JEE Main Practice — Chemistry Set 3

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 element with the most negative electron gain enthalpy among the halogens is

  1. A.F
  2. B.Cl
  3. C.Br
  4. D.I

Correct answer: B

Explanation: Although F is more electronegative, its very small 2p shell causes strong inter-electronic repulsion for the incoming electron, so its electron gain enthalpy is less negative than that of Cl. Cl has the most negative value (≈−349\approx-349 kJ mol−1^{-1}).

Question 2

Chemistry · 4 marks · −1

The correct order of ionic radii of the isoelectronic species O2−\mathrm{O^{2-}}, F−\mathrm{F^-}, Na+\mathrm{Na^+} and Mg2+\mathrm{Mg^{2+}} is

  1. A.Mg2+>Na+>F−>O2−\mathrm{Mg^{2+}>Na^+>F^->O^{2-}}
  2. B.O2−>F−>Na+>Mg2+\mathrm{O^{2-}>F^->Na^+>Mg^{2+}}
  3. C.F−>O2−>Na+>Mg2+\mathrm{F^->O^{2-}>Na^+>Mg^{2+}}
  4. D.O2−>F−>Mg2+>Na+\mathrm{O^{2-}>F^->Mg^{2+}>Na^+}

Correct answer: B

Explanation: All have 10 electrons. Radius decreases as nuclear charge increases (ZZ = 8, 9, 11, 12): O2−>F−>Na+>Mg2+\mathrm{O^{2-}>F^->Na^+>Mg^{2+}}.

Question 3

Chemistry · 4 marks · −1

Which of the following oxides is amphoteric?

  1. A.Na2O\mathrm{Na_2O}
  2. B.SO3\mathrm{SO_3}
  3. C.Al2O3\mathrm{Al_2O_3}
  4. D.MgO\mathrm{MgO}

Correct answer: C

Explanation: Al2O3\mathrm{Al_2O_3} reacts with both acids (→Al3+\to\mathrm{Al^{3+}} salts) and bases (→\to aluminates). Na2O\mathrm{Na_2O} and MgO are basic; SO3\mathrm{SO_3} is acidic.

Question 4

Chemistry · 4 marks · −1

The correct increasing order of first ionisation enthalpy of Mg, Al, Si, P and S is

  1. A.Al < Mg < Si < S < P
  2. B.Mg < Al < Si < P < S
  3. C.Al < Mg < Si < P < S
  4. D.Mg < Al < Si < S < P

Correct answer: A

Explanation: Two anomalies occur in period 3: Al (3p13p^1) < Mg (3s23s^2, filled s subshell; Al's 3p electron is shielded by 3s) and S (3p43p^4) < P (3p33p^3, half-filled). Approximate values (kJ/mol): Al 578, Mg 738, Si 787, S 1000, P 1012.

Question 5

Chemistry · 4 marks · −1

The atomic number of the element placed in period 5, group 16 of the modern periodic table is ______.

Correct answer: 52

Explanation: Period 5 begins at Rb (37). Groups 1–2: Rb 37, Sr 38; groups 3–12: Y 39 to Cd 48; group 13 In 49, 14 Sn 50, 15 Sb 51, 16 Te 52.

Question 6

Chemistry · 4 marks · −1

How many of the following species are isoelectronic with neon: Na+\mathrm{Na^+}, Mg2+\mathrm{Mg^{2+}}, F−\mathrm{F^-}, O2−\mathrm{O^{2-}}, Al3+\mathrm{Al^{3+}}, Cl−\mathrm{Cl^-}, K+\mathrm{K^+}, N3−\mathrm{N^{3-}}?

Correct answer: 6

Explanation: Ne has 10 electrons. Na+\mathrm{Na^+} (10), Mg2+\mathrm{Mg^{2+}} (10), F−\mathrm{F^-} (10), O2−\mathrm{O^{2-}} (10), Al3+\mathrm{Al^{3+}} (10), N3−\mathrm{N^{3-}} (10) qualify; Cl−\mathrm{Cl^-} and K+\mathrm{K^+} have 18. Count =6=6.

