One of the products formed due to the re-action between KMnO4 and HCI is
(A) red liquid
(B) MnO2
(C) greenish yellow gas
(D) HCIO4
Explanation: In redox reactions involving strong oxidizing agents reacting with concentrated Acids, a complex electron transfer process occurs. Potassium permanganate acts as a powerful oxidizer, while hydrochloric Acid provides chloride ions and an acidic medium. During the reaction, manganese in permanganate undergoes reduction to a lower oxidation state, while chloride ions can be oxidized to a diatomic elemental form. Such reactions typically proceed through simultaneous oxidation and reduction steps, making them classic examples of redox Chemistry in Inorganic systems. The products formed depend on the medium and reaction conditions, especially acidity and concentration. The transformation involves changes in oxidation states, gas Evolution, and formation of metal Salts. These processes are commonly studied to understand oxidizing strength and product prediction in qualitative analysis and laboratory reactions.
Option c – greenish yellow gas
Which of the following is not an element?
(A) Graphite
(B) Diamond
(C) 22-Carat gold
(D) Rhombic sulphur
Explanation:Matter in Chemistry is broadly classified into elements, compounds, and mixtures. Elements are pure substances made up of only one kind of Atom and cannot be broken down further by chemical means. However, some materials appear elemental but are actually different forms of the same element arranged in distinct structural forms. These variations arise due to different Bonding and atomic arrangements, leading to unique physical properties while still representing the same basic chemical substance. In contrast, alloys or combinations containing multiple elements are not considered pure elements because they involve more than one type of Atom chemically or physically combined. Understanding this distinction is important in material classification, Metallurgy, and Periodic Table concepts.
Option c – 22-Carat gold
Green vitriol is
(A) FeSO4.7H₂O
(B) ZnSO4.7H₂O
(C) CaSO4 2H₂O
(D) CuSO4.5H₂O
Explanation: Hydrated Salts of transition Metals often exhibit characteristic names based on historical mineralogy and their crystalline forms. These compounds contain a central metal ion surrounded by water molecules in a fixed ratio, which significantly influences their color, stability, and chemical behavior. The presence of ferrous ions in hydrated sulfate structures leads to specific greenish coloration due to electronic transitions within the d-orbitals. Such substances are widely studied in coordination Chemistry because they demonstrate how water of crystallization affects ionic compounds. These hydrated forms are also important in industrial and laboratory processes, including as reducing agents and in analytical Chemistry for detecting oxidation-reduction changes. The study of such compounds helps in understanding oxidation states of Metals and the role of coordinated water molecules in stabilizing crystal structures.
Option a – FeSO4.7H₂O
Wrought iron, pig iron and steel differ in properties due to
(A) carbon content
(B) malleability
(C) conductivity
(D) softness
Explanation: Different forms of iron-based materials are distinguished by variations in their composition and internal structure. The presence of carbon and other impurities significantly affects hardness, brittleness, malleability, and tensile strength. In Metallurgy, controlled variation in impurity levels leads to materials with specific Mechanical Properties suited for different engineering applications. Iron alloys with higher carbon content tend to be harder but more brittle, while those with lower carbon content are more ductile and workable. These differences arise from how carbon atoms are distributed within the metallic lattice, influencing dislocation movement and crystal structure stability. Understanding these variations is crucial in material science for selecting appropriate Metals for construction, tools, and machinery.
Option a – carbon content
The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is.
(A) +4
(B) +6
(C) +2
(D) +3
Explanation: In redox systems involving dichromate ions in acidic medium, chromium undergoes a change in oxidation state as it acts as an oxidizing agent. Iodide ions are typically oxidized while chromium species are reduced in the process. The final chromium-containing product forms after accepting electrons during the reaction sequence. Oxidation states are determined by assigning standard electron accounting rules based on electronegativity and charge balance in compounds. Such transformations are commonly used in volumetric analysis and redox titrations, where color changes help indicate reaction progress. The behavior of chromium in different oxidation states is central to understanding its chemical versatility and reactivity in aqueous solutions.
Option d – +3
How many electrons are involved in the oxidation by KMnO4 in basic medium?
(A) 1
(B) 2
(C) 5
(D) 3
Explanation: Potassium permanganate is a strong oxidizing agent whose reduction pathway depends strongly on the reaction medium. In basic conditions, the reduction of permanganate follows a different electron transfer route compared to acidic media. Electron balance methods are used to determine how many electrons are gained or lost during the redox change. This involves comparing oxidation states of manganese before and after the reaction and applying conservation of charge principles. Such calculations are essential in balancing redox equations and predicting stoichiometric relationships in chemical reactions. The concept is widely applied in analytical Chemistry, especially in titrimetric analysis involving oxidizing agents.
Option a – 1
The number of unpaired electrons in a nickel Atom (ground state) are (At. No. of Ni = 28)
(A) 2
(B) 5
(C) 3
(D) 7
Explanation: Electronic configuration and orbital filling principles determine magnetic and chemical properties of transition metal atoms. Electrons occupy orbitals following the Aufbau principle, Hund’s rule, and Pauli exclusion principle. In partially filled d-subshells, electrons distribute themselves to maximize spin multiplicity before pairing occurs. The number of unpaired electrons directly affects magnetic behavior such as paramagnetism. Transition Metals often exhibit multiple unpaired electrons due to incomplete d-orbital filling. Understanding orbital diagrams helps in predicting Bonding behavior, complex formation, and reactivity of metal ions and atoms.
Option a – 2
Which one of the following compounds does not decolourise an acidified aqueous solution of KMnO4?
(A) Sulphur dioxide
(B) Ferric chloride
(C) Hydrogen peroxide
(D) Ferrous sulphate
Explanation: Acidified potassium permanganate is a strong oxidizing agent that undergoes color change when it reacts with reducing substances. Compounds capable of reducing MnO₄⁻ typically cause decolorization due to reduction of manganese to a lower oxidation state. Substances that are already in a high oxidation state or lack reducing ability do not participate in this electron transfer process. Redox behavior depends on electron donation capacity and stability of oxidation states. Such reactions are used in qualitative Organic and Inorganic analysis to identify reducing agents based on observable color changes.
