Oxidation and Reduction – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance Questions
Q. Consider all possible isomeric ketones including stereoisomers of MW = 100, All these isomers are independently reacted with $\mathrm{NaBH}_{4}$ (NOTE : stereoisomers are also reacted separately).The total number of ketones that give a racemic product(s) is/are. [JEE 2014]

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Sol. (5)M. wt 100 of ketoneSo m. formula = $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}$\ (1 ; 2 ; 3 ; 6 ; 7)

Q. In the reaction,(A)Ethylene(B) Acetyl chloride(C) Acetaldehyde(D) Acetylene [JEE 2014]

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Sol. (A)

Q. The reagent needed for converting is :(A) $\mathrm{H}_{2} /$ Lindlar Cat.(B) Cat. Hydrogenation(D) $\mathrm{Li} / \mathrm{NH}_{3}$ [JEE 2014]

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Sol. (D)

Carbonyl Compound – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance QuestionsParagraph for Question Nos. 1 to 3A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
Q. The structure of the carbonyl compound P is [IIT 2009]

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Sol. (B)

Q. The structure of the products Q and R, respectively, are [IIT 2009]

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Sol. (A)

Q. The structure of the product S is [IIT 2009]

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Sol. (B)

Paragraph for Questions Nos. 4 to 5An acyclic hydrocarbon P, having molecular formula $\mathrm{C}_{6} \mathrm{H}_{10}$, gave acetone as the only organic product through the following sequence of reactions, in the which Q is an intermediate organic compound.
Q. The structure of compound P is – [IIT 2011]

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Sol. (D)

Q. The structure of the compound Q is – [IIT 2011]

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Sol. (B)

Q. The number of aldol reaction(s) that occurs in the given transformation is [IIT 2012]

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Sol. (C)

Q. Among P, Q, R and S, the aromatic compound(s) is / are : [IIT 2013]

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Sol. (A,B,C,D)

Q. After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are)(A) Reaction I : P and Reaction II : P(B) Reaction I : U, acetone and Reaction II : Q acetone(C) Reaction I : T, U, acetone and Reaction II : P(D) Reaction I : R, acetone and Reaction II : S acetone [IIT 2013]

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Sol. (C)In basic medium halogenation dose not stop with replacement of just one hydrogen and poly halogenation takes place because -haloketones are more reactive towards base and haloform reaction takes placeIn above reaction $\mathrm{Br}_{2}$ is limiting agents.

Q. The major product in the following reaction is [IIT 2014]

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Sol. (D)(i) Grignard prefer to give nucleophilic addition on polar $\pi$-bond and form anion intermediate.(ii) In next step anion give intramolecular nucleophilic substitution reaction & form 5 membered ring.

Q. The major product of the following reaction is – [IIT 2015]

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Sol. (A)Mechanism :

Q. In the following reactions, the product S is – [IIT 2015]

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Sol. (A)

Q. The reaction(s) leading to the formation of 1,3,5-trimethylbenzene is (are) [JEE Adv. 2018]

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Sol. (A,B,D)

Q. In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is____.(Atomic weight in g mol–1: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis) [JEE Adv. 2018]

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Sol. 495

Biomolecule – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. SimulatorPrevious Years JEE Advanced Questions
Q. The correct statement(s) about the following sugars X and Y is(are)(A) X is a reducing sugar and Y is a non‑reducing sugar(B) X is a non‑reducing sugar and Y is a reducing sugar(C) The glucosidic linkages in X and Y are $\alpha$ and $\beta$ , respectively.(D) The glucosidic linkages in X and Y are $\beta$ and $\alpha$ , respectively. [JEE 2009]

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Sol. (B,C)

Q. Among cellulose, poly vinyl chloride, nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is(A) Nylon(B) Poly (vinyl chloride)(C) Cellulose(D) Natural Rubber [JEE 2009]

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Sol. (D)Natural rubber is elastomer and has weak vander waal force of attraction and the unit (monomer) is

Q. The following carbohydrate is(A) a ketohexose (B) an aldohexose (C) an $\alpha$-furanose (D) an $\alpha$-pyranose [JEE 2011]

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Sol. (B)

Q. The correct statement about the following disaccharide is –(A) Ring (a) is pyranose with a–glycosidic link(B) Ring (a) is furanose with a–glycosidic link(C) Ring (b) is furanose with a–glycosidic link(D) Ring (b) is pyranose with b–glycosidic link [IIT-2010]

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Sol. (A)The ring is pyranose with glycosidic linkage which means the oxide linkage whichconnect the monosaccharide units in polysaccharides.

Q. The total number of basic groups in the following form of lysine is :(A) 3             (B) 0             (C) 2             (D) 1 [IIT-2010]

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Sol. (C)The two basic groups are $\mathrm{NH}_{2}$ and $\mathrm{COO}^{-}$ groups.

Q. The major product of the following reaction is(A) a hemiacetal          (B) an acetal           (C) an ether            (D) an ester [JEE 2011]

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Sol. (B)

Q. Amongst the compounds given, the one that would form a brilliant coloured dye on treatment with $\mathrm{NaNO}_{2}$ in dil. HCl followed by addition to an alkaline solution of $\beta$ naphthol is – [JEE 2011]

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Sol. (C)

Q. The correct functional group X and the reagent/reaction conditions Y in the following scheme are condensation polymer(A) $\mathrm{X}=\mathrm{COOCH}_{3}, \mathrm{Y}=\mathrm{H}_{2} / \mathrm{Ni} / \mathrm{heat}$(B) $\mathrm{X}=\mathrm{CONH}_{2}, \mathrm{Y}=\mathrm{H}_{2} / \mathrm{Ni} / \mathrm{heat}$(C) $\mathrm{X}=\mathrm{CONH}_{2}, \mathrm{Y}=\mathrm{Br}_{2} / \mathrm{NaOH}$(D) $\mathrm{X}=\mathrm{CN}, \mathrm{Y}=\mathrm{H}_{2} / \mathrm{Ni} / \mathrm{heat}$ [JEE 2011]

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Sol. (A,B,C,D)

Q. The structure of D-(+)-glucose isThe structure of L(–)-glucose is [IIT 2011]

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Sol. (A)The structure of D(+) glucose is

Q. A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is [JEE 2011]

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Sol. 6No. of peptide linkage $=$ No. of water molecules added for complete hydrolysis.= n – 1So, number of molecules of $\mathrm{H}_{2} \mathrm{O}$ added $=9$So total wt. of the product $=$ Mol. wt. of polypeptide $+$ total wt. of $\mathrm{H}_{2} \mathrm{O}$ added.= 796 + (9 × 16)= 796 + 162= 958$\therefore$ wt. of glycine obtained $=958 \times \frac{47}{100} \square 450$No. of units of glycine $=\frac{450}{75}=6$ units

Q. The substitutes $R_{1}$ and $R_{2}$ for nine peptides are listed in the table given below. How many of these peptides are positively charged at $p H=7.0 ?$ [JEE 2012]

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Sol. 4

Q. When the following aldohexose exists in its d-configuration , the total number of stereoisomers in its pyranose form is – [JEE 2012]

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Sol. 8

Q. A tetrapeptide has –COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (Primary structures) with –$\mathrm{NH}_{2}$ group attached to a chiral center is [JEE 2013]

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Sol. 4Considering alanin at one end of tetra peptides structure with $\mathrm{CO}_{2} \mathrm{H}$ group, number of possible combination are as follows:(i) $\&$ (ii) can not be the possible because in those combination $-\mathrm{NH}_{2}$ group is not attached to chiral centre.Hence, answer is (4)

Q. The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is [JEE 2014]

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Sol. 1

Q. Positive Tollen’s test is observed for [JEE – Adv. 2016]

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Sol. (A,B,C)Tollens’s test is given by compounds having aldehyde group. Also -hydroxy carbonyl gives positive tollen’s test.

