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New Simplified Chemistry Class 8 ICSE Objective Solutions for Students
What type of Bonding is found in ferric chloride?
(a) Covalent
(b) Ionic
(c) Coordinate
(d) None of these
Explanation: Ferric chloride is a compound formed between iron and chlorine where the nature of Bonding depends on the electron sharing and transfer between atoms. In such metal–nonmetal combinations, Bonding characteristics may show a mix of ionic and covalent behavior due to polarization effects. Iron in a higher oxidation state tends to strongly attract electron density from surrounding atoms, influencing bond character. Chlorine, being highly electronegative, also contributes to electron sharing tendencies. The compound structure is influenced by the size and charge of the central metal ion, which can distort electron clouds of surrounding ligands. This leads to a Bonding pattern that is not purely ionic or purely covalent but shows directional interaction features. Understanding this requires knowledge of electronegativity difference, Fajan’s rules, and coordination tendencies of transition Metals. Such compounds often show complex lattice or Molecular arrangements depending on conditions like temperature and phase.
Option a – Covalent
Manganese reaches its maximum oxidation state in which compound?
(a) MnO2
(b) Mn2O4
(c) KMnO4
(d) K2MnO4
Explanation: Manganese is a transition element capable of exhibiting multiple oxidation states due to the availability of both 3d and 4s electrons for Bonding and electron loss. Its oxidation behavior depends on the surrounding ligands and oxygen content. In highly oxygen-rich environments, manganese can achieve its highest stable oxidation level. This occurs when it is bonded in a structure where oxygen, being highly electronegative, stabilizes the metal in a very high positive state. Such compounds are typically strong oxidizing agents and are widely used in analytical Chemistry due to their electron-accepting ability. The stability of these high oxidation states is influenced by lattice energy, resonance stabilization, and strong metal–oxygen Bonding. Understanding this concept requires familiarity with transition metal Chemistry, especially variable valency and redox stability trends across compounds of manganese.
Option a – MnO2
What are the typical oxidation states of scandium (Atomic number 21)?
(a) +4
(b) +1
(c) +2 and +3
(d) +4 and +1
Explanation: Scandium is a transition metal located at the beginning of the d-block, where the filling of 3d orbitals has just begun. Its electronic configuration allows limited variability in oxidation behavior compared to heavier transition elements. The stability of oxidation states depends on the energy required to remove electrons from both the outermost s-orbital and the partially filled d-orbitals. In scandium, the removal of electrons is strongly influenced by its relatively small atomic size and higher effective nuclear charge. The involvement of 3d electrons in Bonding is less flexible, making one oxidation condition more stable than others. The general concept involves understanding how transition Metals show multiple oxidation states due to d-electron participation, but early transition elements often prefer a more stable configuration after losing a specific number of electrons that leads to a pseudo-noble gas structure. This stability is also supported by lattice energy in compounds formed by scandium ions in Solids.
Option c – +2 and +3
In addition to zinc and copper, German silver also contains which element?
(a) Tin
(b) Silver
(c) Nickel
(d) Magnesium
Explanation: German silver is an alloy commonly studied in Metallurgy, composed of multiple metallic elements combined to enhance mechanical strength, corrosion resistance, and appearance. Alloys are formed by mixing Metals in specific proportions to modify properties that individual elements alone do not provide. In such combinations, the added element plays a crucial role in improving hardness and resistance to tarnishing. The structure of alloys depends on atomic size similarity, electron distribution, and metallic Bonding strength between constituent atoms. The third component in German silver significantly affects its physical and chemical stability, especially in applications like utensils and decorative items. Understanding alloy composition requires knowledge of substitutional Solid solutions, where atoms of similar size replace each other in a metallic lattice. The presence of this element also influences conductivity and resistance to oxidation, making the alloy more durable than pure copper or zinc-based systems.
Option c – Nickel
What is the correct chemical formula for permanganic Acid?
(a) HMnO4
(b) HMnO5
(c) H2MnO4
(d) M2MnO3
Explanation: Permanganic Acid is an oxyacid derived from a transition metal in a high oxidation state, where manganese is bonded to oxygen atoms forming a tetrahedral oxyanion structure. The stability of such Acids depends on the strong electronegativity of oxygen and the ability of the central metal to stabilize a high positive oxidation state through multiple Bonding interactions. In oxyacids of transition Metals, hydrogen is attached to oxygen, forming acidic protons that can dissociate in aqueous solution. The structure involves resonance stabilization of the oxyanion, distributing electron density over multiple oxygen atoms. Understanding this requires knowledge of Acid formation from oxides, oxidation state stabilization, and coordination of oxygen ligands around manganese. These Acids are typically strong oxidizing agents due to the high oxidation state of the central metal, which makes them readily accept electrons during chemical reactions.
Option a – HMnO4
Why is K2Cr2O7 preferred over Na2Cr2O7 as a primary standard in volumetric analysis?
(a) Na2Cr2O7 absorbs moisture while K2Cr2O7 does not
(b) K2Cr2O7 absorbs moisture while Na2Cr2O7 does not
(c) K2Cr2O7 is pure and Na2Cr2O7 is not
(d) None of these
Explanation: In analytical Chemistry, primary standards must be highly pure, stable, and non-hygroscopic to ensure accurate quantitative measurements. Dichromate compounds are used in redox titrations due to their strong oxidizing nature and well-defined stoichiometry. The choice between potassium and sodium Salts depends on their physical stability and ease of handling. The reliability of a standard solution is influenced by its tendency to absorb moisture from the air, which can alter measured Mass and concentration. A suitable compound must remain unchanged in air and have a consistent Molecular composition during weighing. Additionally, crystalline structure and lattice stability contribute to its suitability in volumetric analysis. Potassium-based Salts often exhibit better crystallinity and lower hygroscopic behavior compared to their sodium counterparts, making them more accurate for preparing standard solutions. This ensures precision in redox titrations involving electron transfer reactions.
Option a – Na2Cr2O7 absorbs moisture while K2Cr2O7 does not
Platinum dissolves in which of the following?
