Caboxylic Acid – 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. Match each of the compound in Column I with its characteristic reaction(s) in Column II.

[IIT 2009]

Sol.


Q. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in Column II.

[IIT 2009]

Sol. ($(A) \rightarrow P, Q, S, T:(B) \rightarrow P, S, T ;(C) \rightarrow P:(D) \rightarrow R$)


Q. The major product of the following reaction is

[IIT 2011]

Sol. ($(\mathrm{A}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{T} ;(\mathrm{B}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{S}, \mathrm{T},(\mathrm{C}) \rightarrow \mathrm{R}, \mathrm{S}, ;(\mathrm{D}) \rightarrow \mathrm{P}$)


Q. With reference the scheme given, which of the given statement(s) about T, U, V & W is (are) correct ?

(A) ‘T’ is soluble in hot aq NaOH

(B) ‘U’ is optically active

(C) mol formula of $\mathrm{W}$ is $\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_{4}$

(D) V gives effervescence with aq NaHCO $_{3}$

[IIT 2012]

Sol. (A)


Q. Identify the binary mixtures (s) that can be separated into the individual compounds, by differential extraction, as shown in the given scheme –

[IIT 2012]

Sol. (A,C,D)


Q. The total number of carboxylic acid groups in the product P is

[IIT 2013]

Sol. (B,D)


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

[IIT 2014]

Sol. 2


Q. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List-I with an appropriate structure from List-II and select the correct answer using the code given below the lists.

[IIT 2014]

Sol. (A)


Q. The major product of the reaction is :

[IIT 2015]

Sol. (A)


Q. Aniline reacts with mixed acid (conc. HNO, and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ ) at 288 K to give P (51%), Q (47%) and R (2%). The major product(s) the following reaction sequence is (are) :-

[JEE Adv. 2018]

Sol. (C)


Q. In the following reaction sequence, the correct structure(s) of X is (are)

[JEE Adv. 2018]

Sol. (D)


Treatment of benzene with CO/HCl in the presence of anhydrous $\mathrm{AlCl}_{3} / \mathrm{CuCl}$ followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with

$\mathrm{Br}_{2} / \mathrm{Na}_{2} \mathrm{CO}_{3}$, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with $\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{C}$, followed by $\mathrm{H}_{3} \mathrm{PO}_{4}$ treatment gives Z as the major product.

(There are two questions based on PARAGRAPH “X”, the question given below is one of them)

Q. The compound Y is :-

[JEE Adv. 2018]

Sol. (B)


Q. The compound Z is :-

[JEE Adv. 2018]

Sol. (C)


An organic acid P $\left(\mathrm{C}_{1} \mathrm{H}_{12} \mathrm{O}_{2}\right)$ can easily be oxidized to a dibasic acid which reacts with ethyleneglycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S.

(There are two questions based on PARAGRAPH “A”, the question given below is one of them)

Q.

[JEE Adv. 2018]

Sol. (A)


Q. The compound R is

(A)

(B)

(C)

(D)

[JEE Adv. 2018]

Sol. (A)


Q. The compound S is

[JEE Adv. 2018]

Sol. (B)


 

Radioactivity – 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 total number of $\alpha$ and $\beta$particles emitted in the nuclear reaction $_{92}^{238} \mathrm{U} \rightarrow_{82}^{214} \mathrm{Pb}$ is

[JEE 2009]

Sol. 8


Q. The number of neutrons emitted when $_{92}^{235} \mathrm{U}$ undergoes controlled nuclear fission to $_{54}^{142} \mathrm{Xe}$ and is –

[JEE 2010]

Sol. 4


Q. Bombardment of aluminium by $\alpha$ -particle leads to its artificial disintegration in two ways,

(i) and (ii) as shown. Products X, Y and Z respectively are :

(A) proton, neutron, positron

(B) neutron, positron, proton

(C) proton, positron, neutron

(D) positron, proton, neutron

[JEE 2011]

Sol. (A)


Q. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group , element X belongs in the periodic table ?

[JEE 2012]

Sol. 8


Q. In the nuclear transmutation

(X, Y) is(are)

$(\mathrm{A})(\gamma, \mathrm{n})$

(B) (p, D)

(C) (n, D)

$(\mathrm{D})(\gamma, \mathrm{D})$

[JEE 2013]

Sol. (AB)


Q. A closed vessel with rigid walls contains 1 mol of 238

92 $\mathrm{U}$ U and 1 mol of air at 298 K. Considering complete decay of $^{238}_{92} \mathrm{U}$ the ratio of the final pressure to the initial pressure of the system at 298 K is –

[JEE 2015]

Sol. 9


Metallurgy – 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.

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Previous Years JEE Advance Questions

Paragraph for questions 1 to 3

Copper is the most nobel of the first row transition metals and occurs in small deposits in several

countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_{4}, 5 \mathrm{H}_{2} \mathrm{O}\right),$ atacamite $\left(\mathrm{Cu}_{2} \mathrm{Cl}(\mathrm{OH})_{3}\right),$ cuprite $\left(\mathrm{Cu}_{2} \mathrm{O}\right),$ copper glance (Cu.S) and malachite $\left(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\right) .$ However, $80 \%$ of the world copper production comes from the ore chalcopyrite (CuFeS_{2} ) . \text { The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.

Q. Partial roasting of chalcopyrite produces :-

(A) $\mathrm{Cu}_{2} \mathrm{S}$ and $\mathrm{FeO}$

(B) $\mathrm{Cu}_{2} \mathrm{O}$ and FeO

(C) $\mathrm{CuS}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(D) $\mathrm{Cu}_{2} \mathrm{O}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$

[JEE-2010]

Sol. (A)


Q. Iron is removed from chalcopyrite as :-

(A) FeO

(B) FeS

(C) $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(D) FeSiO $_{3}$

[JEE-2010]

Sol. (D)


Q. In self-reduction, the reducing species is :-

(A) S

(B) $\mathrm{O}^{2-}$

(C) $\mathrm{S}^{2-}$

(D) $\mathrm{SO}_{2}$

[JEE-2010]

Sol. (C)


Q. Extraction of metal from the ore cassiterite involves –

(A) carbon reduction of an oxide ore

(B) self-reduction of a sulphide ore

(C) removal of copper impurity

(D) removal of iron impurity

[JEE-2011]

Sol. (A,C,D)

$\mathrm{SnO}_{2}+2 \mathrm{C} \rightarrow \mathrm{Sn}+2 \mathrm{CO}$

Cassetrite contains impurity of Wolframite (FeWO,), Which is a ferro magnatic.


Q. Oxidation states of the metal in the minerals haematite and magnetite, respectively, are –

(A) II, III in haematite and III in magnetite

(B) II, III in haematite and II in magnetite

(C) II in haematite and II, III in magnetite

(D) III in haematite and II, III in magnetite

[JEE-2011]

Sol. (D)

Haematite Ore $\Rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3} ;$ Oxidation state $(+3)$ magnetite ore $\Rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}+\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=$ Oxidation

state of $(+2 \&+3)$


Q. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are :

(A) $\mathrm{O}_{2}$ and CO respectively.

