properties, even though they are all made of only carbon atoms.
In simple words: Allotropism is when an element can show up in different physical forms, like carbon in diamond or graphite. These different forms have unique structures.

🎯 Exam Tip: Define allotropism clearly and give carbon's allotropes as a prime example, mentioning their structural differences briefly.

 

Question 3. Give the uses of Borax.
Answer: Borax is a very useful compound with several applications:
1. Borax is used to identify colored metal ions, for example, in the borax bead test.
2. It is important in making optical glass, borosilicate glass, enamels, and glazes for pottery.
3. It acts as a flux in metallurgy, helping to remove impurities, and is also used as a good preservative. Borax helps to lower the melting point of substances in metallurgy.
In simple words: Borax is used to test for metal colors, make special types of glass and pottery coatings, clean metals in industries, and keep things fresh.

🎯 Exam Tip: When listing uses, be specific about the application (e.g., "borax bead test" for identification) and categorize them if possible (e.g., industrial, domestic).

 

Question 4. What is catenation? Describe briefly the catenation property of carbon. (MARCH 2020)
Answer: Catenation is a special property where atoms of an element can form strong covalent bonds with other atoms of the same element, creating long chains or rings. Carbon shows this property more than any other element because of its small size and its ability to form multiple bonds with itself.
The following conditions are necessary for catenation:
1. The element's bonding capacity (valency) must be two or more.
2. The element's atoms must be able to bond strongly with each other.
3. The bonds formed between the same type of atoms must be as strong as or stronger than their bonds with other elements.
4. The long chains or rings formed should be stable and not react easily with other molecules.
5. Carbon has all these qualities, which is why it can form so many different compounds with itself, leading to the vast field of organic chemistry.
In simple words: Catenation is when an element's atoms join together to form long chains. Carbon is great at this because it's small and forms strong bonds with other carbon atoms, creating many types of compounds.

🎯 Exam Tip: For catenation, emphasize carbon's small size, strong C-C bonds, and ability to form multiple bonds. These are the core reasons for its extensive catenation.

 

Question 5. Write a note on Fisher Tropsch synthesis. Fischer Tropsch synthesis: (PTA – 4)
Answer: The Fischer-Tropsch synthesis is a chemical process that turns carbon monoxide and hydrogen into liquid hydrocarbons. This reaction takes place at pressures less than 50 atmospheres and temperatures ranging from 500 to 700 K, with the help of metal catalysts. This process is used to make both saturated (single bonds only) and unsaturated (double or triple bonds) hydrocarbons.
The general reactions are:
\( n \, \text{CO} + (2n+1) \, \text{H}_{2} \rightarrow \text{C}_{n}\text{H}_{2n+2} + n \, \text{H}_{2}\text{O} \)
\( n \, \text{CO} + 2n \, \text{H}_{2} \rightarrow \text{C}_{n}\text{H}_{2n} + n \, \text{H}_{2}\text{O} \)
This synthesis provides a way to produce synthetic fuels and chemicals from coal, natural gas, or biomass. The specific catalyst and reaction conditions determine the type of hydrocarbons formed.
In simple words: Fischer-Tropsch synthesis is a way to make fuels and chemicals from carbon monoxide and hydrogen gas. It uses special metal helpers and heat to change the gases into different types of oil-like substances.

🎯 Exam Tip: Remember that Fischer-Tropsch synthesis is an industrial process for converting syngas (CO and H₂) into liquid hydrocarbons, highlighting the role of metal catalysts and specific temperature/pressure conditions.

 

Question 6. Give the structure of CO and \( \text{CO}_{2} \).
Answer:
Structure of CO:
The carbon monoxide molecule has a triple bond between carbon and oxygen. It is a linear molecule, meaning all atoms are in a straight line. The structure shows resonance and typically has a slight positive charge on oxygen and a negative charge on carbon in some resonance forms. For example:
\( \overset{+}{\text{C}} \equiv \overset{-}{\text{O}} \)
Structure is linear.
Structure of \( \text{CO}_{2} \):
The carbon dioxide molecule has a carbon atom in the center double-bonded to two oxygen atoms. It is also a linear molecule. This arrangement ensures a stable, symmetrical structure. For example:
\( \overset{..}{\text{O}}=\text{C}=\overset{..}{\text{O}} \)
Structure is linear.
In simple words: Both carbon monoxide (CO) and carbon dioxide (\( \text{CO}_{2} \)) are straight-line molecules. CO has a strong bond between carbon and oxygen, and \( \text{CO}_{2} \) has carbon in the middle with two oxygens connected by double bonds.

🎯 Exam Tip: Always remember that both CO and \( \text{CO}_{2} \) molecules are linear. Pay attention to the types of bonds (triple in CO, double in \( \text{CO}_{2} \)) and the number of lone pairs on each atom when drawing their structures.

 

Question 7. Give the uses of silicones.
Answer: Silicones are versatile polymers with a wide range of uses due to their unique properties:
1. They are used as lubricants in low-temperature conditions because they maintain their viscosity.
2. Silicones are also used in vacuum pumps.
3. They are stable at high temperatures, making them suitable for high-temperature oil baths.
4. Silicones are used to make waterproof cloths.
5. They serve as insulating materials for electrical motors and other appliances.
6. When mixed with paints and enamels, silicones make them resistant to high temperatures, sunlight, dampness, and various chemicals. Silicones are also biologically inert, making them useful in medical implants.
In simple words: Silicones are used for many things like lubricants in cold weather, in vacuum pumps, in hot oil baths, to make clothes waterproof, as electrical insulators, and to make paints strong against heat and weather.

🎯 Exam Tip: When listing uses, group them by property (e.g., thermal stability, water repellency, electrical insulation) to make your answer comprehensive and easy to remember.

 

Question 8. Describe the structure of diborane. (PTA – 3)
Answer: Diborane (\( \text{B}_{2}\text{H}_{6} \)) has a unique structure that does not fit simple two-center, two-electron bonding rules. Here's a breakdown:
* In diborane, two \( \text{BH}_{2} \) units are connected by two hydrogen atoms that bridge them. These bridging hydrogens give rise to what are known as "banana bonds" or 3-center-2-electron bonds. This structure includes eight \( \text{B}-\text{H} \) bonds in total.
* The diborane molecule has only 12 valence electrons in total, which are not enough to form normal two-center, two-electron covalent bonds for all atoms.
* The four terminal \( \text{B}-\text{H} \) bonds are normal covalent bonds, each being a 2-center-2-electron bond. These account for 8 electrons.
* The remaining four electrons are used for the two bridged bonds (banana bonds). Each of these is a 3-centered \( \text{B}-\text{H}-\text{B} \) bond and utilizes two electrons. These bridging hydrogen atoms lie in a plane perpendicular to the four terminal hydrogen atoms.
* In diborane, each boron atom is sp³ hybridised.
* Three sp³ hybridized orbitals on each boron atom contain a single electron, while the fourth orbital remains empty. The empty sp³ orbital of one boron overlaps with the half-filled sp³ orbital of the other boron and the 1s orbital of a hydrogen atom to form two bridged 3-center-2-electron \( \text{B}-\text{H}-\text{B} \) bonds.
* These unusual bonds are therefore 3-center-2-electron bonds, crucial for understanding diborane's stability. Diborane's structure can be thought of as two \( \text{BH}_{3} \) units held together by two bridging hydrogen atoms.
In simple words: Diborane has a special structure where two hydrogen atoms connect the two boron atoms, forming "banana bonds" that share two electrons among three atoms. The boron atoms are sp³ hybridized, and there are not enough electrons for all atoms to have regular two-electron bonds.

🎯 Exam Tip: For diborane, always explain the 3-center-2-electron bonds (banana bonds) and the 2-center-2-electron terminal bonds. Mention the sp³ hybridization of boron and the electron deficiency.

 

Question 9. Write a short note on hydroboration.
Answer: Hydroboration is a chemical reaction where diborane (\( \text{B}_{2}\text{H}_{6} \)) adds across the carbon-carbon double or triple bonds in alkenes and alkynes. This reaction usually happens in an ether solvent at room temperature. It is an important reaction in organic chemistry because it allows for the anti-Markovnikov addition of hydrogen and a hydroxyl group (from subsequent oxidation) to the double bond, which is different from typical electrophilic addition reactions.
The reaction proceeds as follows:
\( \text{B}_{2}\text{H}_{6} + 6 \, \text{RCH}=\text{CHR} \rightarrow 2\text{B}(\text{CH}_{2}-\text{CH}_{2}\text{R})_{3} \)
This process is widely used in synthetic organic chemistry to create alcohols or other functionalized compounds from alkenes and alkynes, specifically for anti-Markovnikov addition, where the hydrogen adds to the more substituted carbon. This allows chemists to create specific arrangements of atoms in molecules.
In simple words: Hydroboration is a reaction where diborane adds to carbon double or triple bonds. It helps make new organic compounds, especially when we want to add hydrogen in a special way to a molecule.

🎯 Exam Tip: Remember that hydroboration is a valuable synthetic method for achieving anti-Markovnikov addition products (e.g., primary alcohols from terminal alkenes), and it involves diborane as the reagent in ether solvents.

 

Question 10. Give one example for each of the following:
(i) icosogens
(ii) tetragens
(iii) pnictogen
(iv) chalcogen
Answer:

In simple words: This lists examples for different groups of elements in the periodic table: Boron for Icosagens, Carbon for Tetragens, Nitrogen for Pnictogens, and Oxygen for Chalcogens.

🎯 Exam Tip: Familiarize yourself with the traditional names for groups in the p-block (e.g., Icosagens for Group 13, Tetragens for Group 14, Pnictogens for Group 15, Chalcogens for Group 16, Halogens for Group 17, Aerogens/Noble Gases for Group 18).

 

Question 11. Write a note on metallic nature of p-block elements.
Answer: The metallic nature of elements, also called electropositive character, describes how easily an element loses electrons to form positive ions (cations). This property is closely linked to ionization energy. Here's how it behaves in p-block elements:
* The tendency of an element to form a cation by losing electrons is known as electropositive or metallic character.
* This characteristic mainly depends on the ionization energy; lower ionization energy means higher metallic character.
* Generally, as you move down a group in the p-block, the ionization energy decreases. This happens because the valence electrons are further from the nucleus and are shielded by more inner electrons. So, the metallic character increases down the group.
* In the p-block, the elements located in the lower left portion are metals, while those in the upper right portion are non-metals. Metalloids are found between metals and non-metals.
The metallic nature affects properties like conductivity and luster.
Here is a table summarizing the metallic nature in some p-block groups:

In simple words: The more easily an element loses electrons, the more metallic it is. In the p-block, elements become more metallic as you go down a group because their electrons are less tightly held. Metals are at the bottom-left, and non-metals are at the top-right.

🎯 Exam Tip: Focus on the periodic trends: metallic character increases down a group and decreases across a period. Relate this directly to ionization energy and effective nuclear charge.

 

Question 12. Complete the following reactions:
(a) \( \text{B}(\text{OH})_{3} + \text{NH}_{3} \rightarrow \)
(b) \( \text{Na}_{2}\text{B}_{4}\text{O}_{7} + \text{H}_{2}\text{SO}_{4} + \text{H}_{2}\text{O} \rightarrow \)
(c) \( \text{B}_{2}\text{H}_{6} + 2\text{NaOH} + 2\text{H}_{2}\text{O} \rightarrow \)
(d) \( \text{B}_{2}\text{H}_{6} + 6\text{CH}_{3}\text{OH} \rightarrow \)
(e) \( \text{BF}_{3} + 9\text{H}_{2}\text{O} \rightarrow \)
(f) \( \text{HCOOH} + \text{H}_{2}\text{SO}_{4} \rightarrow \)
(g) \( \text{SiCl}_{4} + \text{NH}_{3} \rightarrow \)
(h) \( \text{SiCl}_{4} + \text{C}_{2}\text{H}_{5}\text{OH} \rightarrow \)
(i) \( \text{B} + \text{NaOH} \rightarrow \)
(j) \( \text{H}_{2}\text{B}_{4}\text{O}_{7} \xrightarrow{\text{Red hot}} \)
Answer:
(a) \( \text{B}(\text{OH})_{3} + \text{NH}_{3} \xrightarrow{\text{800-1200K}} \text{BN} + 3\text{H}_{2}\text{O} \)
Boron nitride is formed.
(b) \( \text{Na}_{2}\text{B}_{4}\text{O}_{7} + \text{H}_{2}\text{SO}_{4} + 5\text{H}_{2}\text{O} \rightarrow \text{Na}_{2}\text{SO}_{4} + 4\text{H}_{3}\text{BO}_{3} \)
Ortho boric acid is formed.
(c) \( \text{B}_{2}\text{H}_{6} + 2\text{NaOH} + 2\text{H}_{2}\text{O} \rightarrow 2\text{NaBO}_{2} + 6\text{H}_{2} \)
Sodium metaborate is formed.
(d) \( \text{B}_{2}\text{H}_{6} + 6\text{CH}_{3}\text{OH} \rightarrow 2\text{B}(\text{OCH}_{3})_{3} + 6\text{H}_{2} \)
Methyl borate is formed.
(e) \( 4\text{BF}_{3} + 3\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{BO}_{3} + 3\text{HBF}_{4} \)
Hydrofluoroboric acid is formed.
(f) \( \text{HCOOH} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{CO} + \text{H}_{2}\text{O} + \text{H}_{2}\text{SO}_{4} \)
Carbon monoxide and water are formed; sulfuric acid acts as a dehydrating agent.
(g) \( \text{SiCl}_{4} + \text{NH}_{3} \xrightarrow{\text{330K, Ether}} \text{Cl}_{3}\text{Si}-\text{NH}-\text{SiCl}_{3} + 2\text{HCl} \)
Chlorosilazane is formed.
(h) \( \text{SiCl}_{4} + 4\text{C}_{2}\text{H}_{5}\text{OH} \rightarrow \text{Si}(\text{OC}_{2}\text{H}_{5})_{4} + 4\text{HCl} \)
Tetraethoxysilane is formed.
(i) \( 2\text{B} + 6\text{NaOH} \rightarrow 2\text{Na}_{3}\text{BO}_{3} + 3\text{H}_{2} \)
Sodium borate is formed.
(j) \( \text{H}_{2}\text{B}_{4}\text{O}_{7} \xrightarrow{\text{Red hot}} 2\text{B}_{2}\text{O}_{3} + \text{H}_{2}\text{O} \)
Boron trioxide is formed.
In simple words: These reactions show how different boron and silicon compounds react with other chemicals. For example, boric acid with ammonia makes boron nitride, and borax with sulfuric acid makes boric acid. Each reaction produces new substances.

