JEE Main Practice — Chemistry Set 2

JEE Main · 2026 · with answers and explanations

questions
25
Total marks
100
Duration
60 minutes
Negative marking
−1

Subjects: Chemistry

Question 1

Chemistry · 4 marks · −1

The bond order of O2+\mathrm{O_2^+} is

  1. A.2
  2. B.2.5
  3. C.1.5
  4. D.3

Correct answer: B

Explanation: O2\mathrm{O_2} has bond order 2 (10 bonding, 6 antibonding valence electrons). Removing one electron from a π∗\pi^* orbital gives 10−52=2.5\frac{10-5}{2}=2.5.

Question 2

Chemistry · 4 marks · −1

Which pair consists of molecules that BOTH have zero dipole moment?

  1. A.CO2\mathrm{CO_2} and BF3\mathrm{BF_3}
  2. B.NH3\mathrm{NH_3} and BF3\mathrm{BF_3}
  3. C.H2O\mathrm{H_2O} and CO2\mathrm{CO_2}
  4. D.CHCl3\mathrm{CHCl_3} and CCl4\mathrm{CCl_4}

Correct answer: A

Explanation: Linear CO2\mathrm{CO_2} and trigonal planar BF3\mathrm{BF_3} have bond dipoles that cancel. NH3\mathrm{NH_3} (pyramidal), H2O\mathrm{H_2O} (bent) and CHCl3\mathrm{CHCl_3} are polar.

Question 3

Chemistry · 4 marks · −1

Which of the following molecules has a square pyramidal shape?

  1. A.XeF4\mathrm{XeF_4}
  2. B.PCl5\mathrm{PCl_5}
  3. C.SF4\mathrm{SF_4}
  4. D.BrF5\mathrm{BrF_5}

Correct answer: D

Explanation: BrF5\mathrm{BrF_5}: 5 bond pairs + 1 lone pair (sp3d2sp^3d^2) ⇒\Rightarrow square pyramidal. XeF4\mathrm{XeF_4} (2 lone pairs) is square planar, PCl5\mathrm{PCl_5} trigonal bipyramidal, SF4\mathrm{SF_4} see-saw.

Question 4

Chemistry · 4 marks · −1

According to molecular orbital theory, which of the following species is diamagnetic?

  1. A.O2\mathrm{O_2}
  2. B.O22−\mathrm{O_2^{2-}}
  3. C.O2−\mathrm{O_2^-}
  4. D.O2+\mathrm{O_2^+}

Correct answer: B

Explanation: π∗\pi^* occupancy: O2+\mathrm{O_2^+} 1, O2\mathrm{O_2} 2 (two unpaired), O2−\mathrm{O_2^-} 3 (one unpaired), O22−\mathrm{O_2^{2-}} 4 (all paired). Only the peroxide ion is diamagnetic.

Question 5

Chemistry · 4 marks · −1

The correct decreasing order of the H–X–H bond angle in CH4\mathrm{CH_4}, NH3\mathrm{NH_3}, H2O\mathrm{H_2O} and H2S\mathrm{H_2S} is

  1. A.CH4>NH3>H2O>H2S\mathrm{CH_4>NH_3>H_2O>H_2S}
  2. B.NH3>CH4>H2O>H2S\mathrm{NH_3>CH_4>H_2O>H_2S}
  3. C.CH4>H2O>NH3>H2S\mathrm{CH_4>H_2O>NH_3>H_2S}
  4. D.CH4>NH3>H2S>H2O\mathrm{CH_4>NH_3>H_2S>H_2O}

Correct answer: A

Explanation: CH4\mathrm{CH_4} 109.5∘109.5^\circ (no lone pair); NH3\mathrm{NH_3} ≈107∘\approx107^\circ (1 lone pair); H2O\mathrm{H_2O} ≈104.5∘\approx104.5^\circ (2 lone pairs). In H2S\mathrm{H_2S} the larger, less electronegative S uses almost pure p orbitals and bond-pair repulsion is small, so the angle is ≈92∘\approx92^\circ.

Question 6

Chemistry · 4 marks · −1

The total number of lone pairs on the central Xe atoms in XeF2\mathrm{XeF_2}, XeF4\mathrm{XeF_4} and XeF6\mathrm{XeF_6} taken together is ______.

Correct answer: 6

Explanation: Xe has 8 valence electrons. XeF2\mathrm{XeF_2}: 2 used, 3 lone pairs. XeF4\mathrm{XeF_4}: 4 used, 2 lone pairs. XeF6\mathrm{XeF_6}: 6 used, 1 lone pair. Total =3+2+1=6=3+2+1=6.

