Question 1

In the history of chemistry, who is widely known as the father of modern chemistry for establishing quantitative laws and a new system of chemical nomenclature?

Accepted Answer

Introduction / Context: Modern chemistry is built on a foundation of careful measurement, conservation laws, and systematic naming of substances. Several scientists made major contributions, but one figure is especially recognised for transforming chemistry from a qualitative art into a quantitative science. This question asks you to identify the person commonly called the father of modern chemistry, a title usually given to the scientist who established the law of conservation of mass in chemical reactions and helped create modern chemical nomenclature. Given Data / Assumptions: Several famous chemists are listed as options, including Priestley, Lavoisier, Dalton, and Mendeleev. We are looking for the one associated with founding modern chemistry as a quantitative science. The title father of modern chemistry is a conventional label used in many textbooks. We assume basic knowledge of the main contributions of each listed scientist. Concept / Approach: Antoine Lavoisier, an eighteenth century French chemist, is widely credited with establishing the law of conservation of mass in chemical reactions, identifying and naming oxygen, and disproving the phlogiston theory. He also helped develop a systematic naming scheme for chemical compounds. These contributions gave chemistry a firm scientific basis and shifted it toward modern quantitative methods. Because of these achievements, he is often described as the father of modern chemistry. The other scientists listed were very important but are associated with other specific advances such as gas discovery, atomic theory, and the periodic table. Step-by-Step Solution: Step 1: Recall that Antoine Lavoisier performed careful experiments on combustion and calcination and concluded that mass is conserved in chemical reactions. Step 2: Note that he identified oxygen role in combustion and respiration and helped give the element its name. Step 3: Remember that Lavoisier contributed to the development of a logical system of chemical nomenclature that replaced older, confusing names with more descriptive ones. Step 4: Compare these foundational contributions with those of Joseph Priestley, John Dalton, and Dmitri Mendeleev, who each made specific but more specialized advances. Step 5: Conclude that Lavoisier is the scientist most commonly called the father of modern chemistry. Verification / Alternative check: Many standard chemistry textbooks and historical accounts present Lavoisier as the figure who transformed chemistry into a modern science. His work Elements of Chemistry is considered one of the earliest modern chemistry textbooks and emphasised measurement, balance use, and systematic terminology. While Priestley discovered several gases, including oxygen, he did not provide the theoretical framework that Lavoisier did. Dalton is known for atomic theory, and Mendeleev for organising the periodic table, both crucial but later developments. This comparison confirms that the title father of modern chemistry is most closely associated with Lavoisier. Why Other Options Are Wrong: Joseph Priestley made important discoveries of gases and experimental techniques, but he did not reform the theoretical foundations of chemistry in the same way. John Dalton introduced atomic theory, which greatly advanced chemistry but came after Lavoisier foundational work and built upon the conservation principles he established. Dmitri Mendeleev is honoured as the father of the periodic table for his organisation of elements, not as the founder of modern chemistry as a whole. Therefore, while all are important chemists, they are not usually given the title specified in the question. Common Pitfalls: Students sometimes confuse the roles of these historical figures, especially when they remember a familiar name such as Mendeleev or Dalton and assume it must go with any major chemistry title. Others may know that Priestley discovered oxygen and think that discovery alone means he is the father of modern chemistry. To avoid such errors, associate Lavoisier with conservation of mass, oxygen theory of combustion, and systematic nomenclature, which together justify the father of modern chemistry label. Final Answer: The scientist widely known as the father of modern chemistry is Antoine Lavoisier.

Question 2

In basic atomic structure, atoms are composed of which combination of fundamental components when we consider the overall structure of the nucleus and the surrounding region?

Accepted Answer

Introduction / Context: Understanding what atoms are made of is one of the first steps in learning chemistry and physics. An atom is not a solid indivisible sphere but a composite object built from smaller particles. This question asks which combination of components best describes the overall structure of an atom when we think in terms of an inner nucleus and an outer region. Choosing the correct answer helps reinforce the basic model of atomic structure used throughout modern science. Given Data / Assumptions: Atoms contain subatomic particles such as electrons, protons, and neutrons. The nucleus is the dense central region containing protons and neutrons. Electrons occupy regions around the nucleus often called shells or orbitals. The options combine these components in different ways, and we must pick the most accurate description of atomic composition. Concept / Approach: In the modern nuclear model, an atom consists of a small, dense nucleus surrounded by electrons. The nucleus contains positively charged protons and neutral neutrons. Electrons are negatively charged and move in a cloud around the nucleus. When we summarise atomic composition at a basic level, we can say that an atom is composed of electrons and a nucleus. The nucleus itself contains protons and neutrons, so the phrase electrons and nuclei correctly summarises all three particle types in two structural units. Options that omit either electrons or the nucleus are incomplete. Step-by-Step Solution: Step 1: Recall that every atom has a central nucleus that contains protons and, in most cases, neutrons. Step 2: Note that electrons are found in regions around the nucleus and are responsible for chemical bonding and charge balance. Step 3: Understand that the combination of electrons plus the nucleus gives a complete picture of atomic structure at the simplest level. Step 4: Identify that the nucleus is not a separate particle from protons; rather, it is made of protons and neutrons together. Step 5: Select the option that explicitly includes electrons and nuclei as the two main structural parts of an atom. Verification / Alternative check: Consider a simple hydrogen atom. Its nucleus consists of one proton, and there is one electron in the region around the nucleus. If we move to a helium atom, the nucleus contains two protons and two neutrons, and there are two electrons around it. In both cases, the atom can be described as electrons plus a nucleus. This pattern extends to all elements: the nucleus always contains protons and usually neutrons, and electrons surround it. Therefore, describing the atom as composed of electrons and nuclei is accurate and complete at this level of detail. Why Other Options Are Wrong: Saying atoms are composed only of electrons and protons leaves out neutrons, which are important components of most nuclei and affect atomic mass and isotopes. The option protons and nuclei is confusing and redundant because nuclei already contain protons and neutrons; it also fails to mention electrons, which are essential. The option only electrons is clearly wrong because an electron cloud without a nucleus would not be an atom and would have very different properties. Thus, these options do not give a correct description of atomic composition. Common Pitfalls: Some learners may focus only on the fact that protons and electrons are charged and therefore believe they alone define the atom. Others may misunderstand the term nucleus and think it is something separate from protons and neutrons. To avoid confusion, remember that the nucleus is the central part made of protons and neutrons, while electrons occupy surrounding space. The simplest structural description is electrons plus nucleus, which naturally includes all three subatomic particle types. Final Answer: Atoms are composed of electrons and nuclei containing protons and neutrons.

Question 3

For gadolinium Gd with atomic number 64, which of the following is the correct outer electronic configuration of the atom?

Accepted Answer

Introduction / Context: Electronic configuration describes how electrons are distributed among the orbitals of an atom. For transition and inner transition elements such as the lanthanides, understanding their electronic configurations can be challenging due to the filling of f orbitals and some exceptions to simple rules. Gadolinium, symbol Gd, is a lanthanide with atomic number 64. This question asks you to identify its correct outer electronic configuration by comparing several possible distributions of electrons in the 4f, 5d, and 6s subshells. Given Data / Assumptions: Gadolinium has atomic number 64, so a neutral atom has 64 electrons. Its electrons fill orbitals in the general order from lower to higher energy, including 4f, 5d, and 6s orbitals for the outer configuration. The options present four combinations of 4f, 5d, and 6s electron counts. We assume the inner core up to xenon is the same, and we only compare the outer part beyond xenon. Concept / Approach: For lanthanides, the general pattern of filling is [Xe] 4f^n 5d^0 or 5d^1 6s^2 with n increasing from cerium to lutetium. Gadolinium is known to have a particularly stable configuration with a half filled 4f subshell. A half filled f subshell corresponds to 4f^7. To keep 64 electrons in total, we consider that xenon accounts for 54 electrons. That leaves 10 electrons to place. A stable arrangement that gives 4f^7 and keeps 6s^2 filled uses the remaining electron in 5d, giving 4f^7 5d^1 6s^2. This configuration benefits from both a half filled 4f subshell and a partially filled 5d subshell, matching experimental observations for gadolinium. Step-by-Step Solution: Step 1: Write the core configuration of xenon, which has 54 electrons, as [Xe]. Step 2: Subtract 54 from the atomic number of gadolinium, 64, to find that 10 electrons remain to be placed in 4f, 5d, and 6s subshells. Step 3: Consider that lanthanides fill the 4f subshell and often have either 5d^0 or 5d^1 along with 6s^2. Step 4: Recognise the special stability of a half filled 4f subshell, which corresponds to 4f^7. Step 5: Allocate 7 electrons to 4f, 2 electrons to 6s, which totals 9, and then place the remaining 1 electron in 5d, giving 4f^7 5d^1 6s^2 as the outer configuration. Verification / Alternative check: Check the options against the requirement for 10 electrons beyond xenon. The configuration 4f^7 5d^1 6s^2 adds up to 7 plus 1 plus 2, which equals 10, matching the required number of electrons. It also gives a half filled 4f subshell, known to be particularly stable due to exchange energy and symmetry considerations. Other options either assign too many or too few electrons or give less stable arrangements, such as 4f^8 with 5d^0 or various combinations that do not reflect known ground state data. Reference tables of lanthanide configurations list gadolinium as [Xe] 4f^7 5d^1 6s^2, confirming this choice. Why Other Options Are Wrong: The option 4f^3 5d^5 6s^2 still has 10 electrons in the outer shells but does not match the pattern of f subshell filling for lanthanides and does not provide the stable half filled 4f^7 arrangement. The option 4f^8 5d^0 6s^2 has 10 electrons but leads to a more crowded f subshell and lacks the known experimental configuration for gadolinium. The option 4f^4 5d^4 6s^2 again gives a strange distribution that is not supported by spectroscopic data. These alternatives ignore the known stability of a half filled f subshell and do not match reference configurations. Common Pitfalls: Students often attempt to apply simple aufbau rules mechanically without considering exceptions in d and f block elements. For lanthanides, it is important to remember that some elements prefer configurations that give half filled or fully filled subshells. Another common mistake is to miscount the number of electrons above xenon and end up with the wrong total in outer shells. To avoid such errors, always subtract the noble gas core electrons, consider the special stability of half filled subshells like 4f^7, and double check that the total number of electrons matches the atomic number. Final Answer: For gadolinium with atomic number 64, the correct outer electronic configuration is 4f^7 5d^1 6s^2.

