NEET 2020, on the exam screen.

The length of the DNA can be calculated using the formula $L = n \times d$, where n is the number of base pairs and d is the distance between two consecutive base pairs. Substituting the values, $L = 6.6 \times 10^{9} \times 0.34 \times 10^{-9}$ m = 2.244 m $\approx$ 2.2 m. Chapter: Molecular Basis of Inheritance.

Platyhelminthes are bilaterally symmetrical and acoelomate, lacking a body cavity. Chapter: Animal Kingdom.

Locusta is a gregarious polyphagous pest; adult Asterias has radial symmetry while its larva is bilateral; scorpions have book lungs; and Ctenoplana is bioluminescent. Thus P-IV, Q-I, R-II, S-III. Chapter: Animal Kingdom.

Golgi bodies are the primary site for the formation of glycoproteins and glycolipids in eukaryotic cells. Chapter: Cell: The Unit of Life.

The QRS complex in a standard electrocardiogram (ECG) represents the depolarisation of the ventricles, which initiates ventricular contraction. Chapter: Body Fluids and Circulation.

Floating ribs do not connect with the sternum; the acromion articulates with the clavicle; the scapula lies between the second and seventh ribs; and the glenoid cavity receives the humeral head. Thus P-IV, Q-III, R-I, S-II. Chapter: Locomotion and Movement.

Thomas Hunt Morgan provided experimental verification of the chromosomal theory of inheritance through his work with Drosophila melanogaster. Chapter: Principles of Inheritance and Variation.

Sapwood is not the innermost secondary xylem; it is the outer, living part of the secondary xylem that is involved in water and mineral conduction. Heartwood is the inner, non-living part that provides mechanical support and is darker in color due to deposits of tannins, resins, and oils. Chapter: Anatomy of Flowering Plants.

Pituitary/antidiuretic hormone (ADH) dysfunction causes diabetes insipidus, thyroid hyperactivity causes Graves' disease, adrenal hypofunction causes Addison's disease, and pancreatic insulin dysfunction causes diabetes mellitus. Thus P-III, Q-I, R-IV, S-II. Chapter: Chemical Coordination and Integration.

Bacillus thuringiensis supplies Cry proteins; Thermus aquaticus supplies thermostable DNA polymerase; Agrobacterium tumefaciens is a plant cloning vector; and Salmonella typhimurium was used in the first recombinant-DNA construction. Thus P-IV, Q-III, R-I, S-II. Chapter: Biotechnology: Principles and Processes.

Inulin contains a glycosidic bond, linking its fructose units. Insulin contains peptide bonds, linking its amino acid residues. Chapter: Biomolecules.

During transcription, DNA-dependent RNA polymerase binds the promoter and locally opens the DNA helix to expose the template strand. DNA helicase is primarily associated with replication, so the keyed answer is RNA polymerase. Chapter: Molecular Basis of Inheritance.

The cockroach can survive for a few days without its head because the major part of its nervous system, including segmental ganglia, is situated along the ventral part of its body. This allows basic functions to continue even without the head. Chapter: Structural Organisation in Animals.

Inspiration requires contraction of the diaphragm and external intercostal muscles. Thoracic and pulmonary volume increase while intrapulmonary pressure falls, so only statements (I) and (II) are correct. Chapter: Breathing and Exchange of Gases.

Hisardale was produced by crossing Bikaneri ewes with Merino rams, so it is a cross-bred sheep variety. Chapter: Strategies for Enhancement in Food Production.

Collagen is the most abundant protein in animals, forming the structural framework of bones, skin, and connective tissues. Chapter: Biomolecules.

Mendel selected 14 true breeding pea plant varieties as pairs, each differing in one character with contrasting traits. Chapter: Principles of Inheritance and Variation.

The body of the ovule is fused to the funicle at the hilum. Chapter: Sexual Reproduction in Flowering Plants.

Viroids are infectious agents composed of short strands of naked RNA without a protein coat. Chapter: Biological Classification.