(B) $\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively.

(C) $\mathrm{HNO}_{3}$ and $\mathrm{Zn}$ dust respectively.

(D) $\mathrm{HNO}_{3}$ and CO respectively.

[JEE-2012]

Sol. (B)

$2 \mathrm{Ag}+4 \mathrm{NaCN}+\mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}(\text { air }) \longrightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+2 \mathrm{NaOH},$ Here $\mathrm{O}_{2}$ is oxidising agent $\& \mathrm{Zn}^{-}$

dust act as reducing agent.


Q. Sulfide ores are common for the metals –

(A) Ag, Cu and Pb

(B) Ag, Cu and Sn

(B) Ag, Cu and Sn

(D) Al, Cu and Pb

[JEE-2013]

Sol. (A)


Q. The carbon-based reduction method is NOT used for the extraction of

(A) $\operatorname{tin}$ from $\mathrm{SnO}_{2}$

(B) Iron from $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(C) aluminium from $\mathrm{Al}_{2} \mathrm{O}_{3}$

(D) magnesium from $\mathrm{MgCO}_{3} . \mathrm{CaCO}_{3}$

[JEE-2013]

Sol. (C,D)

Being a more reactive metal “Al” and “Mg” can be reduced by electrolytic reduction.


Q. Upon heating with $\mathrm{Cu}_{2} \mathrm{S},$ the reagent(s) that give copper metal is/are

(A) CuFeS $_{2}$

(B) CuO

(C) $\mathrm{Cu}_{2} \mathrm{O}$

(D) $\mathrm{CuSO}_{4}$

[JEE Adv. 2015]

Sol. (A,C,D)


Q. Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process is (are) –

(A) Impure Cu strip is used as cathode

(B) Acidified aqueuous $\mathrm{CuSO}_{4}$ is used as electrolyte

(C) Pure Cu deposits at cathode

(D) Impurities settle as anode-mud

[JEE Adv. 2015]

Sol. (B,C,D)

Impure “Cu” act as a Anode . While ” Pure thin strips of Cu act as cathode and aqueous solution of $\mathrm{CuSO}_{4}$ act as electrolyte.


Q. Match the anionic species given in Column-I that are present in the ore(s) given in

Column-II –

[JEE Adv. 2015]

Sol. $(A \rightarrow P, Q, S, B \rightarrow T, C \rightarrow Q, R, D \rightarrow R)$

Name of ores & their metals.


Q. Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of $\mathrm{O}_{2}$ consumed is ______ .

(Atomic weights in $\left.\mathrm{g} \text { mol }^{-1}: \mathrm{O}=16, \mathrm{S}=32, \mathrm{Pb}=207\right)$

[JEE Adv. 2018]

Sol. 6.47


Liquid Solution – 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 Henry’s law constant for the solubility of $\mathrm{N}_{2}$ gas in water at 298 K is 1.0 × $10^{5}$ atm. The mole fraction of N2 in air is 0.8. The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of waterat 298 K and 5 atm pressure is-

(A) $4.0 \times 10^{-4}$

(B) $4.0 \times 10^{-5}$

(C) $5.0 \times 10^{-4}$

(D) $4.0 \times 10^{-5}$

[JEE 2009]

Sol. (A)

$\mathrm{P}_{\mathrm{N}_{2}}=\mathrm{K}_{\mathrm{H}} \mathrm{X}_{\mathrm{N}_{2}}$

$\mathrm{Y}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{T}}=\mathrm{K}_{\mathrm{H}} \times \mathrm{N}_{2}$

$0.8 \times 5=1 \times 10^{5} \times \frac{\mathrm{n}}{\mathrm{n}+10}$

$4=10^{5} \times \frac{\mathrm{n}}{10}$

$\mathrm{n}=4 \times 10^{-4}$


Q. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is-(take $\left.\mathrm{K}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$

(A) 724             (B) 740            (C) 736             (D) 718

[JEE 2011]

Sol. (A)


Q. The freezing point (in °C) of a solution containing 0.1 g of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (Mol. Wt. 329) in

100 g of water $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$ is –

(A) $-2.3 \times 10^{-2}$

(B) $-5.7 \times 10^{-2}$

(C) $-5.7 \times 10^{-3}$

(D)-1.2 \times 10^{-2}

[JEE 2011]

Sol. (A)

$\mathrm{T}_{\mathrm{f}}^{\prime}=\mathrm{T}_{\mathrm{f}}-\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}-\mathrm{i} \mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$

$=0^{\circ} \mathrm{C}-4 \times 1.86 \times \frac{0.1 / 329}{100 / 1000}$

$=-0.023^{\circ} \mathrm{C}=-2.3 \times 10^{2} \mathrm{C}$


Q. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2^{\circ} \mathrm{C}. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is \text { (take }\left.\mathrm{K}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)

(A) 724 (B) 740 (C) 736 (D) 718

[JEE 2012]

Sol. (A)

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}$

$2=0.76 \times \frac{\mathrm{n}}{100 / 1000}$

$\mathrm{n}=0.263 \mathrm{mol}$

$\mathrm{P}_{\mathrm{S}}=\mathrm{P}^{\circ} \mathrm{x}_{\text {solvent }}=760 \times \frac{\frac{100}{18}}{\frac{100}{18}+0.263}=760 \times \frac{5.55}{5.82}=724.7$ torr


Q. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are)

(A) $\Delta \mathrm{G}$ is positive

(B) $\Delta S_{\text {system }}$ is positive

(C) $\Delta \mathrm{S}_{\text {surroundings }}=0$

(D) $\Delta \mathrm{H}=0$

[J-Adv. 2013]

Sol. (B,C,D)


Q. A compound $\mathrm{H}_{2} \mathrm{X}$ with molar weight of 80 g is dissolved in a solvent having density of

$0.4 \mathrm{g} \mathrm{mL}^{-1}$, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

[JEE-Adv. 2014]

Sol. 8

$\mathrm{m}=\frac{3.2 \mathrm{mol}}{0.4 \mathrm{Kg}}=8 \mathrm{mol} / \mathrm{kg}$


Q. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong eletrolyte) is $-0.0558^{\circ} \mathrm{C}$ , the number of chloride (s) in the coordination sphere of the complex is- $\left[\mathrm{K}_{\mathrm{f}} \text { of water }=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right]$

[JEE-Adv. 2015]

Sol. 1

0.0558 = i × 1.86 × 0.01

i = 3

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$


Q. Mixture(s) showing positive deviation from Raoult’s law at $35^{\circ}$C is (are)

(A) carbon tetrachloride + methanol

(B) carbon disulphide + acetone

(C) benzene + toluene

(D) phenol + aniline

[JEE – Adv. 2016]

Sol. (A,C)

(A) H-bonding of methanol breaks when $\mathrm{CCl}_{4}$ is added so bonds become weaker, resulting positive

deviation.