Option b – Ferric chloride
Which of the following contains the maximum number of unpaired electrons
(A) TiCl3
(B) MnCl2
(C) FeSO4
(D) CuSO4
Explanation: The number of unpaired electrons in transition metal compounds depends on their electronic configuration and the nature of ligand field splitting. In crystal field theory, d-orbitals split into different energy levels in the presence of ligands, affecting electron distribution. Weak field ligands allow maximum unpaired electrons, while strong field ligands promote pairing. This directly influences magnetic properties and stability of coordination compounds. Comparing different metal ions requires analyzing their oxidation states and d-electron counts. Such concepts are fundamental in predicting paramagnetism and spectrochemical behavior of coordination complexes.
Option b – MnCl2
In Cu (Z = 29)
(A) 13 electrons have spin in one direction and 16 electrons in other direction
(B) 14 electrons have spin in one direction and 15 electrons in other direction
(C) all the electrons have spin in one direction
(D) none of these
Explanation: Copper, with atomic number 29, exhibits unique electron configuration behavior due to stability associated with completely filled and half-filled subshells. Electron distribution in atoms follows quantum mechanical principles where energy minimization governs configuration. In some transition Metals, slight deviations from expected filling order occur to achieve more stable arrangements. Spin distribution of electrons in orbitals determines magnetic properties and spectral behavior. Understanding electron configuration helps explain chemical reactivity, Bonding tendencies, and formation of complex ions.
Option b – 14 electrons have spin in one direction and 15 electrons in other direction
Explanation: Double Salts are crystalline compounds formed from two different Salts that crystallize together in a fixed stoichiometric ratio. These compounds dissociate into their constituent ions when dissolved in water, unlike coordination complexes. Hydration plays a key role in stabilizing crystal structure and influencing physical properties such as solubility and color. Such compounds are important in analytical Chemistry for standardization and titrimetric analysis. Their behavior in solution helps distinguish between simple ionic compounds and coordination complexes.
In the chemical reaction Ag2O + H₂O + 2e-2Ag + 2OH
(A) water is oxidised
(B) electrons are reduced
(C) silver is oxidised
(D) silver is reduced
Explanation: Redox half-reactions describe electron transfer processes where oxidation and reduction occur simultaneously. In such representations, electrons appear explicitly to balance charge and Mass. Metal oxides in aqueous environments can undergo reduction to elemental Metals while generating hydroxide ions. This reflects changes in oxidation state of the metal species. Understanding half-reactions is essential for balancing electrochemical equations and analyzing electrode processes in electrochemistry. These concepts are widely used in batteries, corrosion science, and electroplating.
Option d – silver is reduced
Which of the following compounds is expected to be coloured?
(A) Ag2SO4
(B) CuF2
(C) MgF2
(D) CuCl
Explanation: Colour in Inorganic compounds is mainly associated with electronic transitions involving d-orbitals in transition metal ions. When electrons absorb visible Light, they get promoted between split d-energy levels, and the remaining reflected or transmitted Light gives the compound its observed colour. This behaviour is especially common in partially filled d-subshells. Compounds with completely filled or empty d-orbitals generally do not show such transitions and often appear colourless. Ligand Environment, oxidation state, and crystal field strength also influence the energy gap between d-orbitals, which in turn affects the wavelength of Light absorbed. Hence, identifying coloured compounds requires understanding electronic configuration and coordination Environment of the central metal ion.
Option b – CuF2
In the preparation of KMnO4, pyrolusite (MnO2) is first converted to potassium manganate (K2MnO4). In this conversion, the oxidation state of manganese changes from
(A) + 1 to +3
(B) +2 to +4
(C) +3 to +5
(D) +4 to +6
Explanation: Oxidation state changes are central to redox chemistry and involve tracking electron transfer during reactions. Manganese compounds can exist in multiple oxidation states depending on reaction conditions and oxidizing or reducing environments. In industrial preparation processes, manganese dioxide undergoes chemical transformation in alkaline conditions where oxygen balance and electron exchange determine the final oxidation state. Such conversions are important in Inorganic synthesis and analytical chemistry. The change in oxidation state reflects whether manganese is oxidized or reduced during the process, and this helps classify the reaction as a redox transformation. Understanding oxidation state progression is essential for predicting reaction pathways and product formation in transition metal chemistry.
Option d – +4 to +6
Ionic character of halides of Metals (3d-transition series) decreases in the order
(A) M-I > M-Br > M-CI > M-F
(B) M-CI > M-Br > M-1 > M-F
(C) M-Br > M-CI > M-F> M-I
(D) M-F> M-CI > M-Br > M-I
Explanation: Bond character in metal halides depends on the extent of electron sharing between metal and halogen atoms. Ionic character is influenced by factors such as size, charge density, and polarizing power of the metal ion. Smaller and more highly charged metal ions tend to polarize the electron cloud of halide ions, increasing covalent character. In contrast, larger halide ions are more easily polarized, affecting bond nature differently. Trends across halides reflect Periodic properties and Fajan’s rules, which explain how ionic and covalent characteristics shift depending on ion interactions. These principles help in predicting Bonding behavior in transition metal compounds.
Option d – M-F> M-CI > M-Br > M-I
Which of the following group of transition Metals is called coinage Metals
(A) Cu, Ag, Au
(B) Ru, Rh, Pd
(C) Fe, Co, Ni
(D) Os, Ir, Pt
Explanation: Transition Metals are classified into groups based on their chemical properties, electronic configuration, and historical usage. Some metals have been widely used in currency due to their resistance to corrosion, malleability, and stability in elemental form. These properties make them suitable for minting coins and ornamental purposes. Their relatively low reactivity compared to other transition metals is due to filled or nearly filled d-subshells, which stabilize the metallic state. Understanding Periodic trends helps explain why certain metals are more suitable for practical applications such as coinage, electrical conductors, and catalytic materials.