In one turn of the citric acid cycle, there is one substrate-level phosphorylation event, which occurs during the conversion of succinyl-CoA to succinate. Chapter: Respiration in Plants.

Nitrogenase catalyzes the reduction of atmospheric nitrogen to ammonia and hydrogen gas. Chapter: Mineral Nutrition.

Typhoid-Salmonella, pneumonia-Haemophilus, filariasis-Wuchereria and malaria-Plasmodium give P-III, Q-IV, R-I, S-II. Chapter: Human Health and Disease.

S.L. Miller's experiment used a mixture of $\mathrm{CH}_{4}$, $\mathrm{H}_{2}$, $\mathrm{NH}_{3}$, and water vapor at a high temperature (800°C) to simulate early Earth conditions and produce amino acids. Chapter: Evolution.

In double-stranded DNA, adenine pairs with thymine through two hydrogen bonds, while guanine pairs with cytosine through three. Complementary base pairing explains the hydrogen-bond count asked here.

Synapsis occurs in zygotene, crossing over in pachytene, chiasmata become visible in diplotene, and terminalization occurs in diakinesis. Thus P-IV, Q-III, R-II, S-I. Chapter: Cell Cycle and Cell Division.

DNA ligase joins DNA fragments by sealing breaks in the sugar-phosphate backbone. It forms phosphodiester bonds between adjacent 3′-hydroxyl and 5′-phosphate ends.

Gonorrhoea, Syphilis, and Genital herpes are all sexually transmitted diseases (STDs). Malaria, Filaria, and Cancer are not STDs. Chapter: Reproductive Health.

Karl Ernst von Baer criticized the idea that embryos of different species pass through the same stages of development, which was a key argument for embryological support of evolution. Chapter: Evolution.

Roots arising in clusters from the base of the stem replace the short-lived primary root and form a fibrous root system, as in grasses. Chapter: Morphology of Flowering Plants.

Separated DNA fragments in gel electrophoresis are visualized using ethidium bromide, which intercalates with DNA and fluoresces under ultraviolet (UV) radiation. Chapter: Biotechnology: Principles and Processes.

High concentrations of estrogen trigger the release of luteinizing hormone (LH), which in turn causes ovulation. Chapter: Human Reproduction.

Goblet cells are modified columnar epithelial cells that secrete mucus to lubricate and protect the lining of the alimentary canal. Chapter: Digestion and Absorption.

Snow-blindness in the Antarctic region is caused by inflammation of the cornea due to high doses of UV-B radiation, which can be reflected off snow surfaces. Chapter: Environmental Issues.

Iron is required for chlorophyll biosynthesis, zinc for indole-3-acetic acid (IAA) biosynthesis, boron for pollen germination, and manganese for water photolysis. Thus P-III, Q-IV, R-II, S-I. Chapter: Mineral Nutrition.

Bt cotton, developed by introducing the toxin gene from Bacillus thuringiensis, is resistant to insect pests. This gene produces a protein that is toxic to certain insects, making Bt cotton effective against insect pests. Chapter: Biotechnology and its Applications.

Ray florets of Asteraceae are epigynous; the other floral whorls arise above the ovary, so the ovary is inferior. Chapter: Morphology of Flowering Plants.

The Montreal Protocol, signed in 1987, aims to control the emission of ozone-depleting substances. Chapter: Environmental Issues.

Each restriction enzyme recognizes and cuts DNA at specific palindromic sequences, not by inspecting the length of a DNA sequence. Chapter: Biotechnology: Principles and Processes.

The infectious stage of Plasmodium that enters the human body is the sporozoite stage. Sporozoites are transmitted through the saliva of an infected female Anopheles mosquito. Chapter: Human Health and Disease.

Meiotic division of the secondary oocyte is completed at the time of fusion of a sperm with an ovum, forming the second polar body. Chapter: Human Reproduction.

During photorespiration, the oxygenation activity of RuBisCo leads to the formation of one molecule of 3-phosphoglycerate (3-C) and one molecule of phosphoglycolate (2-C). Chapter: Photosynthesis in Higher Plants.