Question 4

Which of the following chemicals is commonly used in cloud seeding operations to induce artificial rain from suitable clouds?

Accepted Answer

Introduction / Context: In some regions, meteorologists and environmental agencies attempt to enhance rainfall by a technique called cloud seeding. The goal is to encourage water droplets in clouds to grow large enough to fall as rain. This process involves dispersing certain substances into clouds that act as nuclei for ice or water droplet formation. The question asks which chemical is widely used to induce artificial rain through such cloud seeding operations. Given Data / Assumptions: We consider cloud seeding for artificial rain enhancement. The options include potassium nitrate, ammonium chloride, calcium carbonate, and silver iodide. We assume stratiform or cumuliform clouds that already have sufficient moisture. The correct chemical should act as an effective condensation or ice nucleus under cloud conditions. Concept / Approach: Cloud seeding usually aims to increase precipitation by encouraging the formation of ice crystals or larger water droplets inside supercooled clouds. Silver iodide is commonly used because its crystal structure is similar to that of ice, which makes it an excellent ice nucleating agent. When tiny particles of silver iodide are dispersed into suitable clouds, they provide surfaces on which supercooled water droplets can freeze, grow, and eventually fall as precipitation. Other substances listed in the options do not have the same widely recognised role in standard cloud seeding projects. Step-by-Step Solution: Step 1: Recall that cloud seeding involves dispersing particles into clouds to help form ice crystals or larger droplets. Step 2: Recognise that silver iodide has a crystal lattice that closely resembles the structure of ice crystals. Step 3: Understand that this structural similarity allows silver iodide particles to serve as effective ice nuclei in supercooled clouds. Step 4: When water vapour or supercooled droplets encounter silver iodide particles, ice crystals form, grow, and may trigger precipitation. Step 5: Identify that among the given options, silver iodide is the well known and standard chemical used for inducing artificial rain. Verification / Alternative check: Meteorology and environmental science references describe cloud seeding programs that use aircraft or ground based generators to release silver iodide into cloud systems. Some programmes also use dry ice or hygroscopic materials such as salt, but silver iodide remains one of the most cited agents due to its efficiency as an ice nucleus. Potassium nitrate, ammonium chloride, and calcium carbonate do not have the same reputation or widespread use in this context. Checking multiple sources about artificial rain and cloud seeding consistently shows silver iodide as the standard agent, which confirms the answer. Why Other Options Are Wrong: Potassium nitrate is mainly used in fertilisers and pyrotechnics, not as a typical cloud seeding agent. Ammonium chloride is used in various industrial processes and laboratory demonstrations but is not a standard nucleating agent for artificial rain. Calcium carbonate is a component of limestone and chalk and does not serve as a specialised ice nucleus in cloud seeding. While these compounds may appear in the atmosphere as dust or aerosols, they are not the classic chemicals associated with deliberate artificial rainmaking. Only silver iodide matches that role. Common Pitfalls: Students sometimes confuse artificial rain with other environmental treatments, such as adding chemicals to water bodies, and may choose a compound like potassium nitrate because they have seen it mentioned in agriculture. Another mistake is picking a salt at random, assuming any salt will work. To avoid such errors, remember the key point that the agent used must mimic the structure of ice and act as an ice nucleus. Silver iodide is the well known answer associated with cloud seeding in examination questions. Final Answer: The chemical commonly used to induce artificial rain in cloud seeding operations is silver iodide.

Question 5

In organic chemistry, which of the following substances is commonly known by the old name carbolic acid?

Accepted Answer

Introduction / Context: Many organic chemicals have common or historical names that differ from their modern systematic names. Carbolic acid is an older name that appears in historical literature and some examination questions. Knowing the modern name that corresponds to carbolic acid helps link past and present terminology. This question asks you to identify which listed substance is also known as carbolic acid in older chemical usage. Given Data / Assumptions: The term carbolic acid is the historical name to be identified. The options include sodium hydroxide, sulfuric acid, ethanol, and phenol. We assume the context of basic organic chemistry and industrial history. We look for the compound associated with antiseptic use and a benzene ring with a hydroxyl group. Concept / Approach: Carbolic acid is the traditional name for phenol, which is a compound consisting of a benzene ring directly bonded to a hydroxyl group. Phenol was once widely used as an antiseptic and disinfectant, and in that context it was referred to as carbolic acid. The word carbolic reflects its origin from coal tar and its somewhat acidic nature. Sodium hydroxide is a strong base, sulfuric acid is a mineral acid, and ethanol is an alcohol used in beverages and as a solvent. None of these correspond to carbolic acid in common chemical nomenclature. Step-by-Step Solution: Step 1: Recall that phenol is a simple aromatic compound with formula C6H5OH, where a hydroxyl group is attached directly to a benzene ring. Step 2: Remember that in the nineteenth century phenol was derived from coal tar and used as an antiseptic, and the name carbolic acid was widely used. Step 3: Note that carbolic refers to carbon based origin and acid refers to its weakly acidic character. Step 4: Compare this with sodium hydroxide, which is a strong base, sulfuric acid, which is a strong mineral acid, and ethanol, which is a simple aliphatic alcohol. Step 5: Conclude that the only substance among the options that matches the historical name carbolic acid is phenol. Verification / Alternative check: History of medicine and chemistry sources describe how Joseph Lister used carbolic acid to reduce infection in surgical wounds, and modern accounts clearly state that carbolic acid was phenol. In industrial and laboratory contexts, phenol is still sometimes referred to by the older name, although modern nomenclature prefers the term phenol. No reputable source equates carbolic acid with sodium hydroxide, sulfuric acid, or ethanol, which reinforces the identification of phenol as the correct choice. Why Other Options Are Wrong: Sodium hydroxide, often called caustic soda, is a strong base, not an acid, and is used in soap making and cleaning, not as carbolic acid. Sulfuric acid is a strong mineral acid used in batteries and industrial processes; it has its own specific name and is never called carbolic acid. Ethanol is a two carbon alcohol used in alcoholic beverages and as a solvent, but carbolic acid refers to a benzene derivative, not to ethanol. Therefore, none of these options match the historical name carbolic acid. Common Pitfalls: Some learners may guess sulfuric acid simply because the word acid appears and they recall it as a strong acid. Others might confuse carbolic with carbonic and think of carbonic acid. To avoid such confusion, remember that carbolic acid is linked with coal tar, antiseptic history, and benzene chemistry. Connecting it with the structure C6H5OH and the name phenol will help you quickly answer this type of question in exams. Final Answer: The substance commonly known as carbolic acid is phenol.

Question 6

Steel is an alloy based on iron that typically contains what approximate percentage of carbon by mass?