(B) Mixing of polar and non-polar liquids will produce a solution of weaker interaction, resulting positive deviation

(C) Ideal solution

(D) –ve deviation because stronger H-bond is formed.


Q. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure, Here $\mathrm{x}_{\mathrm{L}}$ and $\mathbf{X}_{\mathrm{M}}$ represent mole fractions of L and M, respectively, in the solution. the correct statement(s) applicable to this system is(are) –

(A) Attractive intramolecular interactions between L–L in pure liquid L and M–M in pure liquid M are stronger than those between L–M when mixed in solution

(B) The point $Z$ represents vapour pressure of pure liquid $\mathrm{M}$ and Raoult’s law is obeyed when $\quad \mathrm{x}_{\mathrm{L}} \rightarrow 0$

(C) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when $\mathrm{x}_{\mathrm{L}} \rightarrow 1$

(D) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from $\mathrm{x}_{\mathrm{L}}=0$ to $\mathrm{x}_{\mathrm{L}}=1$

[JEE – Adv. 2017]

Sol. (A,C)


Q. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg $\mathrm{mol}^{-1}$ . The figures shown below represents plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is 46 g $\mathrm{mol}^{-1}$]

Among the following, the option representing change in the freezing point is –

[JEE – Adv. 2017]

Sol. (D)

Ethanol should be considered non volatile as per given option

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$\Delta \mathrm{T}_{\mathrm{f}}=2 \times \frac{34.5}{46 \times 0.5}=3 \mathrm{K}$


Q. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions $\mathbf{X}_{\mathrm{A}}$ and $\mathrm{x}_{\mathrm{B}}$, respectively, has vapour pressure of 22.5 Torr. The value of $\mathrm{x}_{\mathrm{A}} / \mathrm{x}_{\mathrm{B}}$ in the new solution is____.

(given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

[JEE – Adv. 2018]

Sol. 19

$45=\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \times \frac{1}{2}$$\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}+\mathrm{P}_{\mathrm{B}}^{\circ}=90 \ldots \ldots(1)$

given $\mathrm{P}_{\mathrm{A}}^{\circ}=20$ torr

$\mathrm{P}_{\mathrm{B}}^{\circ}=70 \mathrm{torr}$

$\Rightarrow 22.5$ torr $=20 \mathrm{x}_{\mathrm{A}}+70\left(1-\mathrm{x}_{\mathrm{A}}\right)$

$=70-50 \mathrm{x}_{\mathrm{A}}$

$\mathrm{x}_{\mathrm{A}}=\left(\frac{70-22.5}{50}\right)=0.95$

$\mathrm{x}_{\mathrm{B}}=0.05$

So $\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{B}}}=\frac{0.95}{0.05}=19$


Q. The plot given below shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

On addition of equal number of moles a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ___.

[JEE – Adv. 2018]

Sol. 0.05

From graph


Isomerism – 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 statement(s) about the compound $\mathrm{H}_{3} \mathrm{C}$ (HO)HC – CH = CH – CH(OH) $\mathrm{CH}_{3}$ (X) is (are) :

(A) The total number of stereoisomers possible for X is 6

(B) The total number of diastereomers possible for X is 3

(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2

[iit-2009]

Sol. (A,D)


Q. In the Newman projection for 2,2–dimethylbutane –

X and Y can respectively be –

(A) H and H

(B) $\mathrm{H}$ and $\mathrm{C}_{2} \mathrm{H}_{5}$

(C) $\mathrm{C}_{2} \mathrm{H}_{5}$ and $\mathrm{H}$

(D) $\mathrm{CH}_{3}$ and $\mathrm{CH}_{3}$

[iit-2010]

Sol. (B,D)


Q. Amongst the given option, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are) –

[iit-2011]

Sol. (B,C)


Q. Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?

(A) M and N are non-mirror image stereoisomers

(B) M and O are identical

(C) M and P are enantiomers

(D) M and Q are identical

[JEE-2012]

Sol. (A,B,C)


Q. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are) :

[JEE-2014]

Sol. 3


ElectroChemistry – 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. For the reaction of $\mathrm{NO}_{3}^{-}$ ion in an aqueous solution, $\mathrm{E}^{\circ}$ is +0.96 V. Values of $\mathrm{E}^{\circ}$ for some metal ions are given below

The pair(s) of metal that is(are) oxidised by $\mathrm{NO}_{3}^{-}$ in aqueous solution is(are)

(A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V

[JEE 2009]

Sol. (A,B,D)

(A,B,D) as $\mathrm{E}^{\circ}$ will be positive


Paragraph for Questions 2 to 3

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

$\mathrm{M}(\mathrm{s}) | \mathrm{M}^{+}\left(\mathrm{aq} ; 0.05 \text { molar) } \| \mathrm{M}^{+}(\mathrm{aq} ; 1 \mathrm{molar}) | \mathrm{M}(\mathrm{s})\right.$

For the above electrolytic cell the magnitude of the cell potential $\left|\mathrm{E}_{\mathrm{cell}}\right|$ = 70 mV.

Q. For the above cell :-

(A) $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}>0$

(B) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}<0$

(C) $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}^{0}>0$

(D) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}^{\text {o }}<0$

[JEE 2010]

Sol. (C)

$\mathrm{E}_{1}=-\frac{059}{1} \log \frac{05}{1}=(+) \mathrm{ve} \Rightarrow \mathrm{so}$


Q. If the 0.05 molar solution of $\mathrm{M}^{+}$ is replaced by a 0.0025 molar $\mathrm{M}^{+}$ solution, then the magnitude of the cell potential would be :-

(A) 35 mV              (B) 70 mV             (C) 140 mV              (D) 700 mV

[JEE 2010]

Sol. (B)


Q. Consider the following cell reaction :

$2 \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}(\ell) \quad \mathrm{E}^{\circ}=1.67 \mathrm{V}$

$\mathrm{At}\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_{2}\right)=0.1$ atm and $\mathrm{pH}=3,$ the cell potential at $25^{\circ} \mathrm{C}$ is –

(A) 1.47 V (B) 1.77 V (C) 1.87 V (D) 1.57 V

[JEE 2011]

Sol. (D)


Q. $\mathrm{AgNO}_{3}$ (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. the plot of conductance () versus the volume of $\mathrm{AgNO}_{3}$ is –

(A) (P) (B) (Q) (C) (R) (D) (S)

[JEE 2011]

Sol. (D)


Paragraph for Question 6 and 7

The electrochemical cell shown below is a concentration cell.

$\mathrm{M} | \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt,

$\mathrm{MX}_{2}$) | | $\mathbf{M}^{2+}$ (0.001 mol $\mathrm{dm}^{-3}$) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059V.