Question 7

Chemistry · 4 marks · −1

The IUPAC name of [Co(NH3)5Cl]Cl2\mathrm{[Co(NH_3)_5Cl]Cl_2} is

  1. A.pentaamminechloridocobalt(III) chloride
  2. B.chloridopentaamminecobalt(II) chloride
  3. C.pentaamminechloridocobaltate(III) chloride
  4. D.pentaamminecobalt(III) trichloride

Correct answer: A

Explanation: Ligands in alphabetical order (ammine before chlorido), metal with oxidation state: x+0−1=+2⇒x=+3x+0-1=+2\Rightarrow x=+3. The complex is a cation, so no '-ate' suffix.

Question 8

Chemistry · 4 marks · −1

The oxidation state of iron in K4[Fe(CN)6]\mathrm{K_4[Fe(CN)_6]} is

  1. A.+3
  2. B.+2
  3. C.0
  4. D.+6

Correct answer: B

Explanation: 4(+1)+x+6(−1)=0⇒x=+24(+1)+x+6(-1)=0\Rightarrow x=+2.

Question 9

Chemistry · 4 marks · −1

The spin-only magnetic moment of [Fe(CN)6]3−\mathrm{[Fe(CN)_6]^{3-}} is approximately

  1. A.5.92 BM
  2. B.0 BM
  3. C.3.87 BM
  4. D.1.73 BM

Correct answer: D

Explanation: Fe3+\mathrm{Fe^{3+}} is d5d^5. CN−\mathrm{CN^-} is a strong-field ligand, giving low-spin t2g5t_{2g}^5 with one unpaired electron: μ=1(3)=1.73\mu=\sqrt{1(3)}=1.73 BM.

Question 10

Chemistry · 4 marks · −1

The total number of stereoisomers (geometrical + optical) of [Co(en)2Cl2]+\mathrm{[Co(en)_2Cl_2]^+} is

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

Correct answer: B

Explanation: There are cis and trans forms. The trans isomer has a mirror plane and is optically inactive; the cis isomer is chiral and exists as a pair of enantiomers. Total =1+2=3=1+2=3.

Question 11

Chemistry · 4 marks · −1

The crystal field stabilisation energy of the low-spin complex [Co(NH3)6]3+\mathrm{[Co(NH_3)_6]^{3+}} (ignoring pairing energy) is

  1. A.−0.4Δo-0.4\Delta_o
  2. B.−1.2Δo-1.2\Delta_o
  3. C.−2.4Δo-2.4\Delta_o
  4. D.−1.8Δo-1.8\Delta_o

Correct answer: C

Explanation: Co3+\mathrm{Co^{3+}} is d6d^6; low spin gives t2g6eg0t_{2g}^6e_g^0. CFSE =6(−0.4Δo)=−2.4Δo=6(-0.4\Delta_o)=-2.4\Delta_o.

Question 12

Chemistry · 4 marks · −1

The spin-only magnetic moments (in BM) of [MnBr4]2−\mathrm{[MnBr_4]^{2-}}, [Fe(CN)6]4−\mathrm{[Fe(CN)_6]^{4-}} and [CoF6]3−\mathrm{[CoF_6]^{3-}} respectively are

  1. A.5.92, 4.90, 0
  2. B.1.73, 0, 4.90
  3. C.5.92, 0, 4.90
  4. D.5.92, 0, 0

Correct answer: C

Explanation: [MnBr4]2−\mathrm{[MnBr_4]^{2-}}: Mn2+\mathrm{Mn^{2+}} d5d^5, tetrahedral with weak-field Br−\mathrm{Br^-} (always high spin): 5 unpaired, 5.92 BM. [Fe(CN)6]4−\mathrm{[Fe(CN)_6]^{4-}}: Fe2+\mathrm{Fe^{2+}} d6d^6, strong field, t2g6t_{2g}^6: 0 BM. [CoF6]3−\mathrm{[CoF_6]^{3-}}: Co3+\mathrm{Co^{3+}} d6d^6 with weak-field F−\mathrm{F^-}, high spin t2g4eg2t_{2g}^4e_g^2: 4 unpaired, 24=4.90\sqrt{24}=4.90 BM.

Question 13

Chemistry · 4 marks · −1

The total number of ions produced per formula unit when [Co(NH3)6]Cl3\mathrm{[Co(NH_3)_6]Cl_3}, [Co(NH3)5Cl]Cl2\mathrm{[Co(NH_3)_5Cl]Cl_2} and [Co(NH3)4Cl2]Cl\mathrm{[Co(NH_3)_4Cl_2]Cl} are each dissolved in water, added together, is ______.