Statement I is false because urochordates retain the notochord only in the larval tail. Statement IV is false because Hemichordata is not a chordate subphylum. Statements (II) and (III) are true. Chapter: Animal Kingdom.

Clostridium butylicum produces butyric acid; Trichoderma polysporum produces cyclosporin A; Monascus purpureus produces statins that lower blood cholesterol; and Aspergillus niger produces citric acid. Thus P-II, Q-I, R-IV, S-III. Chapter: Microbes in Human Welfare.

Chlorella and Spirulina are both unicellular algae. Chlorella is a green alga, and Spirulina is a cyanobacterium, both of which are unicellular. Chapter: Plant Kingdom.

Plastoquinone facilitates the transfer of electrons from PS-II to the cytochrome $b_6f$ complex during the light reactions of photosynthesis. Chapter: Photosynthesis in Higher Plants.

Glycosuria and ketonuria together indicate diabetes mellitus: excess blood glucose spills into urine and increased fat use produces urinary ketone bodies. Chapter: Excretory Products and their Elimination.

Secondary metabolites like nicotine, strychnine, and caffeine are produced by plants for defence against herbivores and pathogens. Chapter: Organisms and Populations.

Aldosterone promotes the reabsorption of Na^+ and water from the renal tubules, reducing urine volume and preventing diuresis. Chapter: Excretory Products and their Elimination.

- Haemophilia is X-linked recessive. - Phenylketonuria is autosomal recessive. - Sickle cell anaemia is an autosomal recessive trait linked to chromosome 11. - Thalassemia is autosomal recessive, not X-linked. Chapter: Principles of Inheritance and Variation.

Species interaction is a characteristic of communities, not populations. Chapter: Organisms and Populations.

Inclusion bodies lie free in the cytoplasm, store reserve material and lack a bounding membrane. They are not involved in ingestion of food particles, making that statement incorrect. Chapter: Cell: The Unit of Life.

The described features-scattered vascular bundles, large parenchymatous ground tissue, conjoint and closed vascular bundles, and absence of phloem parenchyma-are characteristic of a monocotyledonous stem. Chapter: Anatomy of Flowering Plants.

Gross primary productivity (GPP) is the total amount of energy fixed by photosynthesis, while net primary productivity (NPP) is GPP minus the energy used by plants for respiration. Therefore, GPP is always more than NPP. Chapter: Ecosystem.

The flowers of water hyacinth and water lily emerge above the water surface and are pollinated by insects or wind, not by water currents. Chapter: Sexual Reproduction in Flowering Plants.

Activated sludge, which contains microorganisms, is added to the anaerobic sludge digester to further break down organic matter in the sewage. Chapter: Microbes in Human Welfare.

The liquid-water loss is guttation. Root pressure supplies the positive force that pushes xylem sap out through hydathodes at night or early morning, so root pressure is the facilitating process. Chapter: Transport in Plants.

Cuboidal epithelium with a brush border of microvilli is found in the proximal convoluted tubule of the nephron, where it increases the surface area for reabsorption. Chapter: Structural Organisation in Animals.

Glucocorticoids, such as cortisol, stimulate gluconeogenesis, the process of glucose synthesis from non-carbohydrate sources. Chapter: Chemical Coordination and Integration.

Gibberellic acid is a plant hormone that promotes seed germination, not dormancy. Abscisic acid, phenolic acid, and para-ascorbic acid are known to inhibit seed germination and maintain dormancy. Chapter: Plant Growth and Development.

Robert May estimated the global species diversity to be around 7 million species. Chapter: Biodiversity and Conservation.

Grass is the first trophic level, rabbit the second, crow the third and vulture the fourth. Thus P-II, Q-III, R-IV, S-I. Chapter: Ecosystem.

Before peptide synthesis begins, each amino acid is activated and attached to its cognate transfer RNA (tRNA) by aminoacyl-tRNA synthetase. This aminoacylation or charging of tRNA is the first phase of translation. Chapter: Molecular Basis of Inheritance.

Strobili or cones are found in Equisetum, a genus of plants commonly known as horsetails. Chapter: Plant Kingdom.

