Question 1

Polythene, also called polyethylene, is one of the most common plastics. It is formed by the polymerisation of which monomer?

Accepted Answer

Introduction / Context: This question focuses on a very common plastic, polythene or polyethylene, and asks you to recall its basic building block monomer. Understanding which monomer gives rise to which polymer is a core idea in polymer chemistry. Polythene is widely used in carry bags, bottles and packaging films, so knowing its monomer helps in understanding its production, structure and properties. Given Data / Assumptions: The polymer of interest is polythene or polyethylene. It is produced by polymerisation of a single type of monomer. The options include ethylene, ethyl alcohol, vinyl chloride and ethanol. All options are common organic compounds but only one is the correct monomer. Concept / Approach: Polythene is an addition polymer formed by the polymerisation of ethylene, which is also called ethene and has the formula CH2=CH2. In addition polymerisation, the carbon carbon double bond in the monomer opens up and many monomer units join to form a long chain. Ethyl alcohol and ethanol are two names for the same compound, CH3CH2OH, which does not undergo simple addition polymerisation to give polythene. Vinyl chloride (CH2=CHCl) is the monomer for polyvinyl chloride, known as PVC, not for polythene. Recognising that ethylene is the correct monomer is the key to answering the question. Step-by-Step Solution: Step 1: Recall that polythene is also called polyethylene and is a polymer of ethene or ethylene. Step 2: Ethylene has a carbon carbon double bond that can open during polymerisation. Step 3: Under high pressure and appropriate conditions, many ethylene molecules join to form a long chain polymer with repeating –CH2–CH2– units. Step 4: Examine the options. Ethyl alcohol and ethanol refer to the same alcohol molecule, which is not used to make polythene. Step 5: Vinyl chloride is the monomer used to produce polyvinyl chloride, a different plastic. Step 6: Therefore, only ethylene matches the known monomer for polythene. Verification / Alternative check: Polymer chapters in textbooks often list common monomer polymer pairs: ethene to polyethene, propene to polypropylene, vinyl chloride to polyvinyl chloride, tetrafluoroethene to polytetrafluoroethene and so on. In all these cases, the monomer is a small unsaturated molecule with a double bond. Ethylene fits this pattern perfectly and is always linked to polythene in such tables. You can also recall that polyethylene is often represented structurally as (CH2–CH2)n, clearly showing that the monomer unit corresponds to CH2=CH2, which is ethylene. Why Other Options Are Wrong: Ethyl alcohol and ethanol are the same compound, an alcohol with an –OH group, and are not used for manufacturing polythene. They may undergo other types of reactions such as oxidation, dehydration or esterification, but not addition polymerisation to form polyethylene. Vinyl chloride, with a chlorine atom attached, is used to make polyvinyl chloride, which has different properties like rigidity and chemical resistance, unlike polythene. Thus these options do not match the well established link between ethylene and polythene. Common Pitfalls: A frequent confusion arises because students may mix up different common plastics and their monomers. For example, it is easy to confuse vinyl chloride and ethylene because both have a carbon carbon double bond and are gaseous at room temperature. Additionally, the presence of both ethyl alcohol and ethanol as options may seem to indicate one of them is correct, but they are actually the same compound and serve as distractors. To avoid such mistakes, memorize a short list of the most important monomer polymer combinations and always focus on the presence of double bonds for addition polymers like polythene. Final Answer: Polythene is obtained by addition polymerisation of the monomer ethylene.

Question 2

In basic chemistry classification, a pure substance can be which of the following?

Accepted Answer

Introduction / Context: This question checks your understanding of fundamental definitions in chemistry, especially the distinction between pure substances and mixtures. Classifying matter correctly is the starting point for more advanced topics like solutions, chemical reactions and stoichiometry. A pure substance has a fixed composition and distinct properties, and this question asks you to identify which categories of matter qualify as pure substances. Given Data / Assumptions: You are given four options describing different types of matter. Option A suggests that only compounds are pure substances. Option B suggests that only elements are pure substances. Option C suggests that both elements and compounds can be pure substances. Option D suggests that a heterogeneous mixture can be a pure substance. Standard school level definitions of pure substances and mixtures apply. Concept / Approach: A pure substance is a form of matter that has a uniform and definite composition and consistent properties throughout the sample. It cannot be separated into other kinds of matter by simple physical processes. Elements are pure substances made up of only one kind of atom, such as oxygen gas or copper metal. Compounds are also pure substances, but they contain two or more elements chemically combined in fixed proportions, such as water or sodium chloride. Mixtures, whether homogeneous or heterogeneous, contain more than one substance and their compositions can vary. Therefore, both elements and compounds are pure, while mixtures are not. Step-by-Step Solution: Step 1: Recall that an element is defined as a pure substance that cannot be broken down into simpler substances by ordinary chemical means. Step 2: A compound is defined as a pure substance composed of two or more elements chemically combined in a fixed ratio. Step 3: Because both elements and compounds have fixed compositions and uniform properties, they are both classified as pure substances. Step 4: Heterogeneous mixtures have non uniform composition and can be separated by physical methods, so they are not pure substances. Step 5: Option A is too restrictive because it excludes elements. Step 6: Option B is also too restrictive because it excludes compounds. Step 7: Therefore, the correct description is that a pure substance can be an element or a compound. Verification / Alternative check: Think of examples commonly used in textbooks. Pure copper is an element and is certainly considered a pure substance. Distilled water is a compound made only of H2O molecules and is also a pure substance. In contrast, sea water contains dissolved salts and gases in varying amounts, so its composition is not fixed and it is a mixture. Similarly, soil, air and alloys like brass are mixtures. This everyday reasoning backs up the classification that both elements and compounds belong to the category of pure substances, while mixtures do not. Why Other Options Are Wrong: Option A claims that only compounds are pure substances, which wrongly excludes elements like hydrogen, oxygen and gold that clearly meet the definition of a pure substance. Option B claims that only elements are pure, ignoring compounds like water that also have fixed composition and homogeneous properties. Option D suggests that a heterogeneous mixture is a pure substance, which contradicts the very definition of heterogeneity, where different parts of the sample have different compositions or properties. Therefore, these options are inconsistent with standard chemical classification. Common Pitfalls: Students sometimes confuse the term pure in everyday language with its scientific meaning. In daily life, people may call tap water pure even though it contains dissolved minerals and gases, so scientifically it is a mixture. Another pitfall is to think that any substance containing more than one element cannot be pure, but compounds like sodium chloride have fixed ratios and thus are pure substances. Remember that purity in chemistry is linked to uniform composition and not to the number of elements present in the substance. Final Answer: A pure substance can be an element or a compound.

Question 3

In atomic physics experiments with discharge tubes, cathode rays produced between electrodes are best described as which of the following?

Accepted Answer

Introduction / Context: This question comes from the early experiments on electrical discharge through gases, where scientists discovered cathode rays in evacuated tubes. Understanding what cathode rays really are was a key step toward discovering the electron and developing modern atomic models. Here you must recall the nature of cathode rays and choose the correct description from the options given. Given Data / Assumptions: Cathode rays are produced in a discharge tube under low pressure and high voltage. They travel from the cathode toward the anode. They cause fluorescence on certain materials and can be deflected by electric and magnetic fields. The options available describe possible physical interpretations: electromagnetic waves, alpha particles, electrons, or unspecified harmful radiation. Concept / Approach: Experiments by J. J. Thomson and others showed that cathode rays have several important properties. They are deflected by electric and magnetic fields, they travel in straight lines, they can cause mechanical motion of small paddles placed in their path and they have negative charge. Based on measurement of the charge to mass ratio, these rays were identified as streams of negatively charged particles with much smaller mass than atoms. These particles were later named electrons. Alpha particles are positively charged and much heavier, and electromagnetic waves such as light are not deflected by electric fields in the same way. Therefore cathode rays must be streams of electrons. Step-by-Step Solution: Step 1: Recall that cathode rays originate from the cathode in a discharge tube and move toward the anode. Step 2: Experimental evidence shows that cathode rays are affected by electric and magnetic fields, indicating that they carry charge. Step 3: The direction of deflection demonstrates that they are negatively charged. Step 4: Thomson measured the charge to mass ratio of the particles making up cathode rays and found it very large, meaning the particles have very small mass. Step 5: These particles were recognised as fundamental constituents of atoms and later called electrons. Step 6: Therefore cathode rays are correctly described as a stream of electrons. Verification / Alternative check: A simple way to check this understanding is to compare cathode rays with other known radiations. Alpha particles are helium nuclei that are positively charged and several thousand times more massive than electrons. They are produced in radioactive decay, not in discharge tubes, and are deflected in the opposite direction compared to electrons. Electromagnetic waves like light or X rays do not consist of charged particles and are not deflected by electric fields in the same way. Since cathode rays respond to electric and magnetic fields exactly as a beam of negative charges would, describing them as a stream of electrons is consistent with all experimental observations. Why Other Options Are Wrong: Electromagnetic waves such as visible light or X rays do not carry net electric charge and therefore are not deflected by uniform electric fields in the way cathode rays are. A stream of alpha particles would be composed of positively charged helium nuclei, which would deflect in the opposite direction from cathode rays in the same field and have a much larger mass. The phrase harmful radiations of unknown type ignores the well established identification of cathode rays as electrons and is historically outdated. Hence only the description as a stream of electrons fits the observed behaviour. Common Pitfalls: Students sometimes confuse cathode rays with X rays because both can be produced in similar apparatus. However, X rays are generated when high energy electrons strike a metal target and are electromagnetic in nature, whereas cathode rays themselves are the electrons. Another source of confusion is between different kinds of radiation (alpha, beta and gamma rays) emitted by radioactive substances. Beta rays are streams of electrons or positrons and are more closely related to cathode rays. To avoid confusion, remember that cathode rays and beta minus rays are both streams of electrons, while alpha rays are helium nuclei and gamma rays are high energy photons. Final Answer: Cathode rays in discharge tubes are best described as a stream of electrons.