🎯 Exam Tip: For reaction completion questions, identify the reactants and conditions to predict the products. Pay close attention to stoichiometry and common reaction types like hydrolysis, acid-base reactions, or dehydration. Knowing common names for the products helps.

 

Question 13. How will you identify borate radical? (PTA – 5)
Answer: The borate radical can be identified using the ethyl borate test, which produces a characteristic green flame.
* When boric acid or a borate salt is heated with ethyl alcohol in the presence of concentrated sulfuric acid, an ester called trialkyl borate (specifically, ethyl borate) is formed.
* The vapors of this ethyl borate then burn with a green-edged flame. This distinct green flame confirms the presence of the borate radical. The concentrated sulfuric acid acts as a dehydrating agent and helps in the formation of the ester. This test is highly sensitive and reliable.
* This is known as the ethyl borate test to identify the borate radical:
\( \text{H}_{3}\text{BO}_{3} + 3\text{C}_{2}\text{H}_{5}\text{OH} \xrightarrow{\text{Conc. H}_{2}\text{SO}_{4}} \text{B}(\text{OC}_{2}\text{H}_{5})_{3} + 3\text{H}_{2}\text{O} \)
The product, \( \text{B}(\text{OC}_{2}\text{H}_{5})_{3} \), is ethyl borate (green-edged flame).
In simple words: To find out if a borate radical is present, you can do a test where you mix boric acid or a borate with alcohol and strong acid, then heat it. If it burns with a green flame, it means the borate is there.

🎯 Exam Tip: Remember the specific reagents (ethyl alcohol, conc. \( \text{H}_{2}\text{SO}_{4} \)) and the key observation (green-edged flame) for the ethyl borate test. The formation of a volatile ethyl borate ester is crucial.

 

Question 15. How will you convert boric acid to boron nitride? (PTA - 3)
Answer: To make boron nitride, you combine urea with boric acid. This reaction happens in an ammonia atmosphere and needs very high temperatures (between 800 K and 1200 K). This method is a common way to synthesize inorganic compounds.
\( \text{B(OH)}_3 + \text{NH}_3 \xrightarrow{\text{800-1200 K}} \text{BN} + 3\text{H}_2\text{O} \)
In simple words: You mix boric acid and urea, heat them up very much with ammonia, and you get boron nitride.

🎯 Exam Tip: Remember the high temperature range and the specific reactants (urea, boric acid, ammonia) for this conversion, as these are key details.

 

Question 16. A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducting agent (C). Identify (A), (B) and ( C) (PTA - 1)
Answer:
(A) is Lithium hydride (\( \text{LiH} \)), which is a hydride of a 2nd period alkali metal.
(B) is Diborane (\( \text{B}_2\text{H}_6 \)), a compound of boron.
(C) is Lithium borohydride (\( \text{LiBH}_4 \)), which is a strong reducing agent formed when \( \text{LiH} \) reacts with \( \text{B}_2\text{H}_6 \) in ether.
The reaction is: \( \text{B}_2\text{H}_6 + 2\text{LiH} \xrightarrow{\text{Ether}} 2\text{LiBH}_4 \). This reaction is important for producing useful reducing agents in synthesis.

In simple words: The alkali metal is lithium hydride. When it reacts with diborane, it makes lithium borohydride, which is a strong chemical that can reduce other substances.

🎯 Exam Tip: When asked to identify compounds in a reaction sequence, always consider the period/group number, properties mentioned (like 'reducing agent'), and specific reactants to narrow down the options.

 

Question 17. A double salt which contains fourth period alkali metal (A) on heating at 500 K gives (B). Aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin. Identify (A) and (B).
Answer: The double salt (A) is Potash alum, \( \text{K}_2\text{SO}_4.\text{Al}_2(\text{SO}_4)_3.24\text{H}_2\text{O} \). Potassium is a fourth-period alkali metal. When heated to 500 K, potash alum loses its water molecules and changes into burnt alum (B), which is \( \text{K}_2\text{SO}_4.\text{Al}_2(\text{SO}_4)_3 \). Burnt alum's solution reacts with barium chloride to form a white precipitate of \( \text{BaSO}_4 \). Also, its solution reacts with alizarin dye to give a red compound, which is a characteristic test for aluminum ions. This identification process uses a series of chemical tests to confirm the substances.
\( \text{K}_2\text{SO}_4.\text{Al}_2(\text{SO}_4)_3.24\text{H}_2\text{O} \xrightarrow{\text{500 K}} \text{K}_2\text{SO}_4.\text{Al}_2(\text{SO}_4)_3 + 24\text{H}_2\text{O} \)
In simple words: (A) is potash alum, a type of salt with potassium. When you heat it, it turns into (B), which is burnt alum. Burnt alum makes a white solid with one chemical and turns red with another, helping us know what it is.

🎯 Exam Tip: For identification questions, list all characteristic reactions and properties described in the problem, like color changes, precipitates, and specific conditions (e.g., heating temperature).

 

Question 18. CO is a reducing agent, justify with an example.
Answer: Carbon monoxide (CO) is a powerful reducing agent. This means it can take oxygen away from other compounds, causing them to be reduced. It is often used in metallurgy to extract metals from their ores. For instance, it can reduce metallic oxides into metals, as seen in the reduction of ferric oxide (\( \text{Fe}_2\text{O}_3 \)) to iron (\( \text{Fe} \)). This property makes carbon monoxide valuable in industrial processes.
\( 3\text{CO} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + 3\text{CO}_2 \)
In simple words: Carbon monoxide is good at removing oxygen from other chemicals. For example, it can turn iron rust into pure iron by taking its oxygen.

🎯 Exam Tip: When justifying chemical properties, always provide a balanced chemical equation as an example to clearly demonstrate the concept.

III. Evaluate Yourself

 

Question 1. Why group 18 elements are called inert gases? Write the general configuraton of group 18 elements.
Answer: Group 18 elements are called inert gases because they are highly stable and do not easily react with other elements. This stability comes from their outer electron shell, which is completely filled with eight electrons (except for helium, which has two). Their general electronic configuration is \( \text{ns}^2\text{np}^6 \). Having a full outer shell makes them chemically unreactive under normal conditions. This full electron shell means they have very low tendency to gain, lose, or share electrons, hence "inert."
In simple words: Group 18 elements are called inert gases because their outermost electron shell is full. This makes them very stable and they don't like to react with other things. Their general electron setup is \( \text{ns}^2\text{np}^6 \).

🎯 Exam Tip: When explaining chemical behavior, always link it back to electron configuration, especially for elements in main groups.

12th Chemistry Guide Chapter 2 p-Block Elements - I Additional Questions and Answers

Part - II - Additional Questions

I. Choose the correct answer

 

Question 1. The general electronic configuration of p-block elements is
(a) \( \text{ns}^1 \)
(b) \( \text{ns}^2 \)
(c) \( \text{ns}^2 \text{np}^{1-6} \)
(d) \( \text{(n-1)s}^2 \text{np}^{1-6} \)
Answer: (c) \( \text{ns}^2 \text{np}^{1-6} \)
In simple words: P-block elements have their outermost electrons in both the 's' and 'p' orbitals, with two electrons in 's' and between one to six electrons in 'p'.

🎯 Exam Tip: Always remember that 'n' refers to the principal quantum number (the main energy level), and for p-block, both 's' and 'p' subshells of that level are involved.

 

Question 2. p-block element consists of the groups
(a) 1 & 2
(b) 3 - 12
(c) 13-17
(d) 13-18
Answer: (d) 13-18
In simple words: The p-block in the periodic table includes groups 13 through 18. These are the elements whose valence electrons are found in the p-orbital.

🎯 Exam Tip: Know the specific group numbers that correspond to each block (s, p, d, f) in the periodic table for quick recall.

 

Question 3. Group 18 elements are inert because of their
(a) unstable incompletely filled orbitals
(b) stable completely filled orbitals
(c) half filled orbitals
(d) stable nucleus
Answer: (b) stable completely filled orbitals
In simple words: Group 18 elements are called inert gases because their electron shells are full, making them very stable and unreactive.

🎯 Exam Tip: The concept of "completely filled orbitals" (octet rule) is fundamental to understanding the stability and reactivity of elements, especially inert gases.

 

Question 4. As we go down the group ionisation energy
(a) decreases
(b) increases
(c) becomes constant
(d) becomes zero
Answer: (a) decreases
In simple words: When you move down a group in the periodic table, the amount of energy needed to remove an electron from an atom becomes less. This is because the outer electrons are further from the nucleus and less attracted to it.

🎯 Exam Tip: Remember the general trend: ionization energy decreases down a group due to increasing atomic size and shielding effect, making it easier to remove outer electrons.

 

Question 5. As we go down the group metallic character
(a) decreases
(b) increases
(c) becomes constant
(d) becomes zero
Answer: (b) increases
In simple words: As you go down a group in the periodic table, atoms become larger and lose electrons more easily, which means their metallic properties become stronger.

🎯 Exam Tip: Metallic character is directly related to the ease of losing electrons, so any factor that makes electron loss easier (like increased atomic radius or shielding) will increase metallic character.

 

Question 6. As ionisation energy decreases, the metallic character of elements
(a) decreases
(b) increases
(c) becomes constant
(d) becomes zero
Answer: (b) increases
In simple words: If it's easier to remove an electron (lower ionization energy), the element acts more like a metal. Metals are known for easily giving up their electrons.

🎯 Exam Tip: Ionization energy and metallic character have an inverse relationship; a lower ionization energy means higher metallic character because metals tend to lose electrons readily.

 

Question 7. In p-block, metals are placed in
(a) upper right part
(b) middle part
(c) lower left part
(d) top of the group
Answer: (c) lower left part
In simple words: In the p-block of the periodic table, you find metals towards the bottom-left side. This is because metallic character increases as you go down and to the left.

🎯 Exam Tip: Metals are generally found on the left and bottom of the periodic table, including the lower-left corner of the p-block, while non-metals are in the upper-right.

 

Question 8. In p-block, non-metals are placed in
(a) upper right part
(b) middle part
(c) lower left part
(d) bottom of the group
Answer: (a) upper right part
In simple words: In the p-block, elements that are not metals are found in the top-right corner. This is where you find the elements that tend to gain electrons.

🎯 Exam Tip: Non-metals are typically located on the right and top of the periodic table, including the upper-right corner of the p-block, reflecting their tendency to gain electrons.

 

Question 9. Which of the following factor is not responsible for the anamolous behaviour of the first member of each group in p-block elements?
(a) small size
(b) high ionisation enthalpy
(c) outer electronic configuration
(d) absence of d-orbitals
Answer: (c) outer electronic configuration
In simple words: The unique behavior of the first element in a p-block group is due to its small size, strong hold on electrons, and lack of d-orbitals, not its general electron setup.

🎯 Exam Tip: Anomalous behavior of the first element is primarily due to its small size, high electronegativity, high ionization enthalpy, and absence of d-orbitals, which allow for unique bonding and coordination numbers.