Option a – Cu, Ag, Au
At 20°C, the vapour pressure of diethyl Ether is 442 mm. When 6.4 g of a non-volatile solute is dissolved in 50 g of solvent, the vapour pressure of the solution becomes 410 mm. The MolecularMass of the solute is
(A) 150
(B) 130.9
(C) 160
(D) 180
Explanation: Vapour pressure lowering is a colligative property that depends on the number of solute particles present in a solution rather than their nature. When a non-volatile solute is added to a solvent, the escaping tendency of solvent molecules decreases, reducing vapour pressure. This effect is proportional to the mole fraction of solute in solution, as described by Raoult’s law. By comparing vapour pressure of pure solvent and solution, one can determine solute molar Mass using Mass-to-mole relationships. Such calculations are widely used in physical chemistry to determine unknown Molecular weights of solutes experimentally.
Option b – 130.9
One molar glucose solution boils at 100.54°C. The relative decrease in vapour pressure of aq. urea solutions are 0.018. The elevation in the boiling point of the urea solution is
(A) 0.18°C
(B) 0.36°C
(C) 0.72°C
(D) 0.54°C
Explanation: Boiling point elevation is a colligative property that arises due to lowering of vapour pressure when a non-volatile solute is added to a solvent. The extent of boiling point increase depends on solute concentration and the ebullioscopic constant of the solvent. Relative lowering of vapour pressure is directly related to mole fraction of solute particles, which can be used to estimate boiling point changes. These relationships are important in solution chemistry for understanding how solutes affect phase transition temperatures. Such principles are widely applied in chemical analysis and industrial processes involving solution behavior.
Option d – 0.54°C
A solution containing 0.1 g of non-volatile solute P (molar Mass 100 g) in 100 g of benzene gives rise in boiling point by 0.2 K. Another solution containing 0.1 g of ‘Q’ non-volatile solute in the same amount of the same solvent gives elevation in B.P. by 0.4 K. The ratio of the molar Mass of P and Q is
(A) 1 : 2
(B) 2 : 1
(C) 1 : 4
(D) 4 : 1
Explanation: Colligative properties such as boiling point elevation depend on the number of solute particles in a given amount of solvent. A smaller molar Mass leads to a greater number of moles for the same Mass of solute, resulting in a larger effect on boiling point. By comparing two solutions under identical conditions, relative molar masses can be deduced using proportional relationships between concentration and observed temperature change. This approach is widely used in determining unknown Molecular masses in physical chemistry. The concept highlights the dependence of solution properties on particle concentration rather than chemical identity.
Option b – 2 : 1
An aq. solution of sucrose-containing 34.2 g has an osmotic pressure of 2.38 atm at 17°C for an aq. solution of glucose to be isotonic with this solution would have Mass
(A) 34.2 g
(B) 17.1 g
(C) 18.0 g
(D) 36.0 g
Explanation: Osmotic pressure is a colligative property that depends on the number of solute particles in a solution and temperature. Isotonic solutions have equal osmotic pressure, meaning they exert the same pressure across a semipermeable membrane. Since different solutes may have different molar masses, their masses must be adjusted to maintain equal particle concentration. This involves using molarity relationships and proportionality between moles of solute and volume of solution. Such principles are important in biological systems and solution chemistry, especially in maintaining cell stability and Fluid balance.
Option c – 18.0 g
The osmotic pressure of 5% (M/V) solution of sugar at 150°C is
(A) 4 atm
(B) 5.07 atm
(C) 3.55 atm
(D) 2.45 atm
Explanation: Osmotic pressure is a colligative property arising from solute particles dissolved in a solvent and depends directly on concentration and temperature. When a solute like sugar dissolves in water, it does not ionize, so each Molecule contributes as a single particle. As temperature increases, Molecularmotion becomes more vigorous, increasing the pressure exerted by solute particles across a semipermeable membrane. The relationship between osmotic pressure, concentration, and temperature is governed by principles similar to gas laws. Such calculations require converting percentage concentration into molarity and considering absolute temperature. This concept is widely applied in biological and chemical systems to understand how solutions behave under different thermal conditions.
Option b – 5.07 atm
30 g of a solute is added to 90 g of water, and the V.P. of the solution is 21.85 mm at 25°C. If 18 g of water is added to the same solution. V.P. of the solution becomes 22.15 mm. The vapour pressure of the solvent is
(A) 21.35 mm
(B) 75.6
(C) 12:35 mm
(D) 35.21 mm
Explanation: Vapour pressure lowering is explained using Raoult’s law, which relates vapour pressure of a solution to the mole fraction of the solvent. When a non-volatile solute is dissolved, the escaping tendency of solvent molecules decreases, reducing vapour pressure. Adding more solvent increases the mole fraction of the solvent, thereby increasing vapour pressure. By comparing two different solution compositions and their vapour pressures, the original vapour pressure of the pure solvent can be determined. This approach is commonly used in physical chemistry to link experimental data with Molecular properties of solutions.
Option a – 21.35 mm
The isotonic solution has equal
(A) Vapour pressure
(B) Osmotic pressure
(C) Solution pressure
(D) Freezing point
Explanation: Isotonic solutions are defined based on their ability to exert equal osmotic pressure when separated by a semipermeable membrane. This property is important in biological systems where cells maintain water balance with their surrounding Environment. When two solutions are isotonic, there is no NET movement of solvent across the membrane. This equilibrium condition depends on solute concentration and temperature, not on the chemical nature of solutes. Understanding this concept is essential in physiology, medicine, and solution chemistry, especially in intravenous Fluid preparation and cell Transport mechanisms.