Question 7

Chemistry · 4 marks · −1

The number of σ\sigma bonds in a molecule of naphthalene (C10H8\mathrm{C_{10}H_8}) is ______.

Correct answer: 19

Explanation: Naphthalene has 11 C–C bonds (two fused six-membered rings share one bond: 6+6−1=116+6-1=11) and 8 C–H bonds, each containing exactly one σ\sigma bond: 11+8=1911+8=19. (It has 5 π\pi bonds in a Kekulé structure.)

Question 8

Chemistry · 4 marks · −1

The units of the rate constant of a second-order reaction are

  1. A.s−1\mathrm{s^{-1}}
  2. B.mol L−1 s−1\mathrm{mol\,L^{-1}\,s^{-1}}
  3. C.L mol−1 s−1\mathrm{L\,mol^{-1}\,s^{-1}}
  4. D.L2 mol−2 s−1\mathrm{L^2\,mol^{-2}\,s^{-1}}

Correct answer: C

Explanation: k=rate[ ]2=mol L−1s−1mol2L−2=L mol−1s−1k=\frac{\text{rate}}{[\,]^2}=\frac{\mathrm{mol\,L^{-1}s^{-1}}}{\mathrm{mol^2L^{-2}}}=\mathrm{L\,mol^{-1}s^{-1}}.

Question 9

Chemistry · 4 marks · −1

A first-order reaction has a half-life of 30 min. The time required for 75% of the reactant to be consumed is

  1. A.45 min
  2. B.60 min
  3. C.90 min
  4. D.120 min

Correct answer: B

Explanation: 75% consumed means 25% left =(12)2=\left(\frac12\right)^2: two half-lives =60=60 min.

Question 10

Chemistry · 4 marks · −1

The rate of a reaction doubles when the temperature is raised from 300 K to 310 K. The activation energy of the reaction is approximately (R=8.314R=8.314 J mol−1^{-1}K−1^{-1}, log⁡2=0.301\log2=0.301)

  1. A.26.8 kJ mol−1^{-1}
  2. B.107 kJ mol−1^{-1}
  3. C.5.36 kJ mol−1^{-1}
  4. D.53.6 kJ mol−1^{-1}

Correct answer: D

Explanation: log⁡k2k1=Ea2.303R(T2−T1T1T2)⇒Ea=2.303×8.314×0.301×300×31010≈5.36×104\log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right)\Rightarrow E_a=\frac{2.303\times8.314\times0.301\times300\times310}{10}\approx5.36\times10^4 J mol−1^{-1} =53.6=53.6 kJ mol−1^{-1}.

Question 11

Chemistry · 4 marks · −1

For the reaction A+B→P\mathrm{A+B\to P}, the following initial-rate data were obtained: [A] = 0.1 M, [B] = 0.1 M, rate =2×10−3=2\times10^{-3} M/s; [A] = 0.2 M, [B] = 0.1 M, rate =8×10−3=8\times10^{-3} M/s; [A] = 0.1 M, [B] = 0.2 M, rate =2×10−3=2\times10^{-3} M/s. The rate law is

  1. A.rate =k[A]2=k[\mathrm A]^2
  2. B.rate =k[A][B]=k[\mathrm A][\mathrm B]
  3. C.rate =k[A]2[B]=k[\mathrm A]^2[\mathrm B]
  4. D.rate =k[A][B]2=k[\mathrm A][\mathrm B]^2

Correct answer: A

Explanation: Doubling [A] (B fixed) multiplies the rate by 4 ⇒\Rightarrow order 2 in A. Doubling [B] (A fixed) leaves the rate unchanged ⇒\Rightarrow order 0 in B. Rate =k[A]2=k[\mathrm A]^2.

Question 12

Chemistry · 4 marks · −1

For a first-order reaction, the ratio of the time required for 99.9% completion to the half-life is approximately

  1. A.2
  2. B.10
  3. C.100
  4. D.3

Correct answer: B

Explanation: t99.9=2.303klog⁡1000=6.909kt_{99.9}=\frac{2.303}{k}\log1000=\frac{6.909}{k} and t1/2=0.693kt_{1/2}=\frac{0.693}{k}. Ratio =6.9090.693≈10=\frac{6.909}{0.693}\approx10.