 

Question 10. The correct order of catenation property in group 14 elements is
(a) C << Si < Ge = Sn < Pb
(b) C >> Si > Ge = Sn > Pb
(c) C >> Si < Ge = Sn < Pb
(d) C < < Si » Ge = Sn > Pb
Answer: (b) C >> Si > Ge = Sn > Pb
In simple words: Carbon is best at forming long chains with itself, much better than silicon, and this ability decreases as you go down the group to germanium, tin, and lead.

🎯 Exam Tip: Catenation is strongest for carbon due to its small size and strong C-C bonds, and this ability generally decreases with increasing atomic size down a group.

 

Question 11. The elements N, O, F readily forms hydrogen bonds due to their high
(a) ionisation energy
(b) electron affinity
(c) electro negativity
(d) atomic radius
Answer: (c) electro negativity
In simple words: Nitrogen, oxygen, and fluorine easily form hydrogen bonds because they are very good at attracting electrons (highly electronegative). This strong pull makes the hydrogen atom slightly positive, allowing it to bond with other electronegative atoms.

🎯 Exam Tip: Hydrogen bonding occurs only when hydrogen is directly attached to a highly electronegative atom like Fluorine, Oxygen, or Nitrogen (FON elements), which creates a strong dipole.

 

Question 12. The most electro negative element is
(a) Flourine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer: (a) Flourine
In simple words: Fluorine is the element that pulls electrons towards itself the strongest in a chemical bond. It has the highest electronegativity value on the periodic table.

🎯 Exam Tip: Fluorine is universally recognized as the most electronegative element, a fact that is critical for understanding bonding and reactivity across the periodic table.

 

Question 13. The element with maximum electron affinity is
(a) Flourine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer: (b) Chlorine
In simple words: Chlorine has the highest electron affinity, meaning it releases the most energy when it gains an electron. While fluorine is the most electronegative, chlorine's larger size reduces electron-electron repulsion, allowing it to accept an electron more readily.

🎯 Exam Tip: Do not confuse electron affinity with electronegativity; while fluorine is most electronegative, chlorine has the highest electron affinity due to less inter-electronic repulsion in its valence shell.

 

Question 14. The most reactive element among halogens is
(a) Flourine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer: (a) Flourine
In simple words: Among the halogens, fluorine reacts the most easily with other substances. This is because it is very small and strongly pulls electrons, making it eager to form bonds.

🎯 Exam Tip: Reactivity in halogens decreases down the group. Fluorine is the most reactive due to its high electronegativity and small atomic size.

 

Question 15. The strongest oxidising agent among halogens is
(a) Flourine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer: (a) Flourine
In simple words: Fluorine is the strongest oxidizing agent among halogens. This means it is best at taking electrons from other atoms, causing them to get oxidized.

🎯 Exam Tip: The oxidizing power of halogens decreases down the group. Fluorine is the strongest oxidizing agent because it has the highest tendency to gain electrons.

 

Question 16. The important property shown by p-block elements is
(a) complex formation
(b) coloured ion formation
(c) inert pair effect
(d) metallic character
Answer: (c) inert pair effect
In simple words: One special thing about heavier p-block elements is the "inert pair effect." This means their two 's' electrons are less likely to be used in bonding, leading to different stable oxidation states.

🎯 Exam Tip: The inert pair effect is a key characteristic for understanding the varying oxidation states and stability trends of heavier p-block elements, especially in groups 13-16.

 

Question 17. In 13th group \( \text{Tl}^{+1} \) ion is more stable than \( \text{Tl}^{3+} \) ion due to
(a) high electronegatively
(b) inert pair effect
(c) high ionisation energy
(d) stable electronic configuration
Answer: (b) inert pair effect
In simple words: For thallium, the \( \text{+1} \) charged ion is more stable than the \( \text{+3} \) charged ion. This happens because the 's' electrons in thallium are held very tightly and don't like to take part in bonding, which is called the inert pair effect.

🎯 Exam Tip: The inert pair effect explains why the lower oxidation state becomes more stable for heavier elements in groups 13, 14, and 15, as their \( \text{ns}^2 \) electrons become less available for bonding.

 

Question 18. Diamond and graphite are of carbon.
(a) Isotopes
(b) Isobars
(c) Isomers
(d) Allotropes
Answer: (d) Allotropes
In simple words: Diamond and graphite are different forms of the same element, carbon. They are called allotropes because they have the same atoms but are arranged in different ways, giving them unique properties.

🎯 Exam Tip: Allotropes are different structural forms of the same element in the same physical state. Remember common allotropic pairs like carbon (diamond, graphite) and oxygen (O2, O3).

 

Question 19. The formula of Borax is
(i) \( \text{Na}_2\text{B}_4\text{O}_7.10\text{H}_2\text{O} \)
(ii) \( \text{Na}_2[\text{B}_4\text{O}_5(\text{OH})_4].8\text{H}_2\text{O} \)
(iii) \( \text{Na}_2[\text{B}_4\text{O}_5(\text{OH})_4].2\text{H}_2\text{O} \)
(a) (i) only
(b) (i) & (ii) only
(c) (i) & (iii) only
(d) (iii) only
Answer: (b) (i) & (ii) only
In simple words: Borax has two common formulas: one that shows it with ten water molecules, and another that details its internal structure with oxygen, boron, and hydroxide groups, along with eight water molecules.

🎯 Exam Tip: Be aware that many compounds, especially complex inorganic ones, can have multiple accepted chemical formulas (e.g., empirical, molecular, or structural) that represent the same substance.

 

Question 20. Ortho boric acid on dehydration at 373K produces mainly (PTA - 3)
(a) metaboric acid
(b) boric anhydride
(c) Boron metal and Oxygen
(d) tetra boric acid
Answer: (a) metaboric acid
In simple words: When you heat ortho boric acid at 373 Kelvin, it loses water and mostly turns into metaboric acid. This is a common way to prepare metaboric acid.

🎯 Exam Tip: Remember that the product of dehydration reactions can vary with temperature. For boric acid, different temperatures yield different products, so specific temperature values are important.

 

Question 21. The formula of colemanite is
(a) \( \text{Na}_2\text{B}_4\text{O}_7 \)
(b) \( \text{Na}_2\text{B}_4\text{O}_7.10\text{H}_2\text{O} \)
(c) \( \text{Ca}_2\text{B}_6\text{O}_{11} \)
(d) \( \text{NaBO}_2 \)
Answer: (c) \( \text{Ca}_2\text{B}_6\text{O}_{11} \)
In simple words: Colemanite is a mineral, and its chemical formula is \( \text{Ca}_2\text{B}_6\text{O}_{11} \). This means it contains calcium, boron, and oxygen atoms in specific amounts.

🎯 Exam Tip: Memorizing the chemical formulas of common minerals and ores is crucial for chemistry exams, as they are often tested directly.

 

Question 22. Which is used as moderator in nuclear reactors?
(a) boron nitride
(b) boron
(c) borax
(d) boric acid
Answer: (b) boron
In simple words: Boron is used in nuclear reactors to slow down neutrons. It does this because boron atoms can absorb neutrons very well, which helps control the nuclear reaction.

🎯 Exam Tip: Boron's high neutron absorption cross-section is a key property that makes it useful in nuclear applications, both as a moderator and in control rods.

 

Question 23. The compound used in eye drops and antiseptics is
(a) boron nitride
(b) boric acid
(c) sodium meta borate
(d) boron tri oxide
Answer: (b) boric acid
In simple words: Boric acid is a mild substance used in eye drops and as a gentle antiseptic to clean wounds. It has properties that make it safe for these medical uses.

🎯 Exam Tip: Many compounds have practical applications, and knowing specific uses (like boric acid in antiseptics) demonstrates a broader understanding of chemistry.

 

Question 24. The compound used as a flux in metallurgy is
(a) boron nitride
(b) boric acid
(c) borax
(d) boron tri oxide
Answer: (c) borax
In simple words: Borax is used as a flux in making metals. It helps clean metal surfaces and removes impurities when heating metals, which is essential for strong, pure metal products.

🎯 Exam Tip: Fluxes are important in metallurgy for cleaning metal surfaces and removing unwanted oxides or impurities, and borax is a common example due to its ability to dissolve metal oxides.

 

Question 25. Boric acid on heating at 413 K gives
(a) meta boric acid
(b) tetra boric acid
(c) boric anhydride
(d) borax
Answer: (b) tetra boric acid
In simple words: When boric acid is heated specifically to 413 Kelvin, it loses some water and forms tetra boric acid. This is another form of boric acid with less water.

🎯 Exam Tip: Different temperatures lead to different dehydration products for boric acid; be precise with the temperature given in the question.

 

Question 26. In ethyl borate test the colour of the flame obtained is
(a) red
(b) yellow
(c) blue
(d) green
Answer: (d) green
In simple words: When you do the ethyl borate test, the flame will turn green. This specific color change is a sign that borate is present in the sample.

🎯 Exam Tip: Flame tests are qualitative analysis techniques; specific colors (like green for borate) are key indicators for identifying particular elements or compounds.

 

Question 27. On hydrolysis BF3 gives Boric acid and converted to fluroboric acid. The fluoroboric acid contains the species. (PTA - 6)
(a) \( \text{H}^+, \text{F}^- \) & \( \text{BF}_3 \)
(b) \( \text{H}^+ \) & \( [\text{BF}_4]^- \)
(c) \( [\text{H BF}_3]^+ \) & \( \text{F}^- \)
(d) \( \text{H}^+, \text{B}^{3+} \) & \( \text{F}^- \)
Answer: (b) \( \text{H}^+ \) & \( [\text{BF}_4]^- \)
In simple words: When \( \text{BF}_3 \) reacts with water, it first forms boric acid. With more \( \text{BF}_3 \), it makes fluoroboric acid, which contains hydrogen ions and tetrafluoroborate ions. This reaction is a key example of how boron trifluoride behaves as a Lewis acid.

🎯 Exam Tip: Remember that \( \text{BF}_3 \) is a strong Lewis acid and its reactions with water (hydrolysis) are crucial. The formation of the tetrafluoroborate ion (\( [\text{BF}_4]^- \)) is characteristic.

 

Question 28. In organic benzene is
(a) diborane
(b) borazole
(c) borax
(d) boric acid
Answer: (b) borazole
In simple words: Borazole is often called "inorganic benzene" because it has a similar ring structure to benzene, but with alternating boron and nitrogen atoms instead of all carbon atoms.

🎯 Exam Tip: Understanding analogous compounds (like borazole being inorganic benzene) helps in relating structures and properties across different areas of chemistry.

 

Question 29. The formula of Inorganic benzene is
(a) \( \text{B}_3\text{N}_3 \)
(b) \( \text{B}_3\text{N}_3\text{H}_3 \)
(c) \( \text{B}_3\text{N}_3\text{H}_6 \)
(d) \( \text{B}_6\text{N}_6\text{H}_6 \)
Answer: (c) \( \text{B}_3\text{N}_3\text{H}_6 \)
In simple words: The chemical formula for inorganic benzene, also known as borazine, is \( \text{B}_3\text{N}_3\text{H}_6 \). It has three boron atoms, three nitrogen atoms, and six hydrogen atoms arranged in a ring.

🎯 Exam Tip: Memorizing common names and their corresponding chemical formulas, especially for important inorganic compounds, is crucial for answering direct recall questions.

 

Question 30. The most stable form of carbon is
(a) graphite
(b) diamond
(c) fullerene
(d) carbon nano tubes
Answer: (a) graphite
In simple words: Graphite is the most stable form of carbon under normal conditions. This means it requires the least amount of energy to exist, making it the preferred form.

🎯 Exam Tip: While diamond is well-known for its hardness and brilliance, graphite is thermodynamically more stable under standard conditions because of its layered structure and extensive pi-electron delocalization.

 

Question 31. The formula of buckminster fullerene is
(a) \( \text{C}_{32} \)
(b) \( \text{C}_{50} \)
(c) \( \text{C}_{60} \)
(d) \( \text{C}_{70} \)
Answer: (c) \( \text{C}_{60} \)
In simple words: Buckminsterfullerene, often called a "buckyball," is a carbon molecule made of 60 carbon atoms. It looks like a soccer ball.

🎯 Exam Tip: Fullerene \( \text{C}_{60} \) is the most common and well-known fullerene, recognized for its distinctive spherical cage structure and widespread applications.

 

Question 32. The number of six membered and five membered rings fused together respectively in buckminster fullerene is
(a) 12 & 20
(b) 20 & 12
(c) 10 & 22
(d) 22 & 10
Answer: (b) 20 & 12
In simple words: A buckminster fullerene molecule has a soccer ball shape, which is made up of 20 rings with six carbon atoms and 12 rings with five carbon atoms joined together.