Option b – Osmotic pressure
The ratio of the value of any colligative property of KCl solution to that of sugar solution is
(A) 1
(B) 0.5
(C) 2
(D) 4
Explanation: Colligative properties depend on the number of particles dissolved in solution. Electrolytes like KCl dissociate into ions in water, increasing the number of particles, while non-electrolytes like sugar do not dissociate. The extent of dissociation is represented by the van’t Hoff factor, which influences the magnitude of colligative effects. Comparing two solutions under identical conditions allows determination of relative particle concentration effects. This principle is widely used in solution chemistry to understand how ionic compounds produce stronger colligative effects than Molecular compounds at the same molarity.
Option c – 2
An aqueous solution containing 1 g urea boils at 100.25°C. The aq. solution containing 3 g of glucose in the same volume will boil at
(A) 100.75°C
(B) 100.50°C
(C) 100.25°C
(D) 100°C
Explanation: Boiling point elevation is a colligative property that depends on the number of solute particles present in a solution. Non-electrolytes such as urea and glucose do not dissociate, so their effect is directly proportional to the number of moles. Increasing solute Mass increases the number of dissolved particles, leading to a higher boiling point. The relationship between mass, molar mass, and boiling point elevation is used to compare different solutions under identical solvent conditions. This concept is important in Thermodynamics of solutions and helps predict boiling point changes based on solute concentration.
Option c – 100.25°C
A compound that acts as a semipermeable membrane is
(A) Calcium sulphate
(B) Nickel phosphate
(C) Barium oxalate
(D) Copper ferrocyanide
Explanation: Semipermeable membranes allow selective passage of solvent molecules while restricting solute particles. This selective permeability is essential in processes like osmosis, where solvent movement occurs across concentration gradients. Certain natural and synthetic materials exhibit this property due to their Molecular structure and pore size. These membranes play a crucial role in biological systems such as cell walls and kidney filtration, as well as in laboratory and industrial processes like dialysis and purification. Understanding membrane behavior is fundamental in biochemistry and physical chemistry for explaining Fluid balance and MolecularTransport.
Option d – Copper ferrocyanide
Ebullioscopy is concerned with
(A) Osmotic pressure of the solution
(B) Boiling point elevation
(C) Relative lowering in vapour pressure
(D) Freezing point depression
Explanation: Ebullioscopy is the study of boiling point elevation in solutions when a non-volatile solute is added to a solvent. It is a colligative property that depends on the number of solute particles rather than their chemical nature. When solute particles reduce the escaping tendency of solvent molecules, a higher temperature is required for boiling. This phenomenon is used to determine Molecular masses of solutes experimentally. Ebullioscopic measurements are widely used in physical chemistry to study solution behavior and thermodynamic properties of liquids.
Option b – Boiling point elevation
The osmotic pressure of 0.2 M urea solution at 27°C is
(A) 4.92 atm
(B) 0.29 atm
(C) 9.24 atm
(D) 2.49 atm
Explanation: Osmotic pressure depends on solute concentration, temperature, and the nature of the solute. Urea does not dissociate in water, so each Molecule contributes as a single particle. The relationship between osmotic pressure and molarity follows gas-like behavior of solute particles in solution. Temperature must be converted to Kelvin for accurate calculation since Molecularmotion depends on absolute temperature. This principle is important in solution Thermodynamics and biological systems, where osmotic balance plays a key role in maintaining cell structure and function.
Option a – 4.92 atm
The osmotic pressure of the solution increase if
(A) Temperature decreases
(B) the Concentration of solution decreases
(C) No. of solute molecules increases
(D) the Volume increases
Explanation: Osmotic pressure is directly related to the number of solute particles per unit volume and the absolute temperature of the solution. Increasing concentration increases the number of solute particles, thereby increasing the tendency of solvent molecules to move across a semipermeable membrane. Temperature also influences molecular motion, affecting osmotic pressure. These relationships are derived from thermodynamic principles similar to ideal gas behavior. Understanding these factors is essential in predicting solution behavior in biological systems, chemical processes, and medical applications involving Fluid balance.
Option c – No. of solute molecules increases
The solution having the highest boiling point is
(A) 0.05 M glucose
(B) 0.01 M KNO3
(C) 0.015 M urea
(D) 0.01 M Na₂SO4
Explanation: Boiling point elevation depends on the number of solute particles present in a solution and is influenced by both concentration and dissociation behavior. Electrolytes that dissociate into multiple ions produce a greater increase in boiling point compared to non-electrolytes at the same molarity. The van’t Hoff factor accounts for this difference in particle count. Comparing different solutions requires analyzing their effective particle concentration in solution. This concept is widely used in colligative property problems to determine which solution exhibits the greatest boiling point elevation under identical conditions.
Option d – 0.01 M Na₂SO4
Which one is known as oil of vitriol?
(A) H₂SO3
(B) H₂SO4
(C) H₂S₂O7
(D) H₂S₂O8
Explanation: Certain Acids and chemical substances have traditional historical names based on their physical appearance, preparation method, or industrial usage. These names are often used in classical chemistry and industrial contexts. The substance referred to here is a highly corrosive mineral Acid widely used in laboratories and industries. It plays a key role in dehydration reactions, fertilizer production, and chemical synthesis. Its name originated from its oily appearance and its production from vitriols (metal sulfates) by heating. Understanding such nomenclature is important for linking modern chemical names with historical terminology used in older literature and examination contexts.
Option b – H₂SO4
In the preparation of sulphuric Acid, V₂O5 is used in the reaction, which is
(A) S + O₂ → SO₂
(B) 2SO2 + O2 →2SO3
(C) SO2 + H₂O → H₂SO3
(D) N₂ + 3H₂ → 2NH3
Explanation: In industrial chemical processes, catalysts are used to speed up reactions without being consumed. In the contact process for sulphuric Acid production, a catalyst facilitates the conversion of sulfur dioxide into sulfur trioxide. This step is crucial because it determines the efficiency and yield of the final product. Catalysts work by providing an alternative reaction pathway with lower activation energy. The reaction involves oxidation of sulfur dioxide in the presence of oxygen, and the catalyst ensures the process occurs efficiently at optimal conditions. Understanding catalytic action is essential in industrial chemistry for large-scale chemical production and process optimization.