(a) Hot concentrated HNO3
(b) Hot concentrated HCl
(c) Hot dilute H2SO4
(d) Mixture of concentrated HCl and HNO3
Explanation: Platinum is a noble metal known for its high resistance to corrosion and chemical attack due to strong metallic Bonding and low reactivity. However, under specific conditions involving highly reactive mixtures, even noble Metals can undergo oxidation and dissolution. This typically occurs when a strong oxidizing Environment is combined with a species capable of forming stable complex ions with the metal. The process involves simultaneous oxidation of the metal surface and stabilization of the resulting ionic form in solution. The presence of chloride ions plays a key role in forming stable coordination complexes, which prevents re-deposition of the metal and drives the reaction forward. Understanding this requires knowledge of redox Chemistry, coordination Chemistry, and the behavior of noble Metals under extreme chemical conditions. Such reactions are widely used in refining and analytical dissolution processes involving resistant Metals.
Option d – Mixture of concentrated HCl and HNO3
Fusing MnO2 with KOH in the presence of air produces which colored compound?
(a) K2MnO4, dark green
(b) KMnO4, purple
(c) Mn2O3, brown
(d) Mn3O4, black
Explanation: Manganese dioxide is a transition metal oxide that can undergo oxidation state changes when heated with strong Bases in the presence of oxygen. During fusion reactions, oxygen from air participates in increasing the oxidation state of manganese, leading to the formation of higher oxidation state manganate species. These compounds are characterized by distinct colors due to d–d electronic transitions and charge transfer effects within the metal–oxygen framework. The reaction involves Solid-state interaction where alkali metal ions stabilize the oxyanion formed from manganese. The color observed is directly related to the electronic structure and oxidation level of manganese in the compound. Understanding this requires knowledge of redox reactions in Solid-state Chemistry, oxidation state changes, and the role of oxygen as an oxidizing agent in alkaline fusion processes. Such reactions are commonly used in qualitative Inorganic analysis to identify manganese-containing species.
Option a – K2MnO4, dark green
Arrange Cu, Ag, and Au in decreasing order of melting points.
(a) Cu > Ag > Au
(b) Cu > Au > Ag
(c) Au > Ag > Cu
(d) Ag > Au > Cu
Explanation: The melting point of Metals is influenced by the strength of metallic bonding, which depends on the number of delocalized electrons, atomic size, and effective nuclear charge. In transition Metals, d-electrons play a significant role in strengthening metallic bonds by contributing to electron density in the lattice. As we move down the group, atomic size increases, but relativistic effects and d-orbital participation can alter bonding strength unexpectedly. The balance between weaker metallic bonding due to larger atomic size and enhanced stability from d-electron interactions determines the final trend. Understanding this requires knowledge of Periodic trends, metallic bonding theory, and relativistic stabilization effects in heavier transition elements. These factors collectively influence how tightly atoms are held together in Solid form, thereby affecting the temperature required to break the lattice structure and convert the metal into liquid state.
Option b – Cu > Au > Ag
Why don’t Group-12 elements (Zn, Cd, Hg) behave as transition Metals?
(a) They act as superconductors
(b) They have high melting points
(c) Their d-subshells are fully filled
(d) They act as semiconductors
Explanation: Transition metal behavior is typically associated with partially filled d-orbitals, which allow variable oxidation states and the formation of colored compounds. Group-12 elements have completely filled d-subshells in their ground state and commonly maintain this filled configuration in their compounds. This limits their ability to exhibit variable oxidation states and reduces their participation in typical transition metal Chemistry. The filled d-orbitals are relatively stable and do not readily participate in bonding or electron exchange processes. As a result, these elements show properties more similar to main group metals rather than classical transition elements. Understanding this requires knowledge of electronic configuration, oxidation state variability, and the role of d-electrons in Chemical Bonding and catalytic behavior. Their chemical behavior is therefore distinct from elements with partially filled d-orbitals, which are responsible for characteristic transition metal properties such as colored ions and complex formation.
Option c – Their d-subshells are fully filled
Group-12 elements are softer than other d-block metals because
(a) Their structure is cage-like
(b) They have high ionization energies
(c) Both s and d electrons participate in bonding
(d) Their d electrons do not take part in metallic bonding
Explanation: The hardness of metals is primarily determined by the strength of metallic bonding, which depends on the availability of delocalized electrons and the extent of overlap between atomic orbitals. In Group-12 elements, the d-subshell is completely filled, which reduces the participation of d-electrons in metallic bonding. This weakens the overall cohesion of the metallic lattice compared to other d-block metals where partially filled d-orbitals contribute significantly to bonding strength. As a result, the metallic bonds are less rigid and more easily disrupted under mechanical force. The concept also involves understanding electron delocalization and how it contributes to structural rigidity in metallic crystals. When fewer electrons are available for bonding interactions, the lattice becomes more deformable, leading to softer physical properties. This behavior contrasts with typical transition metals where stronger electron overlap results in higher hardness and melting points.
Option d – Their d electrons do not take part in metallic bonding
Which ore is used for extracting copper?
(a) Kupfernickel
(b) Dolomite
(c) Malachite
(d) Galena
Explanation: Copper extraction involves identifying suitable ores that contain sufficient metal content for industrial processing. Ores are naturally occurring mineral deposits from which metals can be economically extracted. The efficiency of extraction depends on the concentration of copper compounds and their ease of reduction. Copper commonly occurs in nature in the form of oxides, carbonates, and sulfides. Among these, carbonate-based ores are often used in preliminary extraction steps due to their chemical reactivity and ease of conversion into oxides, which can then be reduced to metallic copper. The process involves multiple stages including concentration, roasting, and reduction, depending on the ore type. Understanding this requires knowledge of Metallurgy, ore classification, and extraction techniques used in hydrometallurgy and pyrometallurgy. The choice of ore significantly affects the efficiency and cost of copper production in industrial processes.
Option c – Malachite
Choose the correct sequence of thermal stability.
(a) Zn > Hg > Cd
(b) Cd > Hg > Zn
(c) Zn > Cd > Hg
(d) Hg > Cd > Zn
Explanation: Thermal stability of compounds is influenced by lattice energy, ionic size, and polarization effects. In metal compounds, stability depends on how strongly ions are held together in the crystal lattice and how easily they decompose under Heat. Smaller, more highly charged ions typically form stronger lattices, increasing thermal stability. As ionic size increases down a group, lattice strength decreases, making compounds less resistant to Heat. Polarizing power of cations and polarizability of anions also play a key role in determining decomposition behavior. Understanding this trend requires knowledge of Periodic properties, ionic bonding strength, and thermodynamic stability of Solid compounds. These factors collectively determine the temperature at which decomposition occurs and the relative stability of similar compounds.
Option c – Zn > Cd > Hg
Regular use of which fertilizer increases soil acidity?
