Question 1

Which one of the following alkali metal elements does not form a stable solid hydrogen carbonate (bicarbonate) salt under normal conditions?

Accepted Answer

Introduction / Context: Hydrogen carbonates, also called bicarbonates, of alkali metals show interesting trends in stability and physical state across the group. Most alkali metals form stable solid hydrogen carbonates that can be isolated as crystalline salts. This question asks which alkali metal is an exception and does not form a stable solid hydrogen carbonate, highlighting periodic trends in properties. Given Data / Assumptions: We consider alkali metals in Group 1: lithium, sodium, potassium, rubidium, and caesium. Hydrogen carbonates have the general formula MHCO3 for an alkali metal M. We focus on whether stable solid hydrogen carbonates exist for each element. Standard conditions of temperature and pressure are assumed. Concept / Approach: As we move down Group 1, the stability of solid hydrogen carbonates generally increases. Solid sodium hydrogen carbonate (baking soda) and potassium hydrogen carbonate are common. However, lithium hydrogen carbonate is not stable as a solid at ordinary conditions and exists only in solution. This is due to the small size and high polarising power of the lithium ion, which destabilises the hydrogen carbonate ion in the solid state. Therefore, lithium is the element that does not form a stable solid hydrogen carbonate, distinguishing it from other alkali metals. Step-by-Step Solution: Step 1: Recall that sodium hydrogen carbonate (NaHCO3) is a well known solid, used as baking soda. Step 2: Remember that potassium hydrogen carbonate (KHCO3) is also known and can exist as a crystalline solid. Step 3: For heavier alkali metals such as rubidium and caesium, their hydrogen carbonates are also more stable compared to lithium. Step 4: Lithium hydrogen carbonate, however, is not obtained as a solid under normal conditions and is stable only in aqueous solution. Step 5: Therefore, lithium is the alkali metal that does not form a stable solid hydrogen carbonate salt. Verification / Alternative check: Periodic trends often show that lithium behaves somewhat differently from other alkali metals, a phenomenon called diagonal relationship, where lithium resembles magnesium in some properties more than its own group. Magnesium also does not form a stable solid hydrogen carbonate. Textbooks on inorganic chemistry emphasise that LiHCO3 cannot be isolated as a solid, whereas NaHCO3 is easily crystallised. This pattern confirms that lithium is the exception in the group. Why Other Options Are Wrong: Sodium: Forms stable solid sodium hydrogen carbonate, widely used in baking and fire extinguishers. Potassium: Forms stable solid potassium hydrogen carbonate. Caesium: Its hydrogen carbonate is more stable and can exist as a solid, though it is less common in school examples. Rubidium: Similar to caesium, it can form solid hydrogen carbonate under appropriate conditions. Common Pitfalls: Students sometimes generalise that all alkali metals behave identically and form similar salts. However, lithium often shows anomalous behaviour due to its small ionic size and high charge density. Another pitfall is to assume that because sodium and potassium hydrogen carbonates are common, lithium must behave the same way. Remember that LiHCO3 is only stable in solution and decomposes on drying, which is why it is not isolated as a solid. Final Answer: The alkali metal that does not form a stable solid hydrogen carbonate is Lithium.

Question 2

In which of the following types of coal is the percentage of carbon content the highest, making it the best grade of coal for energy production?

Accepted Answer

Introduction / Context: Coal is a fossil fuel that exists in several forms differing in appearance, carbon content, and heating value. Understanding the different ranks of coal is important in both environmental science and industrial chemistry. This question asks which type of coal contains the highest percentage of carbon, making it the best grade for energy production. Given Data / Assumptions: Different forms of coal include peat, lignite, sub bituminous, bituminous, and anthracite. As coal matures geologically, its carbon content generally increases. Higher carbon content usually means higher calorific value and less volatile matter. We are comparing the relative ranks of the listed coal types. Concept / Approach: The progression of coal formation starts with peat, followed by lignite, then sub bituminous coal, bituminous coal, and finally anthracite. At each stage, pressure and temperature drive off more volatile components and increase the proportion of fixed carbon. Anthracite is the highest rank, hardest, and contains the highest percentage of carbon, often above 90 percent. Lower rank coals like peat and lignite contain more moisture and volatile matter and less carbon, resulting in lower heating value. Step-by-Step Solution: Step 1: Recall the order of coal ranks: peat, lignite, sub bituminous, bituminous, and anthracite. Step 2: Understand that as the rank increases, carbon content increases and moisture and volatile matter decrease. Step 3: Recognise that peat is the lowest rank, very moist and with low carbon content. Step 4: Lignite and sub bituminous coal are intermediate, with increasing carbon but not the highest. Step 5: Bituminous coal has fairly high carbon content but is still below anthracite. Step 6: Anthracite is the highest rank coal with the maximum carbon percentage and highest calorific value among the listed types. Verification / Alternative check: Approximate values from reference tables show that anthracite coal can contain over 90 percent fixed carbon, bituminous coal around 70 to 85 percent, sub bituminous around 60 to 70 percent, lignite around 50 to 60 percent, and peat much less. These numbers confirm that anthracite has the highest carbon content. Additionally, anthracite burns with a short, blue flame and little smoke, reflecting its high carbon and low volatile matter content. Why Other Options Are Wrong: Bituminous: Has high carbon content but less than anthracite, so it is a lower rank coal. Peat: A very low rank precursor to coal with high moisture and low carbon content. Lignite: A soft brown coal with relatively low carbon content compared to anthracite and bituminous coal. Sub-bituminous coal: Intermediate between lignite and bituminous, with carbon content lower than bituminous and anthracite. Common Pitfalls: Students sometimes confuse bituminous with anthracite because both are used widely as fuels. Another common mistake is to think that the most common type of coal used in power plants must have the highest carbon content, which is not necessarily true. Remember that anthracite is the top grade, but it is less abundant. Learning the standard order of coal ranks helps in avoiding these errors. Final Answer: The form of coal with the highest percentage of carbon is Anthracite.

Question 3

Which one of the following statements about the oxidation number of hydrogen in its compounds is correct, taking into account its behaviour with metals and non-metals?

Accepted Answer

Introduction / Context: Oxidation numbers are useful for understanding redox reactions and the distribution of electrons in compounds. Hydrogen is a special element that can show different oxidation numbers depending on what it is bonded to. This question asks which statement correctly describes the possible oxidation numbers of hydrogen, connecting basic rules of oxidation states with examples of common compounds. Given Data / Assumptions: We consider hydrogen in typical compounds with metals and non metals. Oxidation numbers are assigned using standard rules: elements in free state have zero oxidation number, etc. The options suggest that hydrogen might have a fixed or variable oxidation number. No unusual or exotic compounds are considered, just standard inorganic examples. Concept / Approach: In most compounds with non metals, such as water (H2O), hydrogen chloride (HCl), and methane (CH4), hydrogen has an oxidation number of +1. However, when hydrogen forms hydrides with metals, such as sodium hydride (NaH), calcium hydride (CaH2) or lithium aluminium hydride (LiAlH4), hydrogen has an oxidation number of -1. In its elemental form as H2, the oxidation number of hydrogen is zero. Therefore, hydrogen can have more than one oxidation number, usually +1 or -1, depending on the chemical environment. Step-by-Step Solution: Step 1: Recall that in most covalent compounds with non metals, hydrogen is less electronegative than the other element and is assigned an oxidation number of +1. Step 2: Consider water, H2O. Oxygen is assigned -2, and to balance the molecule, each hydrogen is +1. Step 3: Consider metal hydrides like NaH. Sodium is a Group 1 metal and typically has oxidation number +1. To make the compound neutral, hydrogen must then be -1. Step 4: In elemental hydrogen gas, H2, each hydrogen atom has oxidation number zero because it is in the free state. Step 5: These examples show that hydrogen can have oxidation numbers +1, -1, or 0, so the correct general statement is that hydrogen can have more than one oxidation number. Verification / Alternative check: Checking common textbooks on redox chemistry confirms these rules. They state that hydrogen is usually +1 in compounds but is -1 in metal hydrides and zero in elemental form. Any statement that hydrogen always has the same oxidation number, regardless of compound, is described as incorrect. This confirms that the most accurate general statement is that hydrogen can have more than one oxidation number. Why Other Options Are Wrong: The oxidation number for hydrogen is always zero: This is only true for hydrogen gas, not for its compounds. The oxidation number for hydrogen is always +1: This ignores metal hydrides where hydrogen is -1. The oxidation number for hydrogen is always -1: This only applies to hydrides with metals and is not the general case. Hydrogen never shows a negative oxidation number: This contradicts known examples like NaH and CaH2, in which hydrogen has oxidation number -1. Common Pitfalls: Students often memorise that hydrogen is +1 in all compounds and forget about metal hydrides. Another common mistake is to think that hydrogen behaves exactly like alkali metals in all cases. The best way to avoid errors is to recall that oxidation numbers are formal charges based on electronegativity and compound type. In hydrides with metals, the metal is less electronegative, so hydrogen takes the negative oxidation state. Final Answer: The correct statement is that Hydrogen can have more than one oxidation number.

Question 4

In molecular geometry, match the compounds in List I (CH3F, HCHO, HCN and NH3) with the shapes of their molecules in List II (tetrahedral, trigonal planar, linear and trigonal pyramidal) and select the correct code given below.