Accepted Answer

Introduction / Context: Steel is one of the most important engineering materials and is widely used in construction, machinery, and tools. It is based on iron but gains its useful properties from the addition of carbon and sometimes other alloying elements. The percentage of carbon plays a major role in determining whether the alloy is considered steel or cast iron and also influences hardness, strength, and ductility. This question asks you to identify the typical range of carbon content in steel. Given Data / Assumptions: Steel is an iron based alloy with carbon and possibly other elements. The options describe different ranges of carbon content by mass. We assume standard definitions used in materials science. Cast iron has higher carbon content than steel. Concept / Approach: By conventional definition, steel is an alloy of iron that contains a controlled amount of carbon, generally up to about 2 percent by mass. Low carbon steels may have around 0.1 to 0.3 percent carbon, medium carbon steels around 0.3 to 0.6 percent, and high carbon steels up to about 1.0 or slightly above. All of these fall well below 2 percent carbon. When the carbon content rises significantly above about 2 percent, the alloy is usually classified as cast iron rather than steel. Therefore, the correct description is that steel contains less than about 2 percent carbon by mass. Step-by-Step Solution: Step 1: Recall the definition of steel as an iron carbon alloy with controlled, relatively low carbon content. Step 2: Note that typical structural steels have less than 1 percent carbon, while tool steels may approach or slightly exceed 1 percent. Step 3: Remember that cast irons, in contrast, often contain between about 2 percent and 4 percent carbon. Step 4: Compare these figures with the ranges provided in the options. Step 5: Recognise that the only option matching the standard definition is that steel contains less than about 2 percent carbon. Verification / Alternative check: Materials science texts normally state that steels are iron based alloys with a carbon content up to about 2.0 percent, while cast irons have higher carbon levels. For example, mild steel might have about 0.2 percent carbon, and high carbon spring steel might have around 0.8 percent. None of these approach 10 percent, let alone 29 percent. Zero carbon would mean pure iron, which is relatively soft and not usually called steel in practical engineering. These standard figures confirm that the correct choice is the option describing less than 2 percent carbon. Why Other Options Are Wrong: The option no carbon is incorrect because carbon is essential to harden iron and make steel; pure iron has very different mechanical properties. More than about 10 percent carbon is unrealistic for steel and does not correspond to common metallurgical categories; such an alloy would not be classified as steel. More than about 29 percent carbon is even more extreme and does not match any standard iron carbon alloy; in fact, carbon solubility in iron is limited and such a composition is not meaningful for steel. These options are therefore wrong. Common Pitfalls: Some learners may think that since carbon has a strong effect on properties, steel must contain very high carbon percentages. Others may incorrectly assume that steel and cast iron have similar compositions. To avoid confusion, remember that small changes in carbon content within a relatively low range have big effects on hardness and strength. The key threshold is around 2 percent; below that is considered steel, above that is typically cast iron. Final Answer: Steel is an iron based alloy that usually contains less than about 2 percent carbon by mass.

Question 7

In a neutral atom, which pair of fundamental particles is always present in equal numbers so that the overall electric charge is zero?

Accepted Answer

Introduction / Context: An atom is electrically neutral when its total positive charge balances its total negative charge. The three main subatomic particles are protons, neutrons, and electrons, each with characteristic charge and position in the atom. This question asks which pair of particles must be present in equal numbers to ensure that the net charge of an atom is zero. Understanding this balance is fundamental to atomic theory and basic chemistry. Given Data / Assumptions: Protons carry a positive charge of plus one in elementary charge units. Electrons carry an equal magnitude of negative charge minus one. Neutrons have no electric charge. We consider a neutral atom, not an ion. Concept / Approach: Electric charge in atoms comes from protons and electrons. Protons reside in the nucleus, and electrons are located in regions around the nucleus. Neutrons add mass and contribute to nuclear stability but do not contribute to electric charge. In a neutral atom, the total positive charge from protons must equal the total negative charge from electrons. This means that the number of protons and electrons must be the same. If the numbers differ, the species becomes an ion, either positively or negatively charged. Step-by-Step Solution: Step 1: Recognise that protons provide positive charge and electrons provide negative charge, while neutrons are neutral. Step 2: For the net charge of an atom to be zero, total positive charge and total negative charge must cancel each other. Step 3: Let the number of protons be Z and the number of electrons also be Z in a neutral atom. Step 4: The total positive charge is then plus Z and the total negative charge is minus Z, so the sum is zero. Step 5: Conclude that in any neutral atom, the number of protons and the number of electrons must be equal. Verification / Alternative check: Consider a carbon atom as an example. Carbon has atomic number 6, meaning it has 6 protons in its nucleus. A neutral carbon atom also has 6 electrons around the nucleus. The total positive charge from 6 protons is plus 6, and the total negative charge from 6 electrons is minus 6. Together they give zero net charge. If you remove one electron, you get a positive ion with charge plus one; if you add one electron, you get a negative ion with charge minus one. These simple examples show that equality of proton and electron numbers defines neutrality. Why Other Options Are Wrong: Neutrons and electrons do not have equal charges in magnitude and neutrons are uncharged, so equal numbers of them do not guarantee neutrality. Protons and neutrons may be present in equal numbers in some nuclei, but this has no direct effect on electric charge because neutrons do not contribute to charge; equal numbers of protons and neutrons simply describe some stable isotopes. Protons and positrons both carry positive charge, so equal numbers of them would give a large positive charge and would not produce a neutral atom. Therefore, only protons and electrons must match in number for neutrality. Common Pitfalls: Learners may confuse atomic number and mass number, thinking that equal numbers of protons and neutrons are required for stability and neutrality. While equal proton and neutron counts can make some nuclei stable, neutrons do not affect charge. Another mistake is to overlook electrons when thinking about charge, focusing only on the nucleus. To avoid this, always remember that protons and electrons are the charged particles and that neutrality requires equal numbers of these two in an atom. Final Answer: In a neutral atom, the number of protons and electrons is equal, ensuring zero net electric charge.

Question 8

Which of the following types of radiation does not consist of material particles and instead is a form of electromagnetic radiation?

Accepted Answer

Introduction / Context: Radioactive substances and discharge tubes can emit several kinds of radiation. Some of these radiations are beams of particles with mass and charge, while others are waves of electromagnetic energy without rest mass. This question asks you to identify which listed radiation does not consist of material particles and instead is a form of electromagnetic radiation. Recognising the nature of each type is basic to nuclear physics and radiation safety. Given Data / Assumptions: Alpha rays are associated with helium nuclei emitted from radioactive decay. Beta rays are streams of electrons or positrons from radioactive decay. Gamma rays are high energy electromagnetic waves. Anode rays are positive ion beams observed in discharge tube experiments. Concept / Approach: Each type of radiation can be classified as either particulate or electromagnetic. Alpha radiation consists of alpha particles, which are helium nuclei with two protons and two neutrons, so they are material particles. Beta radiation consists of fast moving electrons or positrons, which are also particles with mass. Anode rays, also called canal rays, are beams of positive ions in gas discharge tubes. Gamma radiation, on the other hand, is a form of electromagnetic radiation, similar to X rays but with higher energy, and consists of photons rather than massive particles. Therefore, gamma rays are the radiation that does not contain material particles. Step-by-Step Solution: Step 1: Recall that alpha rays are composed of alpha particles, which are helium nuclei containing protons and neutrons. Step 2: Recall that beta rays originate from electrons or positrons emitted during nuclear beta decay, which have mass and charge. Step 3: Recall that anode rays, or canal rays, were discovered in gas discharge tubes and consist of positively charged ions moving toward the cathode region. Step 4: Recall that gamma rays are high frequency, high energy electromagnetic waves emitted from excited nuclei. Step 5: Conclude that gamma rays are not beams of material particles but are massless photons, so they do not contain material particles. Verification / Alternative check: Experiments that measure mass and charge show that alpha particles have a definite mass about four times that of a proton and a positive charge of plus two. Beta particles have a mass equal to that of electrons and a negative or positive unit charge. Anode rays are observed to bend in electric and magnetic fields in ways consistent with positive ions of various masses. Gamma rays, in contrast, are affected differently in fields and show wave like properties such as diffraction and interference, with energies in the gamma region of the electromagnetic spectrum. These observations justify classifying gamma rays as electromagnetic radiation without rest mass. Why Other Options Are Wrong: Alpha rays are particulate because they consist of helium nuclei, which are massive and charged. Beta rays are also particulate since they are electrons or positrons, again with definite mass and charge. Anode rays are streams of positive ions in gas discharge tubes, so they are clearly material particles. None of these fit the description of radiation without material particles. Only gamma rays correspond to high energy photons with no rest mass and no electric charge. Common Pitfalls: A common confusion is between beta radiation and gamma radiation, since both can penetrate materials to some degree and are often mentioned together. Learners may also think that because gamma radiation comes from the nucleus, it must involve particles. To avoid confusion, remember that alpha and beta radiation involve ejected particles, while gamma radiation is a pure energy wave emitted when the nucleus moves from an excited state to a lower energy state. The key phrase to remember is that gamma rays are electromagnetic radiation, not matter beams. Final Answer: The radiation that does not contain material particles and is instead electromagnetic in nature is gamma rays.