🎯 Exam Tip: Remember the iconic soccer ball structure of \( \text{C}_{60} \), which consists of interlocking pentagons and hexagons, with a specific count of each type of ring.

 

Question 33. Water gas is a mixture of
(a) \( \text{CO}_2 + \text{H}_2 \)
(b) \( \text{CO} + \text{H}_2\text{O} \)
(c) \( \text{CO} + \text{H}_2 \)
(d) \( \text{CO} + \text{N}_2 \)
Answer: (c) \( \text{CO} + \text{H}_2 \)
In simple words: Water gas is a mixture of carbon monoxide and hydrogen gas. It is often used as a fuel in industries.

🎯 Exam Tip: Differentiate water gas (\( \text{CO} + \text{H}_2 \)) from producer gas (\( \text{CO} + \text{N}_2 \)) and synthesis gas, as their compositions and uses differ significantly.

 

Question 34. Producer gas is a mixture of
(a) \( \text{CO}_2 + \text{H}_2 \)
(b) \( \text{CO} + \text{H}_2\text{O} \)
(c) \( \text{CO} + \text{H}_2 \)
(d) \( \text{CO} + \text{N}_2 \)
Answer: (d) \( \text{CO} + \text{N}_2 \)
In simple words: Producer gas is a mixture of carbon monoxide and nitrogen gas. It is a type of fuel gas made from passing air over hot coke.

🎯 Exam Tip: The main distinction between water gas and producer gas lies in the presence of nitrogen (from air) in producer gas, making it less energy-dense than water gas.

 

Question 35. In the presence of light carbon monoxide reacts with chlorine to form a poisonous gas called
(a) mustard gas
(b) phosgene
(c) phosphine
(d) carbylamine
Answer: (b) phosgene
In simple words: When carbon monoxide and chlorine react with light, they form a very toxic gas known as phosgene. This reaction is important to be aware of due to the dangerous nature of the product.

🎯 Exam Tip: Phosgene (\( \text{COCl}_2 \)) is a highly toxic gas; understanding its formation (reaction of CO with \( \text{Cl}_2 \) under light) is important for safety and chemical reactions knowledge.

 

Question 36. Fischer Tropsch synthesis is used for preparing
(a) Silicones
(b) Boranes
(c) Hydrocarbons
(d) Carbonyls
Answer: (c) Hydrocarbons
In simple words: The Fischer-Tropsch synthesis is a chemical process that takes carbon monoxide and hydrogen and turns them into various liquid hydrocarbons, like fuels and lubricants.

🎯 Exam Tip: The Fischer-Tropsch synthesis is a crucial industrial process for converting synthesis gas (CO and \( \text{H}_2 \)) into liquid fuels and chemicals, especially important for gas-to-liquids technology.

 

Question 37. In metal carbonyls the oxidation state of metals is
(a) 0
(b) +1
(c) +2
(d) +3
Answer: (a) 0
In simple words: In compounds called metal carbonyls, the metal atom is in a zero oxidation state. This is because carbon monoxide (CO) is a neutral ligand and does not carry a charge.

🎯 Exam Tip: A key characteristic of most simple metal carbonyls (e.g., \( \text{Ni(CO)}_4 \), \( \text{Fe(CO)}_5 \)) is that the metal is in a zero oxidation state, reflecting the neutral nature of the CO ligand and strong covalent bonding.

 

Question 38. The structure of CO molecule is
(a) trigonal
(b) tetrahedral
(c) linear
(d) square planar
Answer: (c) linear
In simple words: A carbon monoxide molecule has a straight-line shape. This is because it only has two atoms, carbon and oxygen, connected by a triple bond.

🎯 Exam Tip: Any diatomic molecule (composed of two atoms) will always have a linear geometry because there's only one way for two points to connect.

 

Question 39. The structure of \( \text{CO}_2 \) molecule is
(a) trigonal
(b) tetrahedral
(c) linear
(d) square planar
Answer: (c) linear
In simple words: A carbon dioxide molecule has a straight-line shape. This is because the central carbon atom forms double bonds with two oxygen atoms, and there are no lone pairs on the carbon to push them out of a straight line.

🎯 Exam Tip: For \( \text{AB}_2 \) molecules where the central atom (A) has no lone pairs (like in \( \text{CO}_2 \)), the VSEPR theory predicts a linear geometry with bond angles of 180 degrees.

 

Question 40. The critical temperature of \( \text{CO}_2 \) is
(a) 21 °C
(b) 31°C
(c) 12°C
(d) 13°C
Answer: (b) 31°C
In simple words: The critical temperature of carbon dioxide is 31 degrees Celsius. Above this temperature, \( \text{CO}_2 \) cannot be turned into a liquid, no matter how much pressure is applied.

🎯 Exam Tip: Critical temperature is the temperature above which a gas cannot be liquefied, regardless of the pressure applied. Knowing specific values for common substances like \( \text{CO}_2 \) (around 31°C) is important.

 

Question 41. When \( \text{CO}_2 \) is dissolved in water, the solution is slightly
(a) acidic
(b) basic
(c) amphoteric
(d) neutral
Answer: (a) acidic
In simple words: When carbon dioxide dissolves in water, it forms carbonic acid, which makes the water slightly acidic. This is why carbonated drinks are a little sour.

🎯 Exam Tip: Non-metal oxides typically form acidic solutions when dissolved in water, while metal oxides form basic solutions. \( \text{CO}_2 \) is a classic example of an acidic oxide.

 

Question 42. Which among the following is important for photo synthesis?
(a) O2
(b) N2
(c) CO
(d) CO2
Answer: (d) CO2
In simple words: Carbon dioxide is very important for plants because they use it, along with sunlight, to make their food. This process is called photosynthesis.

🎯 Exam Tip: Remember the basic formula for photosynthesis, which clearly shows carbon dioxide as a key reactant.

 

Question 43. The water repellant property of silicones is due to the presence of
(a) -OH group
(b) -Si group
(c) -R group
(d) -Cl group
Answer: (c) -R group
In simple words: Silicones push water away because they have special -R groups on their outside. These groups are organic and do not like water, making silicones waterproof.

🎯 Exam Tip: Understanding the molecular structure and functional groups of silicones is key to explaining their unique properties.

 

Question 44. The percentage of silicate minerals and silica present in earth's crust is
(a) 75
(b) 85
(c) 95
(d) 100
Answer: (c) 95
In simple words: A very large part, almost 95%, of the Earth's hard outer layer is made up of silicate minerals and silica. These are very common minerals found in rocks and soil.

🎯 Exam Tip: Knowing the approximate abundance of key elements and compounds in the Earth's crust is important for general chemistry knowledge.

 

Question 45. The basic unit present in silicates is
(a) SiO2
(b) \( (SiO_3)^{2-} \)
(c) \( (SiO_4)^{2-} \)
(d) \( (SiO_4)^{4-} \)
Answer: (d) \( (SiO_4)^{4-} \)
In simple words: Silicate minerals are built from tiny parts called silicate tetrahedra. Each of these small units has one silicon atom in the middle and four oxygen atoms around it, making its overall charge \( 4- \).

🎯 Exam Tip: Focus on the tetrahedral structure of silicates, as it is fundamental to understanding their various forms and classifications.

 

Question 46. Talc is an example of
(a) Ino silicates
(b) Phyllo silicates
(c) Tecto silicates
(d) Chain silicates
Answer: (b) Phyllo silicates
In simple words: Talc is a type of silicate mineral that has a layered or sheet-like structure. These are called phyllo silicates because 'phyllo' means 'leaf' or 'sheet'.

🎯 Exam Tip: Associate common minerals with their silicate classification to easily recall examples for each type.

 

Question 47. Quartz is an example of
(a) Ino silicates
(b) Phyllo silicates
(c) Tecto silicates
(d) Chain silicates
Answer: (c) Tecto silicates
In simple words: Quartz is a very common mineral, and it belongs to the tecto silicates family. This means its structure is like a big, strong 3D network where all the oxygen atoms are shared between the silicon atoms.

🎯 Exam Tip: Remember that tecto silicates, like quartz, are known for their strong, interconnected three-dimensional frameworks, which make them very stable.

 

Question 48. The formula of Spodumene is
(a) Sc2Si2O7
(b) Li Ai(SiO3)2
(c) \( [Be_3Al_2(SiO_3)_6] \)
(d) Be2SiO4
Answer: (b) Li Ai(SiO3)2
In simple words: Spodumene is a mineral that contains lithium, aluminium, and silicate units. Its chemical formula is \( LiAl(SiO_3)_2 \), which shows how these elements combine in its structure.

🎯 Exam Tip: For mineral formulas, pay attention to the main elements and the presence of common polyatomic ions like silicate to help recall the correct structure.

 

Question 49. The silicate which is used in the removal of permanent hardness of water is
(a) Feldspar
(b) Quartz
(c) Zeolites
(d) Talc
Answer: (c) Zeolites
In simple words: Zeolites are special minerals that can help clean hard water. They do this by swapping out the hard water minerals for softer ones, which is why they are used to remove permanent hardness from water.

🎯 Exam Tip: Zeolites are known for their ion-exchange properties, making them effective water softeners and catalysts.

 

Question 50. Thermodynamically the most stable form of carbon is (PTA - 4)
(a) Diamond
(b) Fullerenes
(c) graphite
(d) Nano tubes
Answer: (c) graphite
In simple words: Among all the different forms carbon can take, graphite is the most stable one when thinking about how much energy it has. This means it is the most natural form carbon likes to be in under normal conditions.

🎯 Exam Tip: Understand that thermodynamic stability relates to the lowest energy state, even if other forms like diamond are very strong. Graphite's layered structure makes it thermodynamically favored.

II. Pick the Odd Man Out

 

Question 1. W.r.t. their metallic character pick the odd man out.
(a) Ge
(b) Ga
(c) B
(d) As
Answer: (b) Ga
In simple words: Germanium (Ge), Boron (B), and Arsenic (As) are metalloids, meaning they have properties of both metals and non-metals. However, Gallium (Ga) is a true metal. So, Gallium is the odd one out.

🎯 Exam Tip: Familiarize yourself with the periodic table and the classification of elements as metals, non-metals, and metalloids.

 

Question 2. W.r.t. their metallic character pick the odd man out
(a) In
(b) Pb
(c) Cl
(d) Bi
Answer: (c) Cl
In simple words: Indium (In), Lead (Pb), and Bismuth (Bi) are all metals, meaning they are good conductors and have a shiny look. Chlorine (Cl) is a non-metal, a gas that does not conduct electricity. Therefore, Chlorine is the odd one out.

🎯 Exam Tip: Clearly distinguish between the properties of metals and non-metals, especially for elements in the p-block, where the transition is evident.

 

Question 3. Pick the odd man out
(a) Borax
(b) Kernite
(c) Colemanite
(d) Bauxite
Answer: (d) Bauxite
In simple words: Borax, Kernite, and Colemanite are all important sources of boron, a chemical element. Bauxite, however, is a main source for aluminium. So, Bauxite is the different one here because it is an ore of aluminium, not boron.

🎯 Exam Tip: Learn the common ores for important elements in the p-block to identify them correctly.

 

Question 4. W.r.t. to hybridisation pick the odd man out.
(a) Graphite
(b) Diamond
(c) Fullerene
(d) Graphene
Answer: (b) Diamond
In simple words: Diamond is special because its carbon atoms are joined in a \( sp^3 \) way, which makes it very hard. Graphite, fullerenes, and graphene all have carbon atoms joined in a \( sp^2 \) way, giving them different structures and properties.

🎯 Exam Tip: Remember the hybridization of carbon in different allotropes (diamond is \( sp^3 \), graphite/fullerene/graphene are \( sp^2 \)) as this determines their properties.

III. Assertion and Reason

 

Question 1. Assertion (A): Boron shows non metallic character. Reason (R) : Atomic radius of boron is small and its nuclear charge is high.
(i) Both A and R are correct, R explains A
(ii) A is wrong but R is wrong
(iii) A is wrong but R is correct
(iv) Both A and R are correct but R does not explain A
Answer: (i) Both A and R are correct, R explains A
In simple words: Boron acts like a non-metal because it has a very small atomic size and a strong pull from its nucleus. These two reasons make it hold onto its electrons tightly, which is why it behaves like a non-metal.

🎯 Exam Tip: When evaluating Assertion-Reason questions, first check if both statements are true individually, then determine if the reason correctly explains the assertion.

 

Question 2. Assertion (A) : As we move down Boron group the elements show less tendency to exhibit +1 oxidation state rather than +3. Reason (R) : As we move down Boron group the elements show inert pair effect.
(i) Both A and R are correct, R explains A
(ii) A is wrong but R is wrong
(iii) A is wrong but R is correct
(iv) Both A and R are correct but R does not explain A
Answer: (iii) A is wrong but R is correct
In simple words: The assertion is incorrect because as you go down the Boron group, elements actually show a *greater* tendency to have a +1 oxidation state, not less. This is because of something called the inert pair effect, which is correctly stated in the reason. The inert pair effect explains why the outermost s-electrons become harder to remove for heavier elements.