Q. For ‘invert sugar’, the correct statement(s) is (are)(Given : specific rotations of (+)-sucrose, (+)-maltose, L-(–)-glucose and L-(+)-fructose in aqueous solution are $+66^{\circ},+140^{\circ},-52^{\circ}$ and $+92^{\circ}$, respectively)(A) ‘invert sugar’ is prepared by acid catalyzed hydrolysis of maltose(B) ‘invert sugar’ is an equimolar mixture of D-(+) glucose and D-(–)-fructose(C) specific rotation of ‘invert surgar’ is $-20^{\circ}$(D) on reaction with $\mathrm{Br}_{2}$ water, ‘invert sugar’ forms saccharic acid as one of the products [JEE – Adv. 2016]

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Sol. (B,C)Invert sugar is equailmolar mixture of D-glucose and D-fructose which is obtained by hydrolysis of sucroseSpecific rotation of mixture is half of sum of sp. rotation of both components $\frac{+52^{\circ}+\left(-92^{\circ}\right)}{2}$$=-20^{\circ} Paragraph (q. 17 TO q. 18)Treatment of compound O with \mathrm{KMnO}_{4} / \mathrm{H}^{+} gave P, which on heating with ammonia gave Q. The compound Q on treatment with \mathrm{Br}_{2} / \mathrm{NaOH} produced R. On strong heating, Q gave S, which on further treatmenet with ethyl 2-bromopropanoate in the presence of KOH following by acidification, gave a compound T. Q. The compound R is : [JEE – Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A) Q. The compound T is :(A) Glycine (B) Alanine (C) Valine (D) Serine [JEE – Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. The Fischer presentation of D-glucose is given below.The correct structure(s) of \beta-glucopyranose is (are) :- [JEE Adv. 2018] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D) Alcohol & Ether – JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years JEE Advance Questions Q. Amongst the following, the total number of compounds soluble in aqueous NaOH is JEE Advance 2010 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4) Q. In the reaction the products are JEE Advance 2010 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D) Q. The major product in the following reaction is(A) (B) (C) (D) JEE Advance 2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D)(ii) In next step anion give intramolecular nucleophilic substitution reaction & form 5 membered ring. Q. The acidic hydrolysis of ether (X) shown below is fastest when(A) one phenyl group is replaced by a methyl group(B) one phenyl group is replaced by a para-methoxyphenyl group(C) two phenyl groups are replaced by two para-methoxyphenyl group(D) no structural change is made to X JEE Advance 2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C) Q. The correct combination of names for isomeric alcohols with molecular formula \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} is/are-(A) tert-butanol and 2-methylpropan-2-ol(B) tert-butanol and 1, 1-dimethylethan-1-ol(C) n-butanol and butan-1-ol(D) isobutyl alcohol and 2-methylpropan-1-ol JEE Advance 2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D)(A,C,D)The combination of names for isomeric alcohols with molecular formula \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} is/are Q. Consider all possible isomeric ketones including stereoisomers of MW = 100, All these isomers are independently reacted with \mathrm{NaBH}_{4}(NOTE : stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are JEE Advance 2014 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 5 Q. The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one or more appropriate reagents in LIST-II.(given, order of migratory aptitude: aryl > alkyl > hydrogen)The correct option is(A) \mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 2,3 ; \mathrm{R} \rightarrow 1,4 ; \mathrm{S} \rightarrow 2,4(B) \mathrm{P} \rightarrow 1,5 ; \mathrm{Q} \rightarrow 3,4 ; \mathrm{R} \rightarrow 4,5 ; \mathrm{S} \rightarrow 3(C) \mathrm{P} \rightarrow 1,5 ; \mathrm{Q} \rightarrow 3,4 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2,4(D) \mathrm{P} \rightarrow 1,5 ; \mathrm{Q} \rightarrow 2,3 ; \mathrm{R} \rightarrow 1,5 ; \mathrm{S} \rightarrow 2,3 JEE Advance 2018 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D) Q. LIST-I contains reactions and LIST-II contains major products.Match each reaction in LIST-I with one or more product in LIST-II and choose the correct option.(A) \mathrm{P} \rightarrow 1,5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4$$(\mathrm{B}) \mathrm{P} \rightarrow 1,4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 3$(C) $\mathrm{P} \rightarrow 1,4 ; \mathrm{Q} \rightarrow 1,2 ; \mathrm{R} \rightarrow 3,4 ; \mathrm{S} \rightarrow 4$(D) $\mathrm{P} \rightarrow 4,5 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 3,4$ JEE Advance 2018

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Sol. (B)

Hydrocarbon – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.SimulatorPrevious Years JEE Advance Questions
Q. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are(A) $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$(B) $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$(C) $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}$(D) $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$ [IIT-2010]

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Sol. (D)

Q. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure.The correct order of their boiling point is(A) I > II > III(B) III > II > I(C) II > III > I(D) III > I > II [IIT-2014]

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Sol. (B)

Paragraph For Question 3 and 4Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.
Q. The product X is – [IIT-2014]

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Sol. (A)

Q. The correct statement with respect to prodcut Y is –(A) It gives a positive Tollens test and is a functional isomer of X(B) It gives a positive Tollens test and is a geometrical isomer of X(C) It gives a positive Iodoform test and is a functional isomer of X(D) It gives a positive Iodoform test and is a geometrical isomer of X [IIT-2014]

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Sol. (C )

Q. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are) – [IIT-2015]

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Sol. (B,D)

Paragraph For Questions 6 and 7In the following reaction
Q. Compound X is : [IIT-2015]

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Sol. (C)

Q. The major compound Y is : [IIT-2015]

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Sol. (D)

Halogen Derivatives – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years JEE Advance Questions
Q. In the reactionthe products are [IIT 2010]

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Sol. (D)

Q. The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is [IIT 2011]

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Sol. 5

Q. The maximum number of isomers (including stereoisomers) that are possible on mono-chlorination of the following compounds, is [IIT 2011]

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Sol. 8

Q. KI in acetone, undergoes $\mathrm{S}_{\mathrm{N}} 2$ reaction with each of P, Q, R and S. The rates of the reaction vary as – .(A) P > Q > R > S(B) S > P > R > Q(C) P > R > Q > S(D) R > P > S > Q [IIT 2013]

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Sol. (B)

Q. The reactivity of compound Z with different halogens under appropriate conditions is given below-The observed pattern of electrophilic substitution can be explained by –(A) The steric effect of the halogen(B) The steric effect of the tert-butyl group(C) The elctronic effect of the phenolic group(D) The electronic effect of the turt-butyl group [IIT 2014]

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Sol. (A,B,C)The observed pattern of electrophilic substitution can be explained by –(A) The steric effect of the halogen(B) The steric effect of the tert-butyl group(C) The elctronic effect of the phenolic group(D) The electronic effect of the turt-butyl group

General Organic Chemistry – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years JEE Advance Questions
Q. The correct acidity order of the following is :(A) (III) > (IV) > (II) > (I)(B) (IV) > (III) > (I) > (II)(C) (III) > (II) > (I) > (IV)(D) (II) > (III) > (IV) > (I) [IIT-JEE-2009]

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Sol. (A)(A) Acidity $\propto-R /-I \propto \frac{1}{+R /+I}$

Q. The correct stability order of the following resonance structures is [IIT-2009]

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Sol. (B)

Q. Amongst the following, the total number of compounds soluble in aquesous NaOH is: [IIT-JEE-2010]

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Sol. 4(4) Compound which is more acidic than water is soluble in NaOH

Q. The total number of contributing structures showing hyperconjugation (involving C–H bonds) for the following carbocation is. [IIT-2011]

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Sol. 6$6 \propto \mathrm{H}$ ore present.

Q. Among the following compounds, the most acidic is –(A) p-nitrophenol(B) p-hydroxybenzoic acid(C) o-hydroxybenzoic acid(D) p-toluic acid [IIT-JEE-2011]

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Sol. (C)

Q. Which of the following molecules, in pure from , is (are) unstable at room temperature ? [IIT-2012]

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Sol. (B,C)Anti – Aromatic : B,C

Q. The carboxyl functional group (–COOH) is present in –(A) picric acid(B) barbituric acid(C) ascorbic acid(D) aspirin [IIT-JEE-2012]

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Sol. (D)

Q. Identify the binary mixtures (s) that can be separated into the individual compounds, by differential extraction, as shown in the given scheme –(A) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$(B) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$(C) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$(D) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$ [IIT-JEE-2012]

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Sol. (B,D)

Q. The hyperconjugative stbilities of tert-butyl cation and 2-butene, respectively, are due to –(A) $\sigma \rightarrow \mathrm{p}$ (empty) and $\sigma \rightarrow \pi$ electron delocalisations(B) $\sigma \rightarrow \sigma$ and $\sigma \rightarrow \pi$ electron delocalisations(C) $\sigma \rightarrow \mathrm{p}(\text { filled })$ and $\sigma \rightarrow \pi$ electron delocalisations(D) $\mathrm{p}$ (filled) $\rightarrow \sigma$ and $\sigma \rightarrow \pi$ electron delocalisations [IIT-2013]

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Sol. (A)

Q. The total number of lone-pairs of electrons in melamine is – [IIT-2013]

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Sol. 6

Q. The compound that does NOT liberate $\mathrm{CO}_{2}$, on treatment with aqueous sodium bicarbonate solution, is –(A) Benzoic acid(B) Benzenesulphonic acid(C) Salicylic acid(D) Carbolic acid (phenol) [JEE-ADVANCED-2013]

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Sol. (D)(D) Compound which is more acidic than $\mathrm{H}_{2} \mathrm{CO}_{3}$ liberate $\mathrm{CO}_{2}$ gas.