Option b – 2SO2 + O2 →2SO3
The bond angle in H₂S is
(A) 109°28′
(B) 104°5′
(C) 120°
(D) 92.5°
Explanation: Molecular geometry is determined by the arrangement of electron pairs around the central Atom according to VSEPR theory. Lone pair–bond pair repulsions influence bond angles and molecular shape. In hydrides of group 16 elements, the bond angle decreases as atomic size increases due to weaker bond pair repulsions and stronger lone pair effects. The presence of two lone pairs on sulfur leads to a bent structure rather than a linear arrangement. Comparing such molecules helps in understanding how electronegativity, atomic size, and lone pair interactions influence molecular geometry and deviations from ideal tetrahedral angles.
Explanation: Elemental sulfur commonly exists in a cyclic molecular form consisting of eight sulfur atoms bonded together. This structure arises due to sulfur’s ability to form stable S–S single bonds and catenation. The arrangement is not linear or planar but adopts a puckered ring to minimize bond strain and electron repulsion. Such molecular structures are important in understanding allotropy, where an element exists in multiple structural forms. The stability of this configuration contributes to sulfur’s physical properties and its role in various industrial and chemical processes.
Option d – a W-shaped ring structure
The correct order of electron affinity of B, C, Ni, and O is
(A) O> C>N> B
(B) B>N> C>O
(C) O> C> B>N
(D) O> B> C> N
Explanation: Electron affinity refers to the energy change when an electron is added to a gaseous Atom. It depends on atomic size, nuclear charge, and electronic configuration stability. Atoms with half-filled or fully filled subshells tend to resist gaining electrons due to extra stability. Periodic trends show variation in electron affinity across a period due to increasing nuclear charge and decreasing atomic size. However, exceptions occur due to electronic repulsion in paired orbitals. Understanding these trends is essential in predicting chemical reactivity and Bonding tendencies of elements.
Option d – O> B> C> N
The hybridization of sulfur in SO₂ is
(A) sp
(B) sp³
(C) sp³
(D) dsp²
Explanation: Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals suitable for Bonding. The geometry of a Molecule depends on the number of electron regions around the central Atom, including Bonding pairs and lone pairs. In molecules with double bonds and lone pairs, electron distribution leads to specific molecular shapes and bond angles. Resonance and electron delocalization also influence Bonding characteristics. Understanding hybridization helps explain molecular geometry, bond strength, and reactivity in covalent compounds, especially in sulfur oxides and related species.
Option c – sp³
Which of the following compounds contains S=O as well as S = S bonds?
Explanation: Sulfur-containing oxyacids and their derivatives exhibit a variety of Bonding patterns due to sulfur’s ability to expand its valence shell. These compounds may contain both double bonds with oxygen and sulfur–sulfur linkages depending on their structure. Such Bonding diversity arises from sulfur’s flexibility in forming multiple oxidation states and covalent frameworks. Structural analysis of these compounds is important in understanding their reactivity, stability, and redox behavior. These features are commonly studied in Inorganic Chemistry to classify sulfur oxoacids and related species.
Explanation: Molecular geometry and polarity are determined by bond arrangement and electronegativity differences between atoms. A Molecule with lone pairs on the central Atom often adopts a bent shape due to electron pair repulsion. If the bond dipoles do not cancel due to asymmetry, the Molecule exhibits a NET dipole moment, making it polar. The presence of lone pairs on sulfur leads to non-linear geometry, and differences in electronegativity between hydrogen and sulfur contribute to polarity. Understanding molecular shape and dipole moment is essential in predicting physical properties like solubility and boiling point.
Option a – Angular and non-zero
The correct order of O-O bond length (increasing) in O2, H₂O2, and O3 is
(A) H₂O2 < O3 < O₂
(B) O₂ < 03 <H₂O₂
(C) O₂ < 03 < H₂O₂
(D) O3 <H₂O2 < O₂
Explanation: Bond length depends on bond order, atomic interactions, and resonance effects. Higher bond order generally leads to shorter and stronger bonds, while lower bond order results in longer bonds. In molecules involving oxygen, different bonding environments such as single, double, and resonance-stabilized bonds affect bond length. Lone pair repulsions and electron delocalization also influence bond distances. Comparing bond lengths requires analyzing the effective bond order and structural features of each Molecule. These principles are fundamental in understanding molecular structure and stability.
Option b – O₂ < O3
The bond angle is minimum for
(A) H₂O
(B) H₂S
(C) H₂Se
(D) H₂Te
Explanation: Bond angles in hydrides of group 16 elements are influenced by VSEPR theory, lone pair repulsions, and atomic size. As the central Atom becomes larger down the group, bond pair repulsions decrease and bond angles become smaller. Lone pairs on the central Atom exert stronger repulsive effects than bonding pairs, affecting molecular geometry. Comparing such molecules helps in understanding Periodic trends in molecular shape and deviations from ideal geometries. These trends are important in predicting structure-property relationships in covalent hydrides.
Explanation: Chromic acid is formed when certain dichromate compounds react with strong Acids, producing a powerful oxidizing medium. In acidic conditions, dichromate ions shift equilibrium to form a deep-colored species responsible for strong oxidation reactions. This system is widely used in Organic Chemistry for oxidizing Alcohols and other functional groups. The formation depends on combining a dichromate source with a strong mineral acid, which stabilizes the reactive oxidizing species. Such mixtures are important in qualitative analysis because they show characteristic color changes linked to oxidation state changes of chromium. Understanding this equilibrium helps in predicting oxidizing behavior in acidic media and in laboratory preparation of strong oxidizing agents.