(a) Ammonium sulfate
(b) Potassium nitrate
(c) Urea
(d) Superphosphate of lime
Explanation: Soil acidity is influenced by the chemical nature of fertilizers and their behavior after dissolution in water. Fertilizers containing ammonium ions undergo nitrification in soil, a biological process where ammonium is converted into nitrate ions by microorganisms. This process releases hydrogen ions, which increase soil acidity over time. Continuous application of such fertilizers gradually lowers soil pH, affecting nutrient availability and microbial activity. Understanding this requires knowledge of soil chemistry, nitrogen cycle, and biochemical oxidation processes in Agriculture. The long-term impact of fertilizers depends on how their constituent ions interact with soil components and water. Acidification can influence plant growth, nutrient uptake, and soil structure, making fertilizer choice an important aspect of sustainable Agriculture practices.
Option a – Ammonium sulfate
Which non-metal never shows a positive oxidation state?
(a) Chlorine
(b) Iodine
(c) Fluorine
(d) Oxygen
Explanation: Oxidation state depends on an element’s ability to lose, gain, or share electrons during Chemical Bonding. Non-metals generally show both positive and negative oxidation states depending on their bonding partners and electronegativity differences. However, the most electronegative element strongly attracts electrons in all compounds, making it extremely resistant to losing electron density. Because of this, it consistently gains or shares electrons in a way that keeps its oxidation number non-positive. Its high electronegativity and small atomic size make electron loss energetically unfavorable. Understanding this requires knowledge of Periodic trends, electronegativity scale, and oxidation state rules in compounds. This behavior is unique among non-metals and is used as a reference point when assigning oxidation states to other elements in chemical compounds.
Option c – Fluorine
Which element forms the fewest oxyacids?
(a) Nitrogen
(b) Sulfur
(c) Fluorine
(d) Chlorine
Explanation: Oxyacids are Acids that contain oxygen, hydrogen, and another central Atom, usually a non-metal. The number of oxyacids formed by an element depends on its ability to exhibit multiple oxidation states and expand its coordination with oxygen atoms. Elements with higher electronegativity and limited ability to expand their valence shell form fewer stable oxygen-containing Acids. This is because they cannot easily stabilize multiple oxygen attachments or higher oxidation states. The structure of oxyacids depends on the central Atom’s ability to form stable bonds with oxygen while maintaining Acid-forming hydrogen atoms. Understanding this requires knowledge of Periodic trends, valency expansion, and Acid formation through oxoanion stability. Elements with restricted bonding flexibility produce fewer oxyacid varieties compared to those capable of multiple oxidation states.
Option c – Fluorine
Identify the incorrect formula for element X.
(a) X2Cl3
(b) X2O3
(c) X2(SO4)3
(d) XPO4
Explanation: Chemical formula formation depends on the valency and oxidation state of the central element combined with the charge balance of surrounding groups. For a given element X, its compounds must follow the principle of electrical neutrality, where total positive and negative charges balance each other. If the oxidation state of X does not match the charge requirement of attached groups, the formula becomes chemically inconsistent. Phosphate, sulfate, chloride, and oxide combinations require strict stoichiometric balancing based on ionic charges. Understanding this involves knowledge of valency rules, oxidation states, and compound formation principles in Inorganic Chemistry. Incorrect formulas arise when charge neutrality or valence requirements are not properly satisfied. Such errors can be identified by comparing expected oxidation states with the total charge contribution of the surrounding ions in the compound structure.
Option a – X2Cl3
Which of the following statements is not true?
(a) Cl2O7 is an anhydride of perchloric Acid
(b) O3 has a bent Molecular shape
(c) ONF is isoelectronic with NO2
(d) OF2 is an oxide of fluorine
Explanation: Evaluating chemical statements requires understanding structural chemistry, Molecular geometry, and bonding characteristics. Different molecules exhibit distinct shapes and electronic properties based on valence shell electron pair repulsion theory and orbital interactions. Some compounds act as isoelectronic species when they have the same number of electrons and similar structural arrangements, even if composed of different elements. Oxides and fluorine-containing compounds often show unique bonding patterns due to differences in electronegativity and atomic size. Determining incorrect statements involves comparing Molecular geometry, electronic structure, and known chemical behavior of each species. This requires knowledge of Molecular orbital theory, hybridization concepts, and Periodic trends in bonding behavior. Misinterpretations usually arise when structural similarity is confused with chemical equivalence or when oxidation states are incorrectly assigned in Molecular species.
Option d – OF2 is an oxide of fluorine
Choose the correct statement regarding Acid strength.
(a) H3PO3 is more acidic than H2SO3
(b) HF is stronger than HCl in aqueous medium
(c) HClO4 is weaker than HClO3
(d) HNO3 is stronger than HNO2
Explanation: Acid strength is determined by the ability of a compound to donate protons in aqueous solution. This depends on bond polarity, bond strength, and stability of the conjugate Base formed after proton release. Strong Acids dissociate more completely in water due to weaker H–X bonds or highly stable anions formed after dissociation. Oxygen-containing Acids show varying strengths based on the electronegativity of the central Atom and the number of oxygen atoms attached, which stabilize the negative charge through resonance. Understanding Acid strength involves concepts of inductive effect, resonance stabilization, and Periodic trends across elements. The presence of electronegative atoms and multiple oxygen bonds increases Acid strength by stabilizing the conjugate Base, making proton release easier in solution.
Option d – HNO3 is stronger than HNO2
What is the hybridization and geometry of XeF4?
(a) Octahedral, sp3d2
(b) Trigonal bipyramidal, sp3d
(c) Trigonal planar, sp3d3
(d) Square planar, sp3d2
Explanation: The structure of molecules is explained using valence shell electron pair repulsion theory, where electron pairs arrange themselves to minimize repulsion. Central atoms with expanded octets can accommodate more than eight electrons due to available d-orbitals. In such cases, both bonding and non-bonding electron pairs influence Molecular shape. Lone pairs occupy more space than bonding pairs, affecting geometry significantly. The hybridization model combines atomic orbitals to form equivalent hybrid orbitals that explain bonding arrangement in space. The final geometry is determined after considering both bonded atoms and lone pairs around the central Atom. Understanding this requires knowledge of electron pair repulsion, hybridization concepts, and Molecular geometry prediction rules for expanded octet compounds.
Option a – Octahedral, sp3d2
Which property is exclusive to either NO2 or SO2?