Correct answer: 9

Explanation: Only counter-ions outside the coordination sphere ionise: [Co(NH3)6]3++3Cl−\mathrm{[Co(NH_3)_6]^{3+}+3Cl^-} (4 ions), [Co(NH3)5Cl]2++2Cl−\mathrm{[Co(NH_3)_5Cl]^{2+}+2Cl^-} (3 ions), [Co(NH3)4Cl2]++Cl−\mathrm{[Co(NH_3)_4Cl_2]^++Cl^-} (2 ions). Sum =9=9.

Question 14

Chemistry · 4 marks · −1

The spin-only magnetic moment of [Mn(H2O)6]2+\mathrm{[Mn(H_2O)_6]^{2+}} is x\sqrt{x} BM. The value of xx is ______.

Correct answer: 35

Explanation: Mn2+\mathrm{Mn^{2+}} is d5d^5; water is a weak-field ligand, so high spin with 5 unpaired electrons. μ=n(n+2)=35\mu=\sqrt{n(n+2)}=\sqrt{35} BM.

Question 15

Chemistry · 4 marks · −1

Zirconium and hafnium have almost identical atomic radii and very similar chemical properties. This is mainly due to

  1. A.the inert pair effect
  2. B.lanthanoid contraction
  3. C.actinoid contraction
  4. D.their identical electronic configurations

Correct answer: B

Explanation: The 14 lanthanoids (poorly shielding 4f electrons) intervene between Zr (period 5) and Hf (period 6), cancelling the expected size increase down the group — the lanthanoid contraction.

Question 16

Chemistry · 4 marks · −1

The most common and stable oxidation state shown by the lanthanoids is

  1. A.+2
  2. B.+4
  3. C.+3
  4. D.+1

Correct answer: C

Explanation: All lanthanoids show +3 as the characteristic oxidation state; +2 (Eu, Yb) and +4 (Ce, Tb) occur for some, generally tending towards f0f^0, f7f^7 or f14f^{14} configurations.

Question 17

Chemistry · 4 marks · −1

In acidic medium, the number of moles of Fe2+\mathrm{Fe^{2+}} oxidised to Fe3+\mathrm{Fe^{3+}} by one mole of KMnO4\mathrm{KMnO_4} is

  1. A.1
  2. B.2
  3. C.3
  4. D.5

Correct answer: D

Explanation: MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO_4^-+8H^++5e^-\to Mn^{2+}+4H_2O} (Mn: +7 to +2). Each Fe2+\mathrm{Fe^{2+}} loses one electron, so one MnO4−\mathrm{MnO_4^-} oxidises 5 Fe2+\mathrm{Fe^{2+}}.

Question 18

Chemistry · 4 marks · −1

Which set contains only colourless ions in aqueous solution?

  1. A.Sc3+,Zn2+,Cu+\mathrm{Sc^{3+}, Zn^{2+}, Cu^+}
  2. B.Ti3+,Zn2+,Cu2+\mathrm{Ti^{3+}, Zn^{2+}, Cu^{2+}}
  3. C.Sc3+,Cr3+,Cu+\mathrm{Sc^{3+}, Cr^{3+}, Cu^+}
  4. D.V3+,Ni2+,Zn2+\mathrm{V^{3+}, Ni^{2+}, Zn^{2+}}

Correct answer: A

Explanation: Colour in these ions arises from d–d transitions, which need a partly filled d subshell. Sc3+\mathrm{Sc^{3+}} (3d03d^0), Zn2+\mathrm{Zn^{2+}} (3d103d^{10}) and Cu+\mathrm{Cu^+} (3d103d^{10}) are colourless; Ti3+\mathrm{Ti^{3+}} (d1d^1), Cu2+\mathrm{Cu^{2+}} (d9d^9), Cr3+\mathrm{Cr^{3+}} (d3d^3), V3+\mathrm{V^{3+}} (d2d^2), Ni2+\mathrm{Ni^{2+}} (d8d^8) are coloured.

Question 19

Chemistry · 4 marks · −1

A divalent 3d-series ion has a spin-only magnetic moment of 4.90 BM. The ion is

  1. A.Mn2+\mathrm{Mn^{2+}}
  2. B.Fe2+\mathrm{Fe^{2+}}
  3. C.Ni2+\mathrm{Ni^{2+}}
  4. D.Cu2+\mathrm{Cu^{2+}}

Correct answer: B

Explanation: n(n+2)=4.90⇒n=4\sqrt{n(n+2)}=4.90\Rightarrow n=4 unpaired electrons. Fe2+\mathrm{Fe^{2+}} (3d63d^6, free ion) has 4. Mn2+\mathrm{Mn^{2+}} has 5, Ni2+\mathrm{Ni^{2+}} 2, Cu2+\mathrm{Cu^{2+}} 1.