Question 4

Who is credited with the discovery of natural radioactivity in uranium salts, leading to the modern understanding of radioactive phenomena?

Accepted Answer

Introduction / Context: This general knowledge question belongs to the history of atomic and nuclear physics. It asks you to identify which scientist discovered natural radioactivity. The discovery of radioactivity was a landmark event that paved the way for further work by scientists such as Marie and Pierre Curie, Ernest Rutherford and many others. Knowing who first observed this spontaneous emission from uranium salts is a standard fact often tested in examinations. Given Data / Assumptions: The phenomenon being discussed is natural radioactivity, observed in uranium salts. The discovery involved spontaneous emission of radiation that could fog photographic plates. The options list several important scientists associated with nuclear physics and radioactivity. The task is to identify the scientist who first discovered natural radioactivity. Concept / Approach: Radioactivity is the spontaneous emission of particles or electromagnetic radiation from unstable atomic nuclei. Henri Becquerel discovered this phenomenon in 1896 while investigating phosphorescent materials containing uranium. He found that uranium salts emitted rays that could blacken photographic plates even in the absence of sunlight, indicating a new type of penetrating radiation. Later, Marie and Pierre Curie extended this work by discovering polonium and radium. Ernest Rutherford studied the nature of alpha, beta and gamma rays and explained radioactive decay. However, the original discovery of radioactivity is credited to Becquerel. Step-by-Step Solution: Step 1: Recall the historical timeline. In 1896, Henri Becquerel performed experiments with uranium salts placed on wrapped photographic plates. Step 2: He observed that the plates darkened even when kept in the dark, proving that the uranium salts emitted penetrating radiation without any external stimulation. Step 3: This spontaneous emission was later named radioactivity. Step 4: Marie Curie and Pierre Curie followed up by discovering additional radioactive elements, but they did not make the initial discovery. Step 5: Ernest Rutherford contributed pioneering work in classifying radioactive emissions and in nuclear models, but again he did not first discover radioactivity. Step 6: Therefore, the credit for discovering natural radioactivity goes to Henri Becquerel. Verification / Alternative check: Textbook summaries of the discovery of radioactivity almost always mention Becquerel as the starting point. He was awarded the Nobel Prize in Physics in 1903, jointly with Marie and Pierre Curie, for their work on radioactivity. The Nobel citation emphasises his discovery of spontaneous radioactivity in uranium salts. While Marie and Pierre Curie discovered new radioactive elements and studied their properties, their work builds on Becquerel initial finding. Rutherford later used radioactive emissions to probe the structure of the atom. This consistent historical record confirms that Becquerel is the correct answer. Why Other Options Are Wrong: Ernest Rutherford is a central figure in nuclear physics and explained alpha and beta decay and the nuclear model of the atom, but he did not first discover the phenomenon of radioactivity. Marie Curie made major contributions by discovering polonium and radium and coining the term radioactivity, yet she worked after Becquerel discovery. Schmidt is sometimes mentioned as an independent observer of similar effects, but standard credit for the initial discovery is given to Becquerel. Therefore these options are historically inaccurate for the first discovery of radioactivity. Common Pitfalls: Students sometimes confuse the scientist who named or developed the theory of radioactivity with the one who made the original observation. Because Marie Curie is very famous in the context of radioactivity and radiation therapy, some learners mistakenly choose her as the discoverer. Others may select Rutherford because of his nuclear model of the atom. To avoid such confusion, remember the sequence: Becquerel discovered spontaneous radioactivity in uranium salts, the Curies expanded the field by discovering new radioactive elements and Rutherford analysed the nature of the radiation and the nucleus. Final Answer: The discovery of natural radioactivity in uranium salts is credited to Henri Becquerel.

Question 5

In atomic structure, the formation of a cation from a neutral atom occurs by which fundamental process?

Accepted Answer

Introduction / Context: This question checks a basic but crucial idea in atomic structure and chemical bonding: how ions are formed from neutral atoms. Understanding the difference between cations and anions, and the processes that form them, is essential for predicting the behaviour of elements in reactions and in ionic compounds. Here, you must identify which process leads specifically to the formation of a positively charged ion, or cation. Given Data / Assumptions: A neutral atom initially has equal numbers of protons and electrons. A cation is defined as a positively charged ion. The options describe gain or loss of electrons or protons. We assume that changes in the nucleus, such as gain or loss of protons, are not common in ordinary chemical reactions. Concept / Approach: The charge on an atom or ion is determined by the balance between positively charged protons in the nucleus and negatively charged electrons around the nucleus. In chemical processes, the number of protons in the nucleus remains fixed because changing protons would turn one element into another and requires nuclear reactions. However, atoms can readily gain or lose electrons. If an atom loses one or more electrons, it has more protons than electrons and becomes positively charged, forming a cation. If it gains electrons, it becomes negatively charged, forming an anion. Therefore, formation of a cation is associated with loss of electrons, not gain of electrons or change in proton number in ordinary chemistry. Step-by-Step Solution: Step 1: Start with a neutral atom where number of protons equals number of electrons, so overall charge is zero. Step 2: Consider what happens when the atom loses one electron. Now the number of protons exceeds the number of electrons by one. Step 3: Because protons are positively charged and outnumber electrons, the net charge of the particle becomes positive. This species is a cation. Step 4: If instead the atom gained an electron, the electron count would exceed the proton count, yielding a negatively charged anion. Step 5: Gaining or losing protons would change the atomic number and identity of the element, which does not occur in typical chemical processes. Step 6: Therefore, the correct description of cation formation is loss of an electron. Verification / Alternative check: Take the example of a sodium atom, which has 11 protons and 11 electrons in its neutral state. If sodium loses one electron, it now has 11 protons and 10 electrons. The net charge is plus one because there is one more positive charge than negative charge. This ion is written as Na+ and is a cation. Similarly, a magnesium atom with 12 protons and 12 electrons can lose two electrons to become Mg2+, with a net charge of plus two. In both cases, the process involved is loss of electrons with the proton count unchanged. This pattern confirms that cations are formed by electron loss. Why Other Options Are Wrong: Gain of an electron would produce an anion, not a cation, because the species would now have more negative charges than positive charges. Gain of a proton is not a typical process in chemical reactions, and if it occurs in nuclear reactions, it changes the element itself, not just its ionic state. Loss of a proton likewise is a nuclear process and not the general route for forming cations in chemistry. Therefore these options do not correctly describe the usual formation of a cation from a neutral atom. Common Pitfalls: Some students mistakenly think that adding a positive particle like a proton is the way to make a positive ion, but in chemistry courses, ions are usually formed by electron transfer, not by changing the nucleus. Another common error is to mix up cations and anions and forget which is positive and which is negative. A simple mnemonic is that cation contains the letter t, which looks like a plus sign, reminding you that cations carry a positive charge. Remember also that metals tend to form cations by losing electrons, while non metals tend to form anions by gaining electrons. Final Answer: A cation is formed from a neutral atom by the loss of an electron.