🎯 Exam Tip: Pay close attention to keywords like "less" or "greater" in assertions, as a single word can change the truth value of the statement. Understand the inert pair effect and its consequences.

 

Question 3. Assertion (A) : Graphite conducts electricity. Reason (R) : In Graphite, successive carbon sheets are held together by weak Vander Waals force.
(i) Both A and R are correct, R explains A
(ii) A is wrong but R is wrong
(iii) A is wrong but R is correct
(iv) Both A and R are correct but R does not explain A
Answer: (iv) Both A and R are correct but R does not explain A
In simple words: Graphite is good at carrying electricity because it has free-moving electrons that are shared across its layers. The reason given is also true: graphite layers are held by weak Van der Waals forces. However, these weak forces are why graphite is soft and slippery, not why it conducts electricity. So, both statements are correct, but the reason does not explain the assertion.

🎯 Exam Tip: Distinguish between the structural reasons for conductivity (delocalized electrons) and the structural reasons for physical properties like softness (interlayer forces).

 

Question 4. Assertion (A) : Silicones are used for making water proofing clothes. Reason (R) : In silicones the organic side groups which surrounds silicon make the molecule looks like an alkane.
(i) Both A and R are correct, R explains A
(ii) A is wrong but R is wrong
(iii) A is wrong but R is correct
(iv) Both A and R are correct but R does not explain A
Answer: (i) Both A and R are correct, R explains A
In simple words: Silicones are indeed used to make clothes waterproof. This is because the organic parts attached to the silicon atoms in silicones make the molecules behave like alkanes, which repel water. This special structure helps them shed water effectively.

🎯 Exam Tip: Connect the structure and chemical nature of organic groups in silicones to their real-world application as waterproofing agents.

IV. Choose the Correct Statement

 

Question 1. i) Some of the p-block elements show negative oxidation states also. ii) Halogens gain two electrons to give a stable halide ion. iii) Inert gases have \( ns^2np^6 \) configuration and hence more stable. iv) p-block elements have a general electronic configuration \( (n-1)s^2 np^{1-6} \)
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (b) (i) & (iii)
In simple words: Statement (i) is correct because some p-block elements can indeed take on negative oxidation states. Statement (iii) is also correct as inert gases have a full outer shell \( (ns^2np^6) \) which makes them very stable. Statement (ii) is wrong because halogens only need to gain *one* electron to become stable, not two. Statement (iv) is also wrong because the general electronic configuration for p-block elements is \( ns^2np^{1-6} \), not \( (n-1)s^2 np^{1-6} \).

🎯 Exam Tip: Review the electronic configurations and common oxidation states for elements in the p-block and noble gases to avoid common misconceptions.

 

Question 2. i) Boron compounds are electron rich compounds. ii) Boron does not react directly with hydrogen. iii) Borax is sodium salt of metaboric acid. iv) Boric acid is used as an antiseptic,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (ii) & (iv)
(d) (iii) & (iv)
Answer: (c) (ii) & (iv)
In simple words: Statement (ii) is correct; boron usually does not react directly with hydrogen. Statement (iv) is also correct, as boric acid is known for its use as a mild antiseptic. Statement (i) is incorrect because boron compounds are usually electron *deficient*, not rich. Statement (iii) is incorrect as borax is actually the sodium salt of *tetraboric* acid, not metaboric acid.

🎯 Exam Tip: Accurately recall the electron nature of boron compounds and the correct chemical names of common boron-containing compounds.

 

Question 3. i) In graphite carbon atoms are \( sp^3 \) hybridised. ii) A single planar sheet of graphite is known as graphene. iii) in diamond each carbon atom is tetrahedrally surrounded by four other carbon atoms. iv) Carbon nanotubes do not conduct electricity
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: Statement (ii) is correct: a single flat layer of graphite is called graphene. Statement (iii) is also correct because in a diamond, each carbon atom is connected to four other carbon atoms in a pyramid shape. Statement (i) is incorrect as carbon atoms in graphite are \( sp^2 \) hybridised, not \( sp^3 \). Statement (iv) is incorrect because carbon nanotubes *do* conduct electricity.

🎯 Exam Tip: Clearly differentiate between the hybridization, structure, and conductivity of various carbon allotropes like graphite, diamond, graphene, and carbon nanotubes.

 

Question 4. i) Silicones are organo silicon polymers. ii) Hydrolysis of \( R_2SiCl_2 \) yields complex cross linked polymer. iii) Silicones are good thermal and electrical conductors. iv) All silicones are water repellent,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (d) (i) & (iv)
In simple words: Statement (i) is correct; silicones are indeed polymers that contain both organic groups and silicon. Statement (iv) is also correct because all silicones repel water. Statement (ii) is incorrect because the hydrolysis of \( R_2SiCl_2 \) forms a *straight chain* polymer, not a complex cross-linked one. Statement (iii) is incorrect as silicones are generally good thermal and electrical *insulators*, not conductors.

🎯 Exam Tip: Understand that the number of chlorine atoms on the silicon (e.g., \( R_2SiCl_2 \) vs \( RSiCl_3 \)) dictates whether a linear, cyclic, or cross-linked polymer is formed. Also remember that silicones are insulators, not conductors.

V. Choose the Wrong Statement

 

Question 1. i) Boron is a metal. ii) Nitrogen is a metalloid. iii) Oxygen is a non metal. iv) Antimony is a metalloid.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (iii) & (iv)
Answer: (a) (i) & (ii)
In simple words: Statement (i) is wrong because boron is actually a metalloid, not a metal. Statement (ii) is also wrong because nitrogen is a non-metal, not a metalloid. Statements (iii) and (iv) are correct. So, the incorrect statements are (i) and (ii).

🎯 Exam Tip: Master the classification of elements (metals, non-metals, metalloids) in the p-block to correctly identify their properties.

 

Question 2. i) Aluminium chloride is a Lewis acid. ii) Alum is a double salt of potassium aluminium sulphate. iii) Aluminium chloride is used as a styptic agent to arrest bleeding. iv) Alum is used as a catalyst in Friedel Crafts reaction.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (c) (iii) & (iv)
In simple words: Statement (iii) is wrong because *alum* (not aluminium chloride) is used as a styptic agent to stop bleeding. Statement (iv) is also wrong because *anhydrous aluminium chloride* (not alum) is used as a catalyst in Friedel Crafts reactions. Statements (i) and (ii) are correct.

🎯 Exam Tip: Distinguish between the applications of aluminium chloride and alum, as they have different uses despite both containing aluminium.

 

Question 3. Which of the following statement about \( H_3BO_3 \) is not correct? (PTA - 5)
(a) It is a strong tribasic acid
(b) It is prepared by acidifying an aqueous solution of borax.
(c) It is a layer structure in which planer BO3 units are joined by hydrogen bonds.
(d) It does not act as proton donor but acts as a Lewis acid by accepting hydroxyl ion.
Answer: (a) It is a strong tribasic acid
In simple words: Boric acid (\( H_3BO_3 \)) is not a strong tribasic acid; it is actually a weak monobasic Lewis acid. It acts by accepting an hydroxyl ion rather than giving away protons. All the other statements about its preparation and structure are correct.

🎯 Exam Tip: Remember that boric acid is an unusual acid because it functions as a Lewis acid by accepting \( OH^- \) rather than donating \( H^+ \).

 

Question 4. i) Silicates which contain discrete \( [SiO_4]^{4-} \) units are called neso silicates. ii) Beryl is an example for amphiboles. iii) Spodumene is an example for phyllo silicates. iv) Silicates which contain \( [Si_7O_7]^{6-} \) ions are called Soro silicates,
(a) (i) & (ii)
(b) (ii) & (iv)
(c) (ii) & (iii)
(d) (i) & (iv)
Answer: (c) (ii) & (iii)
In simple words: Statement (ii) is wrong because Beryl is an example of *cyclic* silicates, not amphiboles. Statement (iii) is also wrong because Spodumene is an example of *chain* silicates, not phyllo silicates. Statements (i) and (iv) are correct.

🎯 Exam Tip: Carefully review the examples and defining characteristics of different silicate types (neso, cyclic, chain, phyllo, soro) to avoid mixing them up.

VI. Match the Following

 

Question 1.

Answer:In simple words: This matching exercise helps us remember the special names given to different groups in the p-block of the periodic table. For example, Group 13 elements are called Icosagens, and Group 18 elements are known as Inert gases because they are not very reactive.

🎯 Exam Tip: Memorize the common group names for p-block elements, as this is a frequent question type in chemistry exams.

 

Question 2.

Answer:In simple words: This question matches substances with their properties or occurrences. Fluorine is a very strong oxidising agent. Borax is used to find colored metal ions. Aluminium is the most common metal in the Earth's crust. Sulphur is found in volcanic ashes.

🎯 Exam Tip: To do well in matching questions, know the key properties, uses, and natural occurrences of important elements and compounds.

 

Question 3.

Answer:In simple words: This match helps connect boron compounds with their uses. Boron is used as a moderator in nuclear reactors. Amorphous boron can be used as a rocket fuel igniter. Boric acid is used in eye drops. Boric oxide is used to make Pyrex glass.

🎯 Exam Tip: Focus on the industrial and medicinal applications of key p-block compounds, as they often appear in matching questions.

 

Question 4.

Answer:In simple words: This match helps us link different types of silicates with their correct examples. Ortho silicates have single silicon-oxygen units, like Phenacite. Pyro silicates are made of two units joined together, like Thortveitite. Cyclic silicates form rings, like Beryl. Chain silicates are long chains, like Spodumene. Amphiboles are double chains, like Asbestos. Sheet silicates form layers, like Mica. Finally, Tecto silicates have a 3D network, like Quartz.

🎯 Exam Tip: Creating a mental image or drawing simple diagrams of each silicate structure (ortho, pyro, cyclic, chain, sheet, tecto) can greatly aid in recalling their corresponding examples.

VII. 2 Marks Questions

 

Question 1. What are 'p'-block elements? Write their general outer electronic configuration.
Answer: P-block elements are those elements where the last electron enters the p-orbital. They include elements from Group 13 to Group 18 in the periodic table. The general outer electronic configuration for p-block elements is \( ns^2np^{1-6} \), which means they have 2 electrons in the s-orbital and 1 to 6 electrons in the p-orbital of their outermost shell. These elements show a wide range of chemical properties, from metals to non-metals.
In simple words: P-block elements are atoms where the last electron goes into a 'p' shell. They are found in groups 13 to 18. Their outer electron setup is always two 's' electrons and one to six 'p' electrons.

🎯 Exam Tip: To define p-block elements, remember both the definition based on electron configuration and their placement in the periodic table (groups 13-18).

 

Question 2. How are the p-block elements classified.
Answer: P-block elements are classified in two main ways. First, they are grouped from Group 13 to Group 18 based on their outer electronic configuration. This grouping helps understand trends in reactivity and properties. Second, they are also classified based on their nature into non-metals, metalloids, and metals. This highlights how elements in the p-block show a gradual change from non-metallic to metallic character as you go down a group and from right to left across a period.
In simple words: P-block elements are sorted by their outer electron arrangement into groups 13 to 18. They are also sorted by what they are like - non-metals, metalloids (like a mix), or metals.

🎯 Exam Tip: When describing classifications, always provide the basis of classification (e.g., electronic configuration, metallic character) along with the resulting categories.

 

Question 3. Aluminium (III) chloride is stable where as Thallium (III) chloride is unstable. Why? (PTA - 2)
Answer: Aluminium (III) chloride (\( AlCl_3 \)) is stable, but Thallium (III) chloride (\( TlCl_3 \)) is unstable. This happens because of a phenomenon called the inert pair effect, which becomes more noticeable as you go down Group 13. The inert pair effect causes the two \( ns^2 \) electrons in the outermost shell to become less reactive. Because of this, for heavier elements like Thallium, the \( Tl^{3+} \) ion (where all three outer electrons are lost) is less stable, and the \( Tl^{+1} \) ion (where only the \( np^1 \) electron is lost) is more stable. Therefore, \( TlCl_3 \) easily breaks down into \( TlCl \) because Thallium prefers the +1 oxidation state.
In simple words: Aluminium chloride is stable, but thallium chloride is not. This is because of the "inert pair effect" in heavier elements like thallium. It means the two outermost 's' electrons don't easily take part in bonding. So, thallium likes to lose only one electron (+1 state) instead of three (+3 state), making \( TlCl_3 \) unstable.

🎯 Exam Tip: The inert pair effect is a critical concept for explaining oxidation state stability trends in heavier p-block elements; always link it to the decreased participation of \( ns^2 \) electrons in bonding.