Option a – K₂Cr₂O7 and conc. H₂SO4
Carbogen is
(A) Pure from carbon
(B) COCI2
(C) a mixture of CO and CO₂
(D) a Mixture of O2 and CO₂
Explanation: Gas mixtures are often prepared for specific medical and industrial applications based on physiological or chemical requirements. Carbogen is a carefully controlled mixture of gases used in medical settings to influence respiratory activity and oxygen supply. It combines oxygen with another inert or biologically active gas in a fixed proportion to regulate breathing efficiency. Such mixtures are important in respiratory therapy and physiological experiments because they affect oxygen Transport and carbon dioxide elimination in the body. Understanding gas mixtures helps in medical science, particularly in treating respiratory disorders and studying gas exchange mechanisms in biological systems.
Option d – a Mixture of O2 and CO₂
The bleaching action of SO₂ is due to
(A) Reduction
(B) Oxidation
(C) Hydrolysis
(D) Its acidic nature
Explanation: Bleaching processes involve chemical reactions that remove or alter color-causing substances. Some bleaching agents work through oxidation, while others operate through reduction. Sulfur dioxide acts differently from chlorine-based bleaching agents because it changes the chemical structure of colored compounds by reducing them rather than destroying them permanently. This type of bleaching is temporary in many cases, as exposure to air can reverse the effect. Understanding the mechanism of bleaching is important in industrial applications such as paper processing, textile treatment, and preservation of Organic materials. The role of redox reactions is central in explaining how color changes occur at the molecular level.
Option a – Reduction
The trioxides which can exist as the monomeric Molecule is
Explanation: Oxides of elements can exist in different structural forms depending on bonding, molecular size, and intermolecular interactions. Some trioxides form polymeric structures in Solid or liquid states, while others remain as discrete molecular units. The stability of monomeric forms depends on atomic size, bonding preferences, and ability to form multiple bonds with oxygen. Lighter elements in a group are more likely to form discrete molecular oxides, whereas heavier elements tend to form polymeric networks. Understanding these structural differences is important in Inorganic Chemistry for predicting physical state, reactivity, and molecular behavior of oxides.
Option a – SO3 in a gaseous state
The standard Heat of combustion of carbon, hydrogen, and CH4 are -349, -241.8, and -906.7 kJ respectively. The std. the Heat of formation of CH4 is
(A) 174.1 kJ
(B) 274.1 kJ
(C) 374.1 kJ
(D) 74.1 kJ
Explanation: Thermochemical calculations involve using known enthalpy values of combustion and formation to determine unknown reaction enthalpies. Hess’s law states that total enthalpy change depends only on initial and final states, not the reaction path. By combining combustion data of elements and compounds, one can construct a thermochemical cycle to find the Heat of formation of a compound. This approach is widely used in physical chemistry to relate energy changes in chemical reactions. It requires careful balancing of equations and consistent use of enthalpy conventions. Understanding these principles is essential for energy analysis in chemical processes.
Option d – 74.1 kJ
40-joule Heat is supplied to the system and work done by a system is 8 joule. Increase in internal energy of the system
(A) 25 J
(B) 30 J
(C) 32 J
(D) 28 J
Explanation: The first law of Thermodynamics relates Heat, work, and internal energy in a system. When Heat is supplied to a system, part of it may be used to perform work, and the remainder changes the internal energy. Internal energy represents the total microscopic energy stored in the system. work done by the system reduces the energy available for internal storage. By applying the energy conservation principle, one can determine how energy is distributed in thermodynamic processes. This concept is fundamental in understanding energy transformations in physical and chemical systems.
Option c – 32 J
In the isochoric process, the increase in internal energy is equal to
Explanation: In thermodynamic processes, constraints determine how Heat and work are related. In an isochoric process, volume remains constant, meaning no work is done because there is no change in volume. According to the first law of Thermodynamics, internal energy change depends only on heat exchange when work is zero. This makes energy analysis simpler in constant-volume conditions. Such processes are important in understanding heat capacity and energy transfer in closed systems. They are widely used in theoretical Thermodynamics and gas behavior studies.
Option a – Heat absorbed
A system in which there is no exchange of Matter, work, or energy from the surrounding is
(A) Closed
(B) Isolated
(C) Adiabatic
(D) Isothermal
Explanation: Thermodynamic systems are classified based on exchange of Matter and energy with surroundings. Some systems allow both Matter and energy exchange, while others restrict one or both. A system that is completely isolated does not interact with its surroundings in any form. Such idealized systems are useful in theoretical studies of Thermodynamics to apply conservation laws without external influence. Understanding system types helps in analyzing real-world processes like engines, biological systems, and chemical reactions under controlled conditions.
Option b – Isolated
An adiabatic process occurs in
(A) Open system
(B) Closed system
(C) Isolated system
(D) In all system
Explanation: An adiabatic process is one in which no heat exchange occurs between the system and its surroundings. Energy changes in such a process occur solely due to work done on or by the system. These processes are important in gas expansion and compression, where rapid changes prevent heat transfer. Adiabatic conditions are commonly assumed in thermodynamic models of engines and atmospheric processes. Understanding these processes helps in analyzing energy conservation under insulated conditions.
Option c – Isolated system
In the isothermal expansion of a gas
(A) Internal energy increases
(B) Enthalpy decreases
(C) Enthalpy remains unaffected
(D) Enthalpy reduced to zero
Explanation: Isothermal processes occur at constant temperature, meaning internal energy of an ideal gas remains unchanged. During expansion, the system performs work while absorbing heat from the surroundings to maintain temperature. This balance between heat and work ensures thermal equilibrium throughout the process. Such processes are important in thermodynamic cycles and gas behavior studies. They help explain how energy is transferred without changing internal thermal energy, emphasizing the relationship between heat flow and mechanical work.