Question 6

In traditional photography, which nitrogen containing compound is used as a light sensitive material on photographic films and papers?

Accepted Answer

Introduction / Context: This question relates to the chemistry behind traditional photography. Before digital cameras, photographic films and papers relied on light sensitive chemical compounds. Many of these compounds contained nitrogen in their structure and were based on silver salts. The question asks you to identify which nitrogen containing compound is used as a light sensitive material in photographic processes. Given Data / Assumptions: We are considering traditional photographic films and papers, not digital sensors. The active material must be light sensitive, meaning it undergoes chemical change on exposure to light. The options include potassium nitrate, silver nitrate, ammonium carbonate and ammonium bicarbonate. Only one of these serves as a precursor for the silver halides used in photography. Concept / Approach: In classical photography, the light sensitive layer on film or photographic paper is usually a silver halide such as silver bromide, silver chloride or silver iodide embedded in gelatin. These silver halides are prepared from silver nitrate, which contains both silver and nitrogen. When exposed to light, silver halide crystals undergo a photochemical reaction that leads to formation of metallic silver in the exposed regions during development, creating a latent image that is later made visible. Potassium nitrate and ammonium salts are common laboratory and industrial chemicals, but they are not themselves the principal light sensitive materials used in photographic emulsions. Step-by-Step Solution: Step 1: Recall that silver halides like silver bromide are the compounds that darken on exposure to light. Step 2: These silver halides are usually obtained in the laboratory and industry by reacting silver nitrate with suitable halide salts. Step 3: Silver nitrate is a soluble silver salt that contains nitrogen in the nitrate ion. Step 4: On photographic paper, the actual light sensitive species is the silver halide, but silver nitrate is the key starting material in preparing these emulsions. Step 5: Potassium nitrate, ammonium carbonate and ammonium bicarbonate are not primary light sensitive compounds in traditional photographic films. Step 6: Therefore, the nitrogen containing compound associated with photographic light sensitivity is silver nitrate. Verification / Alternative check: You may have seen experiments in school laboratories where a solution of silver nitrate is mixed with sodium chloride to precipitate silver chloride, which darkens when exposed to sunlight. This darkening demonstrates the light sensitivity of silver halides. The same principle operates in photographic films, where tiny grains of silver halides are coated on a support. Silver nitrate is the standard laboratory reagent used to form these light sensitive salts. In contrast, potassium nitrate is more associated with fertilisers and fireworks, while ammonium carbonate and bicarbonate are used in baking and other applications but not as primary photographic chemicals. Why Other Options Are Wrong: Potassium nitrate is a nitrate salt used in fertilisers, explosives and some food preservation processes, but it does not form the core of photographic emulsions. Ammonium carbonate and ammonium bicarbonate are volatile salts used in baking powders and as leavening agents, not as light sensitive materials on films. While all these compounds contain nitrogen, their roles in industry are different from the specialised role of silver nitrate in producing silver halides for photography. Therefore, choosing any of these alternatives would not match the established chemistry of photographic materials. Common Pitfalls: Students sometimes focus only on the presence of nitrogen in the formula and think that any nitrate or ammonium compound could be the answer. However, the key is the presence of silver, which forms the light sensitive halides. Another confusion arises between silver nitrate and silver bromide. While silver bromide is directly light sensitive, the question specifically asks for a nitrogen containing compound, which points to silver nitrate. Remember that silver nitrate is a versatile reagent in inorganic chemistry and an important starting material in photographic technology. Final Answer: The nitrogen containing compound used to prepare light sensitive photographic materials is silver nitrate.

Question 7

The Van der Waals equation is a modified gas equation that explains the behaviour of which type of gases?

Accepted Answer

Introduction / Context: This question belongs to the topic of gaseous state and equations of state. The familiar ideal gas equation PV = nRT assumes point like molecules with no intermolecular forces, which is only an approximation. To describe the actual behaviour of real gases more accurately, Van der Waals proposed a modified equation that corrects for molecular volume and attractive forces. You are asked to identify which type of gases this Van der Waals equation is intended to describe. Given Data / Assumptions: The equation referred to is the Van der Waals equation of state. It introduces correction terms for pressure and volume compared with the ideal gas equation. The options mention ideal gases, real gases, mixtures of gases and diatomic gases only. Standard textbook meaning of ideal and real gases is assumed. Concept / Approach: An ideal gas obeys PV = nRT at all temperatures and pressures, with molecules treated as point particles that do not interact except through elastic collisions. Real gases deviate from this behaviour, especially at high pressure and low temperature, because their molecules occupy finite volume and experience intermolecular attractions and repulsions. The Van der Waals equation modifies the ideal gas equation by introducing a correction term a * n^2 / V^2 to account for intermolecular attraction and a volume correction term b * n to account for finite molecular size. This equation is specifically designed to approximate the behaviour of real gases, not truly ideal ones, and it works for many gases across a range of conditions. Step-by-Step Solution: Step 1: Recall the ideal gas equation PV = nRT, which is based on several simplifying assumptions about gas molecules. Step 2: Recognise that actual gases, such as carbon dioxide or ammonia, deviate from this ideal behaviour, especially at high pressure or low temperature. Step 3: Van der Waals introduced a corrected equation of the form (P + a * n^2 / V^2) * (V - n * b) = n * R * T. Step 4: The correction a * n^2 / V^2 reduces the effective pressure, compensating for attractive forces between molecules. Step 5: The correction V - n * b reduces the effective volume available to the gas, compensating for the finite size of molecules. Step 6: Because these corrections are specifically designed to handle deviations of actual gases from the ideal model, the Van der Waals equation is used for real gases. Verification / Alternative check: If gases truly behaved ideally, the simple PV = nRT equation would be sufficient and no corrections would be needed. In practice, however, measurements of pressure, volume and temperature for gases like carbon dioxide show deviations from ideal values, and the Van der Waals equation gives better agreement with experimental data. The constants a and b are different for different gases, which further confirms that the equation is tailored to describe real gases rather than an abstract ideal gas. Mixtures of gases can also be modelled, but the fundamental concept is still that they are real, not ideal. Why Other Options Are Wrong: Ideal gases are defined as obeying PV = nRT exactly, so there is no need to use the Van der Waals equation for a truly ideal gas. Mixtures of gases may still be treated as ideal or real; the equation in question does not specifically refer to mixtures only. Diatomic gases are not a special category in the context of the Van der Waals equation; they are simply real gases that can be described by it with appropriate constants. Therefore, the only correct and general description is that the Van der Waals equation applies to real gases. Common Pitfalls: One frequent confusion is to think that the Van der Waals equation defines a new ideal gas, whereas it actually corrects for non ideal behaviour. Another mistake is to associate the equation only with certain gases such as carbon dioxide or nitrogen, rather than understanding that it is a general model for many real gases. Some students also forget the physical meaning of constants a and b, which leads to weak conceptual understanding. Remember that the whole purpose of the Van der Waals equation is to improve the description of real gases, especially under conditions where ideal assumptions fail. Final Answer: The Van der Waals equation is used to explain and model the behaviour of real gases.

Question 8

Which of the following carbon compounds has been widely used as a fire extinguisher, especially for liquid and electrical fires?