Accepted Answer

Introduction / Context: This question tests your understanding of basic molecular geometry, as predicted by the valence shell electron pair repulsion (VSEPR) theory. You are asked to match a set of simple molecules and ions, namely CH3F, HCHO, HCN and NH3, with their correct three dimensional shapes such as tetrahedral, trigonal planar, linear and trigonal pyramidal. Knowing how to assign shapes based on the number of electron domains and lone pairs around the central atom is a core skill in introductory chemistry and is frequently used to explain reactivity, polarity and physical properties. Given Data / Assumptions: In this question we consider the following points as given or standard knowledge. CH3F is a molecule with carbon as the central atom bonded to three hydrogen atoms and one fluorine atom. HCHO is methanal or formaldehyde, with carbon double bonded to oxygen and single bonded to two hydrogen atoms. HCN is hydrogen cyanide, where carbon is between hydrogen and nitrogen and forms a triple bond with nitrogen. NH3 is ammonia, with nitrogen bonded to three hydrogen atoms and having one lone pair of electrons. We use standard VSEPR theory to predict shapes from electron domain geometry and lone pairs. Concept / Approach: According to VSEPR theory, the shape of a molecule is determined by the number of electron pairs (bonding plus lone pairs) around the central atom. The electron pair geometry considers all electron domains, while the molecular shape focuses on the positions of atoms. Four electron pairs with no lone pairs around the central atom give a tetrahedral shape. Three electron pairs with no lone pairs give a trigonal planar shape. Two electron pairs with no lone pairs give a linear shape. Three bonding pairs plus one lone pair give a trigonal pyramidal molecular shape. The task is to apply these ideas to each molecule and then write the correct code for the pairings between List I and List II. Step-by-Step Solution: Step 1: For CH3F, carbon is bonded to three hydrogen atoms and one fluorine atom. That means four bonding pairs and no lone pairs on carbon, so the shape around carbon is tetrahedral. Therefore CH3F corresponds to code 2 (tetrahedral). Step 2: For HCHO (formaldehyde), the carbon atom is bonded to two hydrogens and double bonded to oxygen. This gives three regions of electron density around carbon (two single bonds and one double bond) with no lone pairs, which corresponds to a trigonal planar shape. Therefore HCHO corresponds to code 1 (trigonal planar). Step 3: For HCN, the carbon is bonded linearly between hydrogen and nitrogen, and there are two regions of electron density (one single bond and one triple bond), which leads to a linear shape. Therefore HCN corresponds to code 4 (linear). Step 4: For NH3, nitrogen has three N−H bonds and one lone pair of electrons. This gives four electron pairs in total, but one is a lone pair. The electron pair geometry is tetrahedral, while the molecular shape is trigonal pyramidal. Therefore NH3 corresponds to code 3 (trigonal pyramidal). Step 5: Collecting the matches, we get A. CH3F → 2, B. HCHO → 1, C. HCN → 4, D. NH3 → 3, which produces the sequence 2 1 4 3. Verification / Alternative check: Another way to verify the shapes is to consider typical textbook examples. CH4 and CH3F are classic tetrahedral molecules with four sigma bonds around carbon. Formaldehyde HCHO is routinely cited as trigonal planar due to the sp2 hybridisation at carbon. Hydrogen cyanide HCN is used as a standard example of a linear molecule with a triple bond. Ammonia NH3 is a well known trigonal pyramidal molecule due to the lone pair on nitrogen. These standard shapes confirm that the mapping A → 2, B → 1, C → 4 and D → 3 is correct, reinforcing the earlier VSEPR based reasoning. Why Other Options Are Wrong: Option 2 4 1 3 would assign CH3F as tetrahedral but would incorrectly label HCHO as linear and HCN as trigonal planar, which contradicts VSEPR predictions and experimental data. Option 3 4 1 2 mixes up the shapes even more, implying that CH3F is trigonal pyramidal and NH3 is tetrahedral without a lone pair, which is inconsistent with the known structure of ammonia. Option 3 1 4 2 wrongly assigns NH3 as tetrahedral and CH3F as trigonal pyramidal, reversing the correct interpretation of lone pairs on nitrogen versus carbon in these molecules. Common Pitfalls: A common mistake is to confuse electron pair geometry with molecular shape and to forget that lone pairs change the molecular geometry but are not counted as atoms. Students also sometimes miscount double or triple bonds as multiple electron pairs, whereas VSEPR counts them as one domain of electron density. Another pitfall is to assume that any four bonded atoms around a central atom must always produce the same shape, without checking for the presence of lone pairs. Careful counting of electron domains and awareness of lone pairs help to avoid these errors. Final Answer: The correct matching of molecules with their shapes is CH3F tetrahedral, HCHO trigonal planar, HCN linear and NH3 trigonal pyramidal, which corresponds to the code 2 1 4 3.

Question 5

Which one of the following elements cannot be detected by the classical qualitative organic analysis known as the Lassaigne test for elements present in an organic compound?

Accepted Answer

Introduction / Context: This question examines your understanding of qualitative organic analysis, specifically the Lassaigne test, which is used to detect elements such as halogens, nitrogen and sulphur in organic compounds. In the Lassaigne test, the organic compound is fused with metallic sodium to convert covalently bound elements into ionic forms that are then detected by standard inorganic reactions. You are asked to identify which halogen or element in the list cannot be detected by this classical method, which is a favourite concept in introductory organic chemistry courses and competitive examinations. Given Data / Assumptions: We are considering the Lassaigne test applied to organic compounds that may contain halogens or sulphur. I stands for iodine, Cl stands for chlorine, S stands for sulphur and F stands for fluorine. In the Lassaigne test, the compound is fused with sodium metal to form sodium halides or other sodium salts. The aqueous extract obtained after fusion is tested with silver nitrate or other reagents to identify halide ions. We assume standard experimental conditions as commonly described in textbooks. Concept / Approach: The central idea of the Lassaigne test is that organic compounds containing elements such as chlorine, bromine, iodine, nitrogen or sulphur are converted to their corresponding sodium salts when fused with sodium. For halogens, sodium halides like NaCl, NaBr and NaI are formed. These are then treated with nitric acid and silver nitrate solution to give characteristic precipitates of silver halides. However, the behaviour of fluorine is different. Sodium fluoride formed during fusion is extremely soluble in water and does not give the typical insoluble silver halide precipitate under normal test conditions. Therefore, fluorine cannot be detected reliably using the ordinary Lassaigne test procedure, unlike chlorine, bromine or iodine. Step-by-Step Solution: Step 1: Recall that chlorine, bromine and iodine form sodium chloride, sodium bromide and sodium iodide during sodium fusion, all of which can give visible precipitates with silver nitrate in acidic medium. Step 2: For chlorine, the reaction with silver nitrate produces a white precipitate of silver chloride, which is characteristic and confirms chlorine in the organic compound. Step 3: For bromine, the test produces a pale yellow precipitate of silver bromide, while for iodine it produces a yellow precipitate of silver iodide, both of which are relatively insoluble and easy to observe. Step 4: By contrast, fluorine in an organic compound forms sodium fluoride on fusion with sodium. Sodium fluoride is highly soluble in water and does not readily produce a distinct silver fluoride precipitate under normal conditions of the Lassaigne test. Step 5: Because of this high solubility and the absence of a characteristic precipitation reaction using the usual reagents, fluorine cannot be detected by the standard Lassaigne method, whereas iodine, chlorine and sulphur can be detected by appropriate specific tests. Verification / Alternative check: Standard organic chemistry laboratory manuals and textbooks explicitly state that fluorine is an exception in the Lassaigne test. They typically mention that the Lassaigne test detects chlorine, bromine and iodine but not fluorine, and they often provide separate or more specialised procedures if fluorine detection is required. Cross checking with such references confirms that fluorine containing organic compounds do not yield the typical silver halide precipitates and are therefore not detected by ordinary Lassaigne fusion tests. This independent confirmation supports the reasoning used here and validates the choice of fluorine as the correct answer. Why Other Options Are Wrong: Iodine forms sodium iodide during fusion and gives a yellow precipitate of silver iodide with silver nitrate, so it is readily detected by the Lassaigne test and cannot be the correct choice. Chlorine forms sodium chloride, which produces a white precipitate of silver chloride with silver nitrate, so it is also detected reliably by this method and is therefore not the correct answer. Sulphur in organic compounds gives sodium sulphide on fusion, which can be detected by reagents such as lead acetate solution producing a characteristic black precipitate of lead sulphide, so sulphur is detectable and option C is incorrect. Common Pitfalls: Students sometimes assume that all halogens behave identically in qualitative tests and overlook the special case of fluorine, which has a very small ionic radius and forms highly hydrated, strongly solvated fluoride ions. Another common confusion is between the limitations of the Lassaigne test and other halogen detection techniques, leading to the assumption that if a halogen can be detected in general, it must also be detected by this specific test. Careful reading of laboratory procedures and awareness of exceptions like fluorine help avoid such mistakes in conceptual questions and practical examinations. Final Answer: The element that cannot be detected by the ordinary Lassaigne test in qualitative organic analysis is F (fluorine).

Question 6

In the context of structural isomerism in organic chemistry, for which of the following classes of compounds is functional group isomerism not possible?