(a) Causes acid rain
(b) Acts as reducing agent
(c) Soluble in water
(d) Used as a Food preservative
Explanation: Nitrogen dioxide and sulfur dioxide are both polar molecules with distinct electronic structures and oxidation states. Their chemical behavior depends on the central Atom’s ability to stabilize unpaired electrons or participate in redox reactions. Some properties such as solubility or reactivity may be shared, but certain behaviors depend on Molecular structure and bonding differences. NO₂ contains an odd number of electrons, making it paramagnetic and highly reactive, while SO₂ has a bent structure with different resonance stabilization. Understanding exclusivity in properties requires knowledge of molecular orbital theory, electron configuration, and redox behavior of non-metal oxides. Differences in bonding also influence their role in atmospheric chemistry and industrial applications.
Option a – Causes acid rain
What is the correct order of bond dissociation energies for halogens?
(a) Cl2 > Br2 > F2 > I2
(b) Br2 > I2 > F2 > Cl2
(c) F2 > Cl2 > Br2 > I2
(d) I2 > Br2 > Cl2 > F2
Explanation: Bond dissociation energy depends on bond length, atomic size, and electron-electron repulsion within bonded atoms. In halogen molecules, as atomic size increases down the group, bond length increases and bond strength decreases. However, fluorine exhibits an exception due to strong lone pair repulsions between small atoms, weakening the F–F bond unusually. The trend is not strictly linear and requires understanding both Periodic trends and electronic repulsions. Bond strength reflects the energy required to break a covalent bond into gaseous atoms, which is influenced by overlap efficiency between atomic orbitals. Smaller atoms form stronger bonds due to better orbital overlap, but repulsive interactions can modify expected trends.
Option a – Cl2 > Br2 > F2 > I2
Which reaction does NOT show the oxidizing property of concentrated H2SO4?
(a) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
(b) 3S + 2H2SO4 → 3SO2 + 2H2O
(c) C + 2H2SO4 → CO2 + 2SO2 + 2H2O
(d) CaF2 + H2SO4 → CaSO4 + 2HF
Explanation: Concentrated sulfuric acid is a strong oxidizing agent in addition to being an acid and dehydrating agent. Its oxidizing behavior depends on its ability to accept electrons and get reduced to lower oxidation states such as sulfur dioxide. Reactions involving metals, non-metals, or carbon compounds often show redox behavior where sulfuric acid acts as an oxidant. However, some reactions involve only acid-Base or precipitation processes without electron transfer. In such cases, no change in oxidation state occurs, and the oxidizing property is not demonstrated. Understanding this requires knowledge of redox chemistry, oxidation state changes, and reaction classification into acid-Base, redox, and displacement types. The key distinction lies in whether sulfur in the acid undergoes reduction during the reaction process.
Option d – CaF2 + H2SO4 → CaSO4 + 2HF
A sharp rise in reaction rate with temperature suggests
(a) Increased collision frequency
(b) High activation energy
(c) High pre-exponential factor
(d) Increased molecular velocity
Explanation: Reaction rates increase with temperature due to higher molecular kinetic energy, leading to more frequent and effective collisions. According to collision theory, only collisions with sufficient energy and proper orientation result in product formation. A sharp increase in rate indicates that a small temperature change significantly increases the fraction of molecules exceeding activation energy. This behavior is governed by the Arrhenius equation, which relates rate constant to temperature and activation energy. The exponential dependence means that even minor thermal changes can greatly affect reaction speed. Understanding this requires knowledge of activation energy, molecular collision theory, and temperature dependence of reaction kinetics.
Option c – High pre-exponential factor
If activation energy is zero, then the rate constant
(a) Increases with temperature
(b) Depends on activation energy
(c) Is controlled by exponential factor
(d) Rises due to molecular speed
Explanation: The rate constant of a reaction is strongly influenced by temperature and activation energy as described by the Arrhenius equation. Activation energy represents the minimum energy required for reactant molecules to undergo a successful transformation into products. When this energy barrier becomes negligible, almost every collision between reactant molecules becomes effective. In such a case, the exponential factor in the rate expression becomes maximized, meaning the reaction becomes extremely fast and highly sensitive to temperature changes only through molecular motion. Understanding this requires knowledge of collision theory, energy distribution among molecules, and how temperature affects molecular kinetics. The absence of an energy barrier implies that the reaction proceeds almost immediately upon collision, making the process diffusion-controlled rather than energy-limited.
Option d – Rises due to molecular speed
Which statements align with collision theory?
(a) I, III, IV only
(b) II, IV only
(c) I, IV only
(d) I, II, III, IV
Explanation: Collision theory explains how chemical reactions occur based on the frequency and effectiveness of molecular collisions. For a reaction to occur, reactant molecules must collide with sufficient energy and proper orientation. The number of successful collisions determines the reaction rate. Temperature affects reaction speed by increasing molecular motion, thereby increasing collision frequency and energy. Not all collisions lead to product formation because only a fraction of molecules possess energy greater than the activation barrier. Understanding this theory requires knowledge of kinetic molecular behavior, activation energy, and factors affecting collision effectiveness. It provides a molecular-level explanation of how chemical reactions proceed and why certain conditions enhance reaction rates more significantly than others.
Option d – I, II, III, IV
Passing the same charge for 30 minutes deposits 0.375 g of Cu and 0.30 g of Cr. What is the electrochemical equivalent ratio of Cu to Cr?
(a) 0.8
(b) 1.25
(c) 2.5
(d) 1.62
Explanation: Electrochemical equivalent represents the Mass of a substance deposited or liberated when a unit quantity of electric charge passes through an electrolyte. It depends on the molar Mass of the element and the number of electrons involved in the reduction process. When the same amount of charge is passed through different electrolytes, the deposited Mass is directly proportional to their electrochemical equivalents. This allows comparison of deposition efficiency between different metals. Understanding this concept requires knowledge of Faraday’s laws of electrolysis, valency factors, and charge–Mass relationship in redox reactions. The ratio reflects how effectively each metal ion undergoes reduction under identical electrical conditions.
Option b – 1.25
A galvanic cell is constructed using a 100 g zinc rod and 1.0 L of 1.0 M CuSO₄ solution. If it delivers a constant current of 1.0 A, for how long will the cell function? (Atomic Mass of Zn = 65)
(A) 1.1 hr
(B) 46 hr
(C) 53.6 hr
(D) 24.00 hr
Explanation: A galvanic cell generates electrical energy through spontaneous redox reactions, where oxidation occurs at the anode and reduction at the cathode. The duration of operation depends on the limiting reactant available for the redox process. In such systems, zinc acts as the electron donor while copper ions are reduced from solution. The total available charge depends on the amount of reactive species present, which determines how long a constant current can be sustained. Understanding this requires knowledge of Faraday’s laws, mole-to-electron conversion, and stoichiometry of redox reactions in electrochemical cells. The lifetime of the cell is governed by how long the reactants can continue to supply electrons without depletion.