Q. Hydrogen bonding plays a central role in the following phenomena(A) Ice floats in water(B) Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions(C) Formic acid is more acidic than acetic acid(D) Dimerisation of acetic acid in benzene [JEE-ADVANCED-2014]

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Sol. (A,B,D)

Q. The number of resonance structures for N is : [IIT-2015]

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Sol. 9

Aromatic Compounds – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years JEE Advance Questions
Q. In the reaction the intermediate(s) is(are) – [JEE 2009]

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Sol. (A,B,C)

Q. The compounds P, Q and Swere separately subjected to nitration using $\mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}$ mixture. The product formed in each case respectively, is [JEE 2009]

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Sol. (C)(1)

Q. The correct acidity order of the following is :(A) (III) > (IV) > (II) > (I)(B) (IV) > (III) > (I) > (II)(C) (III) > (II) > (I) > (IV)(D) (II) > (III) > (IV) > (I) [JEE 2009]

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Sol. (A)

Q. In the following carbocation, $\mathrm{H} / \mathrm{CH}_{3}$ that is most likely to migrate to the positively charged carbon is :(A) $\mathrm{CH}_{3}$ at $\mathrm{C}-4$(B) H at C – 4(C) $\mathrm{CH}_{3}$ at $\mathrm{C}-2$(D) H at C – 2 [JEE 2009]

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Sol. (D)

Q. The structure of the Product T is : [JEE 2010]

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Sol. (C)

Q. Among the following compounds, the most acidic is(A) p-nitrophenol(B) p-hydroxybenzoic acid(C) o-hydroxybenzoic acid(D) p-toluic acid [JEE 2011]

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Sol. (C)

Q. The maximum number of isomers (including stereoisomers) that are possible on mono-chlorination of the following compounds, is [JEE 2011]

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Sol. 8

Q. Among P, Q, R and S, the aromatic compound(s) is / are :(A) P (B) Q (C) R (D) S [JEE 2013]

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Sol. (A,B,C,D)

Q. The major product(s) of the following reaction is (are) –(A) P (B) Q (C) R (D) S [JEE 2013]

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Sol. (B)

Q. In the following reaction, the product (s) formed is (are)(A) P (major) (B) Q (minor) (C) R (minor) (D) S (major) [JEE 2013]

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Sol. (B,D)

Q. In the following reaction sequences V and W are , respectively – [JEE 2013]

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Sol. (A)

Paragraph for Question 11 and 12 P & Q are isomers of dicarboxylic acid $\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}$. Both decolourize $\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}$, On heating P forms the cyclic anhydride. Upon treatment with dilute alkaline $\mathrm{KMnO}_{4}$, P as well as Q could produce one or more than one from S, T and U. [JEE 2013]
Q. Compounds formed from P and Q are respectively(A) Optically active S and optically active pair (T, U)(B) Optically inactive S and optically inactive pair (T, U)(C) Optically active pair (T, U) and optically active S(D) Optically inactive pair (T, U) and optically inactive S [JEE 2013]

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Sol. (B)

Q. In the reaction shown below, the major product(s) formed is / are : [JEE 2014]

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Sol. (A)

Q. The reactivity of compound Z with different halogens under appropriate conditions is given below-The observed pattern of electrop hilic substitution can be explained by –(A) The steric effect of the halogen(B) The steric effect of the tert-butyl group(C) The electronic effect of the phenolic group(D) The electronic effect of the turt-butyl group [JEE 2014]

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Sol. (A,B,C)

Q. Match the four starting materials (P , Q, R, S) given in List I with the corresponding reaction scheme (I, II, III, IV) provided in List – II and select the correct answer using the code given below in lists. [JEE 2014]

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Sol. (C)

Q. In the reactionthe product E is :- [IIT-2015]

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Sol. (A)

Q. In the following sequence of reactions :(A) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$(B) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$(C) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$(D) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$ [IIT 2015]

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Sol. (B)

Q. The major product U in the following reactions is : [IIT 2015]

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Sol. (C)

Q. In the following reactions, the major product W is : [IIT 2015]

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Sol. (A)

Q. Among the following the number of reaction(s) that produce(s) benzaldehyde is : [IIT 2015]

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Sol. (D)

s-block – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance Questions
Q. The compound(s) formed upon combustion of sodium metal in excess air is (are)(A) $\mathrm{Na}_{2} \mathrm{O}_{2}$(B) $\mathrm{Na}_{2} \mathrm{O}$(C) $\mathrm{NaO}_{2}$(D) $\mathrm{NaOH}$ [JEE 2009]

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Sol. (A,B)Sodium does not form peroxide on reaction with excess air.

Atomic Structure -JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance QuestionsParagraph for questions 1 to 3The hydrogen-like species $\mathbf{L} \mathbf{i}^{2+}$ is in a spherically symmetric state $\mathrm{S}_{1}$ with one radial node. Upon absorbing light the ion undergoes transition to a state $\mathrm{S}_{2}$ The state $\mathrm{S}_{2}$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.
Q. The state S1 is :-(A) 1s (B) 2s (C) 2p (D) 3s [JEE 2010]

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Sol. (B)$\mathrm{S}_{1}=2 \mathrm{s}$$\mathrm{S}_{2}=3 \mathrm{p} Q. Energy of the state \mathrm{S}_{1} in units of the hydrogen atom ground state energy is :-(A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50 [JEE 2010] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C)\mathrm{E}=13.6 \times \frac{3}{4} Q. The orbital angular momentum quantum number of the state \mathrm{S}_{2} is :-(A) 0 (B) 1 (C) 2 (D) 3 [JEE 2010] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. The maximum number of electrons that can have principal quantum number, n=3, and spin quantum number, ms = – 1/2, is [JEE 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 9 Q. The work function (\phi) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is : : [JEE 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 4 Q. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [\mathrm{a}_{0} is Bohr radius](\mathrm{A}) \frac{\mathrm{h}^{2}}{4 \pi^{2} \mathrm{ma}_{0}^{2}}$$(\mathrm{B}) \frac{\mathrm{h}^{2}}{16 \pi^{2} \mathrm{ma}_{0}^{2}}$(C) $\frac{\mathrm{h}^{2}}{32 \pi^{2} \mathrm{ma}_{0}^{2}}$(D) $\frac{\mathrm{h}^{2}}{32 \pi^{2} \mathrm{ma}_{0}^{2}}$ [JEE 2012]

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Sol. (C)

Q. The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of the de Broglie wavelength of He gas at $-73^{\circ} \mathrm{C}$ is “M” times that of the de Broglie wavelength of Ne at $727^{\circ} \mathrm{C}$. M is. [JEE 2013]

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Sol. (5)

Q. In an atom, the total number of electrons having quantum numbers n = $4,\left|\mathrm{m}_{\ell}\right|=1$ and $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2} \mathrm{is}$ [JEE 2014]

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Sol. (6)

Q. P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is $4 \pi \mathrm{r}^{2} \mathrm{dr}$. The qualitative sketch of the dependence of P on r is – [JEE Adv. 2017]

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Sol. (B)

Answer Q.10, Q.11 and Q.12 by appropriately matching the information given in the three columns of the following table.The wave function n, l , m1 is a mathematical function whose value depends upon spherical polar coordinates $(\mathrm{r}, \theta, \phi)$ of the electron and characterized by the quantum numbers n, l and $\mathrm{m}_{1}$. Here r is distance from nucleus,  is colatitude and  is azimuth. In the mathematical functions given in the Table, Z is atomic number $\mathrm{a}_{0}$is Bohr radius.
Q. For the given orbital in column 1, the only CORRECT combination for anyhydrogen – like species is :(A) (IV) (iv) (R)(B) (II) (ii) (P)(C) (III) (iii) (P)(D) (I) (ii) (S) [JEE – Adv. 2017]

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Sol. (B)(A) (IV) (iv) (R)  incorrect, because, $\mathrm{d}_{\mathrm{z}^{2}}$ has no nodal plane.(B) (II) (ii) (P)  correct, because 2s orbtial has 1 radial node.(C) (III) (iii) (P)  incorrect, because probability density for 2p at nucleus is zero.(D) (I) (ii) (S)  incorrect, because 1s orbital has no radial node.

Q. For He+ ion, the only INCORRECT combination is(A) (II) (ii) (Q)(B) (I) (i) (S)(C) (I) (i) (R)(D) (I) (iii) (R) [JEE – Adv. 2017]

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Sol. (D)The option (D) is incorrect because in the wave function of 1s orbital , no angular function should be present.

Q. For hydrogen atom, the only CORRECT combination is(A) (I) (iv) (R)(B) (I) (i) (P)(C) (II) (i) (Q)(D) (I) (i) (S) [JEE – Adv. 2017]

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Sol. (D)We have to select only correct combination hence, the option (D) is correct.