Accepted Answer

Introduction / Context: This question focuses on the concept of functional group isomerism, an important type of structural isomerism in organic chemistry. Functional group isomers are compounds that have the same molecular formula but possess different functional groups, leading to different chemical properties. You are asked to identify the class of compounds among alcohols, aldehydes, alkyl halides and cyanides for which such functional group isomerism cannot occur. Understanding which functional groups can give rise to functional group isomers helps in predicting possible structures from a given molecular formula in many exam problems. Given Data / Assumptions: Alcohols can have functional group isomers that are ethers with the same molecular formula. Aldehydes can have functional group isomers that are ketones with the same molecular formula. Cyanides (nitriles) can have functional group isomers that are isocyanides (isonitriles) with the same molecular formula. Alkyl halides contain a carbon halogen bond but do not have a commonly associated alternative functional group with identical molecular formula that differs only by functional group type. We assume simple monofunctional compounds without additional functional groups. Concept / Approach: Functional group isomerism occurs when the same molecular formula can be written in such a way that the functional group changes, for example alcohol versus ether or aldehyde versus ketone. The key idea is that the molecular formula is the same, but the arrangement of atoms leads to a different functional group. For alcohols, an example is C2H6O, which can be written as ethanol (an alcohol) or as dimethyl ether (an ether). For aldehydes, C3H6O can represent propanal (an aldehyde) or propanone (a ketone). For cyanides, a molecular formula like C2H3N can represent acetonitrile or methyl isocyanide. In contrast, alkyl halides R–X, where X is a halogen, do not have a corresponding set of common functional group isomers with the same simple molecular formula and a different functional group classification. Step-by-Step Solution: Step 1: Consider alcohols. A classic example is C2H6O, which can exist as CH3CH2OH (ethanol, an alcohol) and CH3OCH3 (dimethyl ether, an ether). These are functional group isomers, so alcohols clearly show functional group isomerism. Step 2: Consider aldehydes. For instance, C3H6O can represent CH3CH2CHO (propanal, an aldehyde) or CH3COCH3 (propanone, a ketone). Aldehydes therefore exhibit functional group isomerism with ketones. Step 3: Consider cyanides. A formula like C2H3N can correspond to CH3CN (acetonitrile, a nitrile) or CH3NC (methyl isocyanide, an isonitrile). Thus cyanides show functional group isomerism with isocyanides. Step 4: Now consider alkyl halides, which have the general formula CnH2n+1X, where X is a halogen. There is no common distinct functional group that shares exactly the same simple molecular formula and differs only by a reorganisation that produces a different functional group category. Step 5: Alkyl halides can have chain isomers or position isomers, but they do not exhibit a pair of structures with different functional groups and identical molecular formula in the way that alcohols, aldehydes and cyanides do. Therefore functional group isomerism is not possible for alkyl halides. Verification / Alternative check: To verify, you can systematically write down molecular formulas for each class and attempt to draw another structure with a different functional group but the same formula. For alcohols, ethers serve as the corresponding functional group isomers. For aldehydes, the paired functional group is ketones. For cyanides, the alternative functional group is isocyanides. However, when you take an alkyl halide like C2H5Cl, there is no other stable organic compound with this exact molecular formula that would be placed into a different functional group category, such as an alcohol, ether or nitrile, without changing the overall formula. This confirms that alkyl halides lack functional group isomerism under standard classification schemes. Why Other Options Are Wrong: Alcohols are wrong as an answer because they clearly show functional group isomerism with ethers, for example ethanol and dimethyl ether for C2H6O. Aldehydes are wrong because they exhibit functional group isomerism with ketones, such as propanal and propanone for C3H6O. Cyanides are wrong because they show functional group isomerism with isocyanides, as seen in the nitrile and isonitrile forms of the same molecular formula. Common Pitfalls: Students often confuse different types of isomerism. Chain isomerism, positional isomerism and functional group isomerism may all occur separately, and it is important not to assume that every class of compounds can show every type of isomerism. Another common mistake is thinking that changing the position of the halogen in an alkyl halide is a type of functional group isomerism; in fact, this is positional isomerism, because the functional group carbon halogen bond remains the same. Keeping the definitions of each isomerism type clear helps to answer such questions correctly. Final Answer: Functional group isomerism is not possible for Alkyl halides, so that is the correct choice.

Question 7

With reference to different projection formulas used in stereochemistry (Fischer, Newman and Sawhorse projections), which one of the following statements is not correct?

Accepted Answer

Introduction / Context: This question belongs to the domain of organic stereochemistry and asks about different ways of representing three dimensional structures on a two dimensional plane. Fischer, Newman and Sawhorse projections are widely used projection formulas to visualise conformations and configurations, especially for molecules like alkanes, substituted ethanes and carbohydrates. The task here is to find the statement that is not correct among the given options by recalling how each type of projection relates to molecular conformation and stability. Given Data / Assumptions: Fischer projections are commonly used for carbohydrates and amino acids to represent stereochemistry in a planar form. Newman projections show the view along a particular carbon carbon bond and can depict eclipsed, staggered or skew conformations. Sawhorse projections also show a carbon carbon bond but from an oblique angle, making the relative orientation of substituents clearer. More stable conformations usually correspond to staggered arrangements, while eclipsed conformations are generally less stable due to torsional strain. Concept / Approach: The key idea is to understand what each projection represents. Fischer projections are essentially a convention to show stereochemical relationships and usually correspond to an eclipsed representation when converted to a simple carbon chain. They are convenient but do not necessarily depict the most stable conformation. Newman projections explicitly show the relationship between front and back substituents along a sigma bond and can be drawn in different conformational forms, such as staggered or eclipsed. Sawhorse projections offer a slightly tilted view of the same bond, allowing us to visualise torsional angles and interpret relative positions. The incorrect statement must contradict these standard ideas, particularly concerning whether a Fischer projection represents a most stable conformation. Step-by-Step Solution: Step 1: Evaluate the statement that a Fischer projection represents the molecule in an eclipsed conformation. When simple carbon chains are shown in Fischer projection, the substituents on adjacent carbons are implied to be eclipsed, so this statement is essentially correct. Step 2: Consider the statement that a Newman projection can be represented in eclipsed, staggered and skew conformations. This is true because Newman projections are specifically used to analyse conformations and can show all possible torsional arrangements. Step 3: Examine the statement that a Fischer projection of the molecule is its most stable conformation. Since the Fischer representation corresponds more closely to an eclipsed arrangement, and eclipsed conformations are not the most stable due to torsional strain, this statement is not correct. Step 4: Finally, check the statement about Sawhorse projections. In a Sawhorse projection, the carbon carbon bond is drawn diagonally and the substituent bonds are drawn at approximately 120 degrees to reflect the tetrahedral angle, so this statement is acceptable. Step 5: Based on this analysis, the only incorrect statement is that the Fischer projection corresponds to the most stable conformation of the molecule. Verification / Alternative check: If you take ethane as a simple example, its most stable conformation is the staggered form, which is best shown by a Newman projection. However, drawing ethane in a Fischer style projection would imply that the hydrogen atoms on adjacent carbons are aligned one behind the other, that is an eclipsed arrangement, which is higher in energy. Textbooks often highlight that Fischer projections are a stereochemical shorthand rather than true conformational pictures. This confirms that interpreting the Fischer projection as the most stable conformation is conceptually incorrect, reinforcing our choice of the wrong statement. Why Other Options Are Wrong: Option A is not the incorrect statement because Fischer projections do indeed treat adjacent substituents as if they were in an eclipsed arrangement when converted to simple conformational drawings. Option B is correct because Newman projections are specifically designed to show different conformations and can be drawn for eclipsed, staggered or intermediate skew conformations as the dihedral angle changes. Option D is also correct because Sawhorse projections usually depict the carbon carbon bond at an angle and the substituent bonds roughly at the tetrahedral angle of about 109.5 degrees, which is commonly idealised as 120 degrees in the drawing for clarity. Common Pitfalls: Students often confuse stereochemical representation with conformational stability. A projection that is easy to draw, such as a Fischer formula, is not automatically the lowest energy conformation. Another typical mistake is to assume that all projection formulas show the same conformation, which is not true because Newman and Sawhorse projections are primarily used for conformational analysis, whereas Fischer projections emphasise configuration at stereocentres. Carefully separating these concepts helps avoid misinterpretation in questions about projections and conformations. Final Answer: The statement that is not correct is the claim that Fischer projection of the molecule is its most stable conformation.

Question 8

Which one of the following atmospheric gases is the most important for absorbing heat radiation coming from the Sun as well as reradiated from the Earth surface, thereby contributing significantly to the greenhouse effect?