Cyclostomes have 6-15 pairs of gill slits; Chondrichthyes have a heterocercal caudal fin; Osteichthyes generally have an air bladder; and Trygon bears a poison sting. Thus P-II, Q-III, R-IV, S-I. Chapter: Animal Kingdom.

Cells can exit the cell cycle and enter the quiescent stage (G 0 ) at the end of the G 1 phase. Chapter: Cell Cycle and Cell Division.

The ileum is a highly coiled part of the small intestine, responsible for the absorption of nutrients. Chapter: Digestion and Absorption.

Zygote intra fallopian transfer (ZIFT) transfers an early embryo into the fallopian tube, while intra uterine transfer (IUT) transfers an embryo with more than eight blastomeres into the uterus. Gamete intra fallopian transfer (GIFT) transfers gametes and intra cytoplasmic sperm injection (ICSI) is fertilisation outside the body, so the embryo-transfer pair is ZIFT and IUT. Chapter: Reproductive Health.

Pollen grains inside the anther and embryo sac inside the ovule both consist of two generations, one within the other. Pollen grains contain microspores (haploid) and the embryo sac contains the female gametophyte (haploid) within the ovule (diploid). Chapter: Sexual Reproduction in Flowering Plants.

The synaptonemal complex dissolves during diplotene, leading to the separation of homologous chromosomes while they remain connected at chiasmata. Chapter: Cell Cycle and Cell Division.

The organ of Corti lies on the basilar membrane; the cochlea is the coiled labyrinth; the Eustachian tube joins the middle ear to the pharynx; and the stapes contacts the oval window. Thus P-IV, Q-II, R-I, S-III. Chapter: Neural Control and Coordination.

Plum has a perigynous flower and a half-inferior ovary. Mustard, sunflower and brinjal have superior or inferior ovaries rather than the half-inferior condition asked here. Chapter: Morphology of Flowering Plants.

Lysine is a basic amino acid due to its positively charged amino group on the side chain. Chapter: Biomolecules.

Eosinophils release histaminase and destructive enzymes; basophils release histamine-containing granules; neutrophils are phagocytic; and lymphocytes mediate immune responses. Thus P-III, Q-IV, R-II, S-I. Chapter: Body Fluids and Circulation.

Malonate inhibits catalytic activity competitively; collagen contains peptide bonds; fungal walls contain chitin; and ricin is a secondary metabolite. Thus P-II, Q-IV, R-III, S-I. Chapter: Biomolecules.

During the G1 phase, the cell is metabolically active and grows, but DNA replication does not occur. Chapter: Cell Cycle and Cell Division.

Gibberellins, when sprayed on sugarcane, promote stem elongation and increase the yield. Chapter: Plant Growth and Development.

The gene 'I' has three alleles: $I^A$, $I^B$, and i. A person will have only two of these alleles. When $I^A$ and $I^B$ are present together, they express different types of sugars, not the same type. The allele i does not produce any sugar. Chapter: Principles of Inheritance and Variation.

Active immunity involves the production of antibodies by the host's immune system, which takes time to develop a full response, not immediately. Chapter: Human Health and Disease.

Enterokinase converts trypsinogen into trypsin, an active protease in the small intestine. Chapter: Digestion and Absorption.

Restriction enzyme from Escherichia coli RY 13 (EcoRI) recognizes the palindromic sequence 5' - GAATTC - 3' and its complementary strand 3' - CTTAAG - 5'. Chapter: Biotechnology: Principles and Processes.

Bt cotton is linked to Bacillus thuringiensis, ADA deficiency to gene therapy, RNA interference to cellular defence and polymerase chain reaction (PCR) to human immunodeficiency virus (HIV) detection. Thus P-IV, Q-I, R-II, S-III. Chapter: Biotechnology and its Applications.

Floridean starch has a structure similar to amylopectin and glycogen, both of which are branched polysaccharides. Chapter: Plant Kingdom.

The functional insulin has A and B chains linked together by disulfide bonds, not hydrogen bonds. Chapter: Biotechnology and its Applications.