Option c – Enthalpy remains unaffected
According to Hess’s law, the heat of reaction. depends upon
(A) the Initial state of reactants
(B) the Initial and final state of reactants
(C) the Intermediate path of the reaction
(D) the End point of conditions of the reactant
Explanation: Thermochemical reactions often proceed through multiple steps, but the total energy change remains consistent regardless of the pathway taken. This principle is based on the conservation of energy, which states that energy cannot be created or destroyed, only transformed. In chemical systems, enthalpy change is a state function, meaning it depends only on initial and final states of reactants and products. Intermediate steps or reaction mechanisms do not influence the overall heat change. This allows complex reactions to be broken into simpler steps for easier calculation of enthalpy changes using known data. The concept is widely applied in reaction energetics and industrial chemistry for predicting energy requirements and feasibility of processes.
Option b – the Initial and final state of reactants
Intensive properties among the following are
(A) Specific heat, and temperature
(B) Entropy, density
(C) Enthalpy, mole fraction
(D) Heat, temperature
Explanation: Physical properties of Matter are classified based on whether they depend on the amount of substance present. Some properties remain unchanged regardless of system size, while others vary with quantity. Properties such as temperature and density describe the state of a system independently of its mass or volume. These are useful in identifying substances and understanding equilibrium conditions. In contrast, extensive properties change when the amount of substance changes. Distinguishing between these types is important in Thermodynamics, where system behavior is analyzed using state variables. This classification helps simplify the study of physical and chemical systems.
Option a – Specific heat, and temperature
Which of the following is not a state function? (i) q + W (ii) q (iii) W (iv) HTS
(A) (i), (ii), (iii)
(B) (ii) and (iii)
(C) (i), (iv)
(D) (ii), (iii), (iv)
Explanation: In Thermodynamics, state functions depend only on the current state of a system and not on the path taken to reach that state. Energy forms like heat and work are path-dependent quantities, meaning they vary based on how a process occurs. In contrast, combinations of energy terms can sometimes represent state functions depending on how they are defined. Internal energy is a key example of a state function, while heat and work individually are not. Understanding this distinction is essential for applying the first law of Thermodynamics and analyzing energy changes in physical and chemical processes.
Option b – (ii) and (iii)
Which one of the following statements regarding starch and cellulose is not correct?
(d) Both of them are made up of glucose molecules.
Explanation: Polysaccharides such as starch and cellulose are both composed of glucose units but differ in their structural arrangement and bonding. These structural differences lead to variations in digestibility, function, and chemical behavior. One of them serves as an energy storage Molecule in plants, while the other provides structural support in cell walls. Their interaction with reagents such as iodine also differs due to differences in helical structure. Understanding these differences is important in biochemistry, especially in carbohydrate classification and biological function. The distinction arises from the type of glycosidic linkages and molecular arrangement.
Option c – Both of them give color with iodine
Among the following sweeteners, which one has the lowest sweetness value?
(a) Alitame
(b) Aspartame
(c) Saccharin
(d) Sucralose
Explanation: Artificial sweeteners differ in sweetness intensity compared to natural sugars. Sweetness value refers to how sweet a substance tastes relative to sucrose. Some compounds are many times sweeter than sugar, while others have relatively lower sweetness intensity. These differences arise from molecular structure and how they interact with taste receptors. Artificial sweeteners are widely used in Food industries to provide sweetness with fewer calories. Understanding their comparative sweetness helps in selecting appropriate substitutes in dietary applications and Food formulation.
Option b – Aspartame
Which one of the following artificial sweeteners is modified sugar?
(a) Aspartame
(b) Succharin
(c) Sucralose
(d) Alitame
Explanation: Artificial sweeteners are classified based on their origin and structural modification. Some are synthesized compounds, while others are chemically modified forms of natural sugars. Modified sugars retain the basic carbohydrate framework but have specific functional groups altered to enhance sweetness and reduce caloric value. These modifications change how they are metabolized and how they interact with taste receptors. Such compounds are widely used in low-calorie foods and beverages. Understanding their structure helps in distinguishing between synthetic sweeteners and sugar derivatives.
Option c – Sucralose
Glucose is a source of energy, which one of the following types of Molecule is glucose?
(a) Carbohydrate
(b) Protein
(c) Fat
(d) Nucleic acid
Explanation:Biomolecules are classified based on their chemical composition and biological role. Carbohydrates are Organic compounds that serve as primary energy sources in Living Organisms. They consist of carbon, hydrogen, and oxygen in specific ratios and include simple sugars as well as complex Polymers. Glucose is a simple sugar that plays a central role in cellular Respiration, where it is broken down to release energy. Understanding its classification helps in studying metabolism, energy production, and biological pathways in living systems.
Option a – Carbohydrate
Lactose on hydrolysis gives
(a) glucose
(b) fructose
(c) fructose + glucose
(d) glucose + galactose
Explanation: Disaccharides are carbohydrates composed of two monosaccharide units linked by glycosidic bonds. During hydrolysis, these bonds are broken in the presence of water, often with enzymatic or acidic catalysis. The resulting monosaccharides are simpler sugars that can be directly utilized in metabolic pathways. Lactose is a naturally occurring sugar found in milk and undergoes enzymatic breakdown in the digestive system. Understanding hydrolysis reactions is important in biochemistry for studying Digestion and energy release from Food sources.
Option d – glucose + galactose
Which one of the following carbohydrates is used for the silvering of a mirror?
(a) Sucrose
(b) Maltose
(c) Glucose
(d) Fructose
Explanation: Reducing sugars are capable of donating electrons due to the presence of free aldehyde or ketone groups in solution. This property allows them to reduce metal ions to metallic form. In chemical applications, such reactions are used to deposit a thin metallic layer on glass surfaces, forming reflective coatings. The ability to act as a reducing agent depends on molecular structure and presence of reactive functional groups. This principle is widely used in analytical chemistry and industrial mirror manufacturing processes.