Accepted Answer

Introduction / Context: This question lies at the interface of chemistry, environmental science and general knowledge. It asks which gas in the atmosphere plays a particularly important role in absorbing heat radiation from the Sun and from the Earth surface, a key part of the greenhouse effect. Understanding which gases are greenhouse gases and how they interact with infrared radiation is essential for appreciating global warming, climate change and related environmental issues that appear frequently in competitive examinations and public discussions. Given Data / Assumptions: The options include carbon dioxide, oxygen, carbon monoxide and nitrogen, all of which are common atmospheric gases in varying amounts. We assume typical atmospheric composition where nitrogen and oxygen are the major components by volume, while carbon dioxide is present in smaller but significant amounts. We focus on the ability of these gases to absorb infrared radiation and contribute to greenhouse warming. Water vapour is an important greenhouse gas but is not listed among the options, so it is not considered here. Concept / Approach: Greenhouse gases are those that can absorb and emit infrared radiation due to vibrational transitions in their molecules. Molecules with more than two atoms or with a significant dipole moment often have vibrational modes that interact strongly with infrared light. Carbon dioxide has bending and stretching vibrations that fall in the infrared region relevant for heat radiation emitted by the Earth. In contrast, diatomic molecules like nitrogen and oxygen, which are symmetric and non polar, have much weaker interactions with infrared radiation at ground state conditions. Carbon monoxide is a trace gas and, while it can absorb some infrared radiation, its concentration and impact are much less significant than those of carbon dioxide in the context of the general greenhouse effect. Step-by-Step Solution: Step 1: Recall the main greenhouse gases in the atmosphere. These include water vapour, carbon dioxide, methane, nitrous oxide and ozone, with carbon dioxide being one of the most discussed due to human emissions. Step 2: Compare the molecular structures. Carbon dioxide is a linear triatomic molecule with vibrational modes that can interact with infrared radiation. This makes it efficient at absorbing and reemitting heat energy. Step 3: Consider oxygen and nitrogen, which are diatomic molecules (O2 and N2). These have no permanent dipole moment and only limited interaction with infrared radiation at the relevant wavelengths, so they are not major greenhouse gases. Step 4: Carbon monoxide, although it is a toxic gas and weak greenhouse gas, is present in much lower concentrations than carbon dioxide and does not dominate the greenhouse effect compared to CO2. Step 5: Therefore, among the given options, carbon dioxide is clearly the most important gas for absorption of heat radiated from the Sun and reradiated from the Earth, and it is the principal greenhouse gas listed. Verification / Alternative check: If you consult standard environmental science or atmospheric chemistry texts, they describe carbon dioxide as one of the primary greenhouse gases responsible for trapping heat in the atmosphere. Historical climate data and modern climate models show a strong correlation between rising atmospheric CO2 levels and increases in global average temperature. Nitrogen and oxygen, despite being abundant, are not classified as major greenhouse gases because of their limited infrared absorption. Carbon monoxide is recognised for its role in urban air pollution and smog formation rather than for a dominant greenhouse effect. These independent sources confirm the central role of carbon dioxide in absorbing heat radiation. Why Other Options Are Wrong: Oxygen is incorrect because O2 is a homonuclear diatomic molecule with very limited infrared activity, and it does not significantly contribute to the greenhouse effect under normal atmospheric conditions. Carbon monoxide is incorrect because, although it absorbs some infrared radiation, it is present in much smaller concentrations and is more important as a toxic pollutant than as a major greenhouse gas compared with carbon dioxide. Nitrogen is incorrect because N2, like O2, is a homonuclear diatomic molecule that interacts only weakly with infrared radiation and therefore does not play a leading role in greenhouse warming. Common Pitfalls: Students sometimes assume that because nitrogen and oxygen are the most abundant atmospheric gases, they must also be the most important for every atmospheric process, including heat absorption. This is not true for the greenhouse effect, which depends on the molecular ability to absorb infrared radiation rather than on sheer abundance alone. It is also easy to confuse carbon monoxide with carbon dioxide because of similar names, but their environmental roles are different. Remembering that greenhouse behaviour depends on molecular vibrations and dipole changes helps to pick the correct gas in such questions. Final Answer: The gas that most importantly absorbs heat radiated from the Sun and from the Earth among the given options is Carbon dioxide.

Question 9

In nuclear chemistry, what are isobars with respect to the atomic number and mass number of elements or nuclides?

Accepted Answer

Introduction / Context: This question belongs to the area of nuclear chemistry and atomic structure, where terms such as isotopes, isobars, isotones and isoelectronic species are used to describe relationships between nuclides. The term isobar is often tested in examinations because it reflects how elements can share the same mass number while having different atomic numbers and therefore belong to different chemical elements. Understanding these definitions helps in interpreting nuclear reactions, decay schemes and stability patterns in the periodic table. Given Data / Assumptions: Atomic number is the number of protons in the nucleus and defines the chemical identity of an element. Mass number is the total number of protons and neutrons in the nucleus of a given nuclide. Examples of nuclides often used include 40Ar, 40Ca and 40K, which can be compared by atomic number and mass number. We assume standard definitions used in introductory chemistry and physics texts. Concept / Approach: The key concept is to recall and differentiate the various related terms. Isotopes are nuclides with the same atomic number but different mass numbers. Isobars are nuclides with different atomic numbers but the same mass number. Isotones share the same number of neutrons but have different atomic numbers. Isoelectronic species have the same number of electrons. The question specifically asks about isobars, so we must focus on the relationship involving the mass number being the same while the atomic number differs, meaning the elements are chemically distinct but have nuclei with the same total number of nucleons. Step-by-Step Solution: Step 1: Recall that atomic number (Z) is the number of protons, and mass number (A) is the total of protons plus neutrons. Step 2: For isotopes, Z is the same but A is different, for example 1H, 2H and 3H for hydrogen. This does not match the definition of isobars. Step 3: For isobars, the mass number A is the same for two or more nuclides, but the atomic number Z is different. A classic example is 40Ar (Z = 18), 40Ca (Z = 20) and 40K (Z = 19). Each has mass number 40 but different atomic numbers. Step 4: Thus, the correct description of isobars is that they are elements or nuclides with different atomic numbers but the same mass number. Step 5: Compare this with the options given to find the statement that exactly matches this definition. Verification / Alternative check: To verify, you can use any pair of well known isobars from nuclear data tables. For example, 14C (atomic number 6, mass number 14) and 14N (atomic number 7, mass number 14) are isobars. They clearly have different atomic numbers, they belong to different chemical elements (carbon and nitrogen), but they share the same mass number 14. This is consistent with the standard definition of isobars presented in most textbooks. None of the other definitions, such as isotopes, fits this pair, confirming that isobars must have the same mass number and different atomic numbers. Why Other Options Are Wrong: Elements with the same atomic number but different mass number describe isotopes, not isobars, so option A is incorrect. Elements with different atomic number and different mass number do not define any special relationship and therefore do not correspond to isobars, making option C wrong. Elements with the same atomic number and the same mass number represent identical nuclides and are not given any special relational name in this context, so option D is also incorrect. Common Pitfalls: Students often mix up isotopes and isobars because both terms involve comparisons of atomic and mass numbers. A useful memory aid is that iso in both terms means same, but the second part of the word tells what is the same: in isotopes, tope relates to place in the periodic table, so atomic number is the same; in isobars, bar suggests weight or mass, so mass number is the same. Confusing mass number and atomic number or forgetting which quantity defines the element can also lead to incorrect answers. Keeping examples like 14C and 14N in mind helps to avoid this confusion. Final Answer: Isobars are defined as elements with different atomic number but the same mass number.

Question 10

Haematite, Pitchblende and Monazite are important ores; which one of the following ore–metal pairs listed is incorrect?

Accepted Answer

Introduction / Context: This question checks your general knowledge of common ores and the metals extracted from them, a topic that appears frequently in chemistry and general science examinations. Haematite, Pitchblende and Monazite are well known ore minerals. You are asked to identify which ore metal pair is incorrect among the given statements. Recognising the correct association between ore names and their constituent metals is essential for both mineralogy and metallurgical chemistry questions. Given Data / Assumptions: Haematite is an important ore of iron and is commonly written as Fe2O3. Pitchblende is a uranium ore and is often represented as U3O8 in simplified form. Monazite is a phosphate mineral that is an important ore of thorium and rare earth elements. The option list considers whether each of the pairings Haematite : Iron, Pitchblende : Copper and Monazite : Thorium is correct or not. We assume standard textbook information regarding these ores. Concept / Approach: To solve this question, you need to recall the correct metal associated with each ore name. Haematite is a well known iron ore used widely in iron and steel industries. Pitchblende, also called uraninite, is an ore of uranium and is significant as a source of nuclear fuel. Monazite sand, found in beach sands, is a major source of thorium and some rare earth metals. By checking whether the ore name is matched with the right metal, we can identify which pairing is wrong. Copper is not obtained from Pitchblende, so any statement that pairs Pitchblende with copper is incorrect. Step-by-Step Solution: Step 1: Confirm Haematite. Haematite is iron oxide, primarily Fe2O3, and it is a principal ore of iron used in blast furnace extraction. The pairing Haematite : Iron is therefore correct. Step 2: Check Pitchblende. Pitchblende is another name for uraninite, which is an oxide of uranium. It is mined as a major source of uranium, not copper. Thus Pitchblende : Copper is an incorrect pairing. Step 3: Check Monazite. Monazite is a phosphate mineral containing rare earth elements and thorium. It is well known as an important ore of thorium. Therefore Monazite : Thorium is a correct pairing. Step 4: Since only the Pitchblende : Copper pair is wrong, the correct option must state that only this pairing is incorrect. Step 5: Among the given options, the one that explicitly identifies only Pitchblende : Copper as incorrect is the correct answer. Verification / Alternative check: Standard textbooks and reliable reference sources list Haematite, Magnetite and Siderite as common iron ores. They list Pitchblende or uraninite as uranium ores and mention Monazite sand as an important thorium and rare earth metal source. No authoritative source pairs Pitchblende with copper, since principal copper ores include minerals like chalcopyrite, malachite and cuprite. This independent confirmation from basic mineralogy supports the conclusion that the only incorrect ore metal pair is Pitchblende : Copper. Why Other Options Are Wrong: The option suggesting that Haematite : Iron is incorrect is wrong because Haematite is indeed a primary ore of iron and this pairing is factually correct. Any option that claims two pairings are incorrect, such as including Monazite : Thorium among the wrong pairs, is incorrect because Monazite is genuinely used as an ore for thorium and certain rare earth metals. Options that mark more than one pair as incorrect contradict established textbook data and therefore cannot be accepted as correct answers. Common Pitfalls: A common pitfall is mixing up ore names because many minerals have traditional or less intuitive names. Students sometimes confuse Pitchblende with copper ores such as chalcopyrite due to unfamiliarity with mineral terminology. Another error is forgetting that a single ore can contain more than one metal, leading to the assumption that any metal might be associated with any complex mineral. The best way to avoid these mistakes is to memorise a small list of the most important ores and their corresponding metals, especially those that are frequently tested in examinations. Final Answer: The only incorrect ore metal pair is Pitchblende : Copper, so the option stating this alone is the correct choice.