Flippers of penguins and dolphins are examples of convergent evolution, where unrelated organisms develop similar traits due to similar environmental pressures. Chapter: Evolution.

Herbicide-resistant weeds, drug-resistant eukaryotes and human-created dog breeds reflect anthropogenic selection. Darwin's finches arose through natural environmental selection, so (II), (III) and (IV) are correct. Chapter: Evolution.

Higher $\mathrm{H}^{+}$ concentration in the alveoli does not favour the formation of oxyhaemoglobin; instead, it promotes the release of $\mathrm{O}_{2}$ from haemoglobin. Chapter: Breathing and Exchange of Gases.

The process of growth is maximum during the log phase, when cells divide rapidly and the population increases exponentially. Chapter: Plant Growth and Development.

The Amazon forests are known for the highest species diversity due to their vast area, varied habitats, and tropical climate. Chapter: Biodiversity and Conservation.

The origin of replication (Ori site) controls the copy number of the linked DNA in a vector. This site is essential for the initiation of DNA replication and determines how many copies of the plasmid are made within the host cell. Chapter: Biotechnology: Principles and Processes.

The placenta secretes hCG; zona pellucida is the ovum's glycoprotein layer; bulbo-urethral secretions lubricate the penis; and Leydig cells secrete androgens. Thus P-II, Q-III, R-IV, S-I. Chapter: Human Reproduction.

Sucrose is made of an $\alpha$-D-glucose unit and a $\beta$-D-fructose unit joined through a glycosidic bond. On hydrolysis, this bond breaks to give glucose and fructose. The NEET 2020 keyed pair is $\alpha$-D-glucose and $\beta$-D-fructose.

The elimination reaction of 2-Bromo-pentane to form pent-2-ene is a $\beta$-elimination reaction, follows Zaitsev's rule, and is a dehydrohalogenation reaction. Dehydration involves removing a water molecule, which is not the case here. Chapter: Haloalkanes and Haloarenes.

The reaction for the production of calcium is $\mathrm{Ca}^{2+} + 2\mathrm{e}^- → \mathrm{Ca}$. The number of moles of calcium produced is $\frac{20\ g}{40\ g mol^{-1}} = 0.5\ mol$. Since 2 moles of electrons are required to produce 1 mole of calcium, the number of Faradays required is $0.5 \times 2 = 1\ F$. Chapter: Electrochemistry.

For a body-centered cubic (bcc) structure, the relationship between the atomic radius r and the cell edge a is given by $r = \frac{\sqrt{3}}{4}a$. Substituting $a = 288\ pm$, we get $r = \frac{\sqrt{3}}{4} \times 288\ pm = \frac{288\ pm}{4} \times \sqrt{3}$. Chapter: The Solid State.

Passing $\mathrm{HCl}$ gas increases the concentration of $\mathrm{Cl^-}$ ions and lowers the solubility of the chlorides by the common-ion effect. Under these conditions, $\mathrm{NaCl}$ becomes much less soluble and crystallises, while $\mathrm{MgCl_2}$ and $\mathrm{CaCl_2}$ remain in solution.

The solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in 0.1 M NaOH can be calculated using the solubility product $K_{sp} = 2 \times 10^{-15}$. Let s be the solubility of $\mathrm{Ni}(\mathrm{OH})_2$. The concentration of $\mathrm{OH}^-$ from NaOH is 0.1 M, and from $\mathrm{Ni}(\mathrm{OH})_2$ is 2s. Thus, $K_{sp} = s \times (0.1 + 2s)^2 \approx s \times (0.1)^2 = 2 \times 10^{-15} ⇒ s = 2 \times 10^{-13}\ M$. Chapter: Equilibrium.

The reaction $2\mathrm{Cl}(\text{g}) → \mathrm{Cl}_2(\text{g})$ involves the formation of a bond, which releases energy, making $\Delta_r H < 0$. Additionally, the reaction decreases the number of gas molecules, reducing entropy, so $\Delta_r S < 0$. Chapter: Chemical Thermodynamics.

