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ChemistryMediumMCQ2021 · 27 Jul Shift 2

Q35.Which one of the following set of elements can be detected using sodium fusion extract? (1) Sulfur, Nitrogen, Phosphorus, Halogens (2) Phosphorus, Oxygen, Nitrogen, Halogens (3) Nitrogen, Phosphorus, Carbon, Sulfur (4) Halogens, Nitrogen, Oxygen, Sulfur

What This Question Tests

This question checks the knowledge of Lassaigne's test, which is used for the detection of nitrogen, sulfur, halogens, and phosphorus in organic compounds.

Concepts Tested

Detection of elements in organic compoundsSodium fusion extractPresence of N, S, Halogens, P

📚 NCERT Sections This Tests

4.9Hydrogen Bonding Hydrogen Bond Is Represented By A Dotted

Chemistry Class 11 · Chapter 4

73% match

4.9 Hydrogen Bonding Hydrogen bond is represented by a dotted line (– – –) while a solid line represents theNitrogen, oxygen and fluorine are the highly covalent bond. Thus, hydrogen bond can beelectronegative elements. When they are attached to a hydrogen atom to form covalent defined as the attractive force which binds bond, the electrons of the covalent bond are hydrogen atom of one molecule with the shifted towards the more electronegative electronegative atom (F, O or N) of another atom. This partially positively charged molecule. hydrogen atom forms a bond with the other 4.9.1 Cause of Formation of Hydrogen more electronegative atom. This bond is Bond known as hydrogen bond and is weaker When hydrogen is bonded to stronglythan the covalent bond. For example, in HF electronegative element ‘X’, the electron pairmolecule, the hydrogen bond exists between shared between the two atoms moves farhydrogen atom of one molecule and fluorine away from hydrogen atom. As a result theatom of another molecule as depicted below : hydrogen atom becomes highly electropositive – – – Hδ+–Fδ– – – –Hδ+ – Fδ– – – – Hδ+ – Fδ– with respect to the other atom ‘X’. Since Here, hydrogen bond acts as a bridge between there is displacement of electrons towards two atoms which holds one atom by covalent X, the hydrogen acquires fractional positive bond and the other by hydrogen bond. charge (δ +) while ‘X’ attain fractional negative Reprint 2025-26 132 chemistry charge (δ–). This results in the formation of a H-bond in case of HF molecule, alcohol or polar molecule having electrostatic force of water molecules, etc. attraction which can be represented as: (2) Intramolecular hydrogen bond : It is formed when hydrogen atom is in between Hδ+ – Xδ– – – – Hδ+ – Xδ– – – – Hδ+ – Xδ– the two highly electronegative (F, O, N) The magnitude of H-bonding depends atoms present within the same molecule. For on the physical state of the compound. It is example, in o-nitrophenol the hydrogen is in maximum in the solid state and minimum in between the two oxygen atoms. the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds. 4.9.2 Types of H-Bonds There are two types of H-bonds (i) Intermolecular hydrogen bond (ii) Intramolecular hydrogen bond (1) Intermolecular hydrogen bond : It is formed between two different molecules of the Fig. 4.22 Intramolecular hydrogen bonding in same or different compounds. For example, o-nitrophenol molecule SUMMARY Kössel’s first insight into the mechanism of formation of electropositive and electronegative ions related the process to the attainment of noble gas configurations by the respective ions. Electrostatic attraction between ions is the cause for their stability. This gives the concept of electrovalency. The first description of covalent bonding was provided by Lewis in terms of the sharing of electron pairs between atoms and he related the process to the attainment of noble gas configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols show the number of valence electrons of the atoms of a given element and Lewis dot structures show pictorial representations of bonding in molecules. An ionic compound is pictured as a three-dimensional aggregation of positive and negative ions in an ordered arrangement called the crystal lattice. In a crystalline solid there is a charge balance between the positive and negative ions. The crystal lattice is stabilized by the enthalpy of lattice formation. While a single covalent bond is formed by sharing of an electron pair between two atoms, multiple bonds result from the sharing of two or three electron pairs. Some bonded atoms have additional pairs of electrons not involved in bonding. These are called lone-pairs of electrons. A Lewis dot structure shows the arrangement of bonded pairs and lone pairs around each atom in a molecule. Important parameters, associated with chemical bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity have significant effect on the properties of compounds. A number of molecules and polyatomic ions cannot be described accurately by a single Lewis structure and a number of descriptions (representations) based on the same skeletal structure are written and these taken together represent the molecule or ion. This is a very important and extremely useful concept called resonance. The contributing structures or canonical forms taken together constitute the resonance hybrid which represents the molecule or ion. Reprint 2025-26 Chemical Bonding And Molecular Structure 133 The VSEPR model used for predicting the geometrical shapes of molecules is based on the assumption that electron pairs repel each other and, therefore, tend to remain as far apart as possible. According to this model, molecular geometry is determined by repulsions between lone pairs and lone pairs; lone pairs and bonding pairs and bonding pairs and bonding pairs. The order of these repulsions being : lp-lp > lp-bp > bp-bp The valence bond (VB) approach to covalent bonding is basically concerned with the energetics of covalent bond formation about which the Lewis and VSEPR models are silent. Basically the VB theory discusses bond formation in terms of overlap of orbitals. For example the formation of the H2 molecule from two hydrogen atoms involves the overlap of the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential energy of the system gets lowered as the two H atoms come near to each other. At the equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any attempt to bring the nuclei still closer results in a sudden increase in energy and consequent destabilization of the molecule. Because of orbital overlap the electron density between the nuclei increases which helps in bringing them closer. It is however seen that the actual bond enthalpy and bond length values are not obtained by overlap alone and other variables have to be taken into account. For explaining the characteristic shapes of polyatomic molecules Pauling introduced the concept of hybridisation of atomic