Option c – 53.6 hr
A molten metal Salt is electrolyzed with a 2.0 A current for 5 hours, resulting in the deposition of 22.2 g of the metal. If the atomic Mass of the metal is 177, what is its oxidation state?
(A) +1
(B) +2
(C) +3
(D) +4
Explanation: During electrolysis, metal ions in molten Salts gain electrons at the cathode and are deposited as neutral atoms. The amount of substance deposited is directly related to the total charge passed through the system according to Faraday’s laws of electrolysis. The relationship between Mass deposited, atomic Mass, charge, and number of electrons involved allows determination of the oxidation state of the metal ion. This oxidation state corresponds to the number of electrons required to reduce one ion of the metal. Understanding this requires knowledge of electrochemical stoichiometry, charge quantization, and reduction half-reactions. The oxidation state reflects the charge of the metal ion before reduction in the molten state.
Option c – +3
What quantity of charge must be passed through 200 mL of 0.5 M CuSO₄ to deposit all the copper?
(A) 96500 C
(B) 2 × 9650 C
(C) 2 × 96500 C
(D) 4 × 96500 C
Explanation: Electrolysis involves conversion of ions in solution into neutral atoms through electron transfer at electrodes. The total charge required depends on the number of moles of metal ions present and the number of electrons needed per ion for reduction. In copper sulfate solution, copper exists as Cu²⁺ ions, each requiring a fixed number of electrons to form metallic copper. The total charge is calculated using Faraday’s constant, linking moles of electrons to electrical charge. Understanding this requires knowledge of molarity, stoichiometry, and Faraday’s laws of electrolysis. The process ensures complete reduction of all available metal ions in solution when sufficient charge is supplied.
Option b – 2 × 9650 C
The release of H₂ at the cathode and Cl₂ at the anode during electrolysis indicates that the electrolyte is ( New Simplified Chemistry Class 8 ICSE mcq )
(A) Copper chloride in water
(B) NaCl in water
(C) Ferric chloride in water
(D) AuCl₃ in water
Explanation: During electrolysis, the products formed at electrodes depend on the nature of ions present in solution and their discharge preferences. At the cathode, reduction occurs, typically involving hydrogen ions or metal ions. At the anode, oxidation takes place, often involving halide ions or hydroxide ions depending on their relative discharge potential. The simultaneous formation of hydrogen and chlorine gases suggests a solution containing both hydrogen ions and chloride ions in aqueous medium. Understanding this requires knowledge of electrochemical series, preferential discharge of ions, and behavior of electrolytes in water. The observed gases indicate specific ionic species undergoing redox reactions at respective electrodes.
Option b – NaCl in water
When current flows through an electrolytic solution, positive ions go to the cathode and negative ions go to the anode. If the cathode is removed, what will happen?
(A) Both positive and negative ions will head toward the anode
(B) Positive ions move to the anode, and negative ions stop moving
(C) Negative ions continue to move toward the anode, while positive ions stop
(D) Both types of ions move randomly
Explanation: Electrolysis depends on the presence of both electrodes to complete the electrical circuit and allow redox reactions to occur. The cathode is essential for reduction reactions where cations gain electrons. If the cathode is removed, electron flow cannot be completed, disrupting the reduction process entirely. Although ions may still move in solution due to the Electric Field, the absence of a reduction surface prevents product formation at that electrode. Understanding this requires knowledge of electrochemical circuits, ion migration, and electrode function in redox systems. The continuity of electrolysis relies on both oxidation and reduction occurring simultaneously at separate electrodes.
Option d – Both types of ions move randomly
When 2.5 Faradays of charge is passed through CuSO₄, how many gram equivalents of copper are deposited? ( New Simplified Chemistry Class 8 ICSE mcq )
(A) 1
(B) 2
(C) 2.5
(D) 1.25
Explanation: Faraday’s laws of electrolysis relate the amount of substance deposited at an electrode to the quantity of electric charge passed through the electrolyte. One Faraday corresponds to the charge required to deposit one gram equivalent of a substance. The concept of gram equivalent depends on the molar Mass and the number of electrons involved in the redox reaction. In copper sulfate, copper ions require a specific number of electrons for reduction to metallic copper. Therefore, the amount deposited is directly proportional to the number of Faradays supplied. Understanding this requires knowledge of electrochemical equivalence, charge–mass relationships, and valency factors in redox chemistry. The relationship allows straightforward conversion between electrical charge and chemical quantity.
Option c – 2.5
In an electrolytic cell with platinum electrodes and Ag₂SO₄ solution, how much silver is deposited if 1.6 g of oxygen is released at the anode?
(A) 107.88 g
(B) 1.6 g
(C) 0.8 g
(D) 21.60 g
Explanation: In electrolysis, oxidation occurs at the anode while reduction occurs at the cathode, and both processes involve equal amounts of electric charge transfer. The mass of substances produced at each electrode is directly related through Faraday’s laws. Oxygen Evolution at the anode corresponds to water oxidation, while silver deposition at the cathode results from reduction of silver ions. By relating equivalent masses, one can determine the amount of silver deposited based on oxygen released. Understanding this requires knowledge of electrochemical equivalence, redox balancing, and charge conservation in electrolytic systems. The key idea is that the same quantity of Electricity governs both electrode processes simultaneously.
Option d – 21.60 g
How much chlorine gas is produced when a 2 A current is passed for 30 minutes through an aqueous NaCl solution? ( New Simplified Chemistry Class 8 ICSE mcq )
(A) 66 g
(B) 1.32 g
(C) 33 g
(D) 99 g
Explanation: During electrolysis of aqueous sodium chloride, chloride ions are oxidized at the anode to form chlorine gas. The amount of gas produced depends on the total charge passed through the electrolyte, which is calculated from current and time. Faraday’s laws link electrical charge to the number of moles of electrons involved in the redox reaction. Since each chlorine Molecule formation requires a fixed number of electrons, stoichiometry can be used to determine gas production. Understanding this requires knowledge of electrochemical reactions, gas formation at electrodes, and charge-to-mole conversion using Faraday’s constant. The process demonstrates how electrical energy drives chemical transformation in electrolytic systems.