In the spectrochemical series relevant here, ligand field strength rises as $\mathrm{SCN}^{-}$ < $\mathrm{F}^{-}$ < $\mathrm{C}_{2}\mathrm{O}_{4}^{2-}$ < $\mathrm{CN}^{-}$. Chapter: Coordination Compounds.

The spin-only magnetic moment $\mu$ is given by $\mu = \sqrt{n(n+2)}\ BM$, where n is the number of unpaired electrons. For $\mathrm{Cr}^{2+}$, the electronic configuration is $3d^4$, with 4 unpaired electrons. Substituting n = 4, $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ BM$. Chapter: Coordination Compounds.

Boron trifluoride, beryllium difluoride, and carbon dioxide have symmetrical structures leading to zero net dipole moments. 1,4-dichlorobenzene also has a symmetrical structure, making the net dipole moment zero. Chapter: Chemical Bonding and Molecular Structure.

Potassium ions (K^+) activate many enzymes, participate in the oxidation of glucose to produce adenosine triphosphate (ATP), and, along with sodium ions (Na^+), are crucial for the transmission of nerve signals. Chapter: The s-Block Elements.

Methanal is formed only when ozonolysis cleaves a terminal =CH$_{2}$ group. Allylcyclohexane contains -CH$_{2}$-CH=CH$_{2}$, so its terminal carbon gives HCHO. Chapter: Hydrocarbons.

Why the other options are wrong:

- (1) 3-Propylcyclohexene: its only double bond is inside the ring, so ozonolysis opens the ring into one dialdehyde and gives no methanal.
- (3) 4-Propylcyclohexene: ozonolysis cleaves the ring double bond and gives a single dialdehyde, with no terminal =CH$_{2}$ to give methanal.
- (4) 1-Cyclohexylpropene: ozonolysis splits it into cyclohexanecarbaldehyde and ethanal ($\mathrm{CH}_{3}\mathrm{CHO}$), not methanal.

For a first-order reaction, $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right)$. Substituting, $t = \frac{2.303}{4.606 \times 10^{-3}} \log \left( \frac{2.0}{0.2} \right) = \frac{2.303}{4.606 \times 10^{-3}} \log 10 = \frac{2.303}{4.606 \times 10^{-3}} \times 1 = 500\ s$. Chapter: Chemical Kinetics.

The reaction of acetone with methylmagnesium chloride forms a tertiary alkoxide, which upon hydrolysis yields tert-butyl alcohol. This follows the Grignard reagent mechanism where the nucleophilic attack occurs at the carbonyl carbon, leading to the formation of a tertiary alcohol. Chapter: Aldehydes, Ketones and Carboxylic Acids.

cis-1,4-polyisoprene is the natural polymer found in rubber latex, derived from rubber trees. Other options are synthetic polymers. Chapter: Polymers.

CO$_{2}$ gas is not the refrigerant used here; solid CO$_{2}$ (dry ice) is. C$_{60}$ contains twelve five-membered and twenty six-membered rings, so (Statement II) is false. ZSM-5 converts alcohols to gasoline and CO is colourless and odourless, so only (Statement III) and (Statement IV) are correct. Chapter: p-Block Elements (Groups 13-14).

In a free expansion of an ideal gas under adiabatic conditions, no work is done (w = 0) and no heat is exchanged (q = 0). Since the internal energy of an ideal gas depends only on temperature, $\Delta T = 0$. Chapter: Chemical Thermodynamics.

Peroxodisulphuric acid ($\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}$) contains the peroxide linkage (-O-O-), which is not present in the other oxoacids listed. Chapter: p-Block Elements (Groups 15-18).

Successive side-chain chlorination of toluene under light gives benzal chloride, C$_{6}$H$_{5}$CHCl$_{2}$. Hydrolysis at 373 K converts its gem-dichloride group to -CHO, forming benzaldehyde. Chapter: Hydrocarbons.