 

Question 4. How is boric acid prepared from borax?
Answer: Boric acid (\( H_3BO_3 \)) can be prepared from borax (\( Na_2B_4O_7 \)) by reacting it with a strong acid such as hydrochloric acid (HCl) or sulphuric acid (\( H_2SO_4 \)). When borax is treated with these acids in the presence of water, the borate compound reacts to form boric acid as a product.
\( Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 4H_3BO_3 + 2NaCl \)
\( Na_2B_4O_7 + H_2SO_4 + 5H_2O \rightarrow 4H_3BO_3 + 2Na_2SO_4 \)
In simple words: To make boric acid from borax, you mix borax with water and then add a strong acid like HCl or \( H_2SO_4 \). This reaction makes boric acid, which can then be collected.

🎯 Exam Tip: For preparation questions, remember the reactants, conditions (if any), and the balanced chemical equation, as this is crucial for full marks.

 

Question 5. How is boric acid prepared from Colemanite?
Answer: Boric acid can also be prepared from a mineral called Colemanite (\( Ca_2B_6O_{11} \)). This preparation involves passing sulphur dioxide (\( SO_2 \)) gas through a solution of Colemanite. The reaction between Colemanite, sulphur dioxide, and water yields boric acid as one of the products, along with calcium sulphite.
\( Ca_2B_6O_{11} + 2SO_2 + 9H_2O \rightarrow 2CaSO_3 + 6H_3BO_3 \)
In simple words: Boric acid is made from Colemanite by bubbling sulphur dioxide gas through a solution of Colemanite. This process changes Colemanite into boric acid and calcium sulphite.

🎯 Exam Tip: Recognize that boric acid can be synthesized from different boron-containing minerals; knowing various methods demonstrates a comprehensive understanding.

 

Question 6. What is the action of sodium hydroxide on boric acid?
Answer: When boric acid (\( H_3BO_3 \)) reacts with sodium hydroxide (NaOH), it forms different sodium borate compounds depending on the amount of sodium hydroxide used. With a 1:1 ratio, it forms sodium metaborate. With a 1:2 ratio, it forms sodium tetra borate. Boric acid behaves as a Lewis acid in these reactions, accepting hydroxide ions.
\( NaOH + H_3BO_3 \rightarrow NaBO_2 + 2H_2O \)
\( 2NaOH + 4H_3BO_3 \rightarrow Na_2B_4O_7 + 7H_2O \)
In simple words: Boric acid reacts with sodium hydroxide. Depending on how much sodium hydroxide is used, it can make sodium metaborate or sodium tetraborate.

🎯 Exam Tip: Understand the stoichiometric ratios in reactions and how varying amounts of reactants can lead to different products, as seen with boric acid and NaOH.

 

Question 7. Write the action of water on diborane.
Answer: Diborane (\( B_2H_6 \)) reacts vigorously with water. This reaction leads to the hydrolysis of diborane, producing boric acid (\( H_3BO_3 \)) and hydrogen gas (\( H_2 \)). This is an important chemical reaction that demonstrates diborane's reactivity with protic solvents.
\( B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2 \)
In simple words: When diborane is mixed with water, they react strongly. This makes boric acid and hydrogen gas.

🎯 Exam Tip: For reactions involving hydrides like diborane, always remember to include the byproduct of hydrogen gas if the compound is readily hydrolyzed.

 

Question 8. What is the action of NaOH on diborane.
Answer: Diborane reacts with sodium hydroxide to form sodium metaborate. This reaction creates a useful inorganic compound.
\[ B_2H_6 + 2NaOH + 2H_2O \rightarrow 2NaBO_2 + 6H_2 \]
In simple words: When diborane and sodium hydroxide mix with water, they react to make sodium metaborate and hydrogen gas.

🎯 Exam Tip: Remember to balance all the atoms on both sides of the chemical equation to ensure accuracy, especially for complex reactions.

 

Question 9. What is the action of air on diborane?
Answer: Pure diborane does not react with air or oxygen at normal room temperature. However, impure diborane reacts with air or oxygen, producing boron trioxide and a large amount of heat. This shows how impurities can change a substance's reactivity.
\[ B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O \]
\( \Delta H = -2165 \text{ KJ mol}^{-1} \)
In simple words: Pure diborane is safe in air, but if it's not pure, it can react strongly with air, making heat and boron trioxide.

🎯 Exam Tip: When discussing reactivity, always specify conditions like purity and temperature, as they can significantly alter the outcome of a reaction.

 

Question 10. How does diborane react with methyl alcohol?
Answer: Diborane reacts with methyl alcohol to create trimethyl borate. This is an important reaction in organic synthesis.
\[ B_2H_6 + 6CH_3OH \rightarrow 2B(OCH_3)_3 + 6H_2 \]
In simple words: Diborane mixes with methyl alcohol to make a new substance called trimethyl borate and also hydrogen gas.

🎯 Exam Tip: For reactions involving organic compounds, correctly identifying the products and balancing the equation is crucial for full marks.

 

Question 11. How does diborane react with metal hydrides?
Answer: When diborane is treated with different metal hydrides, it forms metal borohydrides. These compounds are often used as reducing agents in chemistry. The table below summarizes these reactions.
When treated with metal hydrides, diborane forms metal boro hydrides.
\[ B_2H_6 + 2LiH \xrightarrow{Ether} 2LiBH_4 \]
\[ B_2H_6 + 2NaH \xrightarrow{Diglyme} 2NaBH_4 \]

In simple words: Diborane reacts with metal hydrides like LiH or NaH to create new substances called metal borohydrides, which are good at adding hydrogen to other chemicals.

🎯 Exam Tip: Pay attention to the specific catalysts or solvents mentioned (like Ether or Diglyme) as they are crucial conditions for these reactions.

 

Question 12. How does diborane react with ammonia at low temperature?
Answer: When diborane is treated with excess ammonia at a low temperature, it forms diborane diammoniate. This is an addition product formed without major rearrangement. The specific temperature is key to forming this product.
\[ 3B_2H_6 + 6NH_3 \xrightarrow{153K} 3B_2H_6.2NH_3 \]
In simple words: If you mix diborane with a lot of ammonia when it's very cold, they join together to make a new compound called diborane diammoniate.

🎯 Exam Tip: Note the stoichiometric ratio and the low temperature requirement; these are critical for forming the diammoniate product instead of borazine.

 

Question 13. How is inorganic benzene prepared? (PTA – 1)
Answer: Inorganic benzene, also known as borazine or borazole, is prepared by heating diborane with ammonia at higher temperatures. This reaction leads to the formation of a cyclic compound that shares structural similarities with organic benzene.
On heating at higher temperatures with ammonia, diborane forms borazole or borazine.
Borazole or borazine is called as Inorganic benzene.
\[ 3B_2H_6 + 2NH_3 \xrightarrow{\text{High temp / Closed vessel}} B_3N_3H_6 \]
In simple words: You can make inorganic benzene (borazine) by heating diborane with ammonia. It looks a bit like real benzene but has boron and nitrogen atoms instead of just carbon.

🎯 Exam Tip: Distinguish between the low-temperature reaction with ammonia (diborane diammoniate) and the high-temperature reaction (borazine), as temperature dictates the product.

 

Question 14. BF3 acts as a Lewis acid. Give example.
Answer: \( BF_3 \) (Boron trifluoride) acts as a Lewis acid because it is an electron-deficient compound and readily accepts electron pairs. This behavior allows it to form coordinate covalent bonds. For instance, it can accept an electron pair from ammonia or water.
\[ BF_3 + NH_3 \rightarrow F_3B \leftarrow NH_3 \]
\[ BF_3 + H_2O \rightarrow F_3B \leftarrow OH_2 \]
In simple words: \( BF_3 \) is a Lewis acid because it has an empty space for electrons and can take an electron pair from other molecules, like ammonia or water, to form a bond.

🎯 Exam Tip: Remember that Lewis acids are electron pair acceptors, and compounds with an incomplete octet, like \( BF_3 \), are good examples.

 

Question 15. Convert BF3 into hydro fluoro boric acid.
Answer: On hydrolysis, \( BF_3 \) (boron trifluoride) first produces boric acid, which then further reacts to form hydrofluoroboric acid. This reaction demonstrates boron's interaction with water.
\[ 4BF_3 + 3H_2O \rightarrow H_3BO_3 + 3HBF_4 \]
\( 3HBF_4 \) is Hydrofluoroboric acid.
In simple words: When \( BF_3 \) mixes with water, it first makes boric acid, which then quickly turns into hydrofluoroboric acid.

🎯 Exam Tip: Pay attention to the stoichiometry of this reaction, especially the number of \( BF_3 \) molecules needed to produce boric acid and hydrofluoroboric acid.

 

Question 16. Write about McAfee process of manufacturing AlCl3.
Answer: In the McAfee process, aluminum chloride (\( AlCl_3 \)) is manufactured by heating a mixture of alumina (\( Al_2O_3 \)) and coke (carbon) in the presence of a current of chlorine gas. This method is an industrial way to produce aluminum chloride.
\[ Al_2O_3 + 3C + 3Cl_2 \rightarrow 2AlCl_3 + 3CO \]
In simple words: The McAfee process makes aluminum chloride by heating alumina and carbon with chlorine gas.

🎯 Exam Tip: For industrial processes, remember the key reactants and the balanced chemical equation, as well as any special conditions like heating or a current of gas.

 

Question 17. Write the action of NaOH on AlCl3
Answer: When aluminum chloride (\( AlCl_3 \)) reacts with an excess of sodium hydroxide (\( NaOH \)), it forms sodium aluminate. This reaction highlights the amphoteric nature of aluminum compounds.
\[ AlCl_3 + 4NaOH \rightarrow NaAlO_2 + 2H_2O + 3NaCl \]
In simple words: Aluminum chloride reacts with too much sodium hydroxide to make sodium aluminate, water, and sodium chloride.

🎯 Exam Tip: Note that "excess" reactant is specified; this indicates a complete reaction where the product's formation might depend on the amount of reactant.

 

Question 18. Write the uses of aluminium chloride.
Answer: Aluminum chloride (\( AlCl_3 \)) has several important uses in various industries. These applications stem from its properties as a Lewis acid and a halogen carrier.
1. Anhydrous \( AlCl_3 \) is widely used as a catalyst in Friedel-Crafts reactions, which are important in organic synthesis.
2. \( AlCl_3 \) is used in the manufacturing of petrol through the cracking of mineral oils.
3. It also serves as a catalyst in the production of various dyes, drugs, and perfumes.
In simple words: Aluminum chloride helps make many things like petrol, dyes, drugs, and perfumes by speeding up chemical reactions.

🎯 Exam Tip: When listing uses, always try to mention at least two or three distinct applications to show comprehensive knowledge.

 

Question 19. What are alums? Give examples.
Answer: Alums are a type of double salt, typically hydrated double sulfates of aluminum and a monovalent cation. The term "alum" is now used more broadly to refer to all double salts that follow a general formula. These compounds are known for their ability to purify water.
1. Alum is a double salt of potassium aluminum sulfate.
2. Currently, the name alum is used for all double salts that have the general formula \( M'_2SO_4 M''_2(SO_4)_3.24H_2O \), where \( M' \) is a univalent metal ion (or \( NH_4^+ \)) and \( M'' \) is a trivalent metal ion.
Examples: Potash alum \( K_2SO_4.Al_2(SO_4)_3.24H_2O \), Chrome alum \( K_2SO_4.Cr_2(SO_4)_3.24H_2O \).
In simple words: Alums are special salts made from two different sulfates, usually containing aluminum. They are known for helping clean water and have a general chemical pattern.

🎯 Exam Tip: Remember the general formula for alums and be able to provide specific examples like Potash alum and Chrome alum.

 

Question 20. Aqueous solution of carbon di oxide is acidic. Why?
Answer: An aqueous solution of carbon dioxide is slightly acidic because carbon dioxide reacts with water to form carbonic acid. Carbonic acid then dissociates, releasing hydrogen ions (\( H^+ \)), which makes the solution acidic. This process is why carbonated drinks have a slightly tart taste.
\[ CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^- \]
(Carbonic acid)
In simple words: Carbon dioxide mixed in water becomes slightly acidic because it forms carbonic acid, which then releases acidic hydrogen ions.

🎯 Exam Tip: Focus on the formation of carbonic acid and its subsequent dissociation into \( H^+ \) ions as the key reason for acidity.