orbitals. sp, sp2, sp3 hybridizations of atomic orbitals of Be, B, C, N and O are used to explain the formation and geometrical shapes of molecules like BeCl2, BCl3, CH4, NH3 and H2O. They also explain the formation of multiple bonds in molecules like C2H2 and C2H4. The molecular orbital (MO) theory describes bonding in terms of the combination and arrangment of atomic orbitals to form molecular orbitals that are associated with the molecule as a whole. The number of molecular orbitals are always equal to the number of atomic orbitals from which they are formed. Bonding molecular orbitals increase electron density between the nuclei and are lower in energy than the individual atomic orbitals. Antibonding molecular orbitals have a region of zero electron density between the nuclei and have more energy than the individual atomic orbitals. The electronic configuration of the molecules is written by filling electrons in the molecular orbitals in the order of increasing energy levels. As in the case of atoms, the Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals. Molecules are said to be stable if the number of elctrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals. Hydrogen bond is formed when a hydrogen atom finds itself between two highly electronegative atoms such as F, O and N. It may be intermolecular (existing between two or more molecules of the same or different substances) or intramolecular (present within the same molecule). Hydrogen bonds have a powerful effect on the structure and properties of many compounds. EXERCISES 4.1 Explain the formation of a chemical bond. 4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br. 4.3 Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H– 4.4 Draw the Lewis structures for the following molecules and ions : H2S, SiCl4, BeF2, CO32−, HCOOH 4.5 Define octet rule. Write its significance and limitations. Reprint 2025-26 134 chemistry 4.6 Write the favourable factors for the formation of ionic bond. 4.7 Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3 4.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss. 4.9 How do you express the bond strength in terms of bond order ? 4.10 Define the bond length. 4.11 Explain the important aspects of resonance with reference to the CO32− ion. 4.12 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3 ? If not, give reasons for the same. 4.13 Write the resonance structures for SO3, NO2 and NO3−. 4.14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N. 4.15 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment. 4.16 Write the significance/applications of dipole moment. 4.17 Define electronegativity. How does it differ from electron gain enthalpy ? 4.18 Explain with the help of suitable example polar covalent bond. 4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. 4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. 4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ? 4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar. 4.23 Which out of NH3 and NF3 has higher dipole moment and why ? 4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals. 4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl 3  Cl   AlCl 4 Reprint 2025-26 Chemical Bonding And Molecular Structure 135 4.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction? 4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules. 4.28 What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4 4.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s. 4.30 Which hybrid orbitals are used by carbon atoms in the following molecules? CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH 4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type. 4.32 Distinguish between a sigma and a pi bond. 4.33 Explain the formation of H2 molecule on the basis of valence bond theory. 4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. 4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist. 4.36 Compare the relative stability of the following species and indicate their magnetic properties; (superoxide), O22− (peroxide) 4.37 Write the significance of a plus and a minus sign shown in representing the orbitals. 4.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds? 4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces? 4.40 What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–. Reprint 2025-26 Unit 5 Thermodynamics It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown. After studying this Unit, you will be Albert Einstein able to • explain the terms : system and surroundings; • discriminate between close, open and isolated systems; Chemical energy stored by molecules can be released as• explain internal energy, work and heat; heat during chemical reactions when a fuel like methane, • state first law of thermodynamics cooking gas or coal burns in air. The chemical energy may and express it mathematically; also be used to do mechanical work when a fuel burns • calculate energy changes as in an engine or to provide electrical energy through a work and heat contributions in galvanic cell like dry cell. Thus, various forms of energy chemical systems; are interrelated and under certain conditions, these may • explain state functions: U, H. be transformed from one form into another. The study • correlate ∆U and ∆H; of these energy transformations forms the subject matter • measure experimentally ∆U and of thermodynamics. The laws of thermodynamics deal ∆H; with energy changes of macroscopic systems involving• define standard states for ∆H; • calculate enthalpy changes for a large number of molecules rather than microscopic various types of reactions; systems containing a few molecules. Thermodynamics is • state and apply Hess’s law of not concerned about how and at what rate these energy constant heat summation; transformations are carried out, but is based on initial and • differentiate between extensive final states of a system undergoing the change. Laws of and intensive properties; thermodynamics apply only when a system is in equilibrium • define spontaneous and non- or moves from one equilibrium state to another equilibrium spontaneous processes; state. Macroscopic properties like pressure and temperature• e x p l a i n e n t r o p y a s a thermodynamic state function do not change with time for a system in equilibrium state. and apply it for spontaneity; In this unit, we would like to answer some of the important • explain Gibbs energy change (∆G); questions through thermodynamics, like: and How do we determine the energy changes involved in a • establish relationship between chemical reaction/process? Will it occur or not? ∆G and spontaneity, ∆G and equilibrium constant. What drives a chemical reaction/process? To what extent do the chemical reactions proceed? Reprint 2025-26 THERMODYNAMICS 137