Question 11

In polymer chemistry, the synthetic fibre known as Nylon 6 is prepared from which monomer or monomers?

Accepted Answer

Introduction / Context: This question falls under the topic of synthetic polymers and man made fibres. It asks specifically about Nylon 6, which is a widely used polyamide fibre. Knowing which monomer or monomers are used to manufacture different nylons is a standard part of polymer chemistry in school and entrance examinations. Nylon 6 is structurally different from Nylon 6,6, and this difference arises from the nature of the monomer units used in their polymerisation, so identifying the correct monomer is essential to answer the question. Given Data / Assumptions: Nylon 6,6 is a polyamide formed from a diamine and a dicarboxylic acid, typically hexamethylenediamine and adipic acid. Nylon 6 is formed from a single monomer containing six carbon atoms, which undergoes ring opening polymerisation. Urea and formaldehyde are used to produce a different type of polymer, urea formaldehyde resin, which is a thermosetting plastic. Phenol and formaldehyde are used to make phenolic resins such as Bakelite, also a thermosetting polymer. We assume standard industrial processes for the production of these polymers. Concept / Approach: To approach this question, you need to recall the specific monomer structures. Nylon 6,6 is made by condensation polymerisation between hexamethylenediamine and adipic acid, leading to a repeating unit with six carbon atoms on each side of the amide linkage. Nylon 6, however, is made from caprolactam, a cyclic amide with six carbon atoms in the ring. Caprolactam undergoes ring opening polymerisation to form a linear polyamide chain with repeating units derived from aminohexanoic acid. By contrasting these known facts with the options given, we can conclude which monomer corresponds specifically to Nylon 6. Step-by-Step Solution: Step 1: Recall that Nylon 6,6 is produced by condensation of hexamethylenediamine with adipic acid. Therefore, the pair hexamethylenediamine and adipic acid is associated with Nylon 6,6, not Nylon 6. Step 2: Consider caprolactam. Caprolactam is a six membered cyclic lactam that can open up under polymerisation conditions to give a linear chain with amide linkages, forming Nylon 6. Step 3: Urea and formaldehyde react to form urea formaldehyde resin, which is a thermosetting polymer used in adhesives and moulded objects, not a nylon. Step 4: Phenol and formaldehyde react to form phenolic resins such as Bakelite, again a thermosetting polymer and not a polyamide like Nylon 6. Step 5: From these considerations, Nylon 6 is clearly associated with caprolactam as its monomer, so caprolactam is the correct answer. Verification / Alternative check: Industrial chemistry references describe Nylon 6 as polycaprolactam, emphasising that it is obtained by polymerising caprolactam. They also show reaction schemes where caprolactam undergoes ring opening to yield the amide linked polymer chain. Nylon 6,6 is uniquely noted as being derived from hexamethylenediamine and adipic acid. In contrast, urea formaldehyde and phenol formaldehyde resins are listed as separate categories of thermosetting plastics. These independent points confirm that caprolactam is the monomer used to synthesise Nylon 6, validating our choice. Why Other Options Are Wrong: Hexamethylenediamine and adipic acid are incorrectly chosen if linked to Nylon 6, because they are used to manufacture Nylon 6,6, not Nylon 6, so option A is wrong in this context. Urea and formaldehyde form urea formaldehyde resin, which is a thermosetting plastic, not a linear polyamide fibre; therefore option C does not correspond to Nylon 6. Phenol and formaldehyde form phenolic resins like Bakelite and are not used for making nylon type polyamides, so option D is also incorrect for this question. Common Pitfalls: A common confusion arises between Nylon 6 and Nylon 6,6 because both names contain the number six, and students may wrongly assume they come from the same pair of six carbon units. Another pitfall is to assume that any mention of formaldehyde indicates a type of plastic, which is true, but not necessarily a nylon. Remembering that Nylon 6 comes from a single monomer caprolactam, while Nylon 6,6 comes from two monomers, a diamine and a diacid, helps to keep these fibres clearly distinguished in memory. Final Answer: Nylon 6 is manufactured by ring opening polymerisation of the monomer Caprolactam.

Question 12

Within a eukaryotic cell, which one of the following organelles mainly functions as a storehouse of digestive enzymes responsible for intracellular digestion?

Accepted Answer

Introduction / Context: Although this question appears under chemistry, it actually touches on cell biology, which is often included in general science sections of competitive exams. The question asks which cell organelle functions as a storehouse of digestive enzymes and is responsible for intracellular digestion. Understanding the roles of lysosomes, ribosomes, desmosomes and vacuoles is essential in basic biology and helps build a picture of how cells maintain and recycle their own components. Given Data / Assumptions: Desmosomes are specialised structures involved in cell adhesion, particularly in epithelial tissues. Ribosomes are small ribonucleoprotein particles involved in protein synthesis in the cytoplasm or on the rough endoplasmic reticulum. Lysosomes are membrane bound vesicles that contain a variety of hydrolytic enzymes active at acidic pH. Vacuoles are membrane bound compartments mainly involved in storage and turgor maintenance, especially in plant cells. We assume a standard eukaryotic cell with these organelles present and performing their textbook functions. Concept / Approach: The key concept is that lysosomes are often described as the digestive system or waste disposal system of the cell. They contain hydrolytic enzymes capable of breaking down proteins, nucleic acids, lipids and carbohydrates. These enzymes function best at acidic pH, which is maintained inside the lysosome. When cellular components are damaged or obsolete, or when foreign materials are taken up by endocytosis or phagocytosis, lysosomes fuse with the vesicles containing the material and digest it. None of the other organelles listed act as a general storehouse for digestive enzymes in the cytoplasm. Step-by-Step Solution: Step 1: Consider desmosomes. They are intercellular junctions that provide mechanical strength by holding adjacent cells together, especially in tissues subject to stress. They do not contain digestive enzymes for intracellular digestion. Step 2: Consider ribosomes. Ribosomes are the sites of protein synthesis, where messenger RNA is translated into polypeptide chains. They are not storage sites for digestive enzymes. Step 3: Consider lysosomes. Lysosomes are membrane bound organelles that house many hydrolytic enzymes capable of digesting a variety of biomolecules. They are directly responsible for intracellular digestion and recycling of cellular components. Step 4: Consider vacuoles. Vacuoles in plant cells store water, ions and sometimes pigments or waste products. In animal cells, small vacuoles may be involved in storage or transport but are not the main library of hydrolytic enzymes. Step 5: From this comparison, lysosomes clearly fit the description of an organelle that acts as a storehouse of digestive enzymes responsible for intracellular digestion. Verification / Alternative check: Standard biology textbooks describe lysosomes as the suicidal bags or digestive bags of the cell because they contain powerful enzymes that can digest cellular components when released. Experimental evidence shows that if lysosomal membranes are damaged, the enzymes can digest parts of the cell. In contrast, desmosomes are visible under the microscope as thickened regions of the plasma membrane between cells, ribosomes are visible as small granules involved in protein synthesis, and vacuoles are large storage compartments especially in plant cells. These observations reinforce that only lysosomes match the functional description given in the question. Why Other Options Are Wrong: Desmosomes are wrong because they serve as anchoring junctions between cells and do not store or release digestive enzymes. Ribosomes are wrong because they are involved in protein synthesis, not in digestion of biomolecules, and they do not act as enzyme storage vesicles. Vacuoles are wrong because their primary role is storage of substances like water or ions and maintaining turgor pressure, rather than housing a full complement of hydrolytic enzymes for intracellular digestion. Common Pitfalls: Students may confuse vacuoles and lysosomes because both are membrane bound vesicles. In some animal cells, lysosomal functions and vacuolar functions can overlap, but the classical distinction in basic biology is that lysosomes are specifically enriched with digestive enzymes. Another pitfall is to assume that any organelle involved in protein processing must hold digestive enzymes, whereas only lysosomes are specialised for degradation. Keeping clear mental pictures of the functions of each organelle helps avoid such mistakes in multiple choice questions. Final Answer: The cell organelle that mainly functions as a storehouse of digestive enzymes is the Lysosome.

Question 13

The molecule PCl5 has a trigonal bipyramidal structure; what is the hybridisation state of the central phosphorus atom in this molecule?