Accepted Answer

Introduction / Context: This question tests your knowledge of applied chemistry, specifically the use of certain carbon compounds in fire safety. Different fire extinguishing agents are suited to different classes of fire. Some organic liquids were historically used in extinguishers, particularly for electrical and liquid fuel fires. You are asked to identify which carbon compound among the options has been widely used as a fire extinguisher. Given Data / Assumptions: The options are carbon disulphide, chloroform, carbon tetrachloride and methylene chloride. The compound in question should have properties suitable for extinguishing certain types of fire. It should be non flammable, able to interrupt the combustion process and preferably not conduct electricity. Historical usage in fire extinguishers is relevant, even though safety concerns have changed modern practice. Concept / Approach: Carbon tetrachloride (CCl4) is a non flammable, dense liquid that was historically used in fire extinguishers, especially for liquid fuel and electrical fires. When sprayed onto flames, it vaporises, displacing oxygen and interfering with the free radical chain reactions that sustain combustion. It is also a non conductor of electricity, which made it suitable for electrical equipment fires. However, it is toxic and can form harmful phosgene gas at high temperatures, so its use has been discontinued in many countries. The other compounds listed either are flammable or have other uses, not primary fire extinguishing roles. Step-by-Step Solution: Step 1: Recall that carbon tetrachloride is a chlorinated hydrocarbon once used in portable fire extinguishers. Step 2: It is non flammable and can form a heavy vapour blanket over burning liquids, cutting off oxygen supply. Step 3: It also does not conduct electricity, so it was considered suitable for fires involving electrical equipment. Step 4: Carbon disulphide, in contrast, is highly flammable and would not be used as a fire extinguishing agent. Step 5: Chloroform and methylene chloride have roles as solvents and anaesthetics or degreasers, not as standard fire extinguishing liquids. Step 6: Therefore, the correct answer is carbon tetrachloride. Verification / Alternative check: If you look at older descriptions of fire safety equipment, you will find references to carbon tetrachloride extinguishers, which were common before the health risks became widely known. These devices often contained glass bulbs filled with carbon tetrachloride that would burst when thrown at a fire. Modern fire safety guidelines no longer recommend carbon tetrachloride due to toxicity, and safer halogenated compounds or dry chemical powders are used instead. This historical information confirms that among the listed options, carbon tetrachloride is the compound associated with fire extinguishing use. Why Other Options Are Wrong: Carbon disulphide is very flammable and is actually a fire hazard, not a fire extinguisher. Chloroform has been used as an anaesthetic and solvent but is not a standard extinguishing agent. Methylene chloride, also called dichloromethane, is used as a solvent and paint stripper and has its own health risks, but it has not been widely adopted as a fire extinguisher fluid. None of these alternatives meets the historical and practical role of carbon tetrachloride in fire suppression. Common Pitfalls: Students may incorrectly assume that any chlorinated solvent could be a fire extinguishing agent because chlorinated compounds are often non flammable. However, not all such compounds have been used for this purpose, and toxicity and decomposition products must be considered. Another pitfall is to focus on the word carbon in the name and ignore the halogens attached. The correct approach is to recall specific compounds that have documented fire safety applications. Remember that carbon tetrachloride has largely been phased out due to health concerns but remains a classic example in examination questions. Final Answer: The carbon compound widely used in older fire extinguishers is carbon tetrachloride.

Question 9

Which species or particle has 13 protons and 10 electrons associated with it?

Accepted Answer

Introduction / Context: This question focuses on interpreting atomic and ionic composition in terms of protons and electrons. Being able to deduce the identity and charge of a species from its proton and electron counts is essential in atomic structure, chemical bonding and redox chemistry. The question gives you numbers of protons and electrons and asks you to identify which species matches these values. Given Data / Assumptions: The species under consideration has 13 protons. It has 10 electrons. Proton number (atomic number) determines the identity of the element. Only electrons are gained or lost in usual chemical ion formation; proton number remains fixed. The options list an aluminium atom, an aluminium ion Al3+, a nitrogen isotope and none of these. Concept / Approach: The atomic number Z of an element equals the number of protons in its nucleus. If Z is 13, the element is aluminium. A neutral aluminium atom has 13 electrons to balance the 13 protons. When aluminium forms a cation, it commonly loses three electrons to attain a stable electronic configuration, resulting in the ion Al3+, which has 13 protons and 10 electrons. Nitrogen has atomic number 7, so any isotope of nitrogen has 7 protons, not 13. Therefore, a particle with 13 protons and 10 electrons must be the Al3+ ion. Step-by-Step Solution: Step 1: Determine the element from the number of protons. An atomic number of 13 corresponds to aluminium. Step 2: A neutral aluminium atom would have 13 electrons to match the 13 protons, giving zero net charge. Step 3: The species described in the question has only 10 electrons, which is three fewer than a neutral atom would have. Step 4: Losing three electrons gives a net charge of plus three because there are three more protons than electrons. Step 5: An aluminium ion with charge plus three is written as Al3+, with 13 protons and 10 electrons. Step 6: Therefore, the species must be an aluminium ion Al3+. Verification / Alternative check: Check the other options. An aluminium atom Al has 13 protons and 13 electrons in its neutral state, which does not match the given 10 electrons. A nitrogen isotope, regardless of mass number, always has 7 protons, not 13. None of these would only be correct if no listed option matched, but Al3+ clearly does. You can also look at the electron configuration. Neutral aluminium has configuration 2, 8, 3. Removing three valence electrons leaves 2, 8, which is 10 electrons and a stable noble gas like configuration. This is precisely the configuration of Al3+. Why Other Options Are Wrong: The neutral aluminium atom cannot be correct because it would have 13 electrons. A nitrogen isotope contradicts the given proton number of 13. The generic none of these option is inappropriate because one of the options matches both the proton and electron counts perfectly. Thus these choices are inconsistent with the data given in the question. Common Pitfalls: Learners sometimes confuse atomic number with mass number and may think that 13 could represent protons plus neutrons. However, the question clearly refers to protons and electrons separately. Another error is to forget that losing electrons makes an ion positively charged and reduces the electron count. To avoid mistakes, always link atomic number to element identity and remember that only electrons are exchanged in routine chemical ion formation. Calculating charge as protons minus electrons helps verify which ion is being described. Final Answer: A particle with 13 protons and 10 electrons is the aluminium ion Al3+.

Question 10

Which scientific instrument is commonly used to measure and detect levels of radioactivity?

Accepted Answer

Introduction / Context: This question belongs to nuclear physics and applied instrumentation. Radioactivity refers to the spontaneous emission of particles or electromagnetic radiation from unstable nuclei. Measuring and detecting the intensity of this radiation is important in medical applications, radiation protection, nuclear industry and scientific research. The question asks you to identify which device among the options is specifically designed to measure radioactivity. Given Data / Assumptions: The task is to find an instrument that detects and measures radioactive emissions. The options include mass spectrometer, cyclotron, nuclear reactor and Geiger Muller counter. Each of these devices has a specific function in nuclear or atomic science. We assume standard textbook definitions of these instruments. Concept / Approach: A Geiger Muller counter, often simply called a Geiger counter, is a radiation detection instrument that uses a Geiger Muller tube filled with gas. When ionising radiation passes through the tube, it ionises the gas, causing a brief conductive discharge that is detected electronically. The device counts these discharge events to give a measure of radiation intensity. In contrast, a mass spectrometer measures mass to charge ratios of ions, a cyclotron is a particle accelerator used to accelerate charged particles to high energies, and a nuclear reactor is a device where controlled nuclear fission reactions occur. Only the Geiger Muller counter is specifically intended for detection and measurement of radioactivity in the field or laboratory. Step-by-Step Solution: Step 1: Recall the purpose of a Geiger Muller counter. It is designed to detect ionising radiation such as alpha, beta and gamma rays. Step 2: The device works by counting pulses created when radiation ionises gas within the Geiger tube. Step 3: These counts can be converted into a reading of radiation intensity in counts per second or similar units. Step 4: A mass spectrometer, by contrast, is used for analysing the composition of substances and determining the masses of ions, not for routine radiation monitoring. Step 5: A cyclotron is an accelerator that produces high energy charged particles, which is very different from a detection instrument. Step 6: A nuclear reactor is a power generating or research device where nuclear fission reactions are maintained, not a detector. Step 7: Therefore, the correct instrument for measuring radioactivity is the Geiger Muller counter. Verification / Alternative check: Think of images or descriptions from textbooks and media. When people refer to a handheld device that clicks faster near a radioactive source, they are describing a Geiger counter. Radiation safety officers often carry these devices to measure exposure levels. In laboratory exercises on radioactivity, students frequently use Geiger counters to record counts from a radioactive sample. None of the other devices listed is portable or used in this manner for direct radiation measurement. This confirms that the Geiger Muller counter is the correct answer. Why Other Options Are Wrong: A mass spectrometer requires careful vacuum and electromagnetic systems and is used for chemical analysis, not general radiation monitoring. A cyclotron is a large and complex machine used to accelerate particles and is not used as a detector. A nuclear reactor is a site where radioactivity is produced but is not itself a measuring instrument. While all of these devices are connected to nuclear or atomic physics, they do not serve the specific role of measuring ambient radioactivity in the way a Geiger Muller counter does. Common Pitfalls: Students sometimes choose nuclear reactor because they associate it strongly with radiation, forgetting that the question asks for a measuring device, not a source. Others may confuse mass spectrometry with radiation detection because both involve charged particles and detectors. To avoid such confusion, always focus on the primary function of each device. Remember that the Geiger Muller counter is the classic instrument used to detect and measure radioactivity, often represented with characteristic clicking sounds in demonstrations. Final Answer: The instrument commonly used to detect and measure radioactivity is the Geiger–Muller counter.