Chemical Equilibrium – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance Questions
Q. The thermal dissociation equilibrium of $\mathrm{CaCO}_{3}(\mathrm{s})$ is studied under different conditions.$\mathrm{CaCO}_{3}(\mathrm{s}) \square \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$For this equilibrium, the correct statement(s) is(are)(A) $\Delta \mathrm{H}$ is dependent on $\mathrm{T}$(B) $\mathrm{K}$ is independent of the initial amount of $\mathrm{CaCO}_{3}$(C) $\mathrm{K}$ is dependent on the pressure of $\mathrm{CO}_{2}$ at a given $\mathrm{T}$(D) $\Delta \mathrm{H}$ is independent of the catalyst, if any [JEE 2013]

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Sol. (A,B,D) s independent of catalyst but dependent on temperature

Paragraph 1 (2 TO 3)Thermal decomposition of gaseous $\mathrm{X}_{2}$ to gaseous X at 298 K takes place according to the following equation :The standard reaction Gibbs energy, , of this reaction is positive. At the start of thereaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles ofX formed is given is the number of moles of X formed at equilibrium. Thereaction is carried out at a constant total pressure of 2bar. Consider the gases to behave ideally.(Given $\left.: \mathrm{R}=0.083 \mathrm{L} \text { bar } \mathrm{K}^{-1} \mathrm{mol}^{-1}\right)$
Q. The equilibrium constant KP for this reaction at 298 K, in terms of , is [JEE – Adv. 2016]

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Sol. (B)

Q. The INCORRECT statement among the following , for this reaction, is(A) Decrease in the total pressure will result in formation of more moles of gaseous X(B) At the start of the reaction, dissociation of gaseous $\mathrm{X}_{2}$ takes place spontaneously(C) $\beta_{\text {equilibrium }}=0.7$(D) $\mathrm{K}_{\mathrm{C}}<1$ [JEE – Adv. 2016]

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Sol. (C)(A) On decreasing $\mathrm{P}_{\mathrm{T}}\left[\mathrm{Q}=\frac{\mathrm{n}_{x} \mathrm{P}_{\mathrm{T}}}{\mathrm{n}_{\mathrm{x}_{2}} \mathrm{n}_{\mathrm{T}}}\right]$ Q will be less than Kp reaction will move in forward direction(B) At the start of the reaction $\Delta \mathrm{G}=\Delta \mathrm{G}^{0}+\mathrm{RT} \ln \mathrm{Q}$$\mathrm{t}=0, \mathrm{Q}=0 \Rightarrow \Delta_{\mathrm{rxn}} \mathrm{G}=-\mathrm{ve} \quad (spontaneous) Q. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are\Delta_{f} \mathrm{G}^{\circ}[\mathrm{C}(\text { graphite })]=0 \mathrm{kJ} \operatorname{mol}^{-1} \Delta_{f} \mathrm{G}^{\circ}[\mathrm{C}(\text { diamond })]=2.9 \mathrm{kJ} \mathrm{mol}^{-1}The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 \times 10^{-6} \mathrm{m}^{3} \mathrm{mol}^{-1}. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : \left.1 \mathrm{J}=1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2} ; 1 \mathrm{Pa}=1 \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-2} ; 1 \text { bar }=10^{5} \mathrm{Pa}\right](A) 14501 bar (B) 29001 bar (C) 58001 bar (D) 1405 bar [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A) Ionic Equilibrium- JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance Questions Q. The dissociation constant of a substituted benzoic acid at 25^{\circ} \mathrm{C} is 1.0 \times 10^{-4} . The pH of a 0.01 M solution of its sodium salt is [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 8\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\frac{10^{-14}}{10^{-4}}=10^{-10}$$\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{C}_{0}}}=\sqrt{\frac{10^{-10}}{10^{-2}}}=10^{-4}$$\mathrm{pH}=\frac{1}{2}[\mathrm{pKw}+\mathrm{pKa}+\log \mathrm{Co}]$$\mathrm{pH}=\frac{1}{2}\left[14+4+\log 10^{-2}\right]$$\mathrm{pH}=\frac{1}{2}[16]=8 Q. Aqueous solutions of \mathrm{HNO}_{3}, \mathrm{KOH}, \mathrm{CH}_{3} \mathrm{COOH} and \mathrm{CH}_{3} \mathrm{COONa} of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are)(A) \mathrm{HNO}_{3} and \mathrm{CH}_{3} \mathrm{COOH}(B) \mathrm{KOH} and \mathrm{CH}_{3} \mathrm{COONa}(C) \mathrm{HNO}_{3} and \mathrm{CH}_{3} \mathrm{COONa}(D) \mathrm{CH}_{3} \mathrm{COOH} and \mathrm{CH}_{3} \mathrm{COONa} Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C,D) Q. In 1 L saturated solution of AgCl \left[\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.6 \times 10^{-10}\right], 0.1 \mathrm{mol} of \mathrm{CuCl}$$\left[\mathrm{K}_{\mathrm{sp}}(\mathrm{CuCl})=1.0 \times 10^{-6}\right]$ is added.The resultant concentration of $\mathrm{Ag}^{+}$ in the solution is 1.6 × $10^{-x}$. The value of ‘x’ is. [JEE -2011]

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Sol. 7

Q. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is $1 / 100^{\text {th }}$ of that of a strong acid (HX, 1M), at $25^{\circ} \mathrm{C}$. The $\mathrm{K}_{\mathrm{a}}$ of HA is(A) $1 \times 10^{-4}$(B) $1 \times 10^{-5}$(C) $1 \times 10^{-6}$(D) $1 \times 10^{-3}$ [JEE 2013]

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Sol. (A)

Q. The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is $1.1 \times 10^{-12}$ at 298 K. The solubility (in mol/L) of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ in a $0.1 \mathrm{M} \mathrm{AgNO}_{3}$solution is(A) $1.1 \times 10^{-11}$(B) $1.1 \times 10^{-10}$(C) $1.1 \times 10^{-12}$(D) $1.1 \times 10^{-9}$ [JEE 2013]

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Sol. (B)

Q. Dilution process of different aqueous solutions; with water, are given in LIST-I. Theeffects of dilution of the solutions on $\left[\mathrm{H}^{+}\right]$ are given in LIST II.(Note : Degree of dissociation () of weak acid and weak base is << 1; degree of hydrolysis of salt <<1; $\left[\mathrm{H}^{+}\right]$ represents the concentration of $\mathrm{H}^{+}$ ions)Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is$(\mathrm{A}) \mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 1$(B) $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 3$(C) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 3$$(\mathrm{D}) \mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 1 [JEE- Adv. – 2018] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D)correct match : Q-5 Mole Concept – JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Q. Given that the abundances of isotopes ^{54} \mathrm{Fe},^{56} Fe and ^{57} Fe are 5 \%, 90 \% and 5 \%, respectively, the atomic mass of Fe is :(A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05 [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B)\mathrm{M}_{\mathrm{AVg}}=\frac{\mathrm{M}_{1} \mathrm{X}_{1}+\mathrm{M}_{2} \times \mathrm{X}_{2}+\mathrm{M}_{3} \times \mathrm{X}_{3}}{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}$$=\frac{5 \times 54+56 \times 90+5 \times 57}{5+90+5}$= 55.95

Q. Silver (atomic weight $=108 \mathrm{g} \mathrm{mol}^{-1}$ ) has a density of 10.5 g $\mathrm{cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12} \mathrm{m}^{2}$ can be expressed in scientific notation as y ´ $10^{\mathrm{x}}$. The value of x is – [JEE 2010]

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Sol. 7$\mathrm{d}=\frac{\text { mass }}{\mathrm{V}} \Rightarrow 10.5 \mathrm{g} / \mathrm{cm}^{3}$ means in $1 \mathrm{cm}^{3}=10.5 \mathrm{g}$ of $\mathrm{Ag}$No of atom of $\mathrm{Ag}$ in $1 \mathrm{cm}^{3}=\frac{10.5}{108} \times \mathrm{N}_{\mathrm{A}}$in $1 \mathrm{cm}, \mathrm{no}$ of atom of $\mathrm{Ag}=\left(\frac{10.5}{108} \times \mathrm{N}_{\mathrm{A}}\right)^{1 / 3}$in $10^{-12} \mathrm{m}^{2}$ or $10^{-8} \mathrm{cm}^{2},$ No of atom of $\mathrm{Ag}$$=\left(\frac{10.5 \mathrm{N}_{\mathrm{A}}}{108}\right)^{2 / 3} \times 10^{-8}=\left(\frac{1.05 \times 6.022 \times 10^{23}}{108}\right)^{2 / 3} \times 10^{-8}$$=1.5 \times 10^{7}$Hence $\mathrm{x}=7$

Q. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is(A) 1.78 M          (B) 2.00 M             (C) 2.05 M              (D) 2.22 M [JEE 2011]

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Sol. (C)Vol of Solution $=\frac{\text { mass }}{\text { density }}$$=\frac{1000+120}{1.15} \mathrm{mL}$$=\frac{1120}{1.15} \mathrm{ml}$Molarity $=\frac{120 / 60}{\frac{1120}{1.15}} \times 1000=2.05 \mathrm{M}$

Q. A compound $\mathrm{H}_{2} \mathrm{X}$ with molar weight of 80 g is dissolved in a solvent having density of0.4 g /ml, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is. [JEE 2014]