Q. The value of G $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for the given cell is (take If = 96500 C $\mathrm{mol}^{-1}$)

(A) –5.7

(B) 5.7

(C) 11.4

(D) –11.4.

[JEE 2012]

Sol. (B)


Q. The solubility product $\left(\mathrm{K}_{\mathrm{sp}} ; \mathrm{mol}^{3} \mathrm{dm}^{-9}\right)$ of $\mathrm{MX}_{2}$ at 298 K based on the information available for the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)

(A) $1 \times 10^{-15}$

(B) $4 \times 10^{-15}$

(C) $1 \times 10^{-12}$

(D) $1 \times 10^{-12}$

Sol. (D)

$\Delta \mathrm{G}=\frac{-2 \times 96500 \times .059}{1000}=11.4$


Q. The standard reduction potential data at $25^{\circ} \mathrm{C}$ is given below

Match $\mathrm{E}^{\mathrm{o}}$ of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below the lists :

[JEE-Adv. 2013]

Sol. (D)


Q. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List-I. The variation in conductivity of these reactions is given in List-II. Match List-I with List-II and select the correct answer using the code given below the lists :

[JEE-Adv. 2013]

Sol. (A)

$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{N}+\mathrm{CH}_{3} \mathrm{COOH} \Rightarrow$ Weak acid and weak base so conductivity increases and then does not change much so option 3 hence and (a)


Q. In a galvanic cell , the salt bridge –

(A) Does not participate chemically in the cell reaction

(B) Stops the diffusion of ions from one electrode to another

(C) Is necessary for the occurence of the cell reaction

(D) Ensures mixing of the two electrolytic solutions

[JEE-Adv. 2014]

Sol. (A,B)

Fact


Q. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.1 M). If $\lambda_{\mathrm{X}^{-}}^{0} \approx \lambda_{\mathrm{Y}^{-}}^{0}$ the difference in their $\mathrm{pK}_{\mathrm{a}}$ values $, \mathrm{pK}_{\mathrm{a}}(\mathrm{HX})-\mathrm{p} \mathrm{K}_{\mathrm{a}}(\mathrm{HY}),$ is (consider degree of ionization of both acids to be <<1).

[JEE-Adv. 2015]

Sol. 3


Q. All the energy released from the reaction X $\rightarrow \mathrm{Y}, \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-193 \mathrm{kJ} \mathrm{mol}^{-1}$ is used for the oxidizing $\mathrm{M}^{+}$ and $\mathrm{M}^{+} \rightarrow \mathrm{M}^{3+}+2 \mathrm{e}^{-}, \mathrm{E}^{\circ}=-0.25 \mathrm{V}$

Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is $\left[\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$

[JEE-Adv. 2015]

Sol. 4


Q. For the electrochemical cell,

$\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})$

the standard emf of the cell is 2.70 V at 300 K. When the concentration of $\mathrm{Mg}^{2+}$ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is____.

(given, $\frac{\mathrm{F}}{\mathrm{R}}=11500 \mathrm{KV}^{-1}$ where F is the Faraday constant and R is the gas constant, ln(10) = 2.30)

[JEE-Adv. 2018]

Sol. 10


Q. Consider an electrochemical cell: $\mathrm{A}(\mathrm{s})\left|\mathrm{A}^{\mathrm{n}+}(\mathrm{aq}, 2 \mathrm{M}) \| \mathrm{B}^{2 \mathrm{n}+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{B}(\mathrm{s})$. The value of $\Delta \mathrm{H}^{\theta}$ for the cell reaction is twice that of $\Delta \mathrm{G}^{\theta}$ at 300 K. If the emf of the cell is zero, the $\Delta \mathrm{S}^{\theta}$ $\left(\text { in } \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)$ of the cell reaction per mole of B formed at 300 K is___.
(Given : ln (2) = 0.7, R (universal gas constant) = $8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

[JEE-Adv. 2018]

Sol. (-11.62)


Chemical Kinetics – 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. Plots showing the variation of the rate constant (k) with temperature (T) are given

below. The plot that follows Arrhenius equation is –

[JEE 2010]

Sol. (A)

Arrhenius equation : $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

(taking A and $\mathrm{E}_{\mathrm{a}}$ to be constant, differentiating w.r.t. T)

$\frac{\mathrm{d} \mathrm{k}}{\mathrm{d} \mathrm{T}}=\left[\frac{\mathrm{AE}_{\mathrm{a}}}{\mathrm{RT}^{2}}\right] \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

assuming A and $\mathrm{E}_{\mathrm{a}}$ to be constant theoretically, the plot should be

ButNo such option is given. Now experimentally, A and $\mathrm{E}_{\mathrm{a}}$ both vary with temperature and k increases as T increases and become very large at infinite temperature. Hence option (A) is correct.


Q. The concentration of R in the reaction $\mathrm{R} \rightarrow \mathrm{P}$ was measured as a function of time and the following data is obtained :

[JEE 2010]

Sol. 0

Zero order reaction because rate is constant with time.

Alternative solution : By hit and trial method, assuming the reaction is of zero order, putting given data in integrated expression for zero order.

the value of k is same from different data so reaction is zero order reaction.


Q. For the first order reaction

$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$

(A) the concentration of the reactant decreases exponentially with time

(B) the half-life of the reaction decreases with increasing temperature.

(C) the half-life of the reaction depends on the initial concentration of the reactant.

(D) the reaction proceeds to 99.6% completion in eight half-life duration.

[JEE 2011]

Sol. (A,B,D)

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{kt}}$

for option (D)

$\frac{1}{\mathrm{t}_{1 / 2}} \ln \frac{100}{50}=\frac{1}{\mathrm{t}_{99.6 \%}} \ln \frac{100}{0.4}$


Q. An organic compound undergoes first-order decomposition . The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $$

\mathrm{t}_{1 / 8} \text { and } \mathrm{t}_{1 / 10}

$$ What is the value of $$

\frac{\left[\mathrm{t}_{1 / 8}\right]}{\mathrm{t}_{1 / 10}} \times 10 ?\left(\text { take } \log _{10} 2=0.3\right)

$$

[JEE 2012]

Sol. 9


Q. In the reaction :

$$

\mathrm{P}+\mathrm{Q} \longrightarrow \mathrm{R}+\mathrm{S}

$$

the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The

concentration of Q varies with reaction time as shown in the figure. The overall order of the

reaction is –

(A) 2 (B) 3 (C) 0 (D) 1

[JEE 2013]

Sol. 9


Q. For the elementary reaction $$

\mathrm{M} \rightarrow \mathrm{N}$$, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

(A) 4 (B) 3(C) 2(D) 1

[JEE 2014]

Sol. (D)


Q. In dilute aqueous $$

\mathrm{H}_{2} \mathrm{SO}_{4}$$ , the complex diaquodioxalatoferrate(II) is oxidized by $\mathrm{MnO}_{4}^{-}$.For this reaction, the ratio of the rate of change of $$

\left[\mathrm{H}^{+}\right]$$ to the rate of change of [MnO4–] is.