Option b – 1.32 g
What time is needed to deposit 1 millimole of aluminum by passing a current of 9.65 A through a solution containing Al³⁺?
(A) 30 s
(B) 10 s
(C) 30,000 s
(D) 10,000 s
Explanation: Electrolytic deposition depends on the relationship between electric charge, current, and time. The total charge required is determined by the number of moles of ions and the number of electrons needed for reduction. Aluminium ions require multiple electrons to form metallic aluminium, making the charge requirement higher compared to monovalent ions. Once total charge is known, time can be determined using the relation between charge and current flow. This involves understanding Faraday’s laws of electrolysis, stoichiometric electron transfer, and conversion between moles of substance and electrical charge. The process highlights how electrical energy is precisely converted into chemical deposition in controlled electrochemical systems.
Option a – 30 s
How many Coulombs are needed to convert 12.3 g of nitrobenzene into aniline?
(A) 579 C
(B) 5790 C
(C) 57900 C
(D) 579000 C
Explanation: The conversion of nitrobenzene to aniline is a reduction process involving electron transfer in an electrochemical system. The number of electrons required depends on the change in oxidation state of nitrogen-containing functional groups during the reaction. Electrochemical calculations involve determining moles of reactant and relating them to electrons required per Molecule. Faraday’s constant is used to convert mole of electrons into total charge. This requires understanding of Organic electroreduction, stoichiometry of functional group transformation, and charge–mass relationships. The key idea is that the extent of reduction determines the total electrical energy needed for complete conversion of the compound.
Option c – 57900 C
The amount of Electricity required to deposit 108 g of silver from AgNO₃ solution is ( New Simplified Chemistry Class 8 ICSE mcq )
(A) 1 ampere
(B) 1 Coulomb
(C) 1 Faraday
(D) 2 ampere
Explanation: Electroplating of silver involves reduction of silver ions to metallic silver at the cathode. Each silver ion requires a single electron for reduction, making the process directly proportional to the amount of silver deposited. The total charge required can be calculated using the molar mass of silver and Faraday’s constant, which links electrons to electrical charge. Understanding this requires knowledge of electrochemical deposition, molar conversions, and Faraday’s first law of electrolysis. The process illustrates how electrical energy is used to drive reduction reactions in solution, leading to the formation of pure metallic layers on electrodes.
Option c – 1 Faraday
Passing 96500 C through CuSO₄ solution results in the liberation of
(A) 63.5 g of Cu
(B) 31.75 g of Cu
(C) 96500 g of Cu
(D) 100 g of Cu
Explanation: The quantity of Electricity passed in electrolysis determines the amount of substance produced at electrodes. One Faraday corresponds to the charge required to produce one equivalent of a substance. In copper sulfate solution, copper ions are reduced to metallic copper by gaining electrons. The mass of copper deposited depends on its molar mass and valency. Faraday’s laws allow direct conversion between charge and mass of substance deposited. Understanding this requires knowledge of electrochemical equivalence, electron transfer reactions, and stoichiometric relationships in redox processes. The result reflects the direct proportionality between electrical charge and chemical change at the electrode surface.
Option b – 31.75 g of Cu
What is the charge equal to 0.05 Faraday? ( New Simplified Chemistry Class 8 ICSE mcq )
(A) 48250 C
(B) 3776 C
(C) 4825 C
(D) 4285 C
Explanation: Faraday represents a fixed quantity of electric charge associated with one mole of electrons. It is a fundamental constant used in electrochemistry to relate electrical and chemical quantities. When a fraction of a Faraday is given, the total charge is determined by multiplying that fraction with the standard value of Faraday. This relationship is based on the concept that electric charge is quantized and directly proportional to the number of electrons involved in a reaction. Understanding this requires knowledge of electrochemical units, mole concept, and charge quantization. It provides a bridge between microscopic electron transfer and macroscopic measurable Electricity.
Option c – 4825 C
If 3 Faradays are passed through molten Al₂O₃, aqueous CuSO₄, and molten NaCl in separate cells, what is the molar ratio of Al, Cu, and Na deposited? ( New Simplified Chemistry Class 8 ICSE mcq )
(A) 1 mole : 2 mole : 3 mole
(B) 3 mole : 2 mole : 1 mole
(C) 1 mole : 1.5 mole : 3 mole
(D) 1.5 mole : 2 mole : 3 mole
Explanation: Electrolysis of different ionic compounds results in deposition of metals based on the number of electrons required for reduction of each ion. Aluminium requires more electrons per ion compared to copper and sodium due to its higher oxidation state. Since the same amount of charge passes through each system, the amount of metal deposited depends inversely on the number of electrons needed per ion. This requires understanding of valency, Faraday’s laws, and stoichiometric electron transfer in electrolysis. The molar ratio reflects how different ions respond to identical electrical input based on their reduction requirements.
Option c – 1 mole : 1.5 mole : 3 mole
When 1 Faraday of charge is passed through aqueous NiSO₄, the mass of nickel deposited at the cathode is ( New Simplified Chemistry Class 8 ICSE mcq )
(A) 29.35 g
(B) 2.935 g
(C) 0.2935 g
(D) 293.5 g
Explanation: Nickel ions in solution undergo reduction at the cathode during electrolysis, gaining electrons to form metallic nickel. The amount deposited depends on the molar mass of nickel and the number of electrons required per ion. One Faraday corresponds to one mole of electrons, making it a convenient unit for calculating deposition directly. This involves understanding electrochemical equivalence, valency of nickel ions, and Faraday’s laws. The process demonstrates a direct relationship between electrical charge and mass of metal produced in electrolysis.
Option a – 29.35 g
How can the lifespan of a Daniell cell be extended?
(A) By using a larger copper electrode
(B) By using a larger zinc electrode
(C) By reducing the temperature
(D) By lowering the concentration
Explanation: A Daniell cell produces electrical energy through a spontaneous redox reaction between zinc and copper ions. The cell’s lifespan depends on the availability of reactants, particularly the anode metal and electrolyte concentration. Increasing the amount of active material allows the reaction to continue for a longer time before depletion occurs. The efficiency of the cell is also influenced by maintaining proper ion flow and minimizing resistance in the system. Understanding this requires knowledge of electrochemical cell operation, redox stoichiometry, and factors affecting cell capacity. The key idea is that greater availability of reactants extends the duration of sustained electron flow.