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Sol. 8$\mathrm{V}_{\text {solvent }}=\mathrm{V}_{\text {solution }}$3.2 mol present in 1L solution (solvent)3.2 mol / solvent$1000 \times 0.4=400 \mathrm{gm}$$=.4 \mathrm{Kg}Molality =\frac{3.2}{0.4}=8 \mathrm{m} Q. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g \mathrm{cm}^{-3}. The ratio of the molecular weights of the solute and solvent, \left(\frac{\mathrm{MW}_{\text {solute }}}{\mathrm{MW}_{\text {solikent }}}\right), is [JEE – ADV. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 91 mole solution has 0.1 mole solute and 0.9 mole solventLet \mathrm{M}_{1}= Molar mass solute\mathrm{M}_{2}= Molar mass solvent Redox Reaction – JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance Questions Q. Reduction of the metal centre in aqueous permanganate ion involves –(A) 3 electrons in neutral medium(B) 5 electrons in neutral medium(C) 3 electrons in alkaline medium(D) 5 electrons in acidic medium [JEE-2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D) Q. Reaction of \mathrm{Br}_{2} with \mathrm{Na}_{2} \mathrm{CO}_{3} in aqueous solution gives sodium bromide and sodium bromate with evolution of \mathrm{CO}_{2} gas. The number of sodium bromide molecules involved in the balanced chemical equation is. [JEE- 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 5 Q. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen-(A) \mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}(B) \mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{N}_{2}, \mathrm{NH}_{4} \mathrm{Cl}(C) \mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_{2}(D) \mathrm{NO}, \mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2} [JEE- 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} was used to reach the end point. The molarity of the household bleach solution is(A) 0.48 M(B) 0.96 M(C) 0.24 M(D) 0.024 M [JEE- 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C)Eq. of \mathrm{CaOCl}_{2}=\mathrm{Eq}, of \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}25 × M × 2 = 48 × 0.25\left[2 \mathrm{H}^{+}+\mathrm{CaOCl}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}^{2+}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}^{-}\right](nf = 2)M = 0.24 Q. For the reaction\mathrm{I}^{-}+\mathrm{ClO}_{3}^{-}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Cl}^{-}+\mathrm{HSO}_{4}^{-}+\mathrm{I}_{2}The correct statement(s) in the balanced equation is / are :(A) Stoichiometric coefficient of \mathrm{HSO}_{4}^{-} is 6(B) Iodide is oxidized(C) Sulphur is reduced(D) \mathrm{H}_{2} \mathrm{O} is one of the products [JEE- 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,B,D)Oxidation half reaction : Q. In neutral or faintly alkaline solution, 8 moles permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. the magnitude of X is[JEE – Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 6 Q. To measure the quantity of \mathrm{MnCl}_{2} dissolved in an aqueous solution, it was completely converted to \mathrm{KMnO}_{4} using the reaction,\mathrm{MnCl}_{2}+\mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{8}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}+ HCl (equation not balanced).Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 g) was added in portions till the colour of the permanganate ion disappeard. The quantity of MnCl2 (in mg) present in the initial solution is(Atomic weights in \mathrm{g} mol ^{-1}: \mathrm{Mn}=55, \mathrm{Cl}=35.5 ) [JEE – Adv. 2018] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 126 Thermochemistry – JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.SimulatorPrevious Years JEE Advance Questions Q. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ\mathrm{K}^{-1}, the numerical value for the enthalpy of combustion of the gas in kJ \mathrm{mol}^{-1} is [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 9\mathrm{q}=(2.5)(0.45)=1.125 Q. Using the data provided, calculate the multiple bond energy \left(\mathrm{kJ} \operatorname{mol}^{-1}\right) of a C C bond in \mathrm{C}_{2} \mathrm{H}_{2}. That energy is (take the bond energy of a C–H bond as 350 kJ \left.\mathrm{mol}^{-1} .\right)(A) 1165(B) 837(C) 865(D) 815 [JEE 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D) States of Matter – JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years JEE Advance Questions Q. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is – [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 4 Q. The term that corrects for the attractive forces present in a real gas in the Vander Waals’ equation is –(A) nb(B) \frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}$$(\mathrm{C})-\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}$(D) –nb [JEE 2009]

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Sol. (B)Attractive force term of real gas is represented by $\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{v}^{2}}$

Q. To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1 mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm. at $\left.0^{\circ} \mathrm{C}\right)$) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at $0^{\circ} \mathrm{C}$ is close to – [JEE 2011]

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Sol. 7PV = nRT0.32 × V = 0.1 × 0.0821 × 273V = 7 litre

Q. For one mole of a van der Waals’ gas when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the van der Waals’ constant a (atm. litre $^{2} \mathrm{mol}^{-2}$) is(A) 1.0(B) 4.5(C) 1.5(D) 3.0 [JEE 2012]

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Sol. (C)If $\mathrm{b}=0 ;\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right) \times \mathrm{Vm}=\mathrm{RT}$$\mathrm{PVm}+\frac{\mathrm{a}}{\mathrm{Vm}}=\mathrm{RT}$$\mathrm{PVm}+\frac{\mathrm{a}}{\mathrm{Vm}}=\mathrm{RT}$of the form, y = c + mxhere, m = –a$\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{21.6-20.1}{2-3.0}=-1.5$$\therefore \mathrm{a}=1.5 \mathrm{atm} \mathrm{L}^{2} \mathrm{mol}^{-2} Paragraph for Question 5 & 6X and Y are two volatile liquids with molar weights of 10g \operatorname{mol}^{-1} and 40g \operatorname{mol}^{-1} respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. Q. The value of d in cm (shown in the figure), as estimated from Graham’s law, is –(A) 8(B) 12(C) 16(D) 20 [JEE 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C) Q. The experimetnal value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to –(A) Larger mean free path for X as compared to that of Y(B) Larger mean free path for Y as compared to that of X(C) Increased collision frequency of Y with the inert gas as compared to that of X with theinert gas(D) Increased collision frequency of X with the inert gas as compared to that of Y with theinert gas [JEE 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D) Q. One mole of a monoatomic real gas satisfies the equation P(V– b) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by –(A) (B) (C) (D) [JEE 2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C)P(v – b) = RTUsing thermodynamic equation of state, we get Q. The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is [JEE – Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 4Rate of diffusion \propto \lambda \times U_{\text {Avg }}$$\propto \frac{1}{\sqrt{2} \pi \sigma^{2} \mathrm{N}^{*}} \times \mathrm{U}_{\mathrm{Avg}}$$\propto \frac{\mathrm{U}_{\mathrm{Avg}}}{\sqrt{2} \pi \sigma^{2} \mathrm{N}^{*}}$$\propto \frac{\mathrm{U}_{\mathrm{Avg}}(\mathrm{kT})}{\sqrt{2 \pi \sigma^{2} \mathrm{P}}}$Rate of diffusion $\propto \frac{\mathrm{T}^{\frac{3}{2}}}{\mathrm{P}}$$\frac{\mathrm{r}_{\mathrm{final}}}{\mathrm{r}_{\mathrm{initial}}}=\frac{(4)^{\frac{3}{2}}}{2}$$\frac{\mathrm{r}_{\text {final }}}{\mathrm{r}_{\text {inital }}}=4$

Q. A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in $m^{3}$) of the compartment A after the system attains equilibrium is____. [JEE – Adv. 2018]

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Sol. -14.6

Q. The surface of copper gets tarnished by the formation of copper oxide. $\mathrm{N}_{2}$ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below :$2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})$$\mathrm{p}_{\mathrm{H} 2} is the minimum partial pressure of \mathrm{H}_{2} (in bar) needed to prevent the oxidation at 1250 K. The value of ln \left(\mathrm{p}_{\mathrm{H} 2}\right) is ___.(Given : total pressure = 1 bar, R (universal gas constant) = 8 \mathrm{JK}^{-1} \mathrm{mol}^{-1} , ln(10) = 2.3. Cu(s) and \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s}) are mutually immiscible.At 1250 \mathrm{K}: 2 \mathrm{Cu}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s}) ; \Delta \mathrm{G}^{\theta}=-78,000 \mathrm{J} \mathrm{mol}^{-1}$$\left.\mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{G}^{\theta}=-1,78,000 \mathrm{J} \mathrm{mol}^{-1} ; \mathrm{G} \text { is the Gibbs energy }\right)$ [JEE – Adv. 2018]

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Sol. 2.22

Chemical Bonding – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years IIT/JEE Advance Questions
Q. Match each of the diatomic molecules in Column I with its property / properties in Column II. [JEE 2009]

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Sol. $A \rightarrow P, Q, R, S$$B \rightarrow Q, R, S, T$$C \rightarrow P, Q, R$$D \rightarrow P, Q, R, S Q. The nitrogen oxide(s) that contain(s) N–N bond(s) is (are)(A) \mathrm{N}_{2} \mathrm{O}(B) \mathrm{N}_{2} \mathrm{O}_{3}(C) \mathrm{N}_{2} \mathrm{O}_{4}(D) \mathrm{N}_{2} \mathrm{O}_{5} [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,B,C) Q. In the reaction2 \mathrm{X}+\mathrm{B}_{2} \mathrm{H}_{6} \longrightarrow\left[\mathrm{BH}_{2}(\mathrm{X})_{2}\right]^{+}\left[\mathrm{BH}_{4}\right]^{-}the amine(s) X is (are)(A) \mathrm{NH}_{3}(B) \mathrm{CH}_{3} \mathrm{NH}_{2}(C) \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$$(\mathrm{D})\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}$ [JEE 2009]

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Sol. (B,C)

Q. The number of water molecule(s) directly bonded to the metal centre in $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ is [JEE 2009]

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Sol. (4)

Q. The species having pyramidal shape is(A) $\mathrm{SO}_{3}$(B) $\operatorname{BrF}_{3}$(C) $\mathrm{SiO}_{3}^{2-}$(D) $\mathrm{OSF}_{2}$ [JEE 2010]

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Sol. (D)

Q. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule $\mathrm{B}_{2}$ is(A) 1 and diamagnetic(B) 0 and diamagnetic(C) 1 and paramagnetic(D) 0 and paramagnetic [JEE 2010]

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Sol. (A)If Hund’s rule is violated in Boron then unpaired electron of gets paired up and boron becomes diamagnetic.