[JEE 2015]

Sol. (B)


Q. The % yield of ammonia as a function of time in the reaction

If this reaction is conducted at $\left(P, T_{2}\right)$, with $\mathrm{T}_{2}>\mathrm{T}_{1}$, the % yield of ammonia as a function of time is represented by –

[JEE 2015]

Sol. (B)

$\therefore$ % yield will increase in initial stages due to increase in net speed

As time proceeds $\Rightarrow \mathrm{r}_{\mathrm{net}}=\mathrm{k}_{\mathrm{f}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}-\mathrm{k}_{\mathrm{b}}\left[\mathrm{NH}_{3}\right]^{2}$

On increasing temp., $\mathrm{k}_{\mathrm{f}} \& \mathrm{k}_{\mathrm{b}}$ increase

but increase of $\mathrm{k}_{\mathrm{b}}$ is more

so % yield will decrease

% yield will increase in initial stage due to enhance speed but as time proceeds , final yield is governed by thermodynamics due to which yield decrease since reaction is exothermic


Q. For a first order reaction A(g)  2B(g) + C(g) at constant volume and 300 K, the

total pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A

is present with concentration [A]0, and $t_{1 / 3}$ is the time required for the partial pressure

of A to reach $1 / 3^{\mathrm{rd}}$ of its initial value. The correct option(s) is (are) :-

(Assume that all these gases behave as ideal gases)

[JEE Advance 2018]

Sol. (A,D)


Q. Consider the following reversible reaction,

$\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \square \mathrm{AB}(\mathrm{g})$

The activition energy of the backward reaction exceeds that of the forward reaction by 2RT $\left(\text { in } \mathrm{J} \mathrm{mol}^{-1}\right)$. If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta \mathrm{G}^{\theta}$(in J mol–1) for the reaction at 300 K is____.

(Given ; ln (2) = 0.7, RT = 2500 J $\mathrm{mol}^{-1}$ at 300 K and G is the Gibbs energy)

[JEE Advance 2018]

Sol. 8500


Coordination 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. The spin only magnetic moment value (in Bohr magneton units) of \mathrm{Cr}(\mathrm{CO})_{6} is

(A) 0

(B) 2.84

(C) 4.90

(D) 5.92

[JEE 2009]

Sol. (A)


Q. The compound(s) that exhibit(s) geometrical isomerism is (are) :

(A) $\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\right]$

(B) $\left[\mathrm{Pt}(\mathrm{en})_{2}\right] \mathrm{Cl}_{2}$

(C) $\left[\mathrm{Pt}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}$

(D) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$

[JEE 2009]

Sol. (C,D)


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

[JEE 2009]

Sol. 4


Q. The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}$ is –

(A) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{O}_{2} \mathrm{N}\right)\right] \mathrm{Cl}_{2}$

(B) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\left(\mathrm{NO}_{2}\right)$

(C) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}(\mathrm{ONO})\right] \mathrm{Cl}$

(D) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\left(\mathrm{NO}_{2}\right)\right] \cdot \mathrm{H}_{2} \mathrm{O}$

Sol. (B)

Ionisation isomers differ in ions in solution thus, ionisation isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}$ is $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\left(\mathrm{NO}_{2}\right)$. Because

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl} \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \operatorname{Cl}\left(\mathrm{NO}_{2}\right)\right]^{+}+\mathrm{Cl}^{-}$ (Given compound)

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\left(\mathrm{NO}_{2}\right) \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+}+\mathrm{NO}_{2}^{-}$ Ionisation isomer of given compound.


Q. Total number of geometrical isomers for the complex $\left[\mathrm{RhCl}(\mathrm{CO})\left(\mathrm{PPh}_{3}\right)\left(\mathrm{NH}_{3}\right)\right]$ is.

[JEE 2010]

Sol. 3

$\left[\mathrm{RhCl}(\mathrm{Co})\left(\mathrm{PPh}_{3}\right)\left(\mathrm{NH}_{3}\right)\right]$

$\mathrm{dsp}^{2},$ square planar, total 3 geometrical isomer.


Q. The correct structure of ethylenediaminetetraacetic acid (EDTA) is –

[JEE 2010]

Sol. (C)

The correct structure of ethylenediaminetetra acetic acid (EDTA) is


Q. Geometrical shapes of the complexes formed by the reaction of $\mathrm{Ni}^{2+}$ with $\mathrm{Cl}^{-}, \mathrm{CN}$ and $\mathrm{H}_{2} \mathrm{O}$ respectively, are –

(A) octahedral, tetrahedral and square planar

(B) tetrahedral, square planar and octahedral

(C) square planar, tetrahedral and octahedral

(D) octahedral, square planar and octahedral

[JEE 2011]

Sol. (B)


Q. Among the following complexes (K–P)

$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathbf{K}),\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}(\mathrm{L}), \mathrm{Na}_{3}\left[\mathrm{Co}\left(\text { oxalate) }_{3}\right](\mathrm{M}),\left[\mathrm{Ni}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}(\mathrm{N})\right.$$\mathrm{K}_{2}\left[\mathrm{Pt}(\mathrm{CN}) {4}\right](\mathbf{O})$ and $\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}(\mathbf{P})$

The diamagnetic complex are –

(A) K, L, M, N               (B) K, M, O, P               (C) L, M, O, P                (D) L, M, N, O

[JEE 2011]

Sol. (C)


Q. The volume (in mL) of 0.1M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01M solution of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6} \mathrm{Cl}\right] \mathrm{Cl}_{2}$, as silver chloride is close to.

[JEE 2011]

Sol. 6


Q. As per IUPAC nomenclature, the name of the complex $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}_{3}$ is :

(A) Tetraaquadiaminecobalt(III) chloride

(B) Tetraaquadiamminecobalt(III) chloride

(C) Diaminetetraaquacobalt(III) chloride

(D) Diamminetetraaquacobalt(III) chloride

[JEE 2012]

Sol. (D)

$\left[\mathrm{C}_{\alpha}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{3}\right.$

Diamminetetraaquacobalt(III) chloride

$\frac{V I B G Y O R}{\lambda-v^{-} E^{-}}$


Q. The colour of light absorbed by an aqueous solution of $\mathrm{CuSO}_{4}$ is –

(A) orange-red (B) blue-green (C) yellow (D) violet

[JEE 2012]

Sol. (A)


Q. $\mathrm{NiCl}_{2}\left\{\mathrm{P}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)\right\}_{2}$ exhibits temperature dependent magnetic behavior (paramagnetic/diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively :

(A) tetrahedral and tetrahedral

(B) square planar and square planar

(C) tetrahedral and square planar

(D) square planar and tetrahedral

[JEE 2012]

Sol. (C)