Question 11

Which one of the following elements or isotopes is not radioactive?

Accepted Answer

Introduction / Context: This question tests your basic knowledge of radioactivity and the difference between radioactive and stable elements or isotopes. Several elements near the bottom of the periodic table are highly unstable and radioactive, whereas many lighter elements have at least one stable isotope. You are asked to choose which of the given species is not radioactive. Given Data / Assumptions: Astatine and francium are heavy elements in the periodic table. Tritium is an isotope of hydrogen. Zirconium is a transition metal. The question expects you to know which among these do not undergo spontaneous radioactive decay under normal conditions. Concept / Approach: Radioactive substances have unstable nuclei that spontaneously emit particles or radiation. Astatine and francium are both very heavy, highly unstable elements with no stable isotopes; all of their isotopes are radioactive. Tritium is a radioactive isotope of hydrogen with one proton and two neutrons, used in luminous paints and nuclear fusion research. Zirconium, however, is a transition metal with several stable isotopes, such as zirconium 90 and zirconium 91. Ordinary zirconium metal used in industry is not radioactive. Therefore, zirconium is the only species in the list that is not naturally radioactive. Step-by-Step Solution: Step 1: Consider astatine. It is a halogen in group 17 and is very heavy. All its known isotopes are radioactive and it occurs only in trace amounts in nature. Step 2: Consider francium. It is an alkali metal heavier than caesium. All isotopes of francium are also highly radioactive with very short half lives. Step 3: Consider tritium. It is an isotope of hydrogen with atomic number 1 but mass number 3, consisting of one proton and two neutrons, and it is well known to be radioactive. Step 4: Consider zirconium. It is a transition metal used in alloys and nuclear reactors for cladding fuel rods, and it has several stable isotopes. Step 5: Since zirconium has stable forms and is not inherently radioactive, it is the only non radioactive choice among the options. Verification / Alternative check: Textbooks of nuclear chemistry and general chemistry often list astatine and francium among elements with no stable isotopes, meaning that every isotope of these elements is radioactive. Tritium is specifically mentioned as a radioactive hydrogen isotope with a half life of about twelve years. Zirconium, however, is widely used in industry and nuclear technology specifically because of its corrosion resistance and low neutron absorption cross section; its normal isotopic composition is stable. If it were strongly radioactive, it would not be used so commonly in structural applications. This confirms that zirconium is not radioactive in its common forms. Why Other Options Are Wrong: Astatine is always radioactive and decays quickly, making it extremely rare in nature. Francium is another element with only radioactive isotopes, and it decays into other elements rapidly. Tritium emits beta particles and is widely used as a radioactive tracer and in self luminous devices. These three options therefore all describe radioactive species. Zirconium, in contrast, has stable isotopes and is not radioactive under normal conditions, so the other options cannot be correct answers to the question asked. Common Pitfalls: Some students may think that any element associated with nuclear reactors, like zirconium, must be radioactive, but in reality zirconium is chosen precisely because it is non radioactive and stable. Another source of confusion is between tritium and deuterium; deuterium is a stable isotope of hydrogen, while tritium is radioactive. The general rule that very heavy elements near the bottom of the periodic table are often radioactive also helps you remember that astatine and francium do not have stable isotopes. Final Answer: Among the given options, the element that is not radioactive is zirconium.

Question 12

Permanent hardness of water can be removed most effectively by adding which of the following substances?

Accepted Answer

Introduction / Context: This question belongs to the topic of water chemistry and hardness. Hard water contains dissolved salts of calcium and magnesium that interfere with soap action and can form scale in boilers and pipes. Temporary hardness can be removed by boiling, but permanent hardness requires chemical treatment. The question asks which substance is commonly used to remove permanent hardness of water. Given Data / Assumptions: The type of hardness considered is permanent hardness, mainly due to calcium and magnesium sulphates and chlorides. The options include bleaching powder, chlorine, washing soda (sodium carbonate) and potassium permanganate. The chemical added should convert soluble hardness causing salts into insoluble compounds that can be removed. Standard school level methods for softening hard water are assumed. Concept / Approach: Permanent hardness is usually caused by soluble salts like calcium sulphate and magnesium chloride. One common method of removing this hardness is to add washing soda, which is sodium carbonate. Sodium carbonate reacts with the calcium and magnesium ions to form insoluble carbonates, such as calcium carbonate and magnesium carbonate, which precipitate out. This process reduces the concentration of hardness causing ions and softens the water. Bleaching powder and chlorine are used mainly for disinfection. Potassium permanganate is an oxidising agent used for removing iron or organic impurities, but not specifically for permanent hardness. Step-by-Step Solution: Step 1: Identify the main cause of permanent hardness. It is due to calcium and magnesium salts that do not decompose on boiling. Step 2: Recall that one standard method of softening such water is to add sodium carbonate, commonly known as washing soda. Step 3: Sodium carbonate reacts with soluble calcium and magnesium salts to form insoluble carbonates like CaCO3 and MgCO3. Step 4: These insoluble precipitates can be removed by filtration, resulting in softened water. Step 5: Bleaching powder and chlorine are used to kill bacteria and disinfect water rather than to remove hardness. Step 6: Potassium permanganate is a strong oxidising agent useful in some purification processes, but it does not directly remove permanent hardness. Step 7: Therefore, washing soda (sodium carbonate) is the correct choice. Verification / Alternative check: Textbooks on water treatment describe several methods of softening hard water: lime soda process, ion exchange and addition of washing soda. In the lime soda process, both calcium hydroxide (lime) and sodium carbonate are used to precipitate hardness causing ions. For small scale domestic use, washing soda is often mentioned as a convenient chemical for reducing hardness. The description that sodium carbonate converts soluble salts of calcium and magnesium into their insoluble carbonates matches this function. By contrast, bleaching powder and chlorine are always discussed under disinfection rather than softening. Why Other Options Are Wrong: Bleaching powder (calcium hypochlorite) is used to disinfect water and make it safe for drinking by killing microbes. Chlorine gas or chlorine solutions also disinfect water but do not precipitate hardness causing ions. Potassium permanganate is used as an oxidising agent to remove iron and manganese or to control taste and odour in water, but not to remove calcium or magnesium salts that cause permanent hardness. None of these substances directly produce insoluble carbonates of calcium and magnesium, so they are not effective softening agents for permanent hardness. Common Pitfalls: Students sometimes confuse water purification for health, which uses disinfectants, with water softening for industrial or domestic use, which uses chemicals that remove hardness. Seeing familiar water treatment chemicals like chlorine and bleaching powder, they may choose them without recalling the specific processes involved. To avoid this, remember that disinfection kills germs, while softening removes hardness ions, often using sodium carbonate or ion exchange resins. Associating washing soda with the term lime soda process can also help in recalling its role in softening hard water. Final Answer: Permanent hardness of water can be removed by adding washing soda (sodium carbonate).

Question 13

In metallurgical chemistry, which of the following is the most important ore of aluminium?