Why the other options are wrong:

- (1) Benzyl chloride, $\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}_{2}\mathrm{Cl}$: hydrolysis swaps its single Cl for OH and gives benzyl alcohol, not benzaldehyde.
- (3) Benzotrichloride, $\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CCl}_{3}$: hydrolysis turns the $\mathrm{CCl}_{3}$ group into -COOH and gives benzoic acid, not benzaldehyde.
- (4) Chlorobenzene: it has no side-chain carbon to become -CHO, and $\mathrm{Cl}_{2}$ in light chlorinates the side chain, not the ring.

The atomic number (71) gives the number of protons and electrons, and the mass number (175) minus the atomic number gives the number of neutrons: 175 - 71 = 104. Chapter: Structure of Atom.

Chromium is +6 in both CrO$_{4}^{2-}$ and Cr$_{2}$O$_{7}^{2-}$, so the statement that their oxidation states differ is incorrect. The other three statements are valid. Chapter: The d- and f-Block Elements.

Cetyltrimethyl ammonium bromide is a cationic detergent because it has a positively charged quaternary ammonium ion. Sodium lauryl sulphate, sodium stearate, and sodium dodecylbenzene sulphonate are anionic detergents. Chapter: Chemistry in Everyday Life.

The freezing point depression $\Delta T_f$ is given by $\Delta T_f = K_f \times m$. Substituting the values, $\Delta T_f = 5.12 \times 0.078 = 0.40$ K. Chapter: Solutions.

Unnilunium, unniltrium and unnilhexium correctly correspond to elements 101, 103 and 106. The printed 'Unununnium' denotes element 111, whose official name is roentgenium, not darmstadtium; therefore S-IV is the incorrect match. Chapter: Classification of Elements and Periodicity in Properties.

Ethanol and acetone show positive deviation from Raoult's law due to weaker intermolecular interactions in the solution compared to pure components. Chapter: Solutions.

CO is neutral, BaO basic, Al$_{2}$O$_{3}$ amphoteric and Cl$_{2}$O$_{7}$ acidic. Thus P-II, Q-I, R-IV, S-III. Chapter: Classification of Elements and Periodicity in Properties.

To find the number of atoms, use $n = \frac{m}{M} \times N_A$, where m is mass, M is molar mass, and $N_A$ is Avogadro's number. For 1 g of each: - Ag: $\frac{1}{108} \times N_A$ - Mg: $\frac{1}{24} \times N_A$ - $\mathrm{O}_{2}$: $\frac{1}{32} \times 2 \times N_A = \frac{1}{16} \times N_A$ (since $\mathrm{O}_{2}$ has a molar mass of 32 g/mol and two atoms per molecule) - Li: $\frac{1}{7} \times N_A$ Li has the highest number of atoms. Chapter: Some Basic Concepts of Chemistry.

The reaction between benzaldehyde and acetophenone in the presence of dilute NaOH is a cross aldol condensation. This reaction involves the condensation of an aldehyde with a ketone to form a $\beta$-hydroxy ketone. Chapter: Aldehydes, Ketones and Carboxylic Acids.

A tertiary butyl carbocation is more stable than a secondary butyl carbocation due to hyperconjugation, which involves the delocalization of $\sigma$-bond electrons into the empty p-orbital of the carbocation. This effect increases the stability of the carbocation. Chapter: Organic Chemistry: Some Basic Principles and Techniques.

CO reduces oxygen transport by forming carboxyhaemoglobin, which is more stable than oxyhaemoglobin. It is also produced by incomplete combustion, so the claim that carboxyhaemoglobin is less stable is the incorrect statement. Chapter: Environmental Chemistry.

Lysine is a basic amino acid due to the presence of an additional amino group ($-\mathrm{NH}_{2}$) in its side chain, making it positively charged at physiological pH. Chapter: Biomolecules.

Urea hydrolysis ultimately releases NH$_{3}$. Excess ammonia coordinates Cu$^{2+}$ to form the deep-blue tetraamminecopper(II) ion [Cu(NH$_{3}$)$_{4}$]$^{2+}$. Chapter: Coordination Compounds.