Question 13

Chemistry · 4 marks · −1

A substance A decomposes by two parallel first-order reactions: A→B\mathrm{A\to B} with k1=2×10−3 s−1k_1=2\times10^{-3}\ \mathrm{s^{-1}} and A→C\mathrm{A\to C} with k2=6×10−3 s−1k_2=6\times10^{-3}\ \mathrm{s^{-1}}. The half-life of A and the percentage of B in the products are respectively

  1. A.86.6 s and 25%
  2. B.346.5 s and 25%
  3. C.86.6 s and 75%
  4. D.115.5 s and 33%

Correct answer: A

Explanation: Overall k=k1+k2=8×10−3 s−1k=k_1+k_2=8\times10^{-3}\ \mathrm{s^{-1}}, t1/2=0.6938×10−3=86.6t_{1/2}=\frac{0.693}{8\times10^{-3}}=86.6 s. Products form in the ratio k1:k2=1:3k_1:k_2=1:3, so B is 25%.

Question 14

Chemistry · 4 marks · −1

A zero-order reaction has rate constant 0.02 mol L−1 min−10.02\ \mathrm{mol\,L^{-1}\,min^{-1}}. If the initial concentration of the reactant is 0.8 M, the time (in minutes) for the reaction to go to completion is ______.

Correct answer: 40

Explanation: Zero order: [A]=[A]0−kt[\mathrm A]=[\mathrm A]_0-kt. Completion: t=[A]0k=0.80.02=40t=\frac{[\mathrm A]_0}{k}=\frac{0.8}{0.02}=40 min.

Question 15

Chemistry · 4 marks · −1

A first-order reaction is 20% complete in 10 min. The time (in minutes) required for it to be 36% complete is ______.

Correct answer: 20

Explanation: After 10 min, fraction left =0.8=0.8. Fraction left after 36% completion =0.64=0.82=0.64=0.8^2, which takes two such intervals: 20 min. (Check: t=ln⁡(1/0.64)ln⁡(1/0.8)×10=2×10t=\frac{\ln(1/0.64)}{\ln(1/0.8)}\times10=2\times10.)

Question 16

Chemistry · 4 marks · −1

A reaction has ΔH<0\Delta H<0 and ΔS>0\Delta S>0. The reaction is

  1. A.non-spontaneous at all temperatures
  2. B.spontaneous only at high temperatures
  3. C.spontaneous only at low temperatures
  4. D.spontaneous at all temperatures

Correct answer: D

Explanation: ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta S: both terms are negative for every T>0T>0, so ΔG<0\Delta G<0 always.

Question 17

Chemistry · 4 marks · −1

Which of the following is a state function?

  1. A.Heat absorbed, qq
  2. B.Work done, ww
  3. C.Enthalpy, HH
  4. D.qq at constant temperature for an irreversible path

Correct answer: C

Explanation: Enthalpy depends only on the state of the system. Heat and work depend on the path. (qp=ΔHq_p=\Delta H and qV=ΔUq_V=\Delta U are special cases, but qq in general is path dependent.)

Question 18

Chemistry · 4 marks · −1

For the combustion C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\mathrm{C_3H_8(g)+5O_2(g)\to3CO_2(g)+4H_2O(l)} at 298 K, the value of ΔH−ΔU\Delta H-\Delta U is approximately (R=8.314R=8.314 J mol−1^{-1}K−1^{-1})

  1. A.+7.43+7.43 kJ
  2. B.−2.48-2.48 kJ
  3. C.−7.43-7.43 kJ
  4. D.−4.95-4.95 kJ

Correct answer: C

Explanation: Only gases count: Δng=3−(1+5)=−3\Delta n_g=3-(1+5)=-3. ΔH−ΔU=ΔngRT=−3×8.314×298≈−7.43\Delta H-\Delta U=\Delta n_gRT=-3\times8.314\times298\approx-7.43 kJ.

Question 19

Chemistry · 4 marks · −1

Given the standard enthalpies of formation: CH4(g)\mathrm{CH_4(g)} −74.8-74.8, CO2(g)\mathrm{CO_2(g)} −393.5-393.5 and H2O(l)\mathrm{H_2O(l)} −285.8-285.8 kJ mol−1^{-1}. The standard enthalpy of combustion of methane is

  1. A.−890.3-890.3 kJ mol−1^{-1}
  2. B.−604.5-604.5 kJ mol−1^{-1}
  3. C.−1040.0-1040.0 kJ mol−1^{-1}
  4. D.−754.5-754.5 kJ mol−1^{-1}

Correct answer: A

Explanation: CH4+2O2→CO2+2H2O(l)\mathrm{CH_4+2O_2\to CO_2+2H_2O(l)}. ΔcH∘=[−393.5+2(−285.8)]−[−74.8]=−965.1+74.8=−890.3\Delta_cH^\circ=[-393.5+2(-285.8)]-[-74.8]=-965.1+74.8=-890.3 kJ mol−1^{-1}.