 

Question 21. How is silicon tetra choride prepared?
Answer: Silicon tetrachloride (\( SiCl_4 \)) can be prepared in two main ways: by passing dry chlorine over a mixture of silica and carbon at high temperatures, or commercially by reacting silicon with hydrogen chloride gas. These methods efficiently produce the compound for industrial use.
\( SiCl_4 \) is prepared by passing dry chlorine over an intimate mixture of silica and carbon heated to 1675 K in a porcelain tube.
\[ SiO_2 + 2C + 2Cl_2 \rightarrow SiCl_4 + 2CO \]
\( SiCl_4 \) is prepared commercially by the reaction of silicon with hydrogen chloride gas above 600 K.
\[ SiO + 4HCl \rightarrow SiCl_4 + 2H_2 \]
In simple words: Silicon tetrachloride is made by either heating silica and carbon with chlorine gas, or by reacting silicon with hydrogen chloride gas.

🎯 Exam Tip: Note the different reactants and conditions for laboratory versus commercial preparation, including the high temperatures required for both methods.

 

Question 22. Write the uses of silicon tetra chloride.
Answer: Silicon tetrachloride is a versatile compound with several important applications in various industries. Its chemical properties make it valuable for producing specialized materials.
1. It is used in the production of semi-conducting silicon, which is essential for electronic components.
2. It serves as a starting material for synthesizing silica gel, silicic esters, and as a binder for ceramic materials. Silica gel is widely used as a desiccant.
In simple words: Silicon tetrachloride is used to make silicon for electronics and also for making materials like silica gel and ceramic binders.

🎯 Exam Tip: When listing uses, focus on key applications that highlight the unique properties of the compound, like its role in semiconductor production.

 

Question 23. What is water gas equilibrium? (PTA – 5)
Answer: Water gas equilibrium refers to the reversible chemical reaction between carbon dioxide and hydrogen, which produces carbon monoxide and water. This equilibrium reaction has many industrial applications, especially in the production of fuels and chemicals. It's a key process for adjusting the ratio of hydrogen to carbon monoxide.
The equilibrium involved in the reaction between carbon dioxide and hydrogen has many industrial applications and is called water gas equilibrium.
\[ CO_2 + H_2 \rightleftharpoons CO + H_2O \]
In simple words: Water gas equilibrium is a balanced chemical reaction where carbon dioxide and hydrogen can change into carbon monoxide and water, and back again. This reaction is important in factories.

🎯 Exam Tip: Understand that this is a reversible reaction, meaning it can proceed in both directions, and its position of equilibrium can be shifted by changing conditions.

 

Question 1. How is borax prepared from colemanite?
Answer: Borax is prepared from colemanite ore by boiling its solution with sodium carbonate. This process converts the insoluble calcium borate in colemanite into soluble borax. This is an important industrial method for obtaining borax.
When colemanite ore solution is boiled with sodium carbonate solution, borax is obtained.
\[ 2Ca_2B_6O_{11} + 3Na_2CO_3 + H_2O \rightarrow Na_2B_4O_7 + 3CaCO_3 + Ca(OH)_2 \]
In simple words: Borax is made by boiling a solution of colemanite rock with sodium carbonate, which changes the colemanite into borax and other byproducts.

🎯 Exam Tip: Focus on the main reactants (colemanite, sodium carbonate) and the key product (borax) in this preparation method.

 

Question 2. Write the uses of boron.
Answer: Boron is a versatile element with several significant applications, ranging from nuclear technology to everyday products. Its unique properties make it valuable in various fields.
1. Boron-10 absorbs neutrons, making it useful as a moderator in nuclear reactors to control nuclear reactions.
2. Amorphous boron is used as a rocket fuel igniter because of its high reactivity.
3. Boron is an essential element for the cell walls of plants, playing a crucial role in plant structure and growth.
4. Boric acid and borax, boron compounds, are used in eye drops, as antiseptics, and in washing powders.
5. Boric oxide, another boron compound, is used in the manufacture of pyrex glass due to its heat-resistant properties.
In simple words: Boron is used in nuclear reactors, as a rocket fuel starter, and is important for plants. Its compounds like boric acid and borax are found in eye drops and heat-resistant glass.

🎯 Exam Tip: When asked for uses, try to list applications that demonstrate the element's diverse properties (e.g., nuclear, biological, material science).

 

Question 3. Aqueous solution of borax is basic. Why?
Answer: An aqueous solution of borax is basic because, in hot water, borax dissociates into boric acid and sodium hydroxide. Boric acid is a weak acid, while sodium hydroxide is a strong base. The presence of the strong base (sodium hydroxide) makes the overall solution basic. This is an example of hydrolysis of a salt of a strong base and a weak acid.
In hot water, borax dissociates into boric acid and sodium hydroxide.
\[ Na_2B_4O_7 + 7H_2O \rightarrow 4H_3BO_3 + 2NaOH \]
Boric acid is a weak acid, whereas sodium hydroxide is a strong base.
As a result, the resulting solution is basic.
In simple words: Borax water is basic because when borax dissolves, it forms a strong base (sodium hydroxide) and a weak acid (boric acid). The strong base makes the solution basic.

🎯 Exam Tip: To explain why a salt solution is acidic or basic, always identify the nature of the acid and base from which the salt is formed after hydrolysis.

 

Question 4. What is the action of heat on borax?
Answer: When borax is heated, it first loses its water of crystallization. Upon further heating, it decomposes into sodium metaborate and boron trioxide, which appears as transparent glassy beads. This change is visible as the borax melts and forms a glass-like substance.
On heating borax, it loses its water of crystallization first and then decomposes into sodium metaborate and boron trioxide.
Boron trioxide appears as transparent glassy beads.
\[ Na_2B_4O_7.10H_2O \xrightarrow{\Delta \text{, -10 H_2O}} Na_2B_4O_7 \]
\[ Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3 \]
In simple words: Heating borax first removes its water, then it breaks down into sodium metaborate and clear, glassy boron trioxide.

🎯 Exam Tip: Describe both steps of the decomposition: dehydration followed by breakdown into metaborate and trioxide, and mention the physical appearance of the final product.

 

Question 5. What is the action of heat on boric acid?
Answer: The action of heat on boric acid depends on the temperature applied, leading to different products at various heating stages. This thermal decomposition process shows how temperature can control chemical transformations. The table summarizes the products formed at different temperatures.

\[ H_3BO_3 \xrightarrow{373 K} HBO_2 + H_2O \]
\[ 4HBO_2 \xrightarrow{413 K} H_2B_4O_7 + H_2O \]
\[ H_2B_4O_7 \xrightarrow{\text{Red hot}} 2B_2O_3 + H_2O \]
In simple words: Heating boric acid creates different products depending on how hot it gets. It can turn into meta boric acid, then tetra boric acid, and finally a glassy substance called boric anhydride at very high heat.

🎯 Exam Tip: When asked about reactions with varying conditions, remember to specify the temperature for each distinct product formed from boric acid.

 

Question 6. Describe the structure of boric acid.
Answer: Boric acid has a two-dimensional layered structure. It is composed of planar triangular \( [BO_3]^{3-} \) units that are linked to each other by hydrogen bonds. These hydrogen bonds connect the units into extensive sheets, giving boric acid its characteristic properties. This layered arrangement allows the layers to slide over each other.
Boric acid has a two-dimensional structure.
It consists of \( [BO_3]^{3-} \) units.
These units are linked to each other by hydrogen bonds.
In simple words: Boric acid has a flat, sheet-like structure where many triangular \( BO_3 \) parts are connected by hydrogen bonds.

🎯 Exam Tip: Key aspects to mention for boric acid's structure are its two-dimensional nature, the planar \( [BO_3]^{3-} \) units, and the hydrogen bonding between them.

 

Question 7. Write the uses of boric acid.
Answer: Boric acid is a compound with various practical applications due to its mild antiseptic and fluxing properties. These uses span across manufacturing, healthcare, and preservation.
Boric acid is:
1. Used in the manufacture of pottery glazes, glass, enamels, and pigments, where it acts as a flux to lower melting points.
2. Used as an antiseptic, particularly in eye washes and mild disinfectants, due to its ability to kill bacteria.
3. Used as an eye lotion, often in diluted solutions, to cleanse and soothe the eyes.
4. Used as a food preservative, especially for certain food items, to prevent spoilage.
In simple words: Boric acid is used to make glass and pottery coatings, as a mild germ-killer for eyes, and to keep food from spoiling.

🎯 Exam Tip: For uses of a compound, try to categorize them (e.g., industrial, medicinal) and provide specific examples within each category.

 

Question 8. How is diborane prepared?
Answer: Diborane can be prepared through several chemical reactions, usually involving the reduction of boron compounds. Two common laboratory methods include reacting sodium borohydride with iodine in diglyme or heating magnesium boride with hydrochloric acid. These methods allow chemists to synthesize diborane for various research and industrial purposes.
When sodium borohydride in diglyme is reacted with iodine, diborane is obtained.
\[ 2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2 \]
On heating magnesium boride with \( HCl \), a mixture of volatile boranes is obtained.
\[ 2Mg_3B_2 + 12HCl \rightarrow 6MgCl_2 + B_4H_{10} + H_2 \]
\[ B_4H_{10} + H_2 \rightarrow 2B_2H_6 \]
In simple words: Diborane can be made by mixing sodium borohydride with iodine, or by heating magnesium boride with acid, which produces a mix of boron-hydrogen compounds including diborane.

🎯 Exam Tip: Remember the specific reagents and conditions (like solvent "diglyme") for each preparation method, as they are crucial details.

 

Question 9. Write the uses of diborane.
Answer: Diborane is a significant chemical compound with several important applications, primarily due to its strong reducing power and high energy content. These properties make it valuable in specialized fields.
Diborane is:
1. Used as a high-energy fuel for propellants in rockets because of its vigorous combustion.
2. Used as a reducing agent in organic synthesis, helping to add hydrogen to other organic molecules.
In simple words: Diborane is used as a powerful fuel for rockets and as a chemical tool to add hydrogen to other organic compounds.

🎯 Exam Tip: Highlight the "high-energy fuel" and "reducing agent" properties as these are the two primary uses for diborane.

 

Question 10. How is boron trifluoride prepared from boron trioxide?
Answer: Boron trifluoride (\( BF_3 \)) can be prepared from boron trioxide (\( B_2O_3 \)) through reactions with fluoride sources and other reactants. One method involves treating boron trioxide with calcium fluoride in the presence of concentrated sulfuric acid. Another way is by reacting boron trioxide with carbon and fluorine. These reactions are important for synthesizing \( BF_3 \), which is a valuable Lewis acid.
When boron trioxide is treated with calcium fluoride in the presence of conc. sulfuric acid, boron trifluoride is obtained.
\[ B_2O_3 + 3CaF_2 + 3H_2SO_4 \xrightarrow{\Delta} 2BF_3 + 3CaSO_4 + 3H_2O \]
When boron trioxide is reacted with carbon and fluorine, boron trifluoride is obtained.
\[ B_2O_3 + 3C + 3F_2 \rightarrow 2BF_3 + 3CO \]
In simple words: Boron trifluoride is made from boron trioxide by mixing it with calcium fluoride and sulfuric acid, or by reacting it with carbon and fluorine gas.

🎯 Exam Tip: Remember both methods and the specific reactants involved, especially the role of sulfuric acid as a dehydrating agent in the first reaction.

 

Question 11. How is boron trifluoride prepared in the laboratory?
Answer: In the laboratory, pure boron trifluoride (\( BF_3 \)) is prepared by the thermal decomposition of benzene diazonium tetrafluoroborate. This method offers a controlled way to synthesize \( BF_3 \) for experimental purposes. Heating the complex compound leads to the release of nitrogen gas and the desired product.
In the laboratory, pure \( BF_3 \) is prepared by the thermal decomposition of benzene diazonium tetrafluoroborate.
\[ PhN_2BF_4 \xrightarrow{\Delta} BF_3 + PhF + N_2 \]
In simple words: In the lab, pure \( BF_3 \) is made by heating a chemical called benzene diazonium tetrafluoroborate, which breaks down to give \( BF_3 \), a type of fluorine benzene, and nitrogen gas.

🎯 Exam Tip: For laboratory preparations, focus on the specific starting material and the controlled conditions, such as thermal decomposition.

 

Question 12. How is potash alum prepared? (PTA – 4)
Answer: Potash alum can be prepared from alunite (also known as alum stone) by treating it with sulfuric acid, or from a mixture of aluminum sulfate and potassium sulfate. After the reaction, the solution is crystallized to obtain pure potash alum. This process highlights the formation of double salts.
Potash alum is prepared from alunite or alum stone.
When alunite is treated with excess sulfuric acid, aluminum sulfate is formed.
A calculated quantity of potassium sulfate is added.
The solution is crystallized to obtain potash alum.
It is purified by recrystallization.
\[ K_2SO_4.Al_2(SO_4)_3.4Al(OH)_3 + 6H_2SO_4 \rightarrow K_2SO_4 + 3Al_2(SO_4)_3 + 12H_2O \]
\[ K_2SO_4 + Al_2(SO_4)_3 + 24H_2O \rightarrow K_2SO_4.Al_2(SO_4)_3.24H_2O \]
In simple words: Potash alum is made from alunite by adding sulfuric acid, then mixing with potassium sulfate and letting it form crystals.