Q. Consider the following complex ions P, Q and R ,

$\mathbf{P}=\left[\mathrm{FeF}_{6}\right]^{3-}, \mathbf{Q}=\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and $\mathbf{R}=\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is –

(A) R < Q < P (B ) Q < R < P (C) R < P < Q (D) Q < P < R

[JEE 2013]

Sol. (B)


Q. EDTA $^{4}$ is ethylenediaminetetraacetate ion. The total number of $\mathrm{N}-\mathrm{Co}-\mathrm{O}$ bond angles in $[\mathrm{Co}(\mathrm{EDTA})]^{-1}$ complex ion is –

[JEE 2013]

Sol. 8


Q. The pair(s) of coordination complex/ion exhibiting the same kind of isomerism is(are) –

(A) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}$

(B) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$ and $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{+}$

(C) $\left[\mathrm{CoBr}_{2} \mathrm{Cl}_{2}\right]^{2-}$ and $\left[\mathrm{PtBr}_{2} \mathrm{Cl}_{2}\right]^{2-}$

(D)$\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{3}\right)\right] \mathrm{Cl}$ and $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right] \mathrm{Br}$

[JEE 2013]

Sol. (B,D)


Q. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists.

[JEE Adv. 2014]

Sol. (B)

(P) $\left[\mathrm{Cr}^{\mathrm{III}}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}:$

(1) Complex given in (P) is Paramagnetic & show two geometrical

(3 unpaired electrons) isomerism (cis and trans) (does not show ionization isomer)

(Q) $\left[\mathrm{Ti}^{\mathrm{III}}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right]\left(\mathrm{NO}_{3}\right)_{2}$

(2) Complex given in (Q) is paramagnetic show ionization(1 unpaired electrons) isomerism

(R) $\left[\mathrm{Pt}^{\mathrm{Il}}(\mathrm{en})\left(\mathrm{NH}_{3}\right) \mathrm{Cl}\right] \mathrm{NO}_{3}$

(3) Complex given in (R) is diamagnetic and show ionization(1 unpaired electrons) isomerism

(S)$\left[\mathrm{Co}^{\mathrm{III}}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{3}\right)_{2}\right] \mathrm{NO}_{3}$

(4) Complex given in (S) is diamagnetic does not show ionization (0 unpaired electrons) isomerism show geometrical isomerism


Q. A list of species having the formula $\mathrm{XZ}_{4}$ is given below :

$\mathrm{XeF}_{4}, \mathrm{SF}_{4}, \mathrm{SF}_{4}, \mathrm{BF}_{4}^{-}, \mathrm{BrF}_{4}^{-},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},\left[\mathrm{FeCl}_{4}\right]^{2-},\left[\mathrm{CoCl}_{4}\right]^{2-}$ and $\left[\mathrm{PtCl}_{4}\right]^{2-}$.

Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is

[JEE Adv. 2014]

Sol.


Q. The geometries of the ammonia complexes of $\mathrm{Ni}^{2+}, \mathrm{Pt}^{2+}$ and $\mathrm{Zn}^{2+}$ , respectively , are :

(A) octahedral, square planar and tetrahederal

(B) square planar, octahederal and tetrahederal

(C) tetrahederal, square planar and octahederal

(D) octahederal , tetrahederal and square planar

[JEE – Adv. 2016]

Sol. (A)


Q. Among $\left[\mathrm{Ni}(\mathrm{CO}), \mathrm{I},\left[\mathrm{NiCl}_{4}\right]^{2},\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right), \mathrm{Cl}_{2}\right] \mathrm{Cl}, \mathrm{Na}_{3}\left[\mathrm{CoF}_{6}\right], \mathrm{NaO}_{2} \mathrm{and} \mathrm{Co}_{2}\right.$, the total number of paramagnetic compounds is –

(A) 2             (B) 3              (C) 4                (D) 5

[JEE – Adv. 2016]

Sol. (B)


Q. The number of geometric isomers possible for the complex$\left[\mathrm{CoL}_{2} \mathrm{Cl}_{2}\right]^{-}\left(\mathrm{L}=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}\right)$ is

[JEE – Adv. 2016]

Sol. 5


Q. Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\mathrm{MCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ (X) and $\mathrm{NH}_{4} \mathrm{Cl}$ gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.

Among the following options, which statements is(are) correct ?

(A) The hybridization of the central metal ion in Y is d2sp3

(B) Z is tetrahedral complex

(C) Addition of silver nitrate to Y gives only two equivalents of silver chloride

(D) When X and Z are in equilibrium at 0°C, the colour of the solution is pink

[JEE – Adv. 2017]

Sol. (A,B,D)

(A) Hybridisation of $(\mathrm{Y})$ is $\mathrm{d}^{2} \mathrm{sp}^{3}$ as $\mathrm{NH}_{3}$ is strong field ligand

(B) $\left[\mathrm{CoCl}_{4}\right]^{2-}$ have $\mathrm{sp}^{3}$ hybridisation as $\mathrm{Cl}^{-}$ is weak field ligand

When ice is added to the solution the equilibrium shifts right hence pink colour will remain predominant

So, correct answer is (A,B& D)


A

Q. The correct statement(s) regarding the binary transition metal carbonyl compounds is (are)

(Atomic numbers : Fe = 26, Ni = 28)

(A) Total number of valence shell electrons at metal centre in $\mathrm{Fe}(\mathrm{CO})_{5}$ or $\mathrm{Ni}(\mathrm{CO})_{4}$ is 16

(B) These are predominantly low spin in nature

(C) Metal – carbon bond strengthens when the oxidation state of the metal is lowered

(D) The carbonyl C–O bond weakens when the oxidation state of the metal is increased

[JEE – Adv. 2018]

Sol. (B,C)

(A) $\left[\mathrm{Fe}\left(\mathrm{CO}_{5}\right)\right] \&\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ complexes have 18-electrons in their valence shell.

(B) Carbonyl complexes are predominantly low spin complexes due to strong ligand field.

(C) As electron density increases on metals (with lowering oxidation state on metals), the extent of synergic bonding increases. Hence M–C bond strength increases

(D) While positive charge on metals increases and the extent of synergic bond decreases and hence C–O bond becomes stronger.


Q. Among the species given below, the total number of diamagnetic species is____.

H atom, $\mathrm{NO}_{2}$ monomer, $\mathrm{O}_{2}^{-}$ (superoxide), dimeric sulphur in vapour phase,

$\mathrm{Mn}_{3} \mathrm{O}_{4},\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{FeCl}_{4}\right],\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{NiCl}_{4}\right], \mathrm{K}_{2} \mathrm{MnO}_{4}, \mathrm{K}_{2} \mathrm{CrO}_{4}$

[JEE – Adv. 2018]

Sol. (1)


Q. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is___.