Option d – By lowering the concentration
Daniel cell is classified as ( New Simplified Chemistry Class 8 ICSE mcq )
(A) A reversible primary cell
(B) An irreversible primary cell
(C) An irreversible electrochemical cell
(D) A secondary type of cell
Explanation: Electrochemical cells are broadly classified based on whether they generate electrical energy spontaneously or require external energy to operate. A Daniell cell functions through a spontaneous redox reaction between two different metal electrodes immersed in their respective ionic solutions. Electrons flow from the anode to the cathode through an external circuit, generating electrical current. The process is irreversible under normal conditions, meaning the reaction proceeds in one direction only. Understanding this requires knowledge of electrochemical cell types, redox spontaneity, and energy conversion principles. The classification depends on whether the system produces or consumes electrical energy during operation.
Option a – A reversible primary cell
In a Daniel cell, the reduction process occurs at the
(A) Positive terminal
(B) Negative terminal
(C) Zinc electrode
(D) Anode section
Explanation: In electrochemical cells, oxidation and reduction occur at separate electrodes. The electrode where reduction takes place is called the cathode, where electrons are gained by ions in solution. In a Daniell cell, copper ions gain electrons and get deposited as metallic copper. The flow of electrons from the anode to cathode drives the reaction spontaneously. Understanding this requires knowledge of redox reactions, electrode potential, and electron flow in galvanic cells. The location of reduction is fundamental to understanding how electrical energy is generated in such systems.
Option a – Positive terminal
Removing the Salt bridge in a galvanic cell causes the voltage to
(A) Immediately drop to zero
(B) Remain unchanged
(C) Gradually increase
(D) Rise sharply
Explanation: A Salt bridge maintains electrical neutrality in a galvanic cell by allowing ion migration between two half-cells. It prevents charge buildup that would otherwise stop the redox reaction. Without it, excess positive and negative charges accumulate in respective compartments, quickly halting electron flow. This disrupts the continuous operation of the cell and prevents further current generation. Understanding this requires knowledge of ionic balance, electrochemical circuit continuity, and cell potential maintenance. The Salt bridge is essential for sustaining redox reactions by completing the internal ionic circuit.
Option a – Immediately drop to zero
Which of the following is incorrect about a galvanic cell? ( New Simplified Chemistry Class 8 ICSE mcq )
(A) Anode has a negative potential
(B) Cathode carries a positive charge
(C) Reduction takes place at the anode
(D) Reduction occurs at the cathode in the electrochemical system
Explanation: A galvanic cell operates based on spontaneous redox reactions where chemical energy is converted into electrical energy. Oxidation occurs at the anode and reduction at the cathode, and electrons flow through an external circuit. The electrode charges and potentials depend on the nature of the metals and their ionic solutions. Any statement that contradicts the basic principles of electron flow direction, electrode function, or redox behavior would be incorrect. Understanding this requires knowledge of electrochemical conventions, oxidation–reduction processes, and standard electrode potential concepts. The key is consistency with fundamental rules governing galvanic cells.
Option c – Reduction takes place at the anode
The primary role of a Salt bridge is to
(A) Allow ion migration between the two half-cells
(B) Connect both half-cells physically
(C) Ensure the e.m.f. stays positive
(D) Maintain electrical neutrality across the solutions
Explanation: A Salt bridge is an essential component of a galvanic cell that ensures continuous operation by maintaining charge balance between two half-cells. As electrons flow through the external circuit, ions must migrate internally to prevent charge accumulation. The Salt bridge allows movement of inert ions that neutralize excess charges in each compartment. This maintains electrical neutrality and allows redox reactions to continue uninterrupted. Understanding this requires knowledge of ion migration, electrochemical equilibrium, and circuit completion in galvanic systems. It plays a crucial role in sustaining steady current flow in electrochemical cells.
Option d – Maintain electrical neutrality across the solutions
During electrolysis, the number of ions generated is directly related to ( New Simplified Chemistry Class 8 ICSE mcq )
(A) Duration of electrolysis
(B) Electrochemical equivalent value
(C) Total electric charge passed
(D) Mass of the involved electrons
Explanation: Electrolysis involves the decomposition of an electrolyte into ions that move toward respective electrodes under the influence of an Electric Field. The extent of ionic movement and participation in redox reactions depends on the total amount of electric charge passed through the solution. According to electrochemical principles, charge determines how many ions are discharged at electrodes. This is because each ion requires a specific number of electrons for reduction or oxidation, linking ionic conversion directly to electron flow. Understanding this requires knowledge of Faraday’s laws, ion-electron relationships, and conductivity behavior in electrolytic solutions. The process shows a direct proportionality between electrical input and chemical change in ionic systems.
Option c – Total electric charge passed
In a galvanic cell, the energy transformation is from
(A) Chemical form to electrical form
(B) Electrical energy to chemical form
(C) Chemical form to internal energy
(D) Internal energy to chemical energy
Explanation: A galvanic cell operates through spontaneous redox reactions where chemical energy stored in reactants is converted into electrical energy. This occurs due to differences in electrode potentials between two half-cells, causing electrons to flow through an external circuit. The oxidation reaction releases energy, which is then transferred as electrical work. The reduction process simultaneously consumes electrons at the cathode. Understanding this requires knowledge of Thermodynamics, redox chemistry, and electrochemical cell operation. The key principle is that chemical reactions with negative Gibbs free energy can be harnessed to produce usable electrical power in a controlled system.
Option a – Chemical form to electrical form
A typical Salt bridge includes
(A) Saturated KNO₃ with agar-agar
(B) Saturated KCl mixed with agar-agar
(C) Saturated NH₄NO₃ with agar-agar
(D) Any of the above options
Explanation: A Salt bridge is designed to maintain electrical neutrality in electrochemical cells by allowing ion movement between two half-cells without direct mixing of solutions. It is usually filled with an inert electrolyte that does not react with either half-cell components. The electrolyte is often supported in a gel medium to prevent leakage and ensure controlled ion flow. The ions within the Salt bridge migrate to balance charge buildup caused by oxidation and reduction processes occurring at electrodes. Understanding this requires knowledge of ion migration, electrochemical equilibrium, and cell design. The salt bridge ensures continuous electron flow by completing the internal ionic circuit.