9.4Alkynes 1 2 3 4 5

Chemistry Class 12 · Chapter 9

73% match

9.4 Alkynes 1 2 3 4 5 Like alkenes, alkynes are also unsaturated II. H3C–C≡ C– CH2– CH3 Pent–2-yne hydrocarbons. They contain at least one triple 4 3 2 1 bond between two carbon atoms. The number III. H3C–CH–C≡ CH 3-Methyl but–1-yne |of hydrogen atoms is still less in alkynes as CH3compared to alkenes or alkanes. Their general Structures I and II are position isomers formula is CnH2n–2. and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 9.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen corresponding to the 5 th member of gas. Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it all the isomers. What type of isomerism is interesting to study this class of organic is exhibited by different pairs of isomers? compounds. Solution 9.4.1 Nomenclature and Isomerism th 5 member of alkyne has the molecular In common system, alkynes are named as formula C6H10. The possible isomers are: derivatives of acetylene. In IUPAC system, they Table 9.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne Reprint 2025-26 Hydrocarbons 315 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 9.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C different pairs. π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol -1) is more than those 9.4.2 Structure of Triple Bond of C=C bond (bond enthalpy 681 kJ mol –1) Ethyne is the simplest molecule of alkyne and C–C bond (bond enthalpy 348 kJ mol–1). series. Structure of ethyne is shown in The C≡C bond length is shorter (120 pm) Fig. 9.6. than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon Each carbon atom of ethyne has two sp atoms is cylindrically symmetrical about thehybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 9.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating the 1s orbital of each of the two hydrogen calcium carbide with water. Calcium atoms forming two C-H sigma bonds. carbide is prepared by heating quick lime H-C-C bond angle is of 180°. Each carbon with coke. Quick lime can be obtained byhas two unhybridised p orbitals which are heating limestone as shown in the followingperpendicular to each other as well as to the reactions:plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆ CaO + O2 (9.55) Reprint 2025-26 316 chemistry CaO + 3C CaC2 + CO (9.56) the sp hybridised carbon2 atoms whereas they are attached to sp hybridised carbon Calcium 3 atoms in ethene and sp hybridised carbons carbide in ethane. Due to the maximum percentage of CaC2 + 2H2O Ca(OH)2 + C2H2 (9.57) s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have2. From vicinal dihalides : Vicinal dihalides highest electronegativity; hence, these attract on treatment with alcoholic potassium the shared electron pair of the C-H bond of hydroxide undergo dehydrohalogenation. ethyne to a greater extent than that of the One molecule of hydrogen halide is 2 sp hybridised orbitals of carbon in ethene eliminated to form alkenyl halide which 3 and the sp hybridised orbital of carbon in on treatment with sodamide gives alkyne. ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. HC ≡ CH + Na → HC ≡ C–Na++ 1/2 H2 9.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same (9.59) trend of alkenes and alkanes. First three HC ≡ C– Na + Na → Na+ C–Na+ ≡ C–Na++ 1/2 H2members are gases, the next eight are liquids and the higher ones are solids. All alkynes Disodium ethynide are colourless. Ethyene has characteristic (9.60)odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter CH3 – C ≡ C – H + Na+ NH–2 than water and immiscible with water but ↓ soluble in organic solvents like ethers, carbon CH3 – C ≡ C– Na+ + NH3 tetrachloride and benzene. Their melting Sodium propynide (9.61) point, boiling point and density increase with These reactions are not shown by alkenesincrease in molar mass. and alkanes, hence used for distinction Chemical properties between alkynes, alkenes and alkanes. What Alkynes show acidic nature, addition reactions about the above reactions with but-1-yne and and polymerisation reactions as follows : but-2-yne ? Alkanes, alkenes and alkynes A. Acidic character of alkyne: Sodium follow the following trend in their acidic metal and sodamide (NaNH2) are strong behaviour : bases. They react with ethyne to form sodium i) CH ≡ CH > H2C – CH2 > CH3 –CH3acetylide with the liberation of dihydrogen gas. These reactions have not been observed ii) HC ≡ CH > CH3 –C≡ CH >> CH3 –C≡C–CH3in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison B. Addition reactions: Alkynes contain a to ethene and ethane. Why is it so ? Has triple bond, so they add up, two molecules of it something to do with their structures dihydrogen, halogen, hydrogen halides etc. and the hybridisation ? You have read that Formation of the addition product takes place hydrogen atoms in ethyne are attached to according to the following steps. Reprint 2025-26 Hydrocarbons 317 The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions (9.66)are given below: (i) Addition of dihydrogen (iv) Addition of water Pt/Pd/Ni H2 Like alkanes and alkenes, alkynes are alsoHC≡CH+H2 [H2C = CH2] CH3–CH3 immiscible and do not react with water. (9.62) However, one molecule of water adds to alkynes on warming with mercuric sulphate CH3–C≡CH + H2 Pt/Pd/Ni [CH3–CH=CH2] and dilute sulphuric acid at 333 K to form Propyne Propene carbonyl compounds. ↓H2 CH3–CH2–CH3 Propane (9.63) (ii) Addition of halogens (9.67) (9.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (iii) Addition of hydrogen halides (9.68) Two molecules of hydrogen halides (HCl, HBr, (v) Polymerisation HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H–C≡C–H+H–Br [CH2 = CH–Br]→ CHBr2 polyethyne which is a high molecular Bromoethene weight polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —(CH = CH – CH = CH)n— Under special (9.65) conditions, this polymer conducts electricity. Reprint 2025-26 318 chemistry Thin film of polyacetylene can be used as but in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the is retained. However, there are examples of metal conductors. aromatic hydrocarbons which do not contain a (b) Cyclic polymerisation: Ethyne on benzene ring but instead contain other highly unsaturated ring. Aromatic compoundspassing through red hot iron tube at 873K containing benzene ring are known asundergoes cyclic polymerization. Three benzenoids and those not containing amolecules polymerise to form benzene, which benzene ring are known as non-benzenoids.is the starting molecule for the preparation of Some examples of arenes are givenderivatives of benzene, dyes, drugs and large below:number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (9.69) Biphenyl Problem 9.14 How will you convert ethanoic acid into 9.5.1 Nomenclature and Isomerism benzene? The nomenclature and isomerism of aromatic Solution hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:

1.23Suggest The Most Important Type Of Intermolecular Attractive Interaction In

Chemistry Class 11 · Chapter 1

73% match

1.23 Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water (iv) methanol and acetone (v) acetonitrile (CH3CN) and acetone (C3H6O). 1.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

📋 Question Details

Chapter
Qualitative Analysis
Topic
Lassaigne's Test
Year
2021
Shift
27 Jul Shift 2
Q Number
Q35
Type
MCQ
NCERT Ref
Class 11 Chemistry Ch 12: Organic Chemistry - Some Basic Principles and Techniques

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(1) E cell 0 = ϕright −ϕleft (2) E0cell = ϕleft + ϕright (3) E0cell = ϕleft −ϕright (4) E0cell = −(ϕleft + ϕright ) Q118.What will be the emf for the given cell Pt |H2 (P1)|H+(aq)||H2 (P2) ∣Pt (1) RT f log P1P2 (2) RT2f log P1P2 (3) RT f log P2P1 (4) none of these Q119.Which of the following reaction is possible at anode? (1) 2Cr3+ + 7H2O →Cr2O2−7 + 14H+ (2) F2 →2 F− (3) (1/2)O2 + 2H+ →H2O (4) none of these Q120.When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are (1) cathode - pure zinc anode - pure copper (2) cathode - impure sample anode - pure copper (3) cathode - impure zinc anode - impure sample (4) cathode - pure copper anode - impure sample JEE Main 2002 JEE Main Previous Year Paper Q121.Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively (1) sec−1, Msec−1 (2) sec−1, M (3) Msec −1, sec−1 (4) M, sec−1 Q122.For the reaction A + 2B →C , rate is given by R = [A][B]2 then the order of the reaction is (1) 3 (2) 6 (3) 5 (4) 7 Q123.The differential rate law for the reaction H2 + I2 →2HI is (1) −d[H2]dt = −d[I2]dt = −d[HI]dt (2) d[H2]dt = d[I2]dt = 12 1 d[Hl]dt 1 d[H2] (3) 2 dt = 12 d[I2]dt = −d[HI]dt (4) −2 d[H2]dt = −2 d[I2]dt = d[HI]dt Q124.If half-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is (1) 16 grams (2) 2 grams (3) 32 grams (4) 8 grams Q125.The integrated rate equation is Rt = log C0 −log Ct . The straight line graph is obtained by plotting (1) time vs log Ct (2) time1 vs Ct (3) time vs Ct (4) time1 vs Ct1 Q126.The formation of gas at the surface of tungsten due to adsorption is the reaction of order (1) 0 (2) 1 (3) 2 (4) insufficient data Q127.Aluminium is extracted by the electrolysis of (1) bauxite (2) alumina (3) alumina mixed with molten cryolite (4) molten cryolite Q128.The metal extracted by leaching with a cyanide is (1) Mg (2) Ag (3) Cu (4) Na Q129.Cyanide process is used for the extraction of (1) barium (2) aluminium (3) boron (4) silver Q130.When H2 S is passed through Hg2 S we get (1) HgS (2) HgS + Hg2 S (3) Hg2 S (4) Hg2 S2 Q131.In XeF2, XeF4, XeF6 the numebr of lone pairs of Xe are respectively (1) 2, 3, 1 (2) 1, 2, 3 (3) 4, 1, 2 (4) 3, 2, 1 Q132.In case of nitrogen, NCl3 is possible but not NCl5 while in case of phosphorous, PCl3 as well as PCl5 are possible. It is due to JEE Main 2002 JEE Main Previous Year Paper (1) availability of vacant d orbitals in P but not in N (2) lower electronegativity of P than N (3) lower tendency of H - bond formation in P than (4) occurrence of P in solid while N in gaseous state N at room temperature Q133.Number of sigma bonds in P4O10 is (1) 6 (2) 7 (3) 17 (4) 16 Q134.Most common oxidation states of Ce (cerium) are (1) +2, +3 (2) +2, +4 (3) +3, +4 (4) +3, +5 