Question 9

The Bhopal gas tragedy of 1984 was caused by the accidental leakage of which highly toxic industrial gas from a pesticide plant?

Accepted Answer

Introduction / Context: The Bhopal gas tragedy is one of the worst industrial disasters in history. It occurred in December 1984 in Bhopal, India, when a toxic gas leaked from a pesticide manufacturing plant, causing thousands of immediate deaths and long term health problems for many more. This question tests your awareness of the specific chemical involved in this tragedy. Knowing the name of the gas is important in environmental chemistry and general awareness about industrial safety. Given Data / Assumptions: The incident took place at a pesticide plant in Bhopal. A toxic gas used as an intermediate in pesticide production leaked. The options include carbon monoxide, methyl isocyanate, nitrous oxide, and hydrogen sulphide. We assume the context of the 1984 disaster in Bhopal, not another incident. Concept / Approach: The gas that leaked during the Bhopal tragedy was methyl isocyanate, often abbreviated as MIC. It is a highly toxic, volatile, and reactive compound used in the manufacture of carbamate pesticides. At the plant, water entered a storage tank containing methyl isocyanate, causing a runaway reaction, pressure buildup, and subsequent release of the gas into the surrounding area. Carbon monoxide, nitrous oxide, and hydrogen sulphide are toxic or dangerous in other contexts but are not the gases associated with this particular industrial disaster. Step-by-Step Solution: Step 1: Recall that the Bhopal disaster occurred at a pesticide plant operated by a multinational company. Step 2: Note that the main raw material involved was methyl isocyanate, used in the production of certain pesticide formulations. Step 3: Remember that a large amount of methyl isocyanate was stored in tanks at the site and that an uncontrolled reaction led to a massive leak. Step 4: Identify methyl isocyanate among the answer options as a highly toxic industrial chemical, unlike nitrous oxide, which is commonly used as an anaesthetic gas, or carbon monoxide, which is produced from incomplete combustion. Step 5: Conclude that methyl isocyanate is the gas responsible for the Bhopal gas tragedy. Verification / Alternative check: Historical accounts, investigations, and court documents about the Bhopal disaster consistently refer to methyl isocyanate as the leaked gas. Survivors, medical studies, and environmental reports also describe exposure symptoms characteristic of this compound, such as severe respiratory damage and eye irritation. No reliable source states that carbon monoxide, nitrous oxide, or hydrogen sulphide were the primary compounds released in the main accident, although other gases may have been present in minor amounts. This strong and consistent evidence confirms methyl isocyanate as the correct answer. Why Other Options Are Wrong: Carbon monoxide is a toxic gas produced by incomplete combustion of fuels, but it is not the chemical stored in large tanks at Bhopal and is not the gas named in connection with that tragedy. Nitrous oxide is used in medicine and industry and is not associated with the Bhopal accident. Hydrogen sulphide is a poisonous gas associated with sewage and petroleum processing, but the main Bhopal leak was not hydrogen sulphide. These gases can cause poisoning but are not the ones implicated in the 1984 Bhopal disaster. Common Pitfalls: Some learners may confuse the Bhopal gas tragedy with other industrial or environmental incidents and choose a more familiar toxic gas such as carbon monoxide or hydrogen sulphide. Others may recognise the abbreviation MIC but forget the full name methyl isocyanate. To avoid mistakes, associate Bhopal directly with MIC and remember that the tragedy involved a pesticide intermediate, not a combustion product or sewer gas. Final Answer: The Bhopal gas tragedy was caused by the leakage of methyl isocyanate from a pesticide plant.

Question 10

Which one of the following alcohols can be oxidised under suitable conditions to give a corresponding carbonyl compound such as an aldehyde or ketone?

Accepted Answer

Introduction / Context: Oxidation reactions of alcohols are a key topic in organic chemistry. Primary and secondary alcohols can usually be oxidised to aldehydes, ketones, or further to carboxylic acids, while tertiary alcohols are resistant to simple oxidation because they lack a hydrogen atom on the carbon bearing the hydroxyl group. Aromatic compounds like phenol behave differently. This question asks you to identify which given compound can be oxidised to a carbonyl compound under appropriate conditions, highlighting the difference between primary, secondary, tertiary, and phenolic structures. Given Data / Assumptions: 2 methyl 2 hydroxy propane is a tertiary alcohol (common name tert butanol). 2 hydroxy propane is a secondary alcohol, also called isopropanol or propan 2 ol. o nitrophenol is an aromatic compound with nitro and hydroxyl groups on a benzene ring. Phenol is an aromatic compound with a hydroxyl group directly attached to a benzene ring. We are looking for an alcohol that can be oxidised to a simple carbonyl compound, such as a ketone. Concept / Approach: Primary alcohols are oxidised to aldehydes and then to carboxylic acids. Secondary alcohols are oxidised to ketones. Tertiary alcohols generally do not undergo oxidation to carbonyl compounds without breaking carbon carbon bonds because the carbon bearing the hydroxyl group does not have a hydrogen (no alpha hydrogen) to remove. Phenols and substituted phenols undergo different types of reactions such as electrophilic substitution and oxidative coupling rather than simple conversion to carbonyl compounds. Among the options, 2 hydroxy propane is a secondary alcohol and can be oxidised to propanone, a ketone. Step-by-Step Solution: Step 1: Identify the structure of 2 hydroxy propane. It has three carbon atoms and the hydroxyl group on the middle carbon, so it is a secondary alcohol. Step 2: Recall that secondary alcohols can be oxidised to ketones using oxidising agents such as acidified potassium dichromate or potassium permanganate. Step 3: Write the oxidation product of 2 hydroxy propane, which is propanone, a carbonyl compound (a ketone). Step 4: Recognise that 2 methyl 2 hydroxy propane is a tertiary alcohol where the carbon bearing the OH group has no hydrogen, so simple oxidation to a carbonyl compound is not possible without breaking the carbon skeleton. Step 5: Note that phenol and o nitrophenol are aromatic compounds, and their oxidation does not give simple aldehydes or ketones by straightforward mild oxidation; instead they undergo more complex reactions. Verification / Alternative check: In laboratory practice, when a secondary alcohol such as isopropanol is treated with an oxidising agent, it converts to a ketone such as acetone. This is a standard textbook example. In contrast, tertiary alcohols resist oxidation and may require very harsh conditions that break carbon carbon bonds, leading to fragmentation rather than a neat carbonyl compound. Phenol can be oxidised to quinones or other aromatic products, but that is not the same as the simple carbonyl formation described for aliphatic primary and secondary alcohols. Therefore, 2 hydroxy propane is the only option that fits the statement in the question. Why Other Options Are Wrong: 2 methyl 2 hydroxy propane is a tertiary alcohol, and the carbon bearing the OH group has no hydrogen. Mild oxidants cannot convert it to a carbonyl compound without breaking the carbon skeleton, so it does not fit the requirement. o nitrophenol has a hydroxyl group on an aromatic ring and an ortho nitro substituent; its oxidation leads to aromatic products rather than a single simple carbonyl compound. Phenol is similar in that it is not converted to aldehydes or ketones under standard mild oxidation conditions but tends to form complex oxidation products such as quinones. These options therefore do not match the description given in the question. Common Pitfalls: A frequent error is to assume that any compound with an OH group is an alcohol that behaves the same way in oxidation reactions. Students may forget the important difference between aliphatic and aromatic hydroxyl compounds and between primary, secondary, and tertiary alcohols. Another mistake is misidentifying 2 hydroxy propane as a primary alcohol due to its simple name. To avoid these pitfalls, always consider the position of the OH group and the type of carbon it is attached to. Secondary alcohols like 2 hydroxy propane are oxidised to ketones, which are carbonyl compounds. Final Answer: The compound that can be oxidised to a corresponding carbonyl compound is 2 hydroxy propane (a secondary alcohol, isopropanol).

Question 11

During municipal water treatment, coagulation is primarily designed to remove which type of impurity from the water supply?