Accepted Answer

Introduction / Context: This question deals with valence bond theory and hybridisation, concepts used to explain the shapes of molecules. Phosphorus pentachloride, PCl5, is a classic example of a molecule with trigonal bipyramidal geometry. The question asks for the hybridisation of the central phosphorus atom that leads to this specific shape. Understanding how different hybrid orbitals correspond to geometries such as tetrahedral, trigonal bipyramidal and octahedral is a fundamental skill in chemical bonding and molecular structure topics. Given Data / Assumptions: PCl5 has five P–Cl sigma bonds and no lone pairs on the central phosphorus atom in the valence shell model. The molecular geometry of PCl5 is trigonal bipyramidal, which means three chlorine atoms occupy equatorial positions and two occupy axial positions. Hybridisation schemes such as sp2, sp3, dsp2 and dsp3 are common ways of describing the mixing of atomic orbitals to form equivalent hybrid orbitals. We assume standard valence shell electron pair repulsion and valence bond descriptions without invoking advanced molecular orbital theory. Concept / Approach: In valence bond theory, the hybridisation state is chosen to match the number of electron pairs and the observed geometry around the central atom. A trigonal bipyramidal geometry requires five electron domains arranged to minimise repulsions, corresponding to five hybrid orbitals. The notation traditionally used for this is dsp3, indicating that one d orbital, one s orbital and three p orbitals combine to form five equivalent hybrid orbitals. These hybrids are directed toward the five corners of a trigonal bipyramid. Other hybridisation schemes such as sp3 (four hybrids) or sp2 (three hybrids) cannot accommodate five equivalent bonding directions. Step-by-Step Solution: Step 1: Count the sigma bonds and lone pairs on phosphorus in PCl5. Phosphorus forms five sigma bonds with five chlorine atoms and has no lone pairs in the idealised structure. Step 2: This means phosphorus needs five equivalent hybrid orbitals to form these five sigma bonds. Step 3: Consider possible hybridisations. sp2 hybridisation gives three hybrid orbitals, suitable for trigonal planar geometry. sp3 hybridisation gives four hybrid orbitals, suitable for tetrahedral geometry. Step 4: To obtain five hybrid orbitals, one must combine one s orbital, three p orbitals and one d orbital, producing a dsp3 or sp3d set, which corresponds to trigonal bipyramidal geometry. Step 5: Therefore the correct hybridisation description for the central phosphorus atom in PCl5 within this valence bond framework is dsp3. Verification / Alternative check: An alternative way to check this is to match known hybridisation geometries: sp corresponds to linear (2 directions), sp2 to trigonal planar (3 directions), sp3 to tetrahedral (4 directions), dsp3 to trigonal bipyramidal (5 directions) and d2sp3 to octahedral (6 directions). Since PCl5 is experimentally known to have a trigonal bipyramidal shape and five bonded chlorine atoms, its central atom must use a hybridisation that generates five bonding directions, which is dsp3 in the traditional scheme. This pattern appears consistently in textbooks, confirming that dsp3 is the standard answer for PCl5. Why Other Options Are Wrong: sp2 is incorrect because it provides only three hybrid orbitals and produces a trigonal planar geometry, which cannot account for five P–Cl bonds in PCl5. sp3 is wrong because it yields four hybrid orbitals and gives a tetrahedral geometry, again insufficient for bonding to five chlorine atoms. dsp2 is associated with square planar geometry and four coordination sites, so it also cannot describe the trigonal bipyramidal structure of PCl5, making option C incorrect. Common Pitfalls: Students sometimes confuse dsp3 with d2sp3 or assume that any involvement of d orbitals automatically leads to octahedral geometry. Others incorrectly try to fit PCl5 into an sp3 framework because they remember that four electron pairs correspond to sp3 hybridisation but forget that five pairs require a different arrangement. Another common slip is ignoring the actual shape of the molecule and focusing only on the total number of bonds. Remembering the mapping between electron pair number, hybridisation and geometry is essential to avoid these mistakes. Final Answer: For PCl5 with trigonal bipyramidal geometry, the central phosphorus atom is described as having dsp3 hybridisation.

Question 14

Which one of the following statements about histone proteins in eukaryotic cells is correct?

Accepted Answer

Introduction / Context: This question relates to molecular biology and chromatin structure. Histone proteins play a crucial role in packaging DNA within the eukaryotic cell nucleus and regulating gene expression. The question asks you to identify the correct statement about where histones are located and with what they are associated. A basic understanding of chromatin organisation, nucleosomes and the nature of histone DNA interactions is essential to answer this correctly. Given Data / Assumptions: DNA in eukaryotic cells is not naked but is wrapped around proteins to form chromatin. Histones are small, positively charged proteins that interact with negatively charged DNA. Mitochondria have their own DNA, which is generally not packaged with histones in the same way as nuclear DNA. The cytosol contains many proteins, lipids and carbohydrates, but histones are mainly associated with nuclear chromatin. Concept / Approach: The core concept is that in eukaryotic cells, nuclear DNA is organised into nucleosomes, where about 146 base pairs of DNA are wrapped around a histone octamer made of two copies each of H2A, H2B, H3 and H4. Additional histone H1 binds to linker DNA, helping to compact chromatin into higher order structures. These histones are therefore nuclear proteins specifically associated with DNA. They are not membrane proteins, nor are they primarily associated with lipids or carbohydrates in the cytosol. Recognising this role allows us to select the correct statement from the options. Step-by-Step Solution: Step 1: Consider the statement that histones are present in the mitochondrial membrane. Mitochondrial membranes contain proteins involved in electron transport and ATP synthesis, but histones are typically not located there. Step 2: Examine the statement that histones are proteins present in the nucleus in association with DNA. This matches the known role of histones as major components of chromatin in the nucleus. Step 3: Look at the statement that histones are associated with lipids in the cytosol. While some cytosolic proteins associate with lipids, histones do not function this way; they are strongly nuclear. Step 4: Evaluate the statement that histones are associated with carbohydrates in the cytosol. Again, histones are not known for binding carbohydrates in the cytosol as a main function. Step 5: Therefore, the only statement that correctly describes histones is that they are present in the nucleus in association with DNA. Verification / Alternative check: Chromatin structure studies reveal that eukaryotic chromosomes consist of DNA wrapped around histone proteins. Electron micrographs show beads on a string structures, representing nucleosomes composed of histone cores with DNA wrapped around them. Biochemical analysis confirms that histones are abundant nuclear proteins rich in basic amino acids like lysine and arginine, which facilitate their binding to DNA. Mitochondrial DNA, by contrast, is usually associated with different proteins and is less tightly packed. These observations from multiple experimental approaches confirm that histones are nuclear DNA associated proteins. Why Other Options Are Wrong: The option that places histones in the mitochondrial membrane is incorrect because mitochondria have their own non histone associated DNA and their membranes are rich in respiratory chain proteins rather than histones. The option suggesting histones are associated with lipids in the cytosol is wrong because histones are not membrane or lipid binding proteins; they function mainly in the nucleus with DNA. The option that pairs histones with carbohydrates in the cytosol is also incorrect, as histones are not carbohydrate binding structural proteins but DNA associated chromatin proteins. Common Pitfalls: Students sometimes generalise that any abundant cellular protein might be involved in various locations or functions, leading to confusion about where histones reside. Another pitfall is mixing up different DNA binding proteins, such as transcription factors, histones and mitochondrial DNA binding proteins, and assuming they are interchangeable. Keeping a clear picture that histones are specifically nuclear, associated with chromatin, and not general cytosolic proteins helps to avoid these errors. Final Answer: The correct statement is that Histones are proteins that are present in the nucleus in association with DNA.

Question 15

In redox (oxidation–reduction) chemistry, the process of gaining electrons is known as what?

Accepted Answer

Introduction / Context: This question targets a very basic but fundamental idea in redox chemistry. Oxidation and reduction can be defined in several ways, including gain or loss of oxygen, gain or loss of hydrogen, and most generally loss or gain of electrons. You are asked to identify the correct term for the process in which a species gains electrons. This concept is essential in understanding electrochemistry, corrosion, metabolism and many industrial chemical processes. Given Data / Assumptions: Oxidation is commonly associated with loss of electrons, gain of oxygen or loss of hydrogen. Reduction is commonly associated with gain of electrons, loss of oxygen or gain of hydrogen. Radiation in this context refers to emission of energy as electromagnetic waves or particles and is not a synonym for an electron transfer process. The options include a combination that suggests both oxidation and reduction, which must be evaluated carefully. Concept / Approach: The most general and robust definition of oxidation and reduction is in terms of electron transfer. Oxidation is defined as the loss of electrons by a species, increasing its oxidation number. Reduction is defined as the gain of electrons, decreasing its oxidation number. These definitions apply consistently across a wide variety of reactions, including those where no oxygen or hydrogen is involved. Therefore, when the question simply asks about the process of gaining electrons, it is pointing directly at the definition of reduction. Step-by-Step Solution: Step 1: Recall the mnemonic that oxidation is loss of electrons and reduction is gain of electrons. Many students remember this as the phrase oil rig, meaning oxidation is loss, reduction is gain. Step 2: Check the option oxidation. By definition, oxidation involves the loss of electrons, not the gain, so this cannot be correct. Step 3: Check the option reduction. Reduction is defined as the gain of electrons by an atom, ion or molecule, which matches the process described in the question. Step 4: Evaluate the option radiation. Radiation refers to emission of energy and is not used as a term for electron gain in standard redox terminology. Step 5: The option both oxidation and reduction is incorrect because oxidation and reduction are opposite processes in terms of electron transfer, although they always occur together in a redox reaction. The gain of electrons alone cannot be called both at the same time. Verification / Alternative check: Consider the half reaction Cu2+ + 2e minus gives Cu. In this process, cupric ion Cu2 plus gains two electrons to become metallic copper Cu. The oxidation number decreases from +2 to 0, and this half reaction is labelled a reduction half reaction. Similarly, in the reaction Fe3 plus + e minus gives Fe2 plus, the iron ion gains an electron and its oxidation number decreases, which is again defined as reduction. These standard examples confirm that gaining electrons corresponds to reduction, in agreement with the general definition used in electrochemistry and redox theory. Why Other Options Are Wrong: Oxidation is wrong because it refers to loss of electrons, often accompanied by an increase in oxidation number, the opposite of what is described in the question. Radiation is wrong because it does not describe an electron transfer process; instead, it refers to energy emitted as waves or particles, such as in radioactive decay or electromagnetic radiation. The option both oxidation and reduction is wrong because while oxidation and reduction always occur together in a full redox reaction, the specific act of gaining electrons is defined only as reduction, not as both processes simultaneously. Common Pitfalls: A common pitfall is to rely only on the older definition of oxidation as combination with oxygen and reduction as removal of oxygen, which can create confusion in reactions that do not involve oxygen at all. Another error is to mix up the directions of electron flow and words like loss and gain. Using a clear mnemonic and practising with half reactions where electrons are explicitly written on one side of the equation helps to fix the definition that reduction is gain of electrons and oxidation is loss of electrons. Final Answer: The process of gaining electrons in a redox reaction is called Reduction.