Q. Based on VSEPR theory, the number of 90 degree F–Br–F angles in $\mathrm{BrF}_{5}$ is [JEE 2010]

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Sol. 0According to VSEPR theory all bond angle are $<90 \%$

Q. The value of n in the molecular formula $\mathrm{Be}_{\mathrm{n}} \mathrm{Al}_{2} \mathrm{Si}_{6} \mathrm{O}_{18}$ is [JEE 2010]

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Sol. 3

Q. The total number of diprotic acids among the following is [JEE 2010]

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Sol. 6

Q. Among the following, the number of elements showing only one non-zero oxidationstate isO, Cl, F, N, P, Sn, Tl, Na, Ti [JEE 2010]

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Sol. 2Na & F

Q. The difference in the oxidation numbers of the two types of sulphur atoms in $\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$ is. [JEE 2011]

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Sol. 5$\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$

Q. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen-(A) $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$(B) $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{N}_{2}, \mathrm{NH}_{4} \mathrm{Cl}$(C) $\mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_{2}$(D) $\mathrm{NO}, \mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$ [JEE 2012]

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Sol. (B)

Q. In allene $\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)$, the type(s) of hybridisation of the carbon atoms is (are)(A) sp and $\mathrm{sp}^{3}$(B) sp and $\mathrm{sp}^{2}$(C) only $\mathrm{sp}^{2}$(D) $\mathrm{sp}^{2}$ and $\mathrm{sp}^{3}$ [JEE 2012]

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Sol. (B)

Q. The shape of $\mathrm{XeO}_{2} \mathrm{F}_{2}$ molecule is :(A) Trigonal bipyramidal(B) Square planar(C) tetrahedral(D) see-saw [JEE 2012]

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Sol. (D) $\mathrm{sp}^{3} \mathrm{d} \quad$ See-Saw shape

Q. The total number of lone-pairs of electrons in melamine is [JEE Adv. 2013]

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Sol. 6

Q. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is(A) $\mathrm{Be}_{2}$(B) $\mathrm{B}_{2}$(C) $\mathrm{C}_{2}$(D) $\mathrm{N}_{2}$ [JEE Adv. 2014]

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Sol. (C)If 2s–2p is not operative than $\mathrm{C}_{2}$ becomes paramagnetic.

Q. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists. [JEE Adv. 2014]

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Sol. (C)

Q. The compound(s) with TWO lone pairs of electrons on the central atom is(are) –(A) $\mathrm{BrF}_{5}$(B) $\mathrm{ClF}_{3}$(C) $\mathrm{XeF}_{4}$(D) SF $_{4}$ [JEE – ADV. 2016]

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Sol. (B,C)Hence Ans. (B,C)

Q. The crystalline form of borax has(A)Tetranuclear$\left[\mathrm{B}_{4}\mathrm{O}_{5}(\mathrm{OH})_{4}\right]^2-}$ unit(B) All boron atoms in the same plane(C) Equal number of $\operatorname{sp}^{2}$ and $\mathrm{sp}^{3}$ hybridized boronatoms(D) One terminal hydroxide per boron atom [JEE – ADV. 2016]

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Sol. (A,C,D)(A)Having $\left[\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right]^{2-}$ tetranuclear (boron) unit(B) All boron atoms not in same plane(C) Two boron are $\mathrm{Sp}^{2}$ hybridised and two boron are sp3 hybridised(D) One terminal hydroxide per boron atom is present.

Q. According to Molecular Orbital Theory,(A) $\mathrm{C}_{2}^{2-}$ is expected to be diamagnetic(B) $\mathrm{O}_{2}^{2+}$ is expected to have a longer bond length than O2(C) $\mathrm{N}_{2}^{+}$ and $\mathrm{N}_{2}$– have the same bond order(D) $\mathrm{He}_{2}^{+}$ has the same energy as two isolated He atoms [JEE – ADV. 2016]

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Sol. (A,C)In the MO of $\mathrm{C}_{2}^{2-}$ there is no unpaired electron hence it is diamagnatic(B) Bond order of $\mathrm{O}_{2}^{2+}$ is 3 and $\mathrm{O}_{2}$ is 2 therefore bond length of $\mathrm{O}_{2}$ is greater than $\mathrm{O}_{2}^{2+}$(C) The molecular orbital energy configuration of $\mathrm{N}_{2}^{+}$ is(D) $\mathrm{He}_{2}^{+}$ has less energy as compare to two isolated He atoms

Q. The correct statements(s) about the oxoacids, $\mathrm{HClO}_{4}$ and HClO, is (are) –(A) $\mathrm{HClO}_{4}$ is more acidic than HClO because of the resonance stabilization of its anion(B) $\mathrm{HClO}_{4}$ is formed in the reaction between $\mathrm{Cl}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$(C) The central atom in Both $\mathrm{HClO}_{4}$ and HClO is $\mathrm{sp}^{3}$ hybridized(D) The conjugate base of $\mathrm{HClO}_{4}$ is weaker base than $\mathrm{H}_{2} \mathrm{O}$ [JEE – Adv. 2017]

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Sol. (A,C,D)

Q. The colour of the $\mathrm{X}_{2}$ molecules of group 17 elements changes gradually from yellow to violet down the group. This is due to –A) the physical state of $\mathrm{x}_{2}$ at room temperature changes from gas to solid down the group(B) decrease in HOMO-LUMO gap down the group(C) decrease in down the group(D) decrease in ionization energy down the group [JEE – Adv. 2017]

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Sol. (B,C)

Q. Among$\mathrm{H}_{2},\mathrm{He}_{2}^{+},\mathrm{Li}_{2},\mathrm{Be_{2},\mathrm{B}_{2},\mathrm{C}_{2}, \mathrm{N}_{2}, \mathrm{O}_{2}^{-}$ thenumber of diamagnetic species is –(Atomic number) : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8 , f = 9) [JEE – Adv. 2017]

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Sol. 5or6If existence of $\mathrm{Be}_{2}$ is considered in atomic form or very weak bonded higher energetic species having zero bond order then it is diamagnetic , then answer will be 6. But if existence of molecular form of $\mathrm{Be}_{2}$ is not considered then magnetic property can’t be predicted then answer will be 5.

Q. The sum of the number of lone pairs of electrons on each central atom in the following species is.$\left[\mathrm{TeBr}_{6}\right]^{2-},\left[\mathrm{BrF}_{2}\right]^{+}, \mathrm{SNF}_{3}$ and $\left[\mathrm{XeF}_{3}\right]^{-}$[Atomic number : N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54] [JEE – Adv. 2017]

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Sol. 6

Q. The order of the oxidation state of the phosphorus atom in $\mathrm{H}_{3} \mathrm{PO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}$ and $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$ is(A) $\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}$(B) $\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$(C) $\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{4}$(D) $\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$ [JEE – Adv. 2017]

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Sol. (A)

Q. The option(s) with only amphoteric oxides is (are)(A) $\mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{CrO}, \mathrm{SnO}, \mathrm{PbO}$(B) $\mathrm{NO}, \mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{SnO}_{2}$(C) $\mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{BeO}, \mathrm{SnO}, \mathrm{SnO}_{2}$(D) $\mathrm{ZnO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{PbO}_{2}$ [JEE – Adv. 2017]

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Sol. (A,C)

Q. Among the following, the correct statement(s) is are(A) $\mathrm{Al}\left(\mathrm{CH}_{3}\right)_{3}$ has the three-centre two-electronbonds in its dimeric structure(B) $\mathrm{AlCl}_{3}$ has the three-centre two-electron bonds in its dimeric structure(C) BH3 has the three-centre two-electron bonds in its dimeric structure(D) The Lewis acidity of $\mathrm{BCl}_{3}$ is greater than that of $\mathrm{AlCl}_{3}$ [JEE – Adv. 2017]

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Sol. (A,C,D)(A) (B) (C) (D) Lewis acidic strength decreases down the group. The decrease in acid strength occurs because as size increases, the attraction between the incoming electron pair and the nucleus weakens. Hence Lewis acidic strength of $\mathrm{BCl}_{3}$ is more than $\mathrm{AlCl}_{3}$.

Q. Based on the compounds of group 15 elements , the correct statement(s) is (are)(A) $\mathrm{Bi}_{2} \mathrm{O}_{5}$ is more basic than $\mathrm{N}_{2} \mathrm{O}_{5}$(B) $\mathrm{NF}_{3}$ is more covalent than $\mathrm{BiF}_{3}$(C) PH3 boils at lower temperature than $\mathrm{NH}_{3}$(D) The N–N single bond is stronger than the P–P single bond [JEE – Adv. 2018]

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Sol. (A,B,C)(D) Due to small size in N–N single bond l.p. – l.p. repulsion is more than P–P single bond therefore N–N single bond is weaker than the P–P single bond.