Q135.Arrange Ce+3, La+3, Pm+3 and Yb+3 in increasing order of their ionic radii (1) Yb+3 < Pm+3 < Ce+3 < La+3 (2) Ce+3 < Yb+3 < Pm+3 < La+3 (3) Yb+3 < Pm+3 < La+3 < Ce+3 (4) Pm+3 < La+3 < Ce+3 < Yb+3 Q136.Which of the following ions has the maximum magnetic moment ? (1) Mn+2 (2) Fe+2 (3) Ti+2 (4) Cr+2 Q137.Which is the correct order of ionic sizes ? (Atomic Number : Ce = 58, Sn = 50, Yb = 70 and Lu = 71) (1) Ce > Sn > Yb > Lu (2) Sn > Ce > Lu > Yb (3) Lu > Yb > Sn > Ce (4) Sn > Yb > Ce > Lu Q138.When KMnO4 acts as an oxidising agent and ultimately forms [MnO4]−1, MnO2, Mn2O3, Mn+2 then the number of electrons transferred in each case respectively is (1) 4, 3, 1, 5 (2) 1, 5, 3, 7 (3) 1, 3, 4, 5 (4) 3, 5, 7, 1 Q139.A square planar complex is formed by hybridisation of which atomic orbitals ? (1) s, px, py, dyz (2) s, px, py, dx2−y2 (3) s, px, py, dz2 (4) s, py, pz, dxy Q140.The type of isomerism present in nitropentamine chromium (III) chloride is (1) optical (2) linkage (3) ionization (4) polymerisation Q141.The most stable ion is (1) [Fe(OH)3]3− (2) [Fe(Cl)6]3− (3) [Fe(CN)6]3− (4) [Fe(H2O)6]3+ Q142. CH3 −Mg −Br is an organo metallic compound due to (1) Mg −Br bond (2) C - Mg bond (3) C - Br bond (4) C −H bond Q143.What is the product when acetylene reacts with hypochlorous acid ? (1) CH3COCl (2) ClCH2CHO (3) Cl2CHCHO (4) ClCHCOOH JEE Main 2002 JEE Main Previous Year Paper Q144.The reaction: (CH3)3C −Br ⟶(CH3)3H2O −C −OH (1) elimination reaction (2) substitution reaction (3) free radical reaction (4) displacement reaction Q145.On vigorous oxidation by permanganate solution (CH3)2C = CH −CH2 −CHO gives (1) (2) (3) (4) Q146. CH3CH2COOH ⟶Cl2 A alc.⟶KOH B . What is B ? red P (1) CH3CH2COCl (2) CH3CH2CHO (3) CH2 = CHCOOH (4) ClCH2CH2COOH Q147.When primary amine reacts with chloroform in ethanoic KOH then the product is (1) an isocyanide (2) an aldehyde (3) a cyanide (4) an alcohol Q148.Polymer formation from monomers starts by (1) condensation reaction between monomers (2) coordinate reaction between monomers (3) conversion of monomer to monomer ions by (4) hydrolysis of monomers protons Q149. The compound is used as (1) antiseptic (2) antibiotic (3) analgesic (4) pesticide Q150.RNA is different from DNA because RNA contains (1) ribose sugar and thymine (2) ribose sugar and uracil (3) deoxyribose sugar and thymine (4) deoxyribose sugar and uracil JEE Main 2002 JEE Main Previous Year Paper Q151.The functional group, which is found in amino acid is (1) COOH group (2) NH2 group (3) CH3 group (4) both (a) and (b) Q152.If a, b, c are distinct +ve real numbers and a2 + b2 + c2 = 1 then ab + bc + ca is (1) less than 1 (2) equal to 1 (3) greater than 1 (4) any real no. Q153.If α ≠β but α2 = 5α −3 and β2 = 5β −3 then the equation having α/β and β/α as its roots is (1) 3x2 −19x + 3 = 0 (2) 3x2 + 19x −3 = 0 (3) 3x2 −19x −3 = 0 (4) x2 −5x + 3 = 0 Q154.Difference between the corresponding roots of x2 + ax + b = 0 and x2 + bx + a = 0 is same and a ≠b, then (1) a + b + 4 = 0 (2) a + b - 4 = 0 (3) a - b - 4 = 0 (4) a - b + 4 = 0 Q155.Product of real roots of the equation t2x2 + |x| + 9 = 0 (1) is always positive (2) is always negative (3) does not exist (4) none of these Q156.If p and q are the roots of the equation x2 + px + q = 0, then (1) p =1, q = -2 (2) p = 0, q = 1 (3) p = -2, q = 0 (4) p = -2, q = 1 Q157.If 2a + 3b + 6c = 0(a, b, c ∈R) then the quadratic equation ax2 + bx + c = 0 has (1) at least one root in [0, 1] (2) at least one root in [2, 3] (3) at least one root in [4, 5] (4) none of these Q158. z and w are two non zero complex no.s such that |z| = |w| and Arg z + Arg w = π then z equals –– (1) W (2) - W (3) W (4) - W Q159.If |z −4| < |z −2|, its solution is given by (1) Re(z) > 0 (2) Re(z) < 0 (3) Re (z) > 3 (4) Re(z) > 2 Q160.The locus of the centre of a circle which touches the circle |z −z1| = a and |z −z2| = b externally ( z, z1 and z2 are complex numbers) will be (1) an ellipse (2) a hyperbola (3) a circle (4) none of these Q161.Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are (1) 216 (2) 375 (3) 400 (4) 720 Q162.Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed) is JEE Main 2002 JEE Main Previous Year Paper (1) 125 (2) 105 (3) 375 (4) 625 Q163.Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4, 6 and 7 without repetition. Total number of such numbers are (1) 312 (2) 3125 (3) 120 (4) 216 Q164.If 1, log9 (31−x + 2), log3 (4.3x −1) are in A.P. then x equals (1) log3 4 (2) 1 + log3 4 (3) 1 - log3 4 (4) log4 3 Q165.The value of 21/4, 41/8, 81/6 + … … ∞ is (1) 1 (2) 2 (3) 3/2 (4) 4 Q166.Fifth term of a GP is 2, then the product of its 9 terms is (1) 256 (2) 512 (3) 1024 (4) none of these Q167.Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is (1) 5 (2) 3/5 (3) 8/5 (4) 1/5 Q168. 13 −23 + 33 −43 + … . +93 = (1) 425 (2) -425 (3) 475 (4) -475 Q169.The sum of integers from 1 to 100 that are divisible by 2 or 5 is (1) 3000 (2) 3050 (3) 3600 (4) 3250 Q170.If an = √7 + √7 + √7 + … . having n radical signs then by methods of mathematical induciton which is true (1) an > 7∀n ≥1 (2) an > 7∀n ≥1 (3) an < 4∀n ≥1 (4) an < 3∀n ≥1 Q171.The coefficients of xp and xq in the expansion of (1 + x)p+q are (1) equal (2) equal with opposite signs (3) reciprocals of each other (4) none of these Q172.If the sum of the coefficients in the expansion of (a + b)n is 4096 , then the greatest coefficient in the expansion is (1) 1594 (2) 792 (3) 924 (4) 2924 Q173.The positive integer just greater than (1 + 0.0001)10000 is JEE Main 2002 JEE Main Previous Year Paper (1) 4 (2) 5 (3) 2 (4) 3 Q174. r and n are positive integers r > 1, n > 2 and coefficient of (r + 2)th term and 3rth term in the expansion of (1 + x)2n are equal, then n equals (1) 3r (2) 3r + 1 (3) 2r (4) 2r + 1 Q175.The period of sin2 θ is (1) π2 (2) π (3) 2π (4) π/2 Q176.The number of solution of tan x + sec x = 2 cos x in [0, 2π) is (1) 2 (2) 3 (3) 0 (4) 1 Q177.A triangle with vertices (4, 0), (-1, -1), (3, 5) is (1) isosceles and right angled (2) isosceles but not right angled (3) right angled but not isosceles (4) neither right angled nor isoceles Q178.The sides of a triangle are 3x + 4y, 4x+37 and 5x + 57 where x, y > 0 then the triangle is (1) right angled (2) obtuse angled (3) equilateral (4) none of these Q179.If the pair of lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 intersect on the y - axis then (1) 2fgh = bg2 + ch2 (2) bg2 ≠ch2 (3) abc = 2fgh (4) none of these Q180.The point of lines represented by 3ax2 + 5xy + (a2 −2)y2 = 0 and perpendicular to each other for (1) two