Accepted Answer

Introduction / Context: Water treatment plants use several processes to convert raw water from rivers, lakes, or reservoirs into safe drinking water. One important step is coagulation followed by flocculation and sedimentation. Coagulation involves adding chemicals that help very fine suspended particles and colloids clump together so that they can be removed more easily. This question asks what type of impurity coagulation is mainly intended to remove, highlighting the role of this process in water purification. Given Data / Assumptions: Coagulation is part of conventional water treatment processes. It is usually followed by flocculation, sedimentation, and filtration. The options include large particles like sand, bad smells, microorganisms, and suspended particles. We assume a typical municipal drinking water treatment plant. Concept / Approach: Raw water often contains very fine suspended solids and colloidal particles such as clay, organic matter, and microscopic debris. These particles are too small to settle quickly by gravity and may pass through simple filtration. Coagulant chemicals such as alum or ferric salts neutralise the charges on these particles, causing them to aggregate into larger flocs that can settle out or be filtered more easily. This process specifically targets fine suspended and colloidal matter. Sand and gravel are usually removed earlier by screens and grit chambers. Pathogenic microorganisms are reduced later by filtration and disinfection. Bad smells are addressed by other treatments such as activated carbon. Step-by-Step Solution: Step 1: Identify that coagulation is used to treat water that contains very fine suspended and colloidal impurities which do not settle readily. Step 2: Recall that chemicals such as alum are added to destabilise these particles by neutralising their surface charges. Step 3: Recognise that once destabilised, particles clump together during flocculation and form larger flocs. Step 4: Understand that these larger flocs can then be removed by sedimentation and filtration, thereby clearing the water of turbidity. Step 5: Conclude that the primary purpose of coagulation is removal of fine suspended particles and colloidal matter. Verification / Alternative check: Descriptions of conventional water treatment outline steps in order. Screening and grit removal take out large debris such as sticks, leaves, sand, and gravel. Coagulation and flocculation then deal with very small particles that cause turbidity but are too small to settle easily. Sedimentation basins allow the flocs to settle, and filtration removes remaining fine particles and many microorganisms. Disinfection with chlorine or other agents specifically targets pathogens. This flow of processes confirms that coagulation focuses on physical suspended particles and colloids rather than large debris, smells, or microbes directly. Why Other Options Are Wrong: Sand and gravel are largely removed by simple screening and grit removal processes before coagulation, so option sand and gravel is not the main purpose of coagulation. Bad smells and tastes are treated using adsorbents like activated carbon and by controlling organic content, not mainly by coagulation, although coagulation may indirectly help. Pathogenic microorganisms are targeted primarily during filtration and disinfection steps, particularly chlorination or ultraviolet treatment, rather than by coagulation alone. Therefore, these options do not correctly describe the specific function of coagulation in water treatment. Common Pitfalls: Some learners may assume that each water treatment step removes all types of impurities equally and may pick microorganisms, thinking coagulation is a general cleaning process. Others may confuse coagulation with disinfection or aeration. To avoid such mistakes, remember that coagulation is a physical chemical process aimed at improving the removal of particles that cause turbidity. Linking coagulation with floc formation and suspended solids will help you choose the correct answer in similar questions. Final Answer: Coagulation during water treatment is designed mainly to remove fine particles and colloidal matter suspended in water.

Question 12

The law of constant proportions, which states that a chemical compound always contains the same elements in the same fixed ratio by mass, was first clearly enunciated by which scientist?

Accepted Answer

Introduction / Context: One of the foundational ideas in chemistry is that a pure compound always contains the same elements combined in the same fixed proportion by mass, regardless of its source or method of preparation. This principle is called the law of constant proportions or the law of definite proportions. It played a crucial role in the development of atomic theory. The question asks you to identify the scientist who first clearly formulated and enunciated this law. Given Data / Assumptions: The law states that a given chemical compound always has the same composition by mass. Several notable chemists are listed as options: Dalton, Berthelot, Avogadro, and Proust. We assume familiarity with basic historical developments in early chemistry. We must match the correct name to this specific law. Concept / Approach: Joseph Proust, a French chemist, performed careful analyses of compounds such as copper carbonate and found that they always contained the same mass ratio of elements, no matter how they were prepared. From these experiments he concluded that chemical compounds are defined by definite proportions. This became known as Proust law of definite proportions or law of constant composition. John Dalton later used this and other laws to develop atomic theory but is more closely associated with the law of multiple proportions. Avogadro is known for Avogadro hypothesis about equal volumes of gases. Berthelot made contributions in thermochemistry. Therefore, the correct answer is Joseph Proust. Step-by-Step Solution: Step 1: State the law of constant proportions: a given compound always contains the same elements in the same ratio by mass. Step 2: Recall that Joseph Proust experimentally analysed various samples of the same compound prepared in different ways and found identical composition. Step 3: Recognise that he generalised these observations into the law of definite proportions. Step 4: Compare this with John Dalton contributions, which focused on atomic theory and the law of multiple proportions, not constant proportions. Step 5: Identify Proust among the options as the chemist associated with the law of constant proportions. Verification / Alternative check: Historical accounts of the development of chemical laws attribute the law of definite proportions to Proust around the late eighteenth and early nineteenth centuries. Dalton built on Proust and Lavoisier work to propose that atoms combine in simple whole number ratios, explaining these mass relationships. Avogadro later introduced his famous hypothesis relating gas volumes and numbers of molecules. Berthelot contributed to thermochemistry and synthesis of organic compounds. None of these other scientists are typically credited with establishing the law of constant proportions itself, confirming that Proust is the correct choice. Why Other Options Are Wrong: John Dalton did propose several important ideas about atoms and is linked with the law of multiple proportions, which states that when two elements form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other are in simple whole numbers. However, he did not first enunciate the law of constant proportions. Avogadro is remembered for Avogadro law about gas volumes and mole concept, not for constant proportions. Berthelot is associated with thermochemistry and synthetic organic chemistry but not with this law. Thus, these options do not correctly answer the question. Common Pitfalls: Students often confuse the various classical chemists because their work is interrelated and appears in the same chapter of textbooks. It is easy to attach the wrong law to a familiar name such as Dalton or Avogadro. To avoid confusion, it is helpful to link Proust with fixed composition, Dalton with multiple proportions and atomic theory, Avogadro with gas volumes and the mole, and Lavoisier with conservation of mass. These associations make it easier to recall which scientist is linked to which major law. Final Answer: The law of constant proportions was first clearly enunciated by Joseph Proust.

Question 13

Of the following elements, which one almost always has the same oxidation state in all of its compounds because of its very high electronegativity?

Accepted Answer

Introduction / Context: Oxidation state is a useful concept in inorganic chemistry that indicates the apparent charge an atom would have if all bonds were ionic. Many elements can exhibit multiple oxidation states in different compounds. However, a few elements tend to show only one oxidation state in their compounds due to their extreme electronegativity or other factors. This question asks you to identify the element that effectively has the same oxidation state in all of its compounds. Given Data / Assumptions: The elements listed are hydrogen, fluorine, carbon, and oxygen. We focus on oxidation states in compounds, not in the free elemental form. Oxidation states are determined using standard rules based on electronegativity and bonding. We ignore rare or hypothetical compounds and focus on usual chemical behaviour. Concept / Approach: Fluorine is the most electronegative element in the periodic table. In compounds, it always attracts electrons more strongly than any other element. As a result, fluorine virtually always has an oxidation state of minus one in its compounds, because it gains one electron relative to its neutral atom. Hydrogen can be plus one in most compounds and minus one in metal hydrides, so it does not have a single oxidation state. Oxygen is usually minus two but can be minus one in peroxides and also positive states in compounds with fluorine. Carbon shows a wide range of oxidation states from minus four to plus four. Therefore, fluorine is the only element among these that maintains the same oxidation state in all its compounds. Step-by-Step Solution: Step 1: Recall that fluorine is the most electronegative element and tends to gain one electron in bonds. Step 2: Therefore, in almost all of its compounds, fluorine has an oxidation state of minus one. Step 3: Consider hydrogen, which typically has an oxidation state of plus one when bonded to non metals but minus one in metal hydrides such as sodium hydride. Step 4: Consider oxygen, which usually has oxidation state minus two, but in peroxides such as hydrogen peroxide it has minus one, and in compounds with fluorine it can have positive oxidation states. Step 5: Note that carbon has many oxidation states in organic and inorganic compounds, ranging from minus four up to plus four, so it clearly does not have a constant oxidation state. Thus, fluorine is the correct answer. Verification / Alternative check: List some common compounds for each element. Fluorine appears in hydrogen fluoride (oxidation state minus one), sodium fluoride (minus one), and many organic fluorides where it is also minus one. There are no stable compounds in which fluorine has a positive oxidation state because no other element exceeds its electronegativity. Hydrogen appears as plus one in HCl, H2O, and NH3, but as minus one in NaH and CaH2. Oxygen is minus two in H2O and most oxides, minus one in H2O2, and plus two in OF2. Carbon varies widely, for example minus four in methane, zero in elemental carbon, and plus four in carbon dioxide. These examples confirm that only fluorine is fixed at minus one in all its compounds. Why Other Options Are Wrong: Hydrogen does not have a single oxidation state because in metal hydrides it is clearly in the minus one state, while in acids and many other compounds it is plus one. Carbon can have many oxidation states depending on its bonding environment, so it is very flexible and cannot be said to have one fixed oxidation state. Oxygen is usually minus two but has notable exceptions such as peroxides and compounds with fluorine where its oxidation state is different. Therefore, these elements do not meet the requirement of having the same oxidation state in all their compounds. Common Pitfalls: A common error is to choose oxygen because students remember that oxygen is almost always minus two. However, examination questions often test awareness of the exceptions, especially in peroxides and compounds like OF2. Another mistake is to choose hydrogen, forgetting that metal hydrides exist. To avoid these pitfalls, remember the simple rule that fluorine is the most electronegative element and therefore always has oxidation state minus one in its compounds. Final Answer: Among the given elements, fluorine has essentially the same oxidation state of minus one in all of its compounds.