Accepted Answer

Introduction / Context: This question belongs to the topic of metallurgy and ores. An ore is a naturally occurring mineral from which a metal can be profitably extracted. Aluminium is a widely used metal and its main ore is a standard fact in chemistry curricula. The question asks you to identify the most important ore of aluminium from the list provided. Given Data / Assumptions: The metal of interest is aluminium. You are given four minerals: calamine, monazite, magnetite and bauxite. Only one of these is the principal commercial source of aluminium. Standard names and uses of common ores are assumed. Concept / Approach: Aluminium is extracted mainly from bauxite, a hydrated aluminium oxide mineral. Bauxite typically contains aluminum hydroxides along with impurities such as iron oxides and silica. The Bayer process is used to purify bauxite and obtain alumina, which is then reduced by electrolysis in the Hall Heroult process to give aluminium metal. Calamine is an ore of zinc, magnetite is an ore of iron and monazite is a phosphate mineral that is an important source of rare earth elements and thorium. Therefore, only bauxite is correctly associated with aluminium extraction. Step-by-Step Solution: Step 1: Recall the standard list of important ores. For aluminium, the primary ore is bauxite. Step 2: Bauxite consists largely of hydrated aluminium oxides, such as gibbsite and boehmite, along with impurities. Step 3: In the Bayer process, bauxite is digested in sodium hydroxide to dissolve alumina and separate impurities. Step 4: The alumina obtained is then electrolysed in molten cryolite to produce pure aluminium metal. Step 5: Calamine is a zinc ore (zinc carbonate or related minerals), not an aluminium ore. Step 6: Magnetite is an iron oxide used as an iron ore. Step 7: Monazite sands are sources of thorium and rare earths, not aluminium. Verification / Alternative check: Look at typical tables of ores and metals found in textbooks. Against aluminium you will almost always see bauxite listed as the chief ore, sometimes along with cryolite in specific contexts, though cryolite is mostly used as a flux rather than a principal ore. In contrast, zinc is paired with ores like zinc blende and calamine, iron is paired with hematite, magnetite and siderite and thorium with monazite. This cross checking confirms that bauxite is the only appropriate ore listed for aluminium. Why Other Options Are Wrong: Calamine is primarily zinc carbonate and is an ore of zinc, not aluminium. Magnetite is iron oxide Fe3O4 and an important ore of iron, often used in the blast furnace process. Monazite is a phosphate mineral rich in rare earth metals and thorium, used in nuclear fuel and related applications. None of these minerals serves as a major source of aluminium. Therefore, they cannot be the correct answer in a question about the principal ore of aluminium. Common Pitfalls: Students sometimes remember the names of ores but forget which metal they belong to, leading to confusion between bauxite, magnetite and others. Another mistake is to assume that because aluminium is light, its ore might be a light coloured mineral like some carbonates, but this is not a reliable guide. To avoid this, memorise a simple mapping: bauxite for aluminium, hematite and magnetite for iron, zinc blende and calamine for zinc, and monazite for thorium. This mapping simplifies many ore related questions. Final Answer: The most important ore of aluminium is bauxite.

Question 14

Modern periodic tables are based on the atomic number of elements. The fundamental significance of atomic number for arranging elements was demonstrated by which scientist through studies of X ray spectra?

Accepted Answer

Introduction / Context: This general knowledge question relates to the development of the modern periodic table. Early periodic tables arranged elements by atomic weight, but some inconsistencies remained. The concept of atomic number as the fundamental organising principle was established through experimental evidence involving X ray spectra. You are asked to identify which scientist work proved the significance of atomic number in the periodic classification of elements. Given Data / Assumptions: The modern periodic law states that the properties of elements are periodic functions of their atomic numbers. The question mentions that this significance was proved by an experiment involving X rays. The options refer in summary form to work by Moseley, Bragg, Millikan and Roentgen. Standard historical facts from atomic and modern physics are assumed. Concept / Approach: Henry Moseley conducted experiments in which he measured the frequencies of characteristic X rays emitted by different elements. He found that the square root of the frequency of these X rays varied in a systematic way with the atomic number of the element. This relationship, known as Moseley law, showed that atomic number, not atomic weight, is the fundamental quantity determining the position of an element in the periodic table. Bragg worked on X ray diffraction and crystal structure, Millikan performed the oil drop experiment to measure the charge of the electron and Roentgen discovered X rays themselves. None of these other contributions specifically established the importance of atomic number for periodic classification. Step-by-Step Solution: Step 1: Recall that the modern periodic table arranges elements in order of increasing atomic number rather than atomic weight. Step 2: Henry Moseley measured frequencies of X rays emitted when elements were bombarded with high energy electrons. Step 3: He plotted the square root of the frequency against the atomic number and found a straight line relationship, now called Moseley law. Step 4: This proved that atomic number is a fundamental property of an element, directly related to its nuclear charge. Step 5: Bragg contributions relate to X ray diffraction in crystals, which help determine crystal structures but do not define atomic number. Step 6: Millikan oil drop experiment measured the charge of the electron, an important constant but not directly connected to the periodic table arrangement. Step 7: Roentgen discovery of X rays was crucial but did not by itself reveal the link between X ray frequency and atomic number. Verification / Alternative check: In summaries of the development of periodic classification, you will typically see a sequence from Mendeleev periodic table based on atomic weights to Moseley work that refined the table by using atomic numbers. For example, problems with the placement of elements like argon and potassium were resolved when arranged by atomic number rather than weight. The direct experimental link came from Moseley measurements of characteristic X rays, which allowed atomic numbers of elements to be determined unambiguously. The other scientists mentioned made important contributions to physics but are not credited with establishing atomic number as the basis of the periodic table. Why Other Options Are Wrong: Bragg work on X ray diffraction is important for crystallography and for determining interatomic distances in crystals, but it does not address periodic classification based on atomic number. Millikan oil drop experiment measured the elementary charge and gave evidence for quantisation of electric charge, but it is unrelated to the order of elements in the periodic table. Roentgen discovery of X rays opened a new field of research but did not itself link atomic number to X ray frequencies. Thus these options do not match the specific contribution described in the question. Common Pitfalls: Learners sometimes confuse any work involving X rays with Moseley law, leading them to think that Bragg or Roentgen might be the answer. Others may remember Millikan as a key figure in atomic physics but not distinguish his specific contribution. To avoid this confusion, pair each name with a key phrase: Moseley X ray spectra and atomic number, Bragg X ray diffraction and crystal structure, Millikan oil drop and electron charge, Roentgen discovery of X rays. This association makes it easier to select the correct scientist when questions mention the periodic table and atomic number. Final Answer: The significance of atomic number for the modern periodic table was demonstrated by Moseley work on X ray spectra.

Question 15

In nuclear physics, the first clear indication that even a stable atomic nucleus can be broken down into smaller particles was provided by which scientist's experiments?

Accepted Answer

Introduction / Context: This general chemistry and physics question focuses on the early history of nuclear science. It asks which scientist first provided experimental evidence that even a stable atomic nucleus can be broken down or disintegrated. Understanding this historical milestone helps you appreciate how our concept of the atom evolved from an indivisible unit to a complex system that can undergo nuclear reactions. Given Data / Assumptions: The key idea is the first indication that a stable nucleus is not permanent and can be broken down. The options list important scientists: Ernest Rutherford, Madam Curie, Frederick Soddy, Schmidt and J. J. Thomson (in corrected form). Radioactivity itself was known from Becquerel and the Curies, but the question is about deliberate or clear nuclear disintegration of a stable nucleus. We assume standard textbook history of atomic structure and nuclear transmutation. Concept / Approach: Natural radioactivity, discovered by Becquerel and studied extensively by the Curies and Soddy, showed that some nuclei are unstable and break down spontaneously. However, the idea that a stable nucleus can be forced to disintegrate by bombardment was first demonstrated in Rutherford experiments. By firing alpha particles at nitrogen, Rutherford observed the emission of protons and formation of a different nucleus. This was the first artificial nuclear reaction and a direct indication that even a stable nucleus can be broken apart under suitable conditions. Step-by-Step Solution: 1) Becquerel and the Curies discovered spontaneous radioactivity, which involves naturally unstable nuclei but does not directly show that stable nuclei can be broken by external means.2) Frederick Soddy worked with radioactivity and transmutation concepts, but he did not perform the first clear artificial disintegration of a stable nucleus.3) J. J. Thomson studied cathode rays and discovered the electron, focusing on atomic structure outside the nucleus.4) Rutherford carried out famous alpha particle bombardment experiments, including the scattering experiment that led to the nuclear model of the atom.5) In additional work, Rutherford bombarded nitrogen gas with alpha particles and observed ejection of hydrogen nuclei (protons), effectively transforming nitrogen into oxygen. This nuclear reaction showed that a stable nitrogen nucleus can be broken down.6) Therefore, Rutherford experiments provided the first clear indication that a stable nucleus could be disintegrated, making him the correct answer. Verification / Alternative check: Standard nuclear chemistry history describes the reaction: 14N + alpha → 17O + proton. Here, the nitrogen nucleus captures an alpha particle and emits a proton, changing into oxygen. This was the first convincingly demonstrated artificial nuclear transmutation. Although the Curies and Soddy discussed transmutation, they studied natural decay of already unstable elements. Rutherford was the scientist who experimentally forced a stable nucleus to change, demonstrating that the nucleus is not indestructible and can be deliberately broken down. Why Other Options Are Wrong: Madam Curie: She discovered and studied radioactive elements like radium and polonium, but her work focused on natural radioactivity, not the artificial breaking up of stable nuclei. Frederick Soddy: He explained disintegration series and isotopes but did not perform the first artificial disintegration of a stable nucleus. Schmidt: Also worked with early observations of radioactivity but is not credited with the first deliberate nuclear disintegration of a stable nucleus. J. J. Thomson: Discovered the electron and the e/m ratio, which advanced atomic theory but did not involve destruction of the atomic nucleus. Common Pitfalls: Many students confuse the discovery of radioactivity with the first artificial nuclear reaction. It is important to distinguish between spontaneous decay of unstable nuclei and deliberate bombardment that breaks a stable nucleus. Another common error is to choose the Curies because of their fame in radioactivity, but the specific achievement asked here is Rutherford nuclear disintegration of nitrogen, which is a separate historical step. Keeping a simple timeline in mind helps: Becquerel and the Curies for natural radioactivity, Rutherford for nuclear model and first nuclear disintegration, and later scientists for fission and reactors. Final Answer: The first clear indication that a stable nucleus can be broken down was provided by the experiments of Ernest Rutherford.