Hydrogen – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years JEE Advance Questions
Q. Hydrogen peroxide in its reaction with $\mathrm{KlO}_{4}$ and $\mathrm{NH}_{2} \mathrm{OH}$ respectively, is acting as a(A) reducing agent, oxidising agent(B) reducing agent, reducing agent(C) oxidising agent, oxidising agent(D) oxidising agent, reducing agent [JEE Adv. 2014]

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Sol. (A)When $\mathrm{H}_{2} \mathrm{O}_{2}$ react with $\mathrm{KIO}_{4}$ then it act as reducing agent. When $\mathrm{H}_{2} \mathrm{O}_{2}$ react with $\mathrm{NH}_{2} \mathrm{OH}$ then it act as oxidizing agent.

Periodic Table – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. SimulatorPrevious Years JEE Advance Questions
Q. Statement-1 : F atom has a less negative electron gain enthalpy than Cl atom. Statement-2 : Additional electron is repelled more efficiently by 3p electron in Cl atom thanby 2p electron in F atom. (A)Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement 1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement‑1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true. [JEE 2000]

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Sol. (C)

Q. The correct order of radii is:$(\mathrm{A}) \mathrm{N}<\mathrm{Be}<\mathrm{B}$(B) $\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$(C) $\mathrm{Na}<\mathrm{Li}<\mathrm{K}$$(\mathrm{D}) \mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{4+} [JEE 2000] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. The \mathrm{IE}_{1} of Be is greater than that of B. [JEE 2001] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. True Q. The set representing correct order of \mathrm{IP}_{1} is (A) K > Na > Li (B) Be > Mg > Ca (C) B > C > N (D) Fe > Si > C [JEE 2001] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. Identify the least stable ion amongst the following: (A) \mathrm{Li}^{-}(B) \mathrm{Be}^{-}(C) \mathrm{B}^{-}(D) \mathrm{C}^{-} [JEE 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. The increasing order of atomic radii of the following Group 13 elements is(A) Al < Ga < In < Tl(B) Ga < Al < In < Tl(C) Al < In < Ga < Tl(D) Al < Ga < Tl < In [JEE 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B)Increasing order of atomic radii: Ga < Al < In < TlGa is smaller due to poor shielding of d-orbital Q. The option(s) with only amphoteric oxides is(are) :(A) \mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{BeO}, \mathrm{SnO}, \mathrm{SnO}_{2}(B) \mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{CrO}, \mathrm{SnO}, \mathrm{PbO}(C) \quad \mathrm{ZnO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{PbO}_{2}(D) \mathrm{NO}, \mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{SnO}_{2} [JEE 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C)\mathrm{BeO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SnO}, \mathrm{PbO}, \mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{SnO}_{2}, \mathrm{PbO}_{2} are amphoteric oxides.CrO is basic oxide , \mathrm{B}_{2} \mathrm{O}_{3} is acidic oxide, NO is Neutral oxide p-block – JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Simulator Previous Years JEE Advance Questions Q. White phosphorus on reaction with NaOH gives \mathrm{PH}_{3} as one of the products. This is a :-(A) dimerization reaction(B) disproportionation reaction(C) condensation reaction(D) precipitation reaction [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B)\mathrm{P}_{4}+\mathrm{NaOH} \rightarrow \mathrm{PH}_{3}+\mathrm{NaH}_{2} \mathrm{PO}_{2}(Disproportionation reaction) Q. The reaction of P_{4} with X leads selectively to \mathrm{P}_{4} \mathrm{O}_{6}. The X is(A) Dry \mathrm{O}_{2}(B) A mixture of \mathrm{O}_{2}, and \mathrm{N}_{2}(C) Moist \mathrm{O}_{2}(D) \mathrm{O}_{2} in the presence of aqueous \mathrm{NaOH} [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B)\mathrm{P}_{4}+\left(\mathrm{O}_{2}+\mathrm{N}_{2}\right) \stackrel{\Delta}{\longrightarrow} \mathrm{P}_{4} \mathrm{O}_{6}+\mathrm{N}_{2} Q. The nitrogen oxide(s) that contain(s) N–N bond(s) is (are)(A) \mathrm{N}_{2} \mathrm{O}(B) \mathrm{N}_{2} \mathrm{O}_{3} (unsym)(C) \mathrm{N}_{2} \mathrm{O}_{4}(D) \mathrm{N}_{2} \mathrm{O}_{5} [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,B,C) Q. In the reaction, 2 \mathrm{X}+\mathrm{B}_{2} \mathrm{H}_{6} \longrightarrow\left[\mathrm{BH}_{2}(\mathrm{X})_{2}\right]^{+}\left[\mathrm{BH}_{4}\right]^{-} the amine(s) X is (are)(A) \mathrm{NH}_{3}(B) \mathrm{CH}_{3} \mathrm{NH}_{2}(C) \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}(D) \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} [JEE 2009] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,C)2 \mathrm{X}+\mathrm{B}_{2} \mathrm{H}_{6} \longrightarrow\left[\mathrm{BH}_{2}(\mathrm{X})_{2}\right]^{+}\left[\mathrm{BH}_{4}\right]^{-} Q. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type ; the oxidation states of phosphorus in phosphine and the other product are respectively(A) redox reaction ; -3 and –5(B) redox reaction ; +3 and +5(C) disproportionation reaction ; -3 and +1(D) disproportionation reaction ; -3 and +3 [JEE 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C)\mathrm{P}_{4}+\mathrm{NaOH} \rightarrow^{-3} \mathrm{PH}_{3}+\mathrm{NaH}_{2} \mathrm{PO}_{2} Q. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is :(A) \mathrm{Cl}_{2} \mathrm{O}(B) \mathrm{Cl}_{2} \mathrm{O}_{7}(C) \mathrm{ClO}_{2}(D) \mathrm{Cl}_{2} \mathrm{O}_{6} [JEE 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A)Bleaching powder : \mathrm{Ca}(\mathrm{OC} \ell) \mathrm{C} \ell Q. With respect to graphite and diamond, which of the statement(s) given below is (are) correct ?(A) Graphite is harder than diamond. (B) Graphite has higher electrical conductivity than diamond.(C) Graphite has higher thermal conductivity than diamond.(D) Graphite has higher C–C bond order than diamond. [JEE 2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,D) Q. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of –(A) NO(B) \mathrm{NO}_{2}(C) \mathrm{N}_{2} \mathrm{O}(D) \mathrm{N}_{2} \mathrm{O}_{4} [JEE 2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. The correct statement(s) about \mathrm{O}_{3} is(are)(A) O–O bond lengths are equal(B) Thermal decomposition of \mathrm{O}_{3} is endothermic(C) \mathrm{O}_{3} is diamagnetic in nature(D) \mathrm{O}_{3} has a bent structure [JEE 2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D) Comprehension (Q. 10 and 11)The reaction of \mathrm{Cl}_{2} gas with cold dilute and hot concentrated NaOH in water give sodium salt of two (different) oxoacids of chlorine P and Q respectively. The \mathrm{Cl}_{2} gas reacts with \mathrm\mathrm{SO}_{2}gas , in presence of charcoal to give a product R. R reacts with white phosphorous to give a compound S. On hydrolysis, S gives as oxoacid of phosphorous T. Q. R, S and T , respectively are –(A) \mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{PCl}_{5} and \mathrm{H}_{3} \mathrm{PO}_{4}(B) \mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{PCl}_{3} and \mathrm{H}_{3} \mathrm{PO}_{3}(C) \mathrm{SOCl}_{2}, \mathrm{PCl}_{3} and \mathrm{H}_{3} \mathrm{PO}_{2}(D) \mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{PCl}_{5} and \mathrm{H}_{3} \mathrm{PO}_{4} [JEE 2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A) Q. P and Q, respectively, are the sodium salts of – (A) Hypochlorus and chloric acid(B) Hypochlorus and chlorus acid(C) Chloric and perchloric acids(D) Chloric and hypochlorus acids [JEE 2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A) Q. The unbalanced chemical reactions given in List-I show missing reagent or condition (?) which are provided in List-II. Match List-I with List-II and select the correct answer using the code given below the lists :A [JEE 2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D) Q. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is(A) 0 (B) 1 (C) 2 (D) 3 [JEE Adv. 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C) Q. The product formed in the reaction of \mathrm{SOCl}_{2}with white phosphorous is(A) \mathrm{PCl}_{3}(B) \mathrm{SO}_{2} \mathrm{Cl}_{2}(C) \mathrm{SCl}_{2}(D) \mathrm{POCl}_{3} [JEE Adv. 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A)\mathrm{P}_{4}+\mathrm{SOC} \ell_{2} \rightarrow \mathrm{PCl}_{3}+\mathrm{SO}_{2}+\mathrm{S}_{2} \mathrm{Cl}_{2} Q. The correct statements(s) for orthoboric acid is / are –(A) It behaves as a weak acid in water due to self ionization(B) Acidity of its aqueous solution increases upon addition of ethylene glycol(C) It has a three dimensional structure due to hydrogen bonding.(D) It is a weak electrolyte in water [JEE Adv. 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,D) Q. The correct statement(s) regarding, (i) HClO, (ii) \mathrm{HClO}_{2}, (iii) \mathrm{HClO}_{3} and (iv) \mathrm{HClO}_{4}, is (are)(A) The number of Cl=O bonds in (ii) and (iii) together is two(B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three(C) The hybridization of \mathrm{Cl} in (iv) is sp ^{3}(D) Amongst (i) to (iv), the strongest acid is (i) [JEE Adv. 2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,C) Q. When \mathrm{O}_{2} is adsorbed on a metallic surface, electron transfer occurs from the metal to \mathrm{O}_{2}. The TRUE, statement (s) regarding this adsorption is (are)(A) \mathrm{O}_{2} is physisorbed(B) heat is released(C) occupancy of \pi_{2 \mathrm{p}}^{*} of \mathrm{O}_{2} is increased(D) bond length of \mathrm{O}_{2} is increased [JEE Adv. 2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,C,D) Q. Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are(A) \mathrm{CH}_{3} \mathrm{SiCl}_{3} and \mathrm{Si}\left(\mathrm{CH}_{3}\right)_{4}(B) \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2} and \left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiCl}(C) \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2} and \mathrm{CH}_{3} \mathrm{SiCl}_{3}(D) \mathrm{SiCl}_{4} and \left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiCl} [JEE (Adv.) 2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. Three moles of \mathrm{B}_{2} \mathrm{H}_{6} are completely reacted with methanol. The number of moles of boron containing product formed is – [JEE (Adv.) 2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 6 Q. The total number of lone pairs of electrons in \mathrm{N}_{2} \mathrm{O}_{3}is : [JEE Adv. 2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 8 Q. The nitrogen containing compound produced in the reaction of \mathrm{HNO}_{3} with \mathrm{P}_{4} \mathrm{O}_{10}(A) can also be prepared by reaction of \mathrm{P}_{4} and \mathrm{HNO}_{3}(B) is diamagnetic(C) contains one N-N bond(D) reacts with Na metal producing a brown gas [JEE – Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,D) Q. The increasing order of atomic radii of the following group 13 elements is(A) Al < Ga < In < Tl(B) Ga < Al < In < Tl(C) Al < In < Ga < Tl(D) Al < Ga < Tl < In [JEE Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B)The order of radius of 13th group elements is Ga < Al < In < Tl.Reason Due to poor shielding effect of d-orbital, radius of Ga is smallar than Al. Q. The crystalline form of borax has(A) Tetranuclear \left[\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right]^{2-} unit(B) All boron atoms in the same plane(C) Equal number of \mathrm{sp}^{2} and \mathrm{sp}^{3} hybridized boron atoms(D) One terminal hydroxide per boron atom [JEE Adv. 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D)(A) Having \left[\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right]^{2-} tetranuclear (boron) unit(B) All boron atoms not in same plane(C) Two boron are \mathrm{sp}^{2} hybridised and two boron are \mathrm{sp}^{3} hybridised(D) One terminal hydroxide per boron atom is present. Paragraph for Q.24 & 25Upon heating \mathrm{KClO}_{3} in the presence of catalytic amount of\mathrm{MnO}_{2} , a gas W is formed. Excess amount of W reacts with whitephosphorus to give X. The reaction of X with pure \mathrm{HNO}_{3} gives Y and Z. Q. W and X are, respectively(A) \mathrm{O}_{3} and \mathrm{P}_{4} \mathrm{O}_{6}(B) \mathrm{O}_{2} and \mathrm{P}_{4} \mathrm{O}_{10}(C) \mathrm{O}_{3} and \mathrm{P}_{4} \mathrm{O}_{10}(D) \mathrm{O}_{2} and \mathrm{P}_{4} \mathrm{O}_{6} [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B) Q. Y and Z are , respectively(A) \mathrm{N}_{2} \mathrm{O}_{4} and \mathrm{H}_{3} \mathrm{PO}_{3}(B) \mathrm{N}_{2} \mathrm{O}_{4} and \mathrm{HPO}_{3}(C) \mathrm{N}_{2} \mathrm{O}_{5} and \mathrm{HPO}_{3}(D) \mathrm{N}_{2} \mathrm{O}_{3} and \mathrm{H}_{3} \mathrm{PO}_{4} [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C) Q. The correct statements(s) about the oxoacids, \mathrm{HClO}_{4}and \mathrm{HClO}, is (are) –(A) \mathrm{HClO}_{4} is more acidic than HClO because of the resonance stabilization of its anion(B) \mathrm{HClO}_{4} is formed in the reaction between \mathrm{Cl}_{2} and \mathrm{H}_{2} \mathrm{O}(C) The central atom in Both \mathrm{HClO}_{4} and HClO is \mathrm{sp}^{3} hybridized(D) The conjugate base of \mathrm{HClO}_{4} is weaker base than \mathrm{H}_{2} \mathrm{O} [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D) Q. The colour of the \mathrm{X}_{2} molecules of group 17 elements changes gradually from yellow to violet down the group. This is due to –(A) the physical state of \mathrm{X}_{2} at room temperature changes from gas to solid down the group(B) decrease in HOMO-LUMO gap down the group(C) decrease in \pi^{*}-\sigma^{*} down the group(D) decrease in ionization energy down the group [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B,C)Halogens are coloured due to HOMO-LUMO transition of electrons.On moving down the group HOMO-LUMO energy gap decreases so transition of electrons become easier \stackrel{*}{\pi} 2 \mathrm{p} to \stackrel{*}{\sigma} 2 \mathrm{p} therefore colour intensify. Q. The order of the oxidation state of the phosphorus atom in \mathrm{H}_{3} \mathrm{PO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}and \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6} is (A) \mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}(B) \mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}(C) \mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{4}(D) \mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6} [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A) Q. The option(s) with only amphoteric oxides is (are):(A) \mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{CrO}, \mathrm{SnO}, \mathrm{PbO}(B) \mathrm{NO}, \mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{SnO}_{2}(C) \mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{BeO}, \mathrm{SnO}, \mathrm{SnO}_{2}(D) \mathrm{ZnO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{PbO}_{2} [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C,D) Q. Among the following, the correct statement(s) is are(A) \mathrm{Al}\left(\mathrm{CH}_{3}\right)_{3} has the three-centre two-electron bonds in its dimeric structure(B) \mathrm{AlCl}_{3} has the three-centre two-electron bonds in its dimeric structure(C) \mathrm{BH}_{3} has the three-centre two-electron bonds in its dimeric structure(D) The Lewis acidity of \mathrm{BCl}_{3} is greater than that of \mathrm{AlCl}_{3} [JEE – Adv. 2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D)(A)(B)(C)(D) (D) Lewis acidic strength decreases down the group. The decrease in acid strength occurs because as size increases, the attraction between the incoming electron pair and the nucleus weakens. Hence Lewis acidic strength of \mathrm{BCl}_{3} is more than \mathrm{AlCl}_{3}. Q. The compound(s) which generate(s) \mathrm{N}_{2} gas upon thermal decomposition below 300^{\circ} \mathrm{C} is (are)(A) \mathrm{NH}_{4} \mathrm{NO}_{3}(B) \left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(C) \mathrm{Ba}\left(\mathrm{N}_{3}\right)_{2}$$(\mathrm{D}) \mathrm{Mg}_{3} \mathrm{N}_{2}$ [JEE – Adv. 2018]