Question 14

Tetraethyl lead TEL is added to petrol to minimise knocking in internal combustion engines. In terms of reaction rate, TEL mainly acts as which type of catalyst like additive?

Accepted Answer

Introduction / Context: In older petrol engines, knocking or detonation was a serious problem that reduced efficiency and could damage the engine. Tetraethyl lead, abbreviated TEL, was once widely added to petrol as an anti knocking agent to improve fuel performance. Conceptually, its action can be compared to catalytic effects on reaction rates. This question asks you to identify whether TEL behaves like a positive catalyst, a negative catalyst, an auto catalyst, or an induced catalyst with respect to the reactions responsible for knocking. Given Data / Assumptions: Knocking arises from uncontrolled, premature combustion and undesirable chain reactions in the fuel air mixture. Anti knocking agents aim to slow or prevent these undesirable reactions. TEL is added in small amounts to petrol for this purpose. The options describe types of catalysts in terms of whether they speed up or slow down reactions and how they appear in the system. Concept / Approach: A positive catalyst increases the rate of a desired reaction. A negative catalyst, also called an inhibitor or retarder, decreases the rate of a reaction, particularly an undesirable one. Auto catalysts are formed in the reaction mixture itself, and induced catalysts act only in the presence of another catalyst. TEL does not speed up combustion; instead, it suppresses uncontrolled chain reactions that cause knocking, effectively slowing those undesirable side reactions. In this sense, it behaves like a negative catalyst, decreasing the rate of knocking reactions and allowing smoother, controlled combustion of petrol. Step-by-Step Solution: Step 1: Understand that engine knocking results from rapid, uncontrolled combustion and explosive pressure waves in the cylinder. Step 2: Recognise that TEL is added in small quantity to petrol to reduce this knocking, leading to smoother burning of the fuel air mixture. Step 3: Note that TEL does not aim to speed up the combustion reaction; rather, it interferes with the chain processes that lead to premature detonation. Step 4: Relate this role to the concept of a negative catalyst, which retards or inhibits unwanted reactions and stabilises the system. Step 5: Conclude that TEL is best classified, in this context, as a negative catalyst like inhibitor with respect to knocking reactions. Verification / Alternative check: Engine and fuel chemistry texts describe anti knocking agents as substances that raise the octane rating of petrol by making it more resistant to premature ignition. TEL exerts its effect by interacting with radicals and intermediates in the combustion process, thereby slowing down uncontrolled chain branching reactions that cause knocking. This behaviour is analogous to that of inhibitors used in chemical industry to prevent runaway polymerisation or decomposition, which are classic examples of negative catalysts. TEL is not produced automatically by the fuel, so it is not an auto catalyst, and it does not rely on a second catalyst to work, so it is not an induced catalyst in normal classification. Why Other Options Are Wrong: A positive catalyst that speeds up all reactions would actually make knocking worse by accelerating combustion indiscriminately, which is the opposite of what anti knocking agents are intended to do. Auto catalysts are substances formed during the reaction that then catalyse it further, like manganese in some oxidation reactions; TEL is deliberately added from outside and is not generated by petrol combustion. Induced catalysts are catalysts whose activity appears only in the presence of another catalyst; there is no such second catalyst involved in the basic explanation of TEL action. Therefore, these options do not correctly describe TEL role in reducing knocking. Common Pitfalls: Learners sometimes see the word catalyst and assume that any useful additive must be a positive catalyst that speeds reactions up, so they choose that option without considering whether faster reactions would actually help. Another confusion is treating all specialised additives as auto catalytic simply because they are related to the reaction system. To answer correctly, focus on the effect on knocking. Because TEL reduces the rate of undesirable reactions and increases fuel stability, it behaves like a negative catalyst in this context. Final Answer: TEL minimises knocking when mixed with petrol by acting effectively as a negative catalyst that retards undesirable reactions causing knocking.

Question 15

In the balanced chemical equation for the reaction of magnesium with oxygen gas to form magnesium oxide, how should the equation Mg + O2 → MgO be correctly balanced?

Accepted Answer

Introduction / Context: Balancing chemical equations is a fundamental skill in chemistry. A balanced equation ensures that the law of conservation of mass is satisfied, meaning that the number of atoms of each element is the same on both sides of the equation. The reaction between magnesium metal and oxygen gas to form magnesium oxide is a common example used in introductory courses. This question asks you to choose the correctly balanced form of the basic equation Mg + O2 → MgO. Given Data / Assumptions: The unbalanced equation is magnesium plus oxygen gas gives magnesium oxide. Oxygen exists as a diatomic molecule O2 under standard conditions. Magnesium oxide has the empirical formula MgO, with a one to one ratio of Mg and O. We must add coefficients to balance the number of Mg and O atoms on each side. Concept / Approach: To balance the equation, count atoms of each element on both sides. In the unbalanced equation Mg + O2 → MgO, there is one magnesium atom on each side, but two oxygen atoms on the left and only one on the right. We cannot change subscripts in formulas; we can only add coefficients. To balance oxygen, we need two oxygen atoms on the product side, which requires two formula units of MgO. That introduces two magnesium atoms on the product side, so we must also have two magnesium atoms on the reactant side. The balanced equation is therefore 2Mg + O2 → 2MgO. Step-by-Step Solution: Step 1: Write the unbalanced equation: Mg + O2 → MgO. Step 2: Count atoms on the left side: 1 magnesium atom and 2 oxygen atoms. Step 3: Count atoms on the right side: 1 magnesium atom and 1 oxygen atom in MgO. Step 4: To balance oxygen, place a coefficient of 2 in front of MgO, giving 2MgO on the right side. Now there are 2 oxygen atoms on each side but 2 magnesium atoms on the right and only 1 on the left. Step 5: To balance magnesium, place a coefficient of 2 in front of Mg on the left side, giving 2Mg + O2 → 2MgO. Now magnesium and oxygen are balanced. Verification / Alternative check: Check the final balanced equation 2Mg + O2 → 2MgO carefully. On the left, there are 2 magnesium atoms and 2 oxygen atoms. On the right, there are 2 magnesium atoms and 2 oxygen atoms in the two units of MgO. The numbers match for each element, so mass is conserved. Also, the coefficients are the smallest whole numbers that achieve balance, which is the usual convention. Any alternative equation that changes subscripts, such as MgO2, would represent a different compound and would not be correct for magnesium oxide formation. Why Other Options Are Wrong: The option 2Mg + O → 2MgO is wrong because oxygen does not exist as monatomic O under normal conditions; it is diatomic O2. The equation 2Mg + O → MgO2 is incorrect because it changes the product formula to MgO2, which is not the usual magnesium oxide and also leaves atoms unbalanced. The equation 2Mg + O2 → MgO2 is also wrong because the product again is not MgO and the numbers of atoms do not match. These forms violate either the actual formulas of the substances or the conservation of atoms. Common Pitfalls: Beginners sometimes try to balance equations by changing subscripts inside chemical formulas, which changes the identity of the compounds and is not allowed. Others may forget that many gases, including oxygen, exist as diatomic molecules and incorrectly write O instead of O2. To avoid such mistakes, always treat formulas as fixed and balance equations by placing integer coefficients in front of them. For this reaction, remembering the simple pattern metal plus oxygen gives metal oxide, balanced as 2Mg + O2 → 2MgO, can be a helpful memory aid. Final Answer: The correctly balanced equation for the reaction of magnesium with oxygen is 2Mg + O2 → 2MgO.