Option c – Glucose
The reagent which may be used to distinguish between cane sugar and glucose solution is
(a) Molisch’s reagent
(b) Iodine solution
(c) Baeyer’s reagent
(d) Fehling’s solution
Explanation: Different carbohydrates react differently with chemical reagents depending on their structure and functional groups. Reducing sugars can undergo oxidation reactions, while non-reducing sugars do not respond similarly. Specific reagents are used in qualitative analysis to detect the presence of aldehyde groups in solution. These reactions often produce color changes or precipitates, allowing identification of different types of sugars. Understanding these differences is important in carbohydrate chemistry and Food analysis.
Option d – Fehling’s solution
In honey, which one among the following sugars predominates?
(a) Sucrose
(b) Fructose
(c) Galactose
(d) Maltose
Explanation: Natural Food substances often contain mixtures of simple sugars formed through enzymatic breakdown of complex carbohydrates. In honey, enzymatic activity from bees converts sucrose present in nectar into simpler monosaccharides. These monosaccharides differ in structure and sweetness intensity and remain dissolved in the liquid matrix, contributing to viscosity, taste, and osmotic properties. The relative proportions of these sugars are influenced by biological processing and source of nectar. Such compositions also affect crystallization behavior and nutritional properties. Understanding sugar distribution in natural products is important in Food chemistry, as it explains sweetness, stability, and energy availability in biological systems.
Option b – Fructose
The chemical name of vitamin A is
(a) thiamine
(b) axerophthol (retinol)
(c) ascorbic acid
(d) nicothamide
Explanation: Vitamins are Organic compounds required in small amounts for normal physiological functions. They often have specific chemical names based on their molecular structure and biological activity. Fat-soluble vitamins like vitamin A play a crucial role in vision, immune function, and cellular growth. Their structure includes conjugated systems that allow interaction with biological receptors. Deficiency of such compounds leads to well-known Health disorders. Understanding the chemical identity of vitamins helps in studying their metabolic roles and importance in Nutrition science and biochemistry.
Option a – axerophthol (retinol)
The metal present in Vitamin B₁₂ is
(a) Iron
(b) Manganese
(c) Cobalt
(d) Magnesium
Explanation: Some vitamins contain metal ions at their core, forming coordination complexes essential for biological activity. These metals play a key role in enzymatic reactions and metabolic pathways. The central metal ion is coordinated with a complex Organic ring structure, allowing it to participate in redox reactions and biochemical transformations. Such vitamins are crucial for blood formation, neurological function, and DNA synthesis. Understanding metalloproteins and metal-containing vitamins is important in biochemistry, as it explains how trace elements are vital for Life Processes.
Option c – Cobalt
Which one of the following vitamins is essential for the coagulation of blood?
(a) Vitamin-A
(b) Vitamin-B12
(c) Vitamin-K
(d) Vitamin-D
Explanation: Vitamins are classified based on their biological functions, including roles in metabolism, immunity, and physiological regulation. Some vitamins are directly involved in the synthesis of proteins required for blood clotting. These proteins help in forming stable blood clots by activating specific enzymatic pathways in the coagulation cascade. Deficiency in such vitamins leads to impaired clot formation and excessive bleeding tendencies. Understanding vitamin function is essential in physiology and medical science, particularly in Nutrition and Disease prevention.
Option c – Vitamin-K
Which one of the following Acids is also known as Vitamin-C?
(a) Methanoic acid
(b) Ascorbic acid
(c) Lactic acid
(d) Tartaric acid
Explanation: Certain vitamins are OrganicAcids that participate in metabolic reactions as antioxidants and enzyme cofactors. These compounds are water-soluble and play a key role in tissue repair, immune response, and collagen synthesis. They help neutralize free radicals and protect cells from oxidative damage. Their chemical structure includes multiple hydroxyl groups that contribute to their reducing properties. Understanding their biochemical role is important in Nutrition and Health sciences, as deficiency leads to well-known deficiency diseases.
Option b – Ascorbic acid
Which of the following has a coiled helical structure?
(a) Proteins
(b) Lipids
(c) Carbohydrates
(d) Vitamins
Explanation: Biological macromolecules adopt specific three-dimensional structures that determine their function. Protein chains can fold into organized shapes stabilized by hydrogen bonding and other intermolecular forces. One common structural motif is a helical arrangement, where the polypeptide chain coils in a regular pattern. This structure is stabilized by interactions between backbone groups and contributes to the mechanical strength and biological activity of proteins. Such structural organization is fundamental in biochemistry for understanding enzyme function, structural proteins, and molecular recognition processes.
Option a – Proteins
Which one of the following Polymers is made of proteins?
(a) Rubber
(b) Cotton
(c) Wool
(d) Jute
Explanation:Polymers in biological systems are large macromolecules formed by repeating monomer units. Protein-based Polymers are composed of amino Acids linked through specific covalent bonds. These macromolecules serve structural, enzymatic, and functional roles in Living Organisms. Their properties depend on sequence, folding, and intermolecular interactions. Natural fibers and biological materials often fall into this category due to their protein composition. Understanding protein Polymers is essential in biochemistry, material science, and Biotechnology, as they determine biological structure and function.
Option c – Wool
What is the bond between two amino Acids in a protein called?
(a) Ester bond
(b) Glycosidic bond
(c) Peptide bond
(d) Phosphodiester bond
Explanation: Amino Acids are linked together to form proteins through a specific type of covalent linkage formed during a condensation reaction. This bond forms between the carboxyl group of one amino acid and the amino group of another, releasing a molecule of water. The resulting linkage is stable and forms the backbone of protein structure. Such bonds determine the primary structure of proteins, which further folds into higher-order structures. Understanding this bonding is fundamental in biochemistry for studying protein synthesis, structure, and function in biological systems.
Option c – Peptide bond
We covered all the mcqQuestions for Class 11 chemistry chapter 1 above in this post for free so that you can practice well for the exam.
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