Question 20

Chemistry · 4 marks · −1

Which of the following statements is INCORRECT?

  1. A.MnO4−\mathrm{MnO_4^-} is tetrahedral and its intense colour is due to d–d transitions
  2. B.In Cr2O72−\mathrm{Cr_2O_7^{2-}}, each Cr is tetrahedrally surrounded by O atoms and there is one Cr–O–Cr bridge
  3. C.In neutral or faintly alkaline solution MnO4−\mathrm{MnO_4^-} is reduced to MnO2\mathrm{MnO_2}
  4. D.Acidified K2Cr2O7\mathrm{K_2Cr_2O_7} oxidises H2S\mathrm{H_2S} to sulfur

Correct answer: A

Explanation: Mn in MnO4−\mathrm{MnO_4^-} is +7 (3d03d^0), so d–d transitions are impossible; the purple colour arises from ligand-to-metal (O→Mn) charge transfer. B, C and D are correct: dichromate is two corner-sharing CrO4\mathrm{CrO_4} tetrahedra; MnO4−+2H2O+3e−→MnO2+4OH−\mathrm{MnO_4^-+2H_2O+3e^-\to MnO_2+4OH^-}; Cr2O72−+3H2S+8H+→2Cr3++3S+7H2O\mathrm{Cr_2O_7^{2-}+3H_2S+8H^+\to2Cr^{3+}+3S+7H_2O}.

Question 21

Chemistry · 4 marks · −1

The volume (in mL) of 0.02 M KMnO4\mathrm{KMnO_4} required to oxidise completely 20 mL of 0.1 M FeSO4\mathrm{FeSO_4} in acidic medium is ______.

Correct answer: 20

Explanation: Fe2+\mathrm{Fe^{2+}} = 2 mmol. MnO4−\mathrm{MnO_4^-} needed =25=0.4=\frac{2}{5}=0.4 mmol. Volume =0.40.02=20=\frac{0.4}{0.02}=20 mL.

Question 22

Chemistry · 4 marks · −1

The number of unpaired electrons in a ground-state gadolinium atom (Z=64Z=64) is ______.

Correct answer: 8

Explanation: Gd: [Xe] 4f7 5d1 6s2\mathrm{[Xe]}\,4f^7\,5d^1\,6s^2 (half-filled 4f is retained). Unpaired: 7 (4f) + 1 (5d) = 8.

Question 23

Chemistry · 4 marks · −1

For the reaction N2(g)+3H2(g)⇌2NH3(g)\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}, the relation between KpK_p and KcK_c is

  1. A.Kp=Kc(RT)2K_p=K_c(RT)^{2}
  2. B.Kp=KcK_p=K_c
  3. C.Kp=Kc(RT)−2K_p=K_c(RT)^{-2}
  4. D.Kp=Kc(RT)−1K_p=K_c(RT)^{-1}

Correct answer: C

Explanation: Kp=Kc(RT)ΔngK_p=K_c(RT)^{\Delta n_g} with Δng=2−4=−2\Delta n_g=2-4=-2.

Question 24

Chemistry · 4 marks · −1

The pH of a 0.001 M aqueous HCl solution at 25 °C is

  1. A.3
  2. B.1
  3. C.11
  4. D.2

Correct answer: A

Explanation: HCl is a strong acid: [H+]=10−3[\mathrm{H^+}]=10^{-3} M, pH =3=3.

Question 25

Chemistry · 4 marks · −1

The pH of a 0.1 M acetic acid solution (Ka=1.8×10−5K_a=1.8\times10^{-5}) is approximately

  1. A.1.00
  2. B.4.74
  3. C.2.87
  4. D.3.87

Correct answer: C

Explanation: [H+]=Kac=1.8×10−6=1.34×10−3[\mathrm{H^+}]=\sqrt{K_ac}=\sqrt{1.8\times10^{-6}}=1.34\times10^{-3} M. pH =3−log⁡1.34≈2.87=3-\log1.34\approx2.87.

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