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Sol. (B,C)

Q. Based on the compounds of group 15 elements , the correct statement(s) is (are)(A) $\mathrm{Bi}_{2} \mathrm{O}_{5}$ is more basic than $\mathrm{N}_{2} \mathrm{O}_{5}$(B) $\mathrm{NF}_{3}$ is more covalent than $\mathrm{BiF}_{3}$(C) $\mathrm{PH}_{3}$ boils at lower temperature than $\mathrm{NH}_{3}$(D) The $\mathrm{N}-\mathrm{N}$ single bond is stronger than the $\mathrm{P}-\mathrm{P}$ single bond [JEE – Adv. 2018]

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Sol. (A,B,C)(A) $\mathrm{Bi}_{2} \mathrm{O}_{5}$ is metallic oxide but $\mathrm{N}_{2} \mathrm{O}_{5}$ is non metallic oxide therefore $\mathrm{Bi}_{2} \mathrm{O}_{5}$ is basic but $\mathrm{N}_{2} \mathrm{O}_{5}$ is acidic.(B) In $\mathrm{NF}_{3}$ , N and F are non metals but $\mathrm{BiF}_{3}$ , Bi is metal but F is non metal therefore $\mathrm{NF}_{3}$ is more covalent than $\mathrm{BiF}_{3}$.(C) In $\mathrm{PH}_{3}$ hydrogen bonding is absent but in $\mathrm{NH}_{3}$ hydrogen bonding is present therefore $\mathrm{PH}_{3}$ boils at lower temperature than $\mathrm{NH}_{3}$.(D) Due to small size in N–N single bond l.p. – l.p. repulsion is more than P–P single bond therefore N–N single bond is weaker than the P–P single bond.

Q. The total number of compounds having at least one bridging oxo group among the molecules given below is______.$\mathrm{N}_{2} \mathrm{O}_{3}, \mathrm{N}_{2} \mathrm{O}_{5}, \mathrm{P}_{4} \mathrm{O}_{6}, \mathrm{P}_{4} \mathrm{O}_{7}, \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{H}_{5} \mathrm{P}_{3} \mathrm{O}_{10}, \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{3}, \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{5}$ [JEE – Adv. 2018]

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Sol. 5 or 6