Question 16

Besides carbon dioxide, which one of the following gases is also an important greenhouse gas contributing to the greenhouse effect in the Earth atmosphere?

Accepted Answer

Introduction / Context: This question concerns environmental chemistry and climate science. It asks you to identify another greenhouse gas besides carbon dioxide among the listed gases. Greenhouse gases absorb and emit infrared radiation, thereby trapping heat in the Earth atmosphere and contributing to global warming. Knowing which gases are greenhouse gases is important for understanding climate change and for answering general knowledge questions in competitive examinations. Given Data / Assumptions: Carbon dioxide is recognised as a major greenhouse gas, but the question explicitly says besides CO2 and then lists other gases. CH4 denotes methane, a simple hydrocarbon with one carbon and four hydrogen atoms. N2 denotes nitrogen gas, which makes up most of the atmosphere by volume. Ar stands for argon, a noble gas present in small but significant amounts in air. O2 denotes oxygen gas, the second most abundant gas in the atmosphere by volume. Concept / Approach: Greenhouse gases are characterised by their ability to absorb infrared radiation at wavelengths emitted by the Earth surface. Methane, nitrous oxide, ozone and water vapour are some key greenhouse gases in addition to carbon dioxide. Nitrogen and oxygen, despite being abundant, are symmetric diatomic molecules with limited interaction with infrared radiation in the relevant range, and argon is a noble gas that is monoatomic and essentially infrared inactive under normal conditions. Therefore, among the options, methane is the gas that significantly contributes to the greenhouse effect besides carbon dioxide. Step-by-Step Solution: Step 1: Recall the main greenhouse gases: water vapour, carbon dioxide, methane, nitrous oxide and ozone are commonly listed. Step 2: Identify methane among the options. Methane (CH4) is a greenhouse gas that is much more potent than carbon dioxide on a per molecule basis, although it is present in lower concentrations. Step 3: Consider nitrogen. N2 is a homonuclear diatomic molecule with very limited infrared absorption and is not classified as a major greenhouse gas. Step 4: Consider argon. Ar is a noble gas with a closed shell electronic configuration and does not have vibrational modes that would interact strongly with infrared radiation. Step 5: Consider oxygen. O2 is also homonuclear diatomic and similarly has very limited infrared activity, so it is not regarded as a key greenhouse gas compared with methane. Verification / Alternative check: Climate science resources and reports from bodies such as the Intergovernmental Panel on Climate Change list methane as a significant greenhouse gas with a high global warming potential over shorter time scales. They also specify that while nitrogen and oxygen dominate the atmosphere by volume, their contribution to the greenhouse effect is minimal. Argon, being inert and monoatomic, likewise does not contribute significantly to infrared absorption. This independent confirmation from authoritative climate sources supports the conclusion that methane is the correct answer. Why Other Options Are Wrong: Nitrogen is wrong because it is largely infrared inactive and does not significantly trap heat in the atmosphere despite its high concentration. Argon is wrong because as a noble gas it is chemically inert and does not have vibrational modes that interact with infrared radiation in the way greenhouse gases do. Oxygen is wrong because, like nitrogen, it is a homonuclear diatomic molecule and has limited involvement in infrared absorption relevant to the greenhouse effect. Common Pitfalls: Students sometimes assume that the most abundant gases must be the most important for every atmospheric process, including greenhouse warming. This leads to choosing nitrogen or oxygen incorrectly. Another pitfall is confusion between ozone and oxygen; ozone is a greenhouse gas and also protects against ultraviolet radiation, but it is not listed among the options. Remembering that methane is one of the key greenhouse gases and that abundance alone does not determine greenhouse behaviour helps to avoid these errors. Final Answer: Besides carbon dioxide, an important greenhouse gas in the Earth atmosphere is CH4, that is methane.

Question 17

When granite rocks undergo metamorphism under suitable temperature and pressure conditions, they are converted into which of the following metamorphic rocks?

Accepted Answer

Introduction / Context: This question comes from geology and earth science, which are often included under general science or general knowledge. It asks which metamorphic rock is formed when granite, an igneous rock, is subjected to metamorphism. Knowing the parent rock and its metamorphic product is useful in understanding rock cycles, tectonic processes and the classification of rocks. Granite is a coarse grained igneous rock, and metamorphism changes its texture and mineral alignment to produce a distinct metamorphic rock type. Given Data / Assumptions: Granite is an intrusive igneous rock composed mainly of quartz, feldspar and mica. Metamorphism involves changes in mineral structure and texture under heat and pressure without complete melting. Quartzite is a metamorphic rock derived from sandstone rich in quartz. Marble is a metamorphic rock derived from limestone or dolomite rock. Slate is a metamorphic rock derived mainly from shale or mudstone. Concept / Approach: The key concept is the rock cycle and parent rock metamorphic rock relationships. Granite subjected to regional metamorphism, involving high pressure and temperature, can develop foliation, where minerals such as mica align in bands. The resulting rock is called gneiss, which displays characteristic banding and segregation of light and dark minerals. Quartzite, marble and slate each have different sedimentary parent rocks, so they cannot be the direct products of metamorphism of granite. By matching the parent rock granite with the correct metamorphic product, we identify gneiss as the correct answer. Step-by-Step Solution: Step 1: Recall that granite is an igneous rock formed from slow cooling of magma and composed primarily of quartz, feldspar and mica. Step 2: Under regional metamorphic conditions, granite can be subjected to high pressures and temperatures that cause mineral grains to recrystallise and align, creating a banded texture. Step 3: The metamorphic rock that forms from granite in this way is called gneiss, which is distinguished by its coarse banded structure with alternating layers of light and dark minerals. Step 4: Quartzite forms from metamorphism of quartz rich sandstone, not from granite, so it cannot be the product of granitic metamorphism. Step 5: Marble forms from limestone or dolomite rock, and slate forms from shale, so neither is a metamorphosed form of granite, confirming that gneiss is the correct choice. Verification / Alternative check: Geology textbooks and rock classification charts show parent rock metamorphic rock pairs such as sandstone quartzite, limestone marble, shale slate and schist, and granite gneiss. Field studies and thin section analyses confirm that gneiss often has mineral assemblages similar to granite but exhibits strong foliation and banding due to metamorphic processes. In contrast, quartzite is nearly pure quartz, marble is primarily calcite or dolomite and slate has fine grained mica and clay minerals derived from shale. These independent observations confirm that granite metamorphoses into gneiss, not into the other listed rocks. Why Other Options Are Wrong: Quartzite is wrong because it is formed by metamorphism of quartz rich sandstone and does not originate from granitic parent rock. Marble is wrong because it forms from limestone or dolomite rocks under metamorphic conditions and contains calcite or dolomite rather than the feldspar rich composition of granite. Slate is wrong because it is derived from shale or mudstone and is fine grained, whereas granite is coarse grained; the textural and compositional mismatch makes slate an incorrect product of granite metamorphism. Common Pitfalls: Students may incorrectly match rocks based only on name familiarity or on the assumption that all metamorphic rocks can form from any parent. Another pitfall is confusing quartzite and gneiss because both can be hard and resistant, but their textures and origins are different. Remembering a simple mapping of parent to metamorphic rocks, such as granite gneiss and sandstone quartzite, helps to avoid such confusion. Visualising the banded nature of gneiss versus the massive granular nature of quartzite is another useful cue. Final Answer: When granite rocks are metamorphosed, they are converted into the metamorphic rock Gneiss.

Question 18

Which one of the following is considered a commercial source of energy commonly used in households and industries?