Calculate the number of moles of each gas: $n_{N_2} = \frac{7}{28} = 0.25\ \text{mol}$, $n_{Ar} = \frac{8}{40} = 0.2\ \text{mol}$. The total number of moles is $0.25 + 0.2 = 0.45\ \text{mol}$. The partial pressure of $\mathrm{N}_{2}$ is given by $P_{N_2} = \frac{n_{N_2}}{n_{total}} \times P_{total} = \frac{0.25}{0.45} \times 27 = 15\ \text{bar}$. Chapter: States of Matter.

Pig iron can be moulded into a variety of shapes due to its high carbon content and low melting point. Chapter: General Principles and Processes of Isolation of Elements.

The relationship between the equilibrium constant $K_c$ and the standard Gibbs free energy change $\Delta_r G^\circ$ is given by $\Delta_r G^\circ = -RT \ln K_c$. Substituting the values, $\Delta_r G^\circ = -8.314\ J mol^{-1}\ K^{-1} \times 300\ K \times \ln(2 \times 10^{13})$. Chapter: Equilibrium.

$\mathrm{He}_{2}$ does not exist because helium atoms have a complete outer shell and do not form stable bonds with each other. In contrast, $\mathrm{Li}_{2}$, $\mathrm{C}_{2}$, and $\mathrm{O}_{2}$ can form stable molecules. Chapter: Chemical Bonding and Molecular Structure.

An increase in the concentration of reactants leads to a higher collision frequency, as more particles are available to collide. This does not affect activation energy, heat of reaction, or threshold energy. Chapter: Chemical Kinetics.

A Wurtz reaction with one alkyl halide gives a symmetrical alkane with an even total carbon count. Making n-heptane requires two different halides, producing a mixture, so it cannot be obtained in good yield. Chapter: Hydrocarbons.

Only primary amines give the carbylamine test. Aniline is primary, whereas N-methyl-, N-ethyl- and N,N-dimethylaniline are secondary or tertiary. Chapter: Amines.

Why the other options are wrong:

- (1) N-methylaniline is a secondary amine with only one N-H, and an isocyanide needs two N-H hydrogens to form with $\mathrm{CHCl}_{3}$ and alcoholic $\mathrm{KOH}$.
- (2) N,N-dimethylaniline is a tertiary amine with no N-H at all, so no isocyanide can form.
- (3) N-ethylaniline is a secondary amine; its single N-H is not enough to give the foul-smelling isocyanide.

On electrolysis of dilute sulfuric acid with platinum electrodes, oxygen gas is evolved at the anode due to the oxidation of water. The reaction is $2\mathrm{H}_{2}\mathrm{O} → \mathrm{O}_{2} + 4\mathrm{H}^{+} + 4\mathrm{e}^{-}$. Chapter: Electrochemistry.

In $\mathrm{CH}_{4}$, the oxidation number of carbon is -4 (since hydrogen is +1). In $\mathrm{CCl}_{4}$, the oxidation number of carbon is +4 (since chlorine is -1). Thus, the change in oxidation number of carbon is from -4 to +4. Chapter: Redox Reactions.

HI protonates anisole at oxygen, then $\mathrm{I}^{-}$ attacks the methyl group by substitution nucleophilic bimolecular (S$_{\text{N}}$2) cleavage. The products are phenol and methyl iodide (CH$_{3}$I); the aryl C-O bond is not cleaved by S$_{\text{N}}$2. Chapter: Alcohols, Phenols and Ethers.

Why the other options are wrong:

- (1) Iodobenzene and $\mathrm{CH}_{3}\mathrm{OH}$: this needs $\mathrm{I}^{-}$ to attack the ring carbon, but S$_{\text{N}}$2 cannot occur at an sp$^{2}$ aryl carbon, so the aryl C-O bond stays intact.
- (2) Phenol and $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{I}$: anisole is a methyl ether, so the alkyl iodide formed is $\mathrm{CH}_{3}\mathrm{I}$; ethyl iodide would come from phenetole.
- (3) Iodobenzene and $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$: the aryl C-O bond is not broken, and anisole has no ethyl group to give ethanol.