Question 20

Chemistry · 4 marks · −1

For a reaction, ΔH=+30\Delta H=+30 kJ mol−1^{-1} and ΔS=+100\Delta S=+100 J K−1^{-1} mol−1^{-1}, both roughly independent of temperature. The reaction is spontaneous

  1. A.below 300 K
  2. B.above 300 K
  3. C.above 3 K
  4. D.at no temperature

Correct answer: B

Explanation: ΔG<0\Delta G<0 requires T>ΔHΔS=30000100=300T>\frac{\Delta H}{\Delta S}=\frac{30000}{100}=300 K.

Question 21

Chemistry · 4 marks · −1

One mole of an ideal gas expands isothermally at 300 K from 10 L to 100 L, (i) reversibly and (ii) freely into a vacuum. Which statement is correct? (Rln⁡10≈19.1R\ln10\approx19.1 J K−1^{-1})

  1. A.In (i) ΔStotal=0\Delta S_{total}=0; in (ii) ΔSsys=19.1\Delta S_{sys}=19.1 J K−1^{-1} and ΔSsurr=0\Delta S_{surr}=0
  2. B.In (ii) ΔSsys=0\Delta S_{sys}=0 because no heat is absorbed
  3. C.In (i) ΔSsys=0\Delta S_{sys}=0 because the process is reversible
  4. D.In (ii) ΔStotal=0\Delta S_{total}=0 because w=0w=0 and q=0q=0

Correct answer: A

Explanation: Entropy is a state function, so ΔSsys=nRln⁡V2V1=19.1\Delta S_{sys}=nR\ln\frac{V_2}{V_1}=19.1 J K−1^{-1} in both cases. (i) Reversible: the surroundings supply qrev=TΔSsysq_{rev}=T\Delta S_{sys}, so ΔSsurr=−19.1\Delta S_{surr}=-19.1 and ΔStotal=0\Delta S_{total}=0. (ii) Free expansion: q=0q=0, so ΔSsurr=0\Delta S_{surr}=0 and ΔStotal=+19.1\Delta S_{total}=+19.1 J K−1^{-1} (irreversible).

Question 22

Chemistry · 4 marks · −1

The enthalpies of combustion of C(graphite), H2(g)\mathrm{H_2(g)} and C2H6(g)\mathrm{C_2H_6(g)} are −393-393, −286-286 and −1560-1560 kJ mol−1^{-1} respectively (water formed as liquid). The standard enthalpy of formation of ethane is −x-x kJ mol−1^{-1}. The value of xx is ______.

Correct answer: 84

Explanation: 2C+3H2→C2H6\mathrm{2C+3H_2\to C_2H_6}: ΔfH=2(−393)+3(−286)−(−1560)=−786−858+1560=−84\Delta_fH=2(-393)+3(-286)-(-1560)=-786-858+1560=-84 kJ mol−1^{-1}. So x=84x=84.

Question 23

Chemistry · 4 marks · −1

A gas expands against a constant external pressure of 2 bar from 5 L to 15 L. The magnitude of the work done by the gas, in joule, is ______. (1 L bar = 100 J)

Correct answer: 2000

Explanation: ∣w∣=PextΔV=2×(15−5)=20|w|=P_{ext}\Delta V=2\times(15-5)=20 L bar =2000=2000 J (work done on the gas is −2000-2000 J).

Question 24

Chemistry · 4 marks · −1

Among Na, Mg, K and Ca, the element with the largest atomic radius is

  1. A.Na
  2. B.Mg
  3. C.Ca
  4. D.K

Correct answer: D

Explanation: Atomic radius increases down a group and decreases across a period. K (period 4, group 1) is below Na and to the left of Ca, so it is the largest.

Question 25

Chemistry · 4 marks · −1

Among B, C, N and O, the element with the highest first ionisation enthalpy is

  1. A.B
  2. B.C
  3. C.N
  4. D.O

Correct answer: C

Explanation: IE generally increases across period 2, but N (2p32p^3, half-filled, extra stability) has a higher first IE than O (2p42p^4, where removing a paired electron is easier). Order: B < C < O < N.

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