Option d – Any of the above options
When writing a galvanic cell notation, ions in the same phase are separated by
(A) A single vertical bar
(B) Two vertical bars
(C) A comma
(D) A semicolon
Explanation: Cell notation is a symbolic representation of electrochemical cells used to describe redox reactions and electrode arrangements. In this notation, different phases such as Solids, liquids, and gases are separated using specific symbols to indicate phase boundaries. When multiple species exist in the same phase, such as ions dissolved in solution, they are written together using a punctuation separator. This helps distinguish between phase changes and species present in the same medium. Understanding this requires knowledge of electrochemical representation rules and standard conventions used to describe cell structure clearly and concisely. The notation ensures clarity in representing complex redox systems.
Option c – A comma
During electrolysis of an aqueous sodium sulfate solution, the substance released at the cathode is ( New Simplified Chemistry Class 8 ICSE MCQ )
(A) Sodium
(B) Sulfur dioxide
(C) Sulfur trioxide
(D) Hydrogen
Explanation: In aqueous electrolysis, the products formed at electrodes depend on the relative ease of discharge of ions present in solution. At the cathode, reduction occurs, where positively charged species gain electrons. In aqueous sodium sulfate, hydrogen ions from water are more readily reduced compared to sodium ions due to their position in the electrochemical series. As a result, hydrogen gas is typically formed at the cathode. Understanding this requires knowledge of preferential discharge, electrochemical series, and water ionization in electrolytic solutions. The process demonstrates how competing ions determine the actual products of electrolysis in aqueous media.
Option d – Hydrogen
When one end of a metal rod is heated, the other end becomes hot due to
(A) Movement of energized electrons
(B) Slight changes in atomic energy
(C) Electrical resistance of the metal
(D) Movement of metal atoms
Explanation: Heat transfer in Solids occurs primarily through conduction, where thermal energy is passed from one particle to another without the physical movement of the material itself. In metals, conduction is highly efficient due to the presence of free electrons that move throughout the lattice and carry energy rapidly. These electrons transfer kinetic energy from the heated region to cooler regions of the metal rod. This process is faster in metals compared to non-metals due to metallic bonding and electron mobility. Understanding this requires knowledge of thermal conductivity, electron theory of metals, and energy transfer mechanisms in Solids. The phenomenon explains why metals are good conductors of Heat.
Option a – Movement of energized electrons
Electrolysis of dilute sulfuric acid using platinum electrodes yields the following at the anode ( New Simplified Chemistry Class 8 ICSE MCQ )
(A) Hydrogen gas
(B) Oxygen gas
(C) Hydrogen sulfide
(D) Sulfur dioxide
Explanation: During electrolysis of dilute sulfuric acid, water molecules are primarily involved in electrode reactions because sulfuric acid acts mainly as an electrolyte to increase conductivity. At the anode, oxidation occurs where negatively charged species or water molecules lose electrons. Since sulfate ions are difficult to oxidize under normal conditions, water is oxidized instead, producing oxygen gas. Platinum electrodes are inert and do not participate in the reaction, serving only as conductive surfaces. Understanding this requires knowledge of electrochemical oxidation processes, electrode inertness, and preferential discharge of ions in aqueous solutions. The reaction illustrates how water participates in electrolysis when more stable ions are present.
Option b – Oxygen gas
Among the following, which KCl solution exhibits the greatest specific conductivity?
(A) 0.01 M
(B) 0.1 M
(C) 1.0 M
(D) 0.5 M
Explanation: Specific conductivity of an electrolyte solution depends on the number of ions present per unit volume and their mobility. As concentration increases, the number of charge carriers increases, leading to higher conductivity. However, at very high concentrations, ion–ion interactions may slightly reduce mobility, but overall conductivity is still higher compared to dilute solutions. In potassium chloride solutions, conductivity increases with concentration due to greater availability of ions for charge Transport. Understanding this requires knowledge of ionic mobility, concentration effects, and electrolyte conductivity behavior in solution. The balance between ion concentration and mobility determines overall conductivity.
Option c – 1.0 M
The rise in equivalent conductance of a strong electrolyte upon dilution is largely because ( New Simplified Chemistry Class 8 ICSE MCQ )
(A) The ionic mobility of the ions increases
(B) Electrolyte is completely ionized under normal dilution
(C) Both the number of ions and their mobility increase
(D) More ions are formed
Explanation: Equivalent conductance measures the ability of ions to carry electric current in solution per equivalent of electrolyte. In strong electrolytes, ions are already fully dissociated, so dilution does not increase ionization further. Instead, dilution reduces interionic attraction, allowing ions to move more freely. This increases their mobility and thus increases equivalent conductance. Understanding this requires knowledge of ionic Atmosphere theory, interionic forces, and Transport properties of ions in solution. The key idea is that reduced concentration minimizes ionic interactions, enhancing conductivity efficiency per equivalent of solute.
Option a – The ionic mobility of the ions increases
In the electrolysis of acidified water, the volume ratio of hydrogen to oxygen evolved is
(A) 2 : 1
(B) 1 : 2
(C) 8 : 1
(D) 1 : 8
Explanation: Electrolysis of acidified water involves decomposition of water molecules into hydrogen and oxygen gases at cathode and anode respectively. The process follows stoichiometric relationships based on the chemical equation of water splitting. Since water contains hydrogen and oxygen in a fixed atomic ratio, the gases produced reflect this proportion when corrected for gaseous volume relationships under identical conditions. Understanding this requires knowledge of stoichiometry, electrolysis reactions, and gas laws. The relative volumes of gases produced depend on the number of molecules formed at each electrode during the reaction process.
Option a – 2 : 1
According to Kohlrausch’s law, at infinite dilution ( New Simplified Chemistry Class 8 ICSE MCQ )
(A) Each ion contributes independently to the conductance of the electrolyte regardless of the other ion
(B) Each ion contributes a definite value to the equivalent conductance regardless of the other ion
(C) Each ion’s contribution to equivalent conductance depends on the nature of the counter-ion
(D) Conductance depends on both ions even at infinite dilution
Explanation: Kohlrausch’s law states that at infinite dilution, each ion contributes independently to the total conductivity of an electrolyte solution. At this extreme dilution, interionic interactions become negligible, allowing ions to move freely without influencing each other’s mobility. As a result, the molar or equivalent conductance of an electrolyte becomes the sum of individual ionic contributions. This principle helps determine limiting conductance values and understand ionic behavior in solution. It requires knowledge of ionic dissociation, conductivity concepts, and the effect of dilution on electrolyte behavior. The law is fundamental in physical chemistry for studying Transport properties of ions.
Option a – Each ion contributes independently to the conductance of the electrolyte regardless of the other ion
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