Question 16

In the compound potassium dichromate, K2Cr2O7, what is the oxidation number of each chromium (Cr) atom?

Accepted Answer

Introduction / Context: This question checks your ability to assign oxidation numbers in a polyatomic ionic compound, specifically potassium dichromate, K2Cr2O7. Understanding oxidation numbers is essential for redox reactions, balancing equations and predicting oxidising and reducing agents in inorganic chemistry. Given Data / Assumptions: The compound is K2Cr2O7 (potassium dichromate). We need the oxidation number of chromium (Cr) in this compound. Standard oxidation numbers are assumed: potassium as an alkali metal and oxygen in oxides. The overall compound is electrically neutral, so the sum of oxidation numbers equals zero. Concept / Approach: Oxidation number rules are our main tool. In most compounds, potassium (K) has an oxidation number of +1, and oxygen (O) in oxides has an oxidation number of -2. For a neutral compound, the total sum of oxidation numbers of all atoms must be zero. By assigning known oxidation numbers to potassium and oxygen, we can solve for the unknown oxidation number of chromium, assuming both chromium atoms in dichromate have the same value. Step-by-Step Solution: 1) Write down the formula K2Cr2O7 and note the number of each atom: 2 potassium, 2 chromium and 7 oxygen atoms.2) Assign oxidation numbers using standard rules: each K is +1 and each O is -2 in this type of compound.3) Let the oxidation number of chromium be x. Since there are 2 chromium atoms, their total contribution is 2x.4) Write the total sum of oxidation numbers and set it equal to zero for a neutral compound: 2*(+1) + 2*x + 7*(-2) = 0.5) Simplify: 2 + 2x - 14 = 0, which gives 2x - 12 = 0, so 2x = 12 and x = +6.6) Therefore, the oxidation number of each chromium atom in K2Cr2O7 is +6. Verification / Alternative check: We can check by substituting x = +6 back into the expression: total charge = 2*(+1) + 2*(+6) + 7*(-2) = 2 + 12 - 14 = 0. This satisfies the neutrality condition, confirming that +6 is correct. It also matches the fact that dichromate ion Cr2O7^2- is a strong oxidising agent where chromium typically exists in the +6 oxidation state, consistent with many qualitative and quantitative redox reactions in acidic solution. Why Other Options Are Wrong: -5: A negative oxidation state for chromium in dichromate contradicts both the usual chemistry of chromium and the charge balance equation. +3: Chromium in +3 state appears in other compounds such as Cr2O3, but not in the dichromate ion. -4: Again, a negative oxidation state is not consistent with chromium behaviour in this oxygen rich environment. +7: This is characteristic of manganese in permanganate (MnO4^-), but chromium in K2Cr2O7 is not +7; the mathematics of charge balance disproves this value. Common Pitfalls: Students sometimes confuse dichromate with permanganate and guess +7 or mix up the oxidation states of chromium in different oxyanions (chromate and dichromate). Another common error is to forget to multiply the oxidation number of chromium by 2 when writing the total sum. To avoid such mistakes, always count the number of each type of atom carefully and follow the rule that the total sum of oxidation numbers equals the net charge on the species. Final Answer: The oxidation number of chromium in potassium dichromate, K2Cr2O7, is +6 for each chromium atom.

Question 17

In concentrated sulphuric acid, H2SO4, what is the oxidation number of sulphur (S) in the molecule?

Accepted Answer

Introduction / Context: This question assesses your understanding of oxidation numbers in molecular compounds, specifically in sulphuric acid, H2SO4. Oxidation state calculations are crucial in redox reactions, industrial chemistry and in predicting oxidising power of acids such as H2SO4. Given Data / Assumptions: The formula of the compound is H2SO4. We are asked to find the oxidation number of sulphur (S). We use standard oxidation number rules: hydrogen in most compounds is +1, oxygen in oxides is -2. H2SO4 is electrically neutral, so the sum of oxidation numbers must be zero. Concept / Approach: Oxidation number rules state that hydrogen is usually assigned +1 and oxygen is usually assigned -2 in covalent compounds. For a neutral molecule, the algebraic sum of oxidation numbers of all atoms is zero. By assigning known values to hydrogen and oxygen, we can solve for the unknown oxidation number of sulphur in H2SO4. This simple algebraic approach is standard for such problems. Step-by-Step Solution: 1) Write the chemical formula H2SO4 and identify the number of each atom: 2 H, 1 S and 4 O.2) Assign oxidation numbers based on rules: hydrogen = +1 and oxygen = -2 in this typical oxyacid.3) Let the oxidation number of sulphur be x.4) Set up the total sum of oxidation numbers for a neutral molecule: 2*(+1) + x + 4*(-2) = 0.5) Simplify the equation: 2 + x - 8 = 0, which gives x - 6 = 0, so x = +6.6) Therefore, the oxidation number of sulphur in H2SO4 is +6. Verification / Alternative check: Substitute x = +6 back into the sum: total = 2*(+1) + (+6) + 4*(-2) = 2 + 6 - 8 = 0, which satisfies the neutrality condition. This value also matches the known chemistry of sulphuric acid, where sulphur is often described as being in its highest common oxidation state of +6. In redox reactions, concentrated sulphuric acid behaves as an oxidising agent, consistent with a high positive oxidation state for sulphur. Why Other Options Are Wrong: +2: This oxidation state corresponds to sulphur in compounds like H2S, not in H2SO4. -6: A highly negative oxidation state would require sulphur to be combined with elements more electropositive than itself, which is not the case here with oxygen. -2: This is the oxidation state of sulphur in sulphides, such as Na2S or H2S, not in an oxyacid with four oxygens. 0: A zero oxidation state would mean elemental sulphur, which is not what we have in H2SO4. Common Pitfalls: Students often confuse the oxidation state of sulphur in different sulphur containing compounds because sulphur exhibits many valencies. Mixing up H2SO4 with H2S or SO2 leads to incorrect answers. Another mistake is forgetting that there are four oxygen atoms and miscalculating the total contribution from oxygen. Carefully counting atoms and applying the oxidation number rules step by step prevents such errors and makes these problems straightforward. Final Answer: The oxidation number of sulphur in sulphuric acid, H2SO4, is +6.

Question 18

In typical covalent compounds such as ammonia and many organic molecules, how many covalent bonds does a neutral nitrogen atom usually form to complete its valence shell?