Question 16

In a face-centred cubic (fcc) lattice, atom A occupies all corner lattice positions and atom B occupies all face-centred lattice positions. If exactly one B atom is missing from one of the face centres, what is the simplest formula of the resulting …

Accepted Answer

Introduction / Context: This question tests your understanding of solid state chemistry, specifically how to determine the empirical formula of an ionic or atomic solid from occupancy of lattice points in a face centred cubic (fcc) crystal. In such questions, it is essential to understand how many atoms from corners, edges, faces and body centres actually belong to a single unit cell. Here, atom A sits at all corners, atom B sits at all face centres, and one B atom is missing from one of the face centres. You are asked to convert these occupancy details into a simplest whole number formula of the type A x B y for the solid. Given Data / Assumptions: The lattice is face centred cubic (fcc). Atom A occupies all eight corner lattice positions of the cubic unit cell. Atom B occupies all six face centred positions in the ideal case. Exactly one face centre is vacant, so there are B atoms only on five faces. We assume a perfect infinite crystal in which each unit cell is identical apart from this systematic missing B per cell. Concept / Approach: In crystallography, each lattice point can be shared by neighbouring unit cells. Corner atoms are shared by eight cells, so each corner contributes only one eighth to a given cell. Face centred atoms are shared by two neighbouring cells, so each face atom contributes one half to one unit cell. The basic strategy is to calculate the effective number of A and B atoms per unit cell using these fractions, then simplify the ratio to the smallest whole numbers. This ratio gives the empirical formula A x B y of the crystalline solid. Step-by-Step Solution: Step 1: Count the number of A atoms at corners. There are 8 corners, each shared by 8 cells, so effective A atoms per cell = 8 * (1/8) = 1. Step 2: In a normal fcc cell, there are 6 faces and thus 6 face centred atoms. Each face centred atom is shared between 2 unit cells, so normally B atoms per cell = 6 * (1/2) = 3. Step 3: The question states that one face centred B is missing. So there are B atoms only at 5 faces instead of 6. Step 4: Effective number of B atoms per cell = 5 * (1/2) = 2.5. Step 5: Now the ratio of A:B in the crystal is 1 : 2.5. Step 6: To convert 1 : 2.5 into whole numbers, multiply both by 2 to get 2 : 5. Step 7: Hence the empirical formula of the solid is A2B5. Verification / Alternative check: You can verify the result by reversing the process. Start with A2B5 per formula unit. For every 2 A atoms and 5 B atoms, normalising to one A atom gives 1 A and 2.5 B atoms. This matches the effective atoms per unit cell that we calculated directly from the lattice positions. Because no simpler whole number ratio exists than 2:5, A2B5 must be the correct empirical formula. This confirms that the approach of counting fractional contributions from corners and faces is consistent. Why Other Options Are Wrong: Option A2B3 would correspond to a B count of 1.5 per A, which does not match the calculated 2.5 B per A. A3B2 implies B is less abundant than A, contradicting the fact that there are more B atoms than A atoms in the unit cell. A3B4 gives a ratio B:A of 4:3 (about 1.33), again not matching the required 2.5:1. Therefore these alternative formulas do not reflect the actual occupancy of lattice points in the given fcc structure. Common Pitfalls: Many students forget that corner and face atoms are shared between multiple unit cells and incorrectly count them as one full atom per position. Another common error is to ignore the missing face centred B atom and assume 6 B atoms at faces, which would lead to a formula A1B3 instead of A2B5. Some also forget to convert fractional numbers into the simplest whole number ratio, leaving answers like A1B2.5. Always compute effective atoms per cell using sharing fractions, then simplify to whole numbers to get the final formula. Final Answer: The correct empirical formula of the crystalline solid is A2B5.

Question 17

A gas absorbs a photon of wavelength 355 nm and then emits radiation at two longer wavelengths. If one of the emission lines is observed at 680 nm, what is the wavelength of the other emission line?

Accepted Answer

Introduction / Context: This question examines your understanding of basic quantum theory and the relationship between photon energy and wavelength. When a gas absorbs a photon, an electron in an atom or molecule is promoted to a higher energy level. It can then return to lower energy levels in one or more steps, emitting photons of specific wavelengths. The sum of the energies of emitted photons must equal the energy of the absorbed photon. Using the relation between energy and wavelength, you can determine an unknown emission wavelength when the absorbed wavelength and one emission wavelength are known. Given Data / Assumptions: The gas absorbs a photon with wavelength lambda_abs = 355 nm. The gas emits two photons when returning to the ground state or a lower state. One emitted photon has wavelength lambda_1 = 680 nm. The other emitted photon has wavelength lambda_2, which is unknown. Energy levels are assumed to be fixed and there is no loss of energy as heat. The energy of a photon is given by E = h * c / lambda. Concept / Approach: Photon energy is inversely proportional to wavelength. For a photon, E = h * c / lambda, where h is Planck constant and c is speed of light. When an atom absorbs a photon, it gains energy E_abs. When it relaxes in two steps and emits two photons, their energies are E_1 and E_2. By conservation of energy, E_abs = E_1 + E_2. Because h and c are constants, you can work conveniently with inverse wavelengths (called wave numbers). The relationship becomes 1 / lambda_abs = 1 / lambda_1 + 1 / lambda_2. Solving this equation for lambda_2 gives the missing wavelength. Step-by-Step Solution: Step 1: Write the conservation of energy using wave numbers: 1 / lambda_abs = 1 / lambda_1 + 1 / lambda_2. Step 2: Substitute lambda_abs = 355 nm and lambda_1 = 680 nm. Step 3: Compute 1 / lambda_2 = 1 / 355 nm minus 1 / 680 nm. Step 4: Evaluate 1 / 355 nm approximately as 0.0028169 nm^-1 and 1 / 680 nm approximately as 0.0014706 nm^-1. Step 5: Subtract to get 1 / lambda_2 approximately equal to 0.0028169 minus 0.0014706 which is about 0.0013463 nm^-1. Step 6: Take the reciprocal: lambda_2 is about 1 / 0.0013463 nm which is very close to 743 nm. Step 7: Compare the calculated value with the options and select 743 nm. Verification / Alternative check: As a quick check, convert the three wavelengths into relative energies. The smallest wavelength 355 nm should correspond to the highest energy. The emission wavelengths must be longer than 355 nm and their energies must add up to the absorbed energy. A 680 nm photon clearly has less energy than a 355 nm photon. The second photon must therefore also have energy smaller than that of the absorbed photon but large enough so that E_1 + E_2 equals E_abs. A wavelength of 743 nm is longer than 680 nm, so its energy is slightly lower. If you compute E_355 minus E_680, the difference should be nearly equal to E_743. This confirms that 743 nm is consistent with energy conservation, while the other options are not. Why Other Options Are Wrong: An emission wavelength of 518 nm would have energy too high when combined with the energy of a 680 nm photon, exceeding the absorbed energy at 355 nm. A wavelength of 325 nm would have even more energy than the absorbed photon and cannot be an emission wavelength in this process. A wavelength of 1035 nm would have a very low energy; combined with the 680 nm photon it would not account for the full absorbed energy, leaving an energy gap. Hence these options violate energy conservation in the transition scheme. Common Pitfalls: Common mistakes include adding wavelengths directly instead of working with energies or inverse wavelengths. Some students forget that energy is proportional to 1 / lambda, not to lambda itself. Others mistakenly use 1 / lambda_2 = 1 / lambda_1 minus 1 / lambda_abs, reversing the relation. Rounding too aggressively in intermediate steps can also lead to slightly incorrect values that do not match any option. To avoid these issues, always start from energy conservation, convert it to an equation in inverse wavelengths and maintain enough significant figures until the final step. Final Answer: The wavelength of the second emitted photon is 743 nm.

Question 18

In basic environmental chemistry, recycled plant matter that has decomposed into a soil conditioner is commonly known by which of the following names?