Accepted Answer

Introduction / Context: This question lies in the area of sources of energy and environmental science. It distinguishes between commercial and non commercial sources of energy. Commercial energy sources are those that are bought and sold in the market and form the backbone of modern energy supply systems. Non commercial sources are typically traditional or freely available sources often used in rural settings. You are asked to identify which option is a commercial source of energy from the list given. Given Data / Assumptions: Agricultural waste refers to crop residues and biomass left over from farming activities. Dried dung is animal dung used as fuel after drying, often in rural households. The Sun is the ultimate natural source of solar energy, but direct sunlight is not usually bought or sold as a fuel in the traditional sense. Natural gas is a fossil fuel extracted, processed and distributed through pipelines or in cylinders for domestic and industrial use. We assume standard textbook definitions of commercial and non commercial energy sources. Concept / Approach: Commercial energy sources are characterised by organised production, distribution and sale, and include fossil fuels such as coal, petroleum and natural gas, as well as electricity. Non commercial sources like firewood, agricultural waste and dried dung are often collected locally and used without entering formal markets. The Sun provides solar energy, which can be harnessed commercially through photovoltaic systems, but the direct resource itself is not typically treated as a purchased fuel. Natural gas, on the other hand, is clearly traded, billed and delivered as a commercial energy source, used for cooking, heating and power generation. Step-by-Step Solution: Step 1: Evaluate agricultural waste. It is often used as a traditional form of biomass fuel in rural areas and is usually not purchased on a large commercial scale; it is considered non commercial energy. Step 2: Evaluate dried dung. Like agricultural waste, dried dung is traditionally used in rural households and is generally not part of formal commercial energy markets, so it is also non commercial. Step 3: Evaluate the Sun. The Sun is the source of solar energy, but sunlight itself is not purchased as a fuel. While solar power installations are commercial, the direct energy source is not classified as a commercial fuel in the same sense as gas or coal. Step 4: Evaluate natural gas. Natural gas is extracted from wells, processed, and sold through companies to consumers and industries, making it a clear example of a commercial energy source. Step 5: Therefore, among the given options, natural gas is the correct answer as a commercial source of energy. Verification / Alternative check: Energy resource classifications in textbooks and economic reports typically list coal, petroleum products, natural gas and electricity as commercial energy sources because they are traded and priced in organised markets. In contrast, firewood, agricultural waste and animal dung are labelled as non commercial or traditional sources. Solar energy is recognised as a renewable commercial energy form when harnessed by solar panels and sold as electricity, but the direct Sun option in this question is not meant in that technological sense. These standard classifications confirm that natural gas is the commercial energy source among the choices given. Why Other Options Are Wrong: Agricultural waste is incorrect because it is generally collected and used locally without a formal commercial market, so it is considered a non commercial energy source. Dried dung is wrong because, like agricultural waste, it is mostly used in rural households without entering the organised energy market and is thus non commercial. The Sun is wrong in the context of this question because, although solar energy can be harnessed commercially, the Sun listed as a source here refers to the natural resource, not a traded fuel like natural gas. Common Pitfalls: Students may be tempted to choose the Sun because solar energy is increasingly important and modern. However, the question emphasises commercial sources of energy in the conventional sense of fuels and supply systems. Another pitfall is failing to distinguish between local traditional fuels and large scale commercial energy resources. Remembering that commercial energy is typically bought and sold through companies for domestic and industrial use helps to identify natural gas as the correct option. Final Answer: The commercial source of energy among the options given is Natural gas.

Question 19

Which one of the following chemical elements is used as a precise timekeeper in modern atomic clocks?

Accepted Answer

Introduction / Context: This question focuses on an application of atomic physics and physical chemistry in timekeeping technology. Atomic clocks are the most accurate clocks available and define the standard for the second in the International System of Units. The question asks which element is used as the timekeeper in these atomic clocks. Knowing that the frequency of radiation associated with specific transitions in certain atoms forms the basis of atomic timekeeping is an important piece of scientific general knowledge. Given Data / Assumptions: Atomic clocks rely on the frequency of transitions between specific energy levels of atoms or ions. The SI second is defined in terms of a particular transition frequency of a specific isotope of one element. The options include potassium, caesium, calcium and magnesium, all of which can have atomic transitions but only one is used for the official definition of the second. We assume standard modern definitions as used in metrology. Concept / Approach: The official definition of the second is based on the transition between two hyperfine levels in the ground state of the caesium 133 atom. The second is defined as a fixed number of periods of the radiation corresponding to this transition. Although other elements like rubidium and hydrogen are used in various types of atomic clocks, the primary standard and SI definition involves caesium. Therefore, among the listed options, caesium is the element used as the timekeeper in atomic clocks for defining the second. Step-by-Step Solution: Step 1: Recall that the SI second is defined by the caesium 133 standard, which uses a specific microwave transition in its ground state. Step 2: Check the options. Potassium, calcium and magnesium are not used in the official SI definition of the second. Step 3: Identify caesium among the options. Caesium is a soft, alkali metal whose isotope caesium 133 has a well defined hyperfine transition used in atomic clocks. Step 4: Atomic clocks use the frequency of this transition as a reference, making caesium the timekeeping element in these devices. Step 5: Therefore the correct answer is caesium. Verification / Alternative check: The formal definition of the second states that it is the duration of a specific number of periods (currently 9192631770 periods) of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom. This definition is widely cited in physics and metrology references. Other elements like rubidium are used for different atomic clock designs, but the SI definition explicitly names caesium. Since potassium, calcium and magnesium do not play this defining role, the independent official definition confirms that caesium is the correct element. Why Other Options Are Wrong: Potassium is wrong because, although it has atomic transitions, it is not the element used in the SI definition of the second or in the main caesium standard atomic clocks. Calcium is wrong because it is not the standard timekeeping element for defining the second, even though it may have useful transitions for other optical clock experiments. Magnesium is wrong because, like calcium, it is not the basis for the internationally agreed definition of the second in atomic clock technology. Common Pitfalls: Students may confuse caesium with other elements like rubidium, which are also used in some atomic clocks, or they may recall that optical clocks can use ions of elements such as strontium. However, the key point here is that the SI definition of the second is based on the caesium 133 atom. Remembering this official definition helps to distinguish caesium from other elements involved in precision timekeeping research. Final Answer: The element used as a precise timekeeper in modern atomic clocks is Caesium.

Question 20

Which one of the following gases dissolves in water to form an acidic solution due to the formation of an acid in aqueous medium?

Accepted Answer

Introduction / Context: This question belongs to the topic of acid base chemistry and environmental science. It asks which gas, when dissolved in water, produces an acidic solution. Gases can form acids or bases in water depending on their chemical behaviour. Understanding which gases cause acidification of water bodies is important for interpreting phenomena such as acid rain and ocean acidification, as well as for general knowledge about aqueous solutions. Given Data / Assumptions: Carbon dioxide can react with water to form carbonic acid, H2CO3, which is a weak acid. Oxygen, nitrogen and hydrogen are relatively inert diatomic gases under normal conditions in water and do not form strong acids or bases simply by dissolving. The question focuses on natural dissolution in water rather than special high energy or catalytic conditions. Concept / Approach: When carbon dioxide dissolves in water, it reacts to form carbonic acid according to the equilibrium CO2 + H2O reversible arrow H2CO3. Although carbonic acid is weak and only partially dissociates, its presence increases the hydrogen ion concentration and makes the solution slightly acidic. Oxygen dissolves in water mainly to support aquatic life but does not form an acid. Nitrogen and hydrogen also dissolve to a small extent but do not form acids or bases under normal environmental conditions. Therefore, the gas that clearly leads to an acidic solution is carbon dioxide. Step-by-Step Solution: Step 1: Consider carbon dioxide. When CO2 dissolves in water, it forms carbonic acid, which partially dissociates into hydrogen ions and bicarbonate ions, making the solution acidic. Step 2: Consider oxygen. O2 dissolves in water but remains as molecular oxygen and does not react significantly to form an acid or base in ordinary conditions, so the solution remains essentially neutral. Step 3: Consider nitrogen. N2 is even less reactive than oxygen under normal conditions and dissolves only slightly in water without forming an acid. Step 4: Consider hydrogen. H2 can dissolve in water but does not form an acid merely by dissolving; it requires specific conditions to react, such as in catalytic hydrogenation processes. Step 5: Therefore, among the options, only carbon dioxide clearly produces an acidic solution when dissolved in water due to carbonic acid formation. Verification / Alternative check: Environmental chemistry explains that ocean acidification and the slight acidity of natural rainwater are largely due to the dissolution of atmospheric carbon dioxide forming carbonic acid. Measured pH values of pure water in equilibrium with CO2 are slightly below 7, reflecting this weak acidity. In contrast, oxygen and nitrogen do not significantly affect the pH of water when they dissolve at normal atmospheric partial pressures. These observations confirm that carbon dioxide is the gas that creates an acidic solution. Why Other Options Are Wrong: Oxygen is wrong because it does not react substantially with water to form an acid under normal conditions and mainly serves as a dissolved gas for aquatic respiration. Nitrogen is wrong because it is chemically inert in this context and does not form acids or bases simply by dissolving in water. Hydrogen is wrong because, although reactive under certain controlled conditions, it does not form an acid with water under ordinary dissolution conditions and does not significantly change the pH. Common Pitfalls: Students may think that any gas that dissolves in water must change the pH, which is not true. Another pitfall is confusing the role of oxygen in oxidation reactions with acid formation. It is also possible to forget that carbon dioxide is acidic in water because it is often thought of as neutral in air. Remembering that CO2 forms carbonic acid in water and contributes to phenomena like ocean acidification helps maintain the correct association. Final Answer: The gas that dissolves in water to give an acidic solution is Carbon dioxide.

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