(Atomic weights in g $\mathrm{mol}^{-1}$: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

(A) It has two geometrical isomers

(B) It will have three geometrical isomers if bidentate ‘en’ is replaced by two cyanide ligands

(C) It is paramagnetic

(D) It absorbs light at longer wavelength as compared to $\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right]^{3+}$

[JEE – Adv. 2018]

Sol. 2992

Total mass = 12 × 172 + 4 × 232 = 2992 g


Q. The correct option(s) regarding the complex $\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}:-$

$\left(\mathrm{en}=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)$ is (are)

[JEE – Adv. 2018]

Sol. (A,B,D)


Q. Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II.

[JEE – Adv. 2018]

Sol. (C)


Solid State – 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 statement(s) regarding defects in solid is (are)

(A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and

anion.

(B) Frenkel defect is a dislocation defect

(C) Trapping of an electron in the lattice leads to the formation of F‑center.

(D) Schottky defects have no effect on the physical properties of solids.

[JEE 2009]

Sol. (B,C)


Q. The packing effeciency of the two-dimensional square unit cell shown below is

(A) 39.27%         (B) 68.02%            (C) 74.05%             (D) 78.54%

[JEE-2010]

Sol. (D)


Q. The number of hexagonal faces that present in a truncated octahedron is.

[JEE-2011]

Sol. 8


Q. A compound $\mathrm{M}_{\mathrm{p}} \mathrm{X}_{\mathrm{q}}$ has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is :

(A) MX

(B) $\mathrm{MX}_{2}$

(C) $\mathrm{M}_{2} \mathrm{X}$

(D) $\mathrm{M}_{5} \mathrm{X}_{14}$

[JEE-2012]

Sol. (B)


Q. The arrangement of $\mathrm{X}^{-}$ ions around $\mathrm{A}^{+}$ ion in solid AX is given in the figure (not drawn to scale). If the radius of $\mathrm{X}^{-}$ is 250 pm, the radius of $\mathrm{A}^{+}$ is –

(A) 104 pm         (B) 125 pm          (C) 183 pm         (D) 57 pm

[JEE-2013]

Sol. (A)


Q. The correct statement(s) for cubic close packed (ccp) three dimensional

structure is (are)

(A) The number of the nearest neighbours of an atom present in the topmost layer is 12

(B) The efficiency of atom packing is 74%

(C) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively

(D) The unit cell edge length is $2 \sqrt{2}$ times the radius of the atom

[JEE – Adv. 2016]

Sol. (B,C,D)

CCP is ABC ABC ….. type packing

(A) In topmost layer, each atom is in contact with 6 atoms in same layer and 3 atoms below

this layer.


Q. A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of

400 pm. If the density of the substance in the crystal is 8g $\mathrm{cm}^{-3}$, then the number of atoms present in 256g of the crystal is $\mathrm{N} \times 10^{24}$. The value of N is

[JEE – Adv. 2017]

Sol. 2


Q. Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.

(i) Remove all the anions (X) except the central one

(ii) Replace all the face centered cations (M) by anions (X)

(iii) Remove all the corner cations (M)

(iv) Replace the central anion (X) with cation (M)

[JEE – Adv. 2017]

Sol. 3


Surface 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.

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Previous Years JEE Advance Questions

Q. Among the electrolytes $\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{CaCl}_{2}, \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ and $\mathrm{NH}_{4} \mathrm{Cl}$, the most effective coagulation agent for $\mathrm{Sb}_{2} \mathrm{S}_{3}$ sol is

(A) $\mathrm{Na}_{2} \mathrm{SO}_{4}$

(B) $\mathrm{CaCl}_{2}$

(C) $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

(D) $\mathrm{NH}_{4} \mathrm{Cl}$

JEE 2009

Sol. (C)

$\mathrm{Sb}_{2} \mathrm{S}_{3}$ forms negatively charged sol. Hence cation with

greatest charge density is most effective in bringing about coagulation C.F. Hardy sulz rule.


Q. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is

(are) –

(A) Adsorption is always exothermic

(B) Physisorption may transform into chemisorption at high temperature

(C) Physisorption increases with increasing temperature but chemisorption decreases with

increasing temperature

(D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation

JEE 2011

Sol. (A,B,D)

general properties.


Q. Choose the correct reason(s) for the stability of the lyophobic colloidal particle.

(A) Preferential adsorption of ions on their surface from the solution

(B) Preferential adsorption of solvent on their surface from the solution

(C) Attraction between different particles having opposite charges on their surface

(D) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles

JEE 2012

Sol. (A,D)

Theory Based


Q. The given graphs / data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct ?

(A) I is physisorption and II is chemisorption

(B) I is physisorption and III is chemisorption

(C) IV is chemisorption and II is chemisorption

(D) IV is chemisorption and III is chemisorption

JEE 2012

Sol. (A,C)

For physical adsorption $\rightarrow$ favourable conditions is decrease in temp.

for chemical adsorption $\rightarrow$ chemical bond

Occurs $\rightarrow$ PE $\downarrow$ with bonding and increase in temp..


Q. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at $25^{\circ} \mathrm{C}$. For this process, the correct statement is

(A) The adsorption requires activation at $25^{\circ} \mathrm{C}$

(B) The adsorption is accompanied by a decrease in enthalpy

(C) The adsorption increases with increase of temperature

(D) The adsorption is irreversible

JEE Adv. 2013

Sol. (B)

Explanation for physical adsorption

• Activation energy is very low

• Physical adsorption is an exothermic process

• Physical adsorption decreases with increase in temperature

• Physical adsorption is reversible


Q. The qualitative sketches I , II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{11} \mathrm{OSO}_{3}^{-} \mathrm{Na}^{+}$ at room temperature. The correct assignment of the sketches is –

JEE Adv. 2016

Sol. (D)

Water has large surface tension due to very strong interaction. Generally adding organic

derivatives to water decreases its surface tension due to hydrophobic interaction.

In case III, hydrophobic interaction is stronger than case I causing surface tension to decrease more rapidly.


Q. The correct statement(s) about surface properties is (are)

(A) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion

medium

(B) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system.

(C) Brownian motion of colloidal particles does not depend on the size of the particles but

depends on viscosity of the solution.

(D) The critical temperatures of ethane and nitrogen and 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature.

JEE Adv. 2017

Sol. (B,D)

(A) Emulsion is liquid in liquid type colloid.

(B) For adsorption, $\Delta \mathrm{H}<0 \& \Delta \mathrm{S}<0$

(C) Smaller the size and less viscous the dispersion medium, more will be the brownian motion.

(D) Higher the $\mathrm{T}_{\mathrm{C}}$ , greater will be the extent of adsorption.


Polymer – 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.