Accepted Answer

Introduction / Context: This general knowledge question connects simple chemistry with environmental science and agriculture. It focuses on the term used for recycled plant matter, such as leaves, kitchen waste and garden clippings, that has undergone biological decomposition and is returned to the soil as a nutrient rich conditioner. Knowing this term is important for understanding sustainable practices like organic farming, waste management and home gardening, where recycling organic waste reduces landfill load and improves soil fertility. Given Data / Assumptions: Plant matter, such as leaves, grass clippings and food scraps, is collected and allowed to decompose. This decomposition is largely biological, driven by microorganisms and sometimes aided by earthworms and other organisms. The final product is a dark, crumbly, soil like material used to enrich soil. The question asks for the standard term used to describe this recycled and decomposed plant material. Concept / Approach: In environmental chemistry and gardening terminology, decomposed organic matter used to improve soil is called compost. Composting is the controlled aerobic decomposition of organic waste. Microorganisms break down complex organic molecules such as carbohydrates, proteins and fats into simpler compounds, carbon dioxide, water and humus like material. The resulting compost is rich in nutrients such as nitrogen, phosphorus and potassium and improves soil structure, water holding capacity and biological activity. Recognising the meaning of compost and distinguishing it from unrelated chemical terms like carbide is the key to answering this question correctly. Step-by-Step Solution: Step 1: Recall that recycling plant matter by biological decomposition in a pit or bin is usually called composting. Step 2: The product obtained after sufficient time, when the organic waste has fully broken down, is called compost. Step 3: Examine the options. Decomposed matter is a generic phrase, not a standard technical term used in agriculture. Step 4: Carbide refers to a class of inorganic compounds of carbon with metals or metalloids, for example calcium carbide, and is unrelated to recycled plant waste. Step 5: None of these would only be correct if no option matched the standard term, which is not the case here. Step 6: Therefore the correct and specific term is compost. Verification / Alternative check: As a simple check, think of how gardeners and municipal waste managers label bins and pits for organic waste. They often refer to them as compost bins or compost pits, and the process as composting. The enriched material that is dug out and spread on fields or gardens is called compost manure or simply compost. School level environmental science chapters also describe ways to make compost from vegetable peels and dry leaves. No textbook uses the term carbide or a vague phrase like decomposed matter to name this product. This supports the choice of compost as the correct answer. Why Other Options Are Wrong: Decomposed matter is too general and could refer to any material that has undergone decomposition, including plastics or dead animals, so it is not a precise technical term in this context. Carbide refers to compounds such as calcium carbide that are manufactured industrially and used to generate acetylene gas or for other chemical processes, not to organic soil conditioners. None of these is incorrect because compost is indeed listed as one of the options and is the widely accepted term. Hence these options do not correctly capture the standard environmental term for recycled plant matter used in soil improvement. Common Pitfalls: Learners may be tempted to choose a vague sounding option like decomposed matter because it seems to match the idea of decayed plant material, but exam questions usually prefer proper technical terms. Another mistake is confusion between compost and manure. Manure is usually animal waste, whereas compost comes primarily from plant waste, although both enrich the soil. Some students may not have encountered carbide before and thus ignore it without understanding that it is unrelated to soil conditioning. Remembering key environmental terms such as compost, humus, manure and vermicompost helps avoid such confusion. Final Answer: Recycled plant matter that has decomposed into a soil conditioner is known as compost.

Question 19

In thermodynamics, which of the following statements about the sign of work W when energy is transferred between a system and its surroundings is or are correct?

Accepted Answer

Introduction / Context: This question assesses your understanding of sign conventions for work in thermodynamics. When studying the first law of thermodynamics, it is important to choose a consistent sign convention for heat and work. In chemistry and many physics courses, work done on the system is taken as positive and work done by the system is taken as negative. This question presents two statements involving energy transfer as work between a system and its surroundings and asks you to identify which one correctly matches this sign convention. Given Data / Assumptions: W denotes work in the energy balance for a thermodynamic system. The convention being tested is that used in chemistry: work done on the system is positive, work done by the system is negative. Statement A describes energy transferred to the system from the surroundings as work. Statement B describes energy transferred from the system to the surroundings as work. No heat transfer details are given; only the sign of work is in focus. Concept / Approach: The common convention in chemistry is: when the surroundings compress the system (for example by decreasing its volume), they do work on the system and W is positive. When the system expands and pushes against the surroundings, the system does work on the surroundings and W is negative. This fits with the first law written as delta U = q + W, where q is heat absorbed by the system and W is work done on the system. The key is to relate phrases like to the system and from the system to the sign of W in this convention. Step-by-Step Solution: Step 1: Analyse statement A. It says that energy is transferred to the system from the surroundings as work, and that the surroundings do work on the system. Step 2: In the chemistry convention, work done on the system increases its internal energy and is taken as positive. Therefore, W should be positive when the surroundings perform work on the system. Step 3: Statement A also explicitly says that the value of W is positive in this case. This matches the convention and is therefore correct. Step 4: Analyse statement B. It says that energy is transferred from the system to the surroundings as work, and that work is done on the surroundings. Step 5: When the system does work on the surroundings, energy leaves the system as mechanical work, and W is negative in the chosen convention. Step 6: Statement B explicitly says that the value of W is negative in this case. This again matches the convention and is correct. Step 7: Therefore both statements A and B correctly describe the sign of W for their respective situations. Verification / Alternative check: A helpful way to verify the sign convention is to consider a gas in a cylinder with a movable piston. If an external force pushes the piston inwards, compressing the gas, the surroundings are doing work on the gas. The internal energy of the gas can increase and W is positive. If instead the gas expands, pushing the piston outwards against an external pressure, the gas does work on the surroundings. This reduces the energy available to remain in the system and W is negative. Applying this model directly to statements A and B shows that both are in agreement with the physical behaviour of expanding or compressing gases and the standard sign convention used in chemistry. Why Other Options Are Wrong: Option A alone is incomplete because it ignores the fact that statement B is also correct. Option B alone is also incomplete for the same reason. Option C, which states that both A and B are correct, matches the analysis. Option D claims that none of the statements are correct, which contradicts the standard definitions of work done on and work done by the system. Therefore, the only fully correct choice is that both A and B are correct statements about the sign of work W. Common Pitfalls: A common pitfall is confusing different sign conventions used in physics and engineering texts. Some textbooks define W as work done by the system, resulting in a sign convention opposite to the chemistry convention used here. Another mistake is to ignore the phrase to the system or from the system and focus only on the direction of energy transfer without relating it to the convention chosen. To avoid confusion in examinations, always check how the first law is written in the course material and apply the same convention consistently when interpreting statements about work and heat. Final Answer: Both given statements correctly describe the sign of work in this convention, so the correct choice is Both (A) and (B) are correct.

Question 20

In polymer chemistry, Orlon is the trade name for a synthetic fibre that is a polymer of which monomer?

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Introduction / Context: This question belongs to the topic of synthetic polymers and trade names of common fibres. Examinations often test whether you can link a commercial name to its underlying monomer and polymer type. Orlon is a well known trade name used for an acrylic fibre. Knowing its monomer helps you categorise it as a type of addition polymer and understand its properties and uses in fabrics and textiles. Given Data / Assumptions: The polymer in question is marketed under the trade name Orlon. Orlon is described as a synthetic fibre used in textiles. You are given several possible monomers or related polymer types such as Dacron, acrylonitrile and polyamide nylon. The task is to identify the correct monomer from which the polymer known as Orlon is produced. Concept / Approach: Orlon is an acrylic fibre. Acrylic fibres are typically polymers of acrylonitrile, a monomer with the structure CH2=CHCN. During polymerisation, the double bond opens and forms a long chain polyacrylonitrile. Trade names like Orlon are used by companies for marketing, but examination questions expect you to connect them with their chemical monomers. Dacron, on the other hand, is a trade name for polyethylene terephthalate, a polyester formed from ethylene glycol and terephthalic acid. Polyamide nylon is a condensation polymer formed from diamines and dicarboxylic acids. Therefore only one option matches the known chemistry of Orlon. Step-by-Step Solution: Step 1: Recall that Orlon is a trade name associated with acrylic fibres. Step 2: Acrylic fibres are made primarily from the monomer acrylonitrile, which has a carbon carbon double bond and a nitrile group. Step 3: Through addition polymerisation, many acrylonitrile molecules join together to form polyacrylonitrile, the main component of Orlon. Step 4: Examine the options. Dacron is a completely different polymer, a polyester, and not related directly to Orlon. Step 5: Polyamide nylon refers to a family of condensation polymers, such as nylon 6 and nylon 6,6, based on amide linkages. Step 6: None of the above is not correct because acrylonitrile is a listed option and is in fact the correct monomer. Step 7: Hence, Orlon is a polymer of acrylonitrile. Verification / Alternative check: You can confirm this by checking typical examples from polymer chapters in textbooks. Common trade names are often grouped: Teflon from tetrafluoroethene, Orlon from acrylonitrile, Dacron from terephthalic acid and ethylene glycol, Bakelite from phenol and formaldehyde. Orlon is always placed under acrylic fibres and described as polyacrylonitrile. Its uses include sweaters, blankets and carpets where wool like properties are desired. This matches the information that Orlon is a synthetic acrylic fibre derived from acrylonitrile monomer through chain growth polymerisation. Why Other Options Are Wrong: Dacron is not a monomer at all but a trade name for a polyester polymer, so it cannot be the monomer of Orlon. Polyamide nylon is also a class of polymers, not a monomer, and is based on repeating amide linkages from diamines and dicarboxylic acids. None of the above would only be appropriate if none of the other options matched the known chemistry of Orlon, but acrylonitrile clearly does. Therefore these options do not correctly describe the monomer used to prepare Orlon. Common Pitfalls: Learners sometimes confuse trade names and polymer classes, for example mixing up Dacron and Orlon or thinking that any synthetic fibre must be a type of nylon. Another mistake is to believe that terms like Dacron or Orlon represent monomers rather than polymers. To avoid such confusion, make a small table linking trade names, monomers and polymer classes while studying. Remember that Orlon is associated with acrylic fibres and acrylonitrile, whereas Dacron is linked with polyester and nylon with polyamides. Final Answer: Orlon is a polymer formed from the monomer acrylonitrile.

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