values of a (2) ∀a (3) for one value of a (4) for no values of a Q181.Locus of mid point of the portion between the axes of x cos α + y sin α = p where p is constant is (1) x2 + y2 = 4 (2) x2 + y2 = 4p2 p2 (3) 1 + 1 = 2 (4) 1 + 1 = 4 x2 y2 p2 x2 y2 p2 Q182.If the chord y = mx + 1 of the circle x2 + y2 = 1 subtends an angle of measure 450 at the major segment of the circle then value of m is (1) 2 ± √2 (2) −2 ± √2 (3) −1 ± √2 (4) none of these Q183.The centres of a set of circles, each of radius 3 , lie on the circle x2 + y2 = 25. The locus of any point in the set is (1) 4 ≤x2 + y2 ≤64 (2) x2 + y2 ≤25 (3) x2 + y2 ≥25 (4) 3 ≤x2 + y2 ≤9 Q184.The centre of the circle passing through (0, 0) and (1, 0) and touching the circle x2 + y2 = 9 is JEE Main 2002 JEE Main Previous Year Paper , −√2) (1) ( 21 , 12 ) (2) ( 12 (3) ( 23 , 12 ) (4) ( 12 , 32 ) Q185.Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are (1) x = ±(y + 2a) (2) y = ±(x + 2a) (3) x = ±(y + a) (4) y = ±(x + a) Q186. (1) 1 (2) -1 (3) zero (4) does not exist Q187. (1) e4 (2) e2 (3) e3 (4) 1 Q188.Let f(x) = 4 and f ′(x) = 4. Then Limx→2 xf(2)−2f(x)x−2 is given by (1) 2 (2) -2 (3) -4 (4) 3 Q189. 1p + 2p + 3p + … + np Limn→∞ np+1 is (1) 1 (2) 1 p+1 1−p (3) p 1 − p−11 (4) p+21 denotes greatest integer less than or equal to x) Q190. Limx→0 log xn−[x][x] , n ∈N([x] (1) has value -1 (2) has value 0 (3) has value 1 (4) does not exist Q191.If f(1) = 1, f ′(1) = 2, then Limx→1 √f(x)−1 is √x−1 (1) 2 (2) 4 (3) 1 (4) 1/2 Q192.In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? (1) 73 (2) 65 (3) 68 (4) 74 Q193.The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is JEE Main 2002 JEE Main Previous Year Paper (1) x2 + y2 = 9a2 (2) x2 + y2 = 16a2 (3) x2 + y2 = 4a2 (4) x2 + y2 = a2 Q194.In a triangle with sides a, b, c, r1 > r2 > r3 (which are the ex-radii) then (1) a > b > c (2) a < b < c (3) a > b and b < c (4) a < b and b > c Q195. log l p 1 l, m, n are the pth , qth and rth term of a G.P. all positive, then log m q 1 equals log n r 1 (1) -1 (2) 2 (3) 1 (4) 0 Q196. a b ax + b If a > 0 discriminant of ax2 + 2bx + c is -ve, then b c bx + c is ax + b bx + c 0 (1) +ve (2) (ac −b2) (ax2 + 2bx + c) (3) -ve (4) 0 Q197. cot−1(√cos α) = tan−1(√cos α) = x, then sin x = (1) tan2 ( α2 ) (2) cot2 ( α2 ) (3) tan α (4) cot ( α2 ) Q198.The domain of sin−1 [log3(x/3)] is (1) [1, 9] (2) [-1,9] (3) [-9, 1] (4) [-9, -1] Q199.Which one is not periodic (1) |sin 3x| + sin2 x (2) cos √x + cos2 x (3) cos 4x + tan2 x (4) cos 2x + sin x Q200.If f(x + y) = f(x) ⋅f(y)∀x ⋅y and f(5) = 2, f ′(0) = 3 then f ′(5) is (1) 0 (2) 1 (3) 6 (4) 2 Q201.f is defined in [-5, 5] as f(x) = x if x is rational and = -x is irrational. Then (1) f(x) is continuous at every x, except x = 0 (2) f(x) is discontinuous at every x, except x = 0 (3) f(x) is continuous everywhere (4) f(x) is discontinuous everywhere n d2y dy (1 + x2) dx2 + x dx is Q202.If y = (x + √1 + x2) , then (1) n2y (2) −n2y (3) −y (4) 2x2y Q203.The maximum distance from origin of a point on the curve x = a sin t −b sin ( atb ) y = a cos t −b cos ( atb ), both a, b > 0 is (1) a - b (2) a + b (3) √a2 + b2 (4) √a2 −b2 JEE Main 2002 JEE Main Previous Year Paper Q204. ∫10π0 | sin x|dx is (1) 20 (2) 8 (3) 10 (4) 18 xdx then Limn→∞n [In + In−2] equals Q205. In = ∫π/40 tann (1) 1/2 (2) 1 (3) ∞ (4) zero is Q206. ∫ 0√2 [x2]dx (1) 2 −√2 (2) 2 + √2 (3) √2 −1 (4) √2 −2 Q207. ∫π−π 2x(1+sin1+cos2 xx) dx is (1) π2 (2) π2 4 (3) zero (4) π 2 Q208.If y = f(x) makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then ∫20 xf ′(x)dx is (1) 3/2 (2) 1 (3) 5/4 (4) -3/4 Q209.The area bounded by the curves y = ln x, y = ln |x|, y = | ln x| and y = | ln ||x| is (1) 4 sq. units (2) 6 sq. units (3) 10 sq. units (4) none of these d3y Q210.The order and degree of the differential equation 2/3 are + 3 dx = 4 dx3 (1 dy ) (1) (1, 32 ) (2) (3, 1) (3) (3, 3) (4) (1, 2) Q211.The solution of the equation d2y = e−2x dx2 (1) e−2x (2) e−2x 4 4 + cx + d (3) 4 1 e−2x + cx2 + d (4) 14 e−4x + cx + d Q212. f(x) and g(x) are two differentiable functions on [0, 2] such that f ′′(x) −g′′(x) = 0 f ′(1) = 2g′(1) = 4f(2) = 3g(2) = 9 then f(x) −g(x) at x = 3/2 is (1) 0 (2) 2 (3) 10 (4) 5 Q213.If |→a| = 4, |→b| = 2 and the angle between →a and →b is π/6 then (→a × →b)2 = 2 is equal to (1) 48 (2) 16 (3) →a (4) none of these Q214. If →a,→b, →c are vectors such that |→a→b→c| = 4 then (1) 16 (2) 64 (3) 4 (4) 8 JEE Main 2002 JEE Main Previous Year Paper Q215.If →a,→b, →c are vectors such that →a + →b + →c = 0 and |→a| = 7, |→b| = 5, |→c| = 3 then angle between vector →b and →c is (1) 60∘ (2) 30∘ (3) 45∘ (4) 90∘ Q216.If |a| = 5, |b| = 4, |c| = 3 thus what will be the value of |a ⋅b + b. c + c. a| , given that →a + →b + →c = 0 (1) 25 (2) 50 (3) -25 (4) -50 Q217. 3λ→c + 2μ(→a × →b) = 0 then (1) 3λ + 2μ = 0 (2) 3λ = 2μ (3) λ = μ (4) λ + μ = 0 Q218. →a = 3^i −5^j and →b = 6^i + 3^j are two vectors and →c is a vector such that →c = →a × →b then |→a| : |→b| : |→c| (1) √34 : √45 : √39 (2) √34 : √45 : 39 (3) 34 : 39 : 45 (4) 39 : 35 : 34 Q219.If →a × →b = →b × →c = →c × →a then →a + →b + →c = (1) abc (2) -1 (3) 0 (4) 2 Q220.The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are (1) 13, 5 (2) 12, 6 (3) 14, 4 (4) 11, 7 Q221.A plane which passes through the point (3, 2, 0) and the line x−41 = y−75 = z−44 is (1) x - y + z = 1 (2) x + y + z = 5 (3) x + 2y - z = 1 (4) 2x - y + z = 5 Q222.The d.r. of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle π/4 with plane x + y = 3 are (1) 1, √2, 1 (2) 1, 1, √2 (3) 1, 1, 2 (4) √2, 1, 1 Q223.A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is 2 1 , 13 and 14 . Probability that the problem is solved is (1) 3 (2) 1 4 2 (3) 2 (4) 1 3 3 Q224. A and B are events such that P(A ∪B) = 3/4, P(A ∩B) = 1/4, P(¯A) = 2/3 then P(¯A ∩B) is (1) 5/12 (2) 3/8 (3) 5/8 (4) 1/4 Q225.A die is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is JEE Main 2002 JEE Main Previous Year Paper (1) 8/3 (2) 3/8 (3) 4/5 (4) 5/4 JEE Main 2002 JEE Main Previous Year Paper

2002
Hard

Q40.The correct order of increasing basicity of the given conjugate bases (R = CH3) is –––––– (1) RCOO < HC = C < R < NH2 (2) R < HC ≡C < RCOO < NH2 ––––– (3) RCOO < NH2 < HC ≡C < R (4) RCOO < HC ≡C < NH2 < R

2010
Medium

Q43.Copper wire test for halogens is known as (1) Duma's Test (2) Beilstein's Test (3) Liebig's Test (4) Lassigne's Test

2012
Easy

Q45.Beilstein test is used for the estimation of which one of the following elements? (1) S (2) Cl (3) C and H (4) N

2012
Easy
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