 

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Previous Years JEE Advanced Questions

Q. The correct functional group X and the reagent/reaction conditions Y in the following scheme are

[JEE 2011]

Sol. (A,B,C,D)


Q. On complete hydrogenation, natural rubber produces

(A) ethylene-propylene copolymer

(B) vulcanised rubber

(C) polypropylene

(D) polybutylene

[JEE – ADV. 2016]

Sol. (A)


JEE Advanced Previous Year Papers
JEE Advanced Previous Year Papers Chapter-wise with Solution

Here you will find JEE Advanced Previous Year Papers which are divided chapter-wise. All the JEE Advanced Previous Year papers are provided with complete and detailed Solution. You can download them in the PDF format or you can also read them online. eSaral also provides JEE Main Topic-wise Previous Year Question Papers with Solutions.

  1. Physics – JEE Advanced Previous Year Questions with Solutions
  2. Chemistry – JEE Advanced Previous Year Questions with Solutions
  3. Maths – JEE Advanced Previous Year Questions with Solutions

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Physics JEE Advanced Papers

Chemistry JEE Advanced Papers

Maths JEE Advanced Papers

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Chemistry Topic-wise JEE Advanced Previous Year Question with Solutions

Practicing JEE Advanced Previous Year Question Papers will help you in many ways in your Exam preparation. It will help you to boost your confidence level. Students can check where they are lagging through practicing these previous year question papers. Here you will get the JEE Advanced previous year question papers from year  2009 to 2019 along with solutions.
These JEE Advanced Previous Year Questions for Chemistry plays an important role in IIT-JEE preparation. We are providing IIT-JEE Advanced Previous Year Questions with detailed Solution.

We have tried our best to provide you last 10 years question with solutions.
This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.
While preparing for the IIT-JEE exam, aspirants should be aware about the question paper structure and the format of questions to be asked in this exam. This will help to make an effective preparation strategy for the exam. JEE Advanced Previous Year Question Papers are the best resources to prepare for exam. This will help an individual to understand the exam pattern of JEE. This will also enhance your level of preparation. JEE aspirants must solve multiple sample papers and analyse their performances in order to recognize their strengths and weaknesses.

 

Here are the Chemistry Topic-wise Previous year question for JEE Advanced:

 

Click here to Download Physics Topic-wise JEE Advanced Previous Year Questions with Solutions

Click Here to Download Mathematics Topic-wise JEE Advanced Previous Year Question with Solutions 

 

Click Here to Download Topic-wise JEE Main Previous Year Questions with Solutions

We have tried our best to provide you last 10 years question with solutions.
This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.

 

  

Chemistry Topic-wise JEE Main Previous Year Question with Solutions
Chemistry Topic-wise JEE Main Previous Year Question with Solutions

Practicing JEE Main previous Year Question Papers will help you in many ways in your Exam preparation. It will help you to boost your confidence level. Students can check where they are lagging through practicing these previous year question papers. Here you will get the JEE-Mains previous year question papers from year  2009 to 2019 along with solutions.
These JEE Main previous Year Question for Chemistry plays an important role in IIT-JEE preparation. We are providing IIT-JEE Mains Previous Year Question Papers with detailed Solution.

We have tried our best to provide you last 10 years question with solutions.
This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning.
The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam.
While preparing for the IIT-JEE exam, aspirants should be aware about the question paper structure and the format of questions to be asked in this exam. This will help to make an effective preparation strategy for the exam. JEE Main Previous Year Question Papers are the best resources to prepare for exam. This will help an individual to understand the exam pattern of JEE. This will also enhance your level of preparation. JEE aspirants must solve multiple sample papers and analyse their performances in order to recognize their strengths and weaknesses.

 

Here are the Chemistry Topic-wise Previous year question for JEE Main:

 

In the last few months of JEE Main 2020, candidates should go through the sample papers and mock tests to test their preparation level and get hands-on practice in solving questions.
One of the most effective ways to prepare for the JEE Advanced, after the IIT-JEE Previous Year Papers, is to take a JEE mock test. At eSaral we provide almost all types of test series which includes topic wise tests, full length mock tests, etc.

Click Here to Download Math Topic-wise JEE Main Previous Year Question with Solutions

Click Here to Download Physics Topic-wise JEE Main Previous Year Question with Solutions

Click Here to Download JEE Advanced Previous Year Questions with solutions. 

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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]

Sol. (5)

M. wt 100 of ketone

So 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]

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]

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.

 

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Previous Years JEE Advance Questions

Paragraph for Question Nos. 1 to 3

A 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]

Sol. (B)


Q. The structure of the products Q and R, respectively, are

[IIT 2009]

Sol. (A)


Q. The structure of the product S is

[IIT 2009]

Sol. (B)


Paragraph for Questions Nos. 4 to 5

An 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]

Sol. (D)


Q. The structure of the compound Q is –

[IIT 2011]

Sol. (B)


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

[IIT 2012]

Sol. (C)


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

[IIT 2013]

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]

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 place

In above reaction $\mathrm{Br}_{2}$ is limiting agents.


Q. The major product in the following reaction is

[IIT 2014]

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]

Sol. (A)

Mechanism :


Q. In the following reactions, the product S is –

[IIT 2015]

Sol. (A)


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

[JEE Adv. 2018]

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]

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.

 

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Previous 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]

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]

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]

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]

Sol. (A)

The ring is pyranose with glycosidic linkage which means the oxide linkage which

connect 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]

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]

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]

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]

Sol. (A,B,C,D)


Q. The structure of D-(+)-glucose is

The structure of L(–)-glucose is

[IIT 2011]

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]

Sol. 6

No. of peptide linkage $=$ No. of water molecules added for complete hydrolysis.

= n – 1

So, 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]

Sol. 4


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

[JEE 2012]

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]

Sol. 4

Considering 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]

Sol. 1


Q. Positive Tollen’s test is observed for

[JEE – Adv. 2016]

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]

Sol. (B,C)

Invert sugar is equailmolar mixture of D-glucose and D-fructose which is obtained by hydrolysis of sucrose

Specific 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]

Sol. (A)


Q. The compound T is :

(A) Glycine               (B) Alanine               (C) Valine              (D) Serine

[JEE – Adv. 2016]

Sol. (B)


Q. The Fischer presentation of D-glucose is given below.

The correct structure(s) of $\beta$-glucopyranose is (are) :-

[JEE Adv. 2018]

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

Sol. (4)


Q. In the reaction

the products are

JEE Advance 2010

Sol. (D)


Q. The major product in the following reaction is

(A)

(B)

(C)

(D)

JEE Advance 2014

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

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

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

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

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

Sol. (B)


Easy & Short Trick to Learn Periodic Table

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Easy & Short Trick to Learn Periodic Table

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.

Simulator

Previous 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]

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]

Sol. (B)


Paragraph For Question 3 and 4

Schemes 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]

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]

Sol. (C )


Q. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are) –

[IIT-2015]

Sol. (B,D)


Paragraph For Questions 6 and 7

In the following reaction

Q. Compound X is :

[IIT-2015]

Sol. (C)


Q. The major compound Y is :

[IIT-2015]

Sol. (D)