Accepted Answer

Introduction / Context: This question examines basic valence concepts in covalent bonding, focusing on nitrogen. Knowing how many covalent bonds an atom typically forms is essential for drawing Lewis structures, understanding molecular shapes and predicting the formulas of simple inorganic and organic compounds that contain nitrogen. Given Data / Assumptions: The atom of interest is a neutral nitrogen atom in its common valence state. We are asked how many covalent bonds it typically forms in ordinary molecules. We consider classic examples like ammonia, amines and many organic nitrogen compounds. We assume basic knowledge of electronic configuration and the octet rule. Concept / Approach: Nitrogen has atomic number 7 and an electronic configuration of 1s2 2s2 2p3. This means it has five valence electrons in the second shell. To achieve a stable octet (eight valence electrons), nitrogen usually shares three of its valence electrons in covalent bonds and retains one lone pair. This is seen in many common compounds such as ammonia (NH3), where nitrogen forms three N–H covalent bonds and has one lone pair of electrons. Step-by-Step Solution: 1) Write the valence shell configuration of nitrogen: 2s2 2p3, giving a total of five valence electrons.2) According to the octet rule, atoms tend to achieve eight electrons in their outer shell for stability.3) Nitrogen already has five valence electrons, so it needs three more electrons to reach an octet.4) In covalent bonding, it can obtain these three additional electrons by forming three shared pairs (three covalent bonds) with other atoms such as hydrogen or carbon.5) After forming three covalent bonds, nitrogen has three bonding pairs and one lone pair, giving eight electrons around it and a typical valency of 3. Verification / Alternative check: Consider ammonia, NH3. Each hydrogen atom contributes one electron, and nitrogen contributes one electron to each of the three N–H bonds. This creates three shared pairs (three covalent bonds). Counting electrons around nitrogen: two from its lone pair plus six from three bonding pairs gives eight electrons, satisfying the octet rule. This same pattern appears in many amines and organic nitrogen compounds, confirming that three covalent bonds is the standard for neutral nitrogen in simple molecules. Why Other Options Are Wrong: 2: Two covalent bonds would give nitrogen only seven electrons around it (four from bonds and three lone electrons), which is less common for a stable neutral nitrogen atom in simple molecules. 4: Nitrogen can appear as four coordinate in species such as ammonium ion (NH4+), but the fourth bond is a coordinate (dative) bond and the atom carries a positive charge, not a typical neutral covalent bonding pattern. 1: One covalent bond would leave nitrogen far from the octet and is not representative of common stable compounds. 5: Five covalent bonds around nitrogen would exceed the octet and is not typical for second period elements. Common Pitfalls: Some students confuse the usual three covalent bonds of neutral nitrogen with the four bonds seen in ammonium ion, forgetting that the latter involves a positive charge and a coordinate bond. Others assume that any group 15 element can routinely expand its octet like heavier elements, which is not true for nitrogen due to the absence of available d orbitals in the second period. Always think of nitrogen as having five valence electrons and needing three more to complete an octet, leading naturally to three standard covalent bonds. Final Answer: A neutral nitrogen atom in typical covalent compounds usually forms three covalent bonds to complete its valence shell.

Question 19

In food technology, which gas is commonly used for the artificial ripening of harvested green fruits such as bananas and mangoes?

Accepted Answer

Introduction / Context: This general chemistry and biology question deals with post harvest food technology. It asks which gas is used for the artificial ripening of green fruits. Understanding this concept is important for agriculture, storage, transport and the food industry because ripening control affects both quality and marketability of fruits. Given Data / Assumptions: The context is artificial ripening of green fruits such as bananas, mangoes and tomatoes. Options list several substances: acetone, carbon dioxide, ethylene, ethyl alcohol and ammonia. We are looking specifically for a gas that can trigger and regulate ripening. We assume modern, safer technology rather than outdated chemical practices. Concept / Approach: Ripening of many climacteric fruits is controlled by plant hormones, among which ethylene is the most important natural ripening hormone. Ethylene is a simple hydrocarbon gas (C2H4) produced by plants. In post harvest technology, controlled exposure of fruits to small amounts of ethylene accelerates ripening in a uniform and predictable manner. While some older practices used calcium carbide, which generates acetylene gas, modern recommended practice is to use ethylene itself because it is safer and more closely mimics natural processes. Step-by-Step Solution: 1) Identify which listed substances are gases under ordinary conditions and known to influence plant processes.2) Ethylene (C2H4) is a gaseous plant hormone that promotes ripening, flower opening and leaf abscission.3) In artificial ripening chambers, fruits are exposed to controlled concentrations of ethylene to trigger and speed up the natural ripening process.4) Acetone and ethyl alcohol are liquids used as solvents and antiseptics, not standard ripening agents.5) Carbon dioxide and ammonia are industrial gases but do not function as primary ripening hormones for fruit; high carbon dioxide levels can even slow respiration.6) Therefore, the correct gas used for artificial ripening of green fruits is ethylene. Verification / Alternative check: Horticulture and food science materials describe ripening facilities where ethylene generators or cylinders maintain ethylene levels around a few parts per million to ripen bananas and other fruits in storage rooms. The process is carefully monitored to avoid over ripening. Regulatory bodies in many countries discourage or restrict the use of calcium carbide because of health hazards, further encouraging direct use of ethylene gas for controlled ripening. This confirms ethylene as the correct answer. Why Other Options Are Wrong: Acetone: An organic solvent used in laboratories and industry; it is not a ripening hormone for fruits. Carbon dioxide: Plays roles in respiration and atmosphere control but is not the primary ripening gas; high levels can even suppress ripening. Ethyl alcohol: Used as a disinfectant and solvent; it does not function as a standard ripening gas in post harvest technology. Ammonia: A pungent gas used in fertilisers and cleaning agents, not suitable or safe as a ripening hormone. Common Pitfalls: Some students confuse ethylene with acetylene because of the use of calcium carbide in older ripening practices. Calcium carbide reacts with water to release acetylene, which can have ripening like effects but poses health risks due to impurities. Modern exam questions typically expect the scientifically correct answer, ethylene, which is the natural plant hormone. Remember that ethylene is the key plant hormone for ripening, and its name appears in many plant physiology topics and competitive exam questions. Final Answer: The gas commonly used for the artificial ripening of green fruits is ethylene.

Question 20

In atomic physics, which scientist is credited with the discovery of the neutron as a neutral particle inside the atomic nucleus?

Accepted Answer

Introduction / Context: This question belongs to the history of atomic structure. It asks who discovered the neutron, the neutral particle inside the atomic nucleus. Knowing which scientists are associated with the electron, proton and neutron helps you understand how our model of the atom developed over time. Given Data / Assumptions: We are looking for the scientist who first identified the neutron. Options include J. J. Thomson, James Chadwick, Ernest Rutherford, Joseph Priestley and Niels Bohr. The neutron is an electrically neutral particle found in atomic nuclei, except for the most common hydrogen isotope. We assume standard historical information from physics and chemistry textbooks. Concept / Approach: The neutron was discovered significantly later than the electron and proton because it carries no electric charge and therefore is harder to detect with early instruments that relied on electric and magnetic fields. James Chadwick performed experiments in 1932 using beryllium bombarded with alpha particles, which emitted a penetrating radiation. By careful analysis, he showed that this radiation consisted of neutral particles with mass roughly equal to that of the proton, and he named them neutrons. This work completed the basic picture of the atomic nucleus as composed of protons and neutrons. Step-by-Step Solution: 1) Recall that J. J. Thomson is famous for the discovery of the electron through cathode ray experiments.2) Ernest Rutherford is known for the discovery of the atomic nucleus and the proton through alpha scattering and related experiments.3) Joseph Priestley discovered oxygen gas and made contributions to early chemistry, not directly to nuclear particles.4) Niels Bohr proposed the planetary model of the atom with quantised orbits but did not discover the neutron.5) James Chadwick, in 1932, bombarded beryllium with alpha particles and detected neutral particles knocking protons out of paraffin wax.6) From the behaviour of this radiation, Chadwick concluded that it consisted of neutrally charged particles with mass similar to that of protons and named them neutrons.7) Therefore, the discovery of the neutron is credited to James Chadwick. Verification / Alternative check: Standard physics histories state that Rutherford had proposed the existence of a neutral nuclear particle earlier, to explain the discrepancy between atomic number and atomic mass, but he did not observe it experimentally. Chadwick experiment provided the direct evidence needed. The discovery of the neutron allowed better understanding of isotopes and nuclear reactions, and for this work Chadwick received the Nobel Prize in Physics in 1935. These details confirm that the correct answer is James Chadwick, not the other listed scientists. Why Other Options Are Wrong: J. J. Thomson: Discovered the electron and measured its charge to mass ratio, but he did not identify the neutron. Ernest Rutherford: Proposed the nuclear model and identified the proton but only hypothesised the existence of neutrons. Joseph Priestley: Worked in gaseous chemistry and discovered oxygen, unrelated to nuclear particles. Niels Bohr: Developed the quantum model of the hydrogen atom, explaining spectral lines, but did not discover the neutron. Common Pitfalls: Students sometimes confuse Rutherford name with every nuclear concept because he is strongly associated with the nucleus. However, while Rutherford suggested that a neutral nuclear particle should exist, it was Chadwick who actually discovered it experimentally. Another error is to choose Thomson by mistakenly associating all subatomic particle discoveries with one scientist. To avoid confusion, link each scientist to one major discovery: Thomson for electrons, Rutherford for the nucleus and proton, and Chadwick for neutrons. Final Answer: The neutron was discovered by James Chadwick.

Chemistry Questions