🎯 Exam Tip: Remember the starting materials (alunite, sulfuric acid, potassium sulfate) and the final purification step (recrystallization) for preparing potash alum.

 

Question 13. Write the uses of alum.
Answer: Alum, particularly potash alum, has numerous applications across various industries and in daily life, mainly due to its coagulant and astringent properties. These uses demonstrate its versatility.
Alum is used:
1. For the purification of water, where it acts as a coagulant to settle suspended impurities.
2. For waterproofing fabrics and textiles, by forming a water-resistant layer.
3. In dyeing, paper, and leather tanning industries, as a mordant to fix dyes, in papermaking to improve paper strength, and in tanning to preserve leather.
4. As a styptic agent to arrest bleeding by coagulating blood, often used in shaving cuts.
In simple words: Alum is used to clean water, make clothes waterproof, help dyes stick, and stop bleeding from small cuts.

🎯 Exam Tip: When listing uses, focus on the practical applications and the underlying property of alum (e.g., coagulant for water purification, styptic for bleeding).

 

Question 14. Write the uses of carbon monoxide.
Answer: Carbon monoxide (\( CO \)) is a gas with several important industrial applications, primarily as a reducing agent and a component in synthesis gas. Its properties make it valuable in metallurgy and chemical production.
Carbon monoxide is used:
1. As a reducing agent, capable of reducing many metal oxides to their respective metals, especially in blast furnaces.
2. As an important ligand, forming metal carbonyls, which are used as catalysts in various chemical processes.
3. A mixture of \( CO \) and \( H_2 \) is known as water gas, and a mixture of \( CO \) and \( N_2 \) is called producer gas. Both are important industrial fuels and raw materials for chemical synthesis.
In simple words: Carbon monoxide is used to extract metals from their ores, to make special metal compounds, and in gas mixtures (like water gas and producer gas) that are used as fuels in industries.

🎯 Exam Tip: Remember the role of carbon monoxide as a reducing agent and its significance as a component of industrial fuel gases like water gas and producer gas.

 

Question 15. Write the uses of carbon dioxide.
Answer: Carbon dioxide (\( CO_2 \)) is an essential gas with diverse applications across various sectors, from industrial processes to biological systems. Its chemical and physical properties contribute to its wide range of uses.
Carbon dioxide is used:
1. To produce an inert atmosphere for chemical processing, preventing unwanted reactions with oxygen or moisture.
2. By plants in photosynthesis, where it is converted into sugars, forming the base of most food chains.
3. As a fire extinguisher, especially for electrical fires, because it is non-combustible and heavier than air, displacing oxygen.
4. As a propellant gas in aerosol cans and in carbonated beverages to create fizz.
5. In the production of carbonated beverages, adding its characteristic effervescence.
6. In the production of foam, such as in certain types of plastics and building materials.
In simple words: Carbon dioxide is used in factories to create a safe atmosphere, by plants for food, to put out fires, to make fizzy drinks, and in some foams.

🎯 Exam Tip: For \( CO_2 \), highlight its roles in photosynthesis, fire suppression, and carbonation, as these are its most recognized applications.

 

Question 16. Write note on Boron Neutron Capture Therapy (BNCT).
Answer: Boron Neutron Capture Therapy (BNCT) is an advanced experimental cancer treatment that uses boron compounds and neutrons to selectively destroy cancer cells while sparing healthy tissue. It works on the principle of a nuclear reaction where boron-10 captures neutrons. This treatment targets specific areas in the body where boron compounds accumulate. The basic mechanism involves the following points:
1. Boron compounds are injected into a brain tumor patient, and these compounds preferentially collect in the tumor cells.
2. The tumor area is then irradiated with thermal neutrons. When boron-10 captures a neutron, it undergoes a nuclear reaction.
3. This nuclear reaction results in the release of highly energetic alpha particles and a lithium particle, which cause damage locally.
4. This alpha-particle damages the tissue in the tumor each time a boron-10 nucleus captures a neutron, effectively destroying the cancer cells.
5. The damage can be limited preferentially to the tumor, leaving the normal brain tissue less affected, due to the short range of the alpha particles.
6. BNCT has been studied as a treatment for several other tumors beyond the brain, including those of the head and neck, breast, prostate, bladder, and liver.
In simple words: BNCT is a cancer treatment where special boron chemicals are put into tumors. When these tumors are hit with neutrons, the boron reacts to release tiny damaging particles that kill only the cancer cells, leaving healthy cells mostly untouched.

🎯 Exam Tip: Focus on the mechanism: boron compound accumulation in tumor, neutron capture by boron-10, release of destructive alpha particles, and localized damage.

 

Question 1. How is higher boranes obtained from diborane.
Answer: Higher boranes are obtained from diborane through a process of thermal decomposition and rearrangement reactions at different high temperatures, liberating hydrogen gas. The specific borane produced depends on the heating conditions. This allows for the synthesis of various complex boron hydrides.
At high temperatures, diborane forms higher boranes liberating hydrogen.
\[ 5B_2H_6 \xrightarrow{\text{388 K, U-tube}} 2B_5H_{11} + 4H_2 \]
\[ 2B_2H_6 \xrightarrow{\text{198-373 K}} B_4H_{10} + H_2 \]
\[ 5B_2H_6 \xrightarrow{\text{373 K, sealed tube}} B_{10}H_{14} + 8H_2 \]
\[ 5B_2H_6 \xrightarrow{\text{473-523 K}} 2B_5H_9 + 6H_2 \]
\[ 10B_2H_6 \xrightarrow{\text{523 K}} 2B_9H_{15} + 2B_{10}H_{14} + 11H_2 \]
\[ B_2H_6 \xrightarrow{\text{Red hot}} 2B + 3H_2 \]
In simple words: You can get larger boron-hydrogen compounds (higher boranes) from diborane by heating it at different temperatures, which causes it to break apart and re-form, releasing hydrogen gas.

🎯 Exam Tip: Remember that temperature is the key factor determining which higher borane is formed. Note at least two specific examples with their temperatures and products.

 

Question 2. Explain the allotropes of carbon.
Answer: Carbon is unique because it exists in many different structural forms called allotropes. These allotropes have distinct physical and chemical properties even though they are all made of only carbon atoms. The most common and important allotropes are graphite, diamond, fullerenes, and carbon nanotubes. This diversity arises from different ways carbon atoms can bond together.
Carbon exists in many allotropic forms.
Graphite and diamond are the most common allotropes.
Graphene, fullerenes, and carbon nanotubes are other important allotropes of carbon.

**Graphite:**
It is the most stable allotrope of carbon at normal temperature and pressure.
It is made of flat, two-dimensional hexagonal sheets of \( sp^2 \) hybridised carbon atoms. The C-C bond length is \( 1.41 \, \text{Å} \), which is close to the C-C bond distance in benzene (\( 1.40 \, \text{Å} \)).
Each carbon atom forms three sigma bonds with three neighboring carbon atoms using three of its valence electrons. The fourth electron, present in the unhybridized p-orbital, forms a pi-bond.
These pi-electrons are delocalized over the entire sheet, which allows graphite to conduct electricity.
Successive carbon sheets are held together at a distance of \( 3.40 \, \text{Å} \) by weak Van der Waals forces. This weak bonding between layers makes graphite soft, slippery, and useful as a lubricant.

**Diamond:**
Carbon atoms in diamond are \( sp^3 \) hybridised.
Each carbon atom is bonded tetrahedrally with four other carbon atoms by strong sigma bonds. The C-C bond length is \( 1.54 \, \text{Å} \). This strong, rigid network makes diamond very hard.
Since all four valence electrons of carbon are involved in bonding, there are no free electrons, meaning diamond is not a conductor of electricity.
Being the hardest known substance, diamond is used for sharpening hard tools, cutting glass, making bores, and rock drilling.

**Fullerenes:**
These are newly synthesized allotropes of carbon.
Unlike graphite and diamond, fullerenes are discrete molecules of carbon, such as \( C_{32}, C_{50}, C_{60}, C_{70}, C_{76} \), and so on.
They have cage-like structures.
Buckminsterfullerene, or "buckyball," has a soccer ball-like structure with the formula \( C_{60} \).
It has a fused ring structure with 20 six-membered rings and 12 five-membered rings.
Each carbon is \( sp^2 \) hybridised and forms three sigma bonds and one delocalized pi-bond, giving it aromatic character.
The C-C bond distance is \( 1.44 \, \text{Å} \) and the C=C bond distance is \( 1.38 \, \text{Å} \).

**Carbon nanotubes:**
This is another recently discovered allotrope of carbon.
They have graphite-like tubes with fullerene ends.
Along the axis, carbon nanotubes are stronger than steel and can conduct electricity.
They have many applications in nanoscale electronics, catalysis, polymers, and medicine.
In simple words: Carbon can take many forms called allotropes, like soft graphite (used in pencils), hard diamond, cage-like fullerenes, and strong carbon nanotubes. Each form has different uses because their carbon atoms are arranged and bonded in unique ways.

🎯 Exam Tip: For each allotrope, remember its hybridization, bonding pattern, unique structure (e.g., layered for graphite, tetrahedral for diamond, cage-like for fullerene), and one key property or use (e.g., conductor/insulator, hardness).

 

Question 3. Write about the preparation and structure of silicones.
Answer: Silicones, also known as polysiloxanes, are special types of organosilicon polymers. Their basic empirical formula is \((R_2SiO)\), which looks similar to ketones \((R_2CO)\), leading to their name "silicones". These compounds can have either a straight chain (linear) or a linked (cross-linked) structure. They are highly stable even at high temperatures, making them useful as high-temperature polymers. Silicones are crucial materials in many industries due to their versatility and stability.
**Types of Silicones:**
* **Linear Silicones:** These are made by reacting dialkyl or diaryl dichlorosilanes (\(R_2SiCl_2\) or \(Ar_2SiCl_2\)) with water (hydrolysis), followed by a process called condensation.
\(R_2SiCl_2 + 2H_2O \rightarrow HO-Si(R)_2-OH + 2HCl\)
\(n(HO-Si(R)_2-OH) \rightarrow (-Si(R)_2-O-)_n + nH_2O\)
Examples include silicone rubbers, which are joined by methylene groups, and silicone resins, formed by mixing silicones with organic resins like acrylic esters.
The structure of linear silicones shows a long chain:
\(HO-Si(R)_2-O-Si(R)_2-O-Si(R)_2-OH\)
* **Cyclic Silicones:** These are also formed by hydrolyzing \(R_2SiCl_2\). Linear silicones can become cyclic if water molecules are removed from their ends.
An example of a cyclic structure (for R=Me, methyl group):
\(Me-Si-O-Si-Me\)
\(\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,|\)
\(\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,O\)
\(\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,|\)
\(Me-Si-O-Si-Me\)
* **Cross-linked Silicones:** When \(RSiCl_3\) is hydrolyzed, it creates a very complex, branched (cross-linked) polymer structure.
A general representation of a cross-linked structure:
\(-O-Si(R)-O-Si(R)-O-\)
\(\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\)
\(\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\)
\(\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\)
\(-O-Si(R)-O-Si(R)-O-\)
**Preparation:**
Silicones are prepared by passing the vapor of alkyl or aryl halides (RCl or ArCl) over silicon at 570 K, using a copper catalyst. This reaction produces dialkyl or diaryl dichlorosilanes.
\(2RCl + Si \xrightarrow{Cu/570K} R_2SiCl_2\)
In simple words: Silicones are man-made materials with silicon, oxygen, and carbon parts, used in many things because they are strong and can handle heat well. They are made by taking certain silicon compounds, mixing them with water, and then joining them together into long chains or networks.

🎯 Exam Tip: Remember to clearly differentiate between the empirical formula, the types of silicones, and their general preparation method. Including basic reaction equations helps in scoring full marks.

 

Question 4. Explain various types of silicates.
Answer: Silicates are minerals made from silicon and oxygen atoms. Their basic building block is a tetrahedral unit of \([SiO_4]^{4-}\). These units link together in different ways, creating various types of silicate structures. Understanding these structures helps explain the properties of many common rocks and minerals around us.
Here are the different types of silicates based on how their \([SiO_4]^{4-}\) units are linked:

In simple words: Silicates are rock-forming minerals that have silicon and oxygen. They come in different kinds based on how their basic building blocks, called \([SiO_4]^{4-}\) units, connect to each other. Some are single units, some form chains, and some make rings or complex layers.

🎯 Exam Tip: When describing silicate types, always mention the basic structural unit \([SiO_4]^{4-}\) and how these units link together. Providing an example for each type is crucial for full marks.