JEE Main Practice — Chemistry Set 1

JEE Main · 2026 · with answers and explanations

questions
25
Total marks
97
Duration
60 minutes
Negative marking
−1

Subjects: Chemistry

Question 1

Chemistry · 4 marks · −1

The number of radial nodes in a 4d orbital is

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

Correct answer: D

Explanation: Radial nodes =n−l−1=4−2−1=1=n-l-1=4-2-1=1. (Angular nodes =l=2=l=2; total nodes =n−1=3=n-1=3.)

Question 2

Chemistry · 4 marks · −1

Which of the following sets of quantum numbers (n,l,ml,ms)(n, l, m_l, m_s) is NOT allowed?

  1. A.(3,2,−2,+12)(3, 2, -2, +\frac12)
  2. B.(2,2,0,−12)(2, 2, 0, -\frac12)
  3. C.(4,0,0,+12)(4, 0, 0, +\frac12)
  4. D.(3,1,−1,−12)(3, 1, -1, -\frac12)

Correct answer: B

Explanation: ll can take values 00 to n−1n-1. For n=2n=2, l=2l=2 is impossible. All the other sets satisfy 0≤l≤n−10\le l\le n-1 and ∣ml∣≤l|m_l|\le l.

Question 3

Chemistry · 4 marks · −1

The energy of the electron in the second Bohr orbit of He+\mathrm{He^+} is (energy of the H atom in its ground state =−13.6=-13.6 eV)

  1. A.−3.4-3.4 eV
  2. B.−54.4-54.4 eV
  3. C.−13.6-13.6 eV
  4. D.−6.8-6.8 eV

Correct answer: C

Explanation: En=−13.6Z2n2E_n=-13.6\frac{Z^2}{n^2} eV =−13.6×44=−13.6=-13.6\times\frac{4}{4}=-13.6 eV.

Question 4

Chemistry · 4 marks · −1

If the position of an electron is known to within 1 A˚1\ \mathrm{\AA}, the minimum uncertainty in its velocity is approximately (h=6.626×10−34h=6.626\times10^{-34} J s, me=9.1×10−31m_e=9.1\times10^{-31} kg)

  1. A.5.8×1055.8\times10^{5} m/s
  2. B.1.2×1061.2\times10^{6} m/s
  3. C.7.3×1067.3\times10^{6} m/s
  4. D.5.8×1045.8\times10^{4} m/s

Correct answer: A

Explanation: Δx mΔv≥h4π⇒Δv=6.626×10−344π×9.1×10−31×10−10≈5.8×105\Delta x\,m\Delta v\ge\frac{h}{4\pi}\Rightarrow\Delta v=\frac{6.626\times10^{-34}}{4\pi\times9.1\times10^{-31}\times10^{-10}}\approx5.8\times10^5 m/s.

Question 5

Chemistry · 4 marks · −1

In a sample of hydrogen atoms, all excited to the n=5n=5 level, electrons return to the n=2n=2 level (but not below) by all possible routes. The maximum number of distinct spectral lines emitted is

  1. A.10
  2. B.6
  3. C.4
  4. D.3

Correct answer: B

Explanation: Levels involved: 5, 4, 3, 2 (Δn=3\Delta n=3). Number of lines =Δn(Δn+1)2=3×42=6=\frac{\Delta n(\Delta n+1)}{2}=\frac{3\times4}{2}=6 (5→4, 5→3, 5→2, 4→3, 4→2, 3→2).

Question 6

Chemistry · 4 marks · −1

An electron in a hydrogen atom occupies an orbital that has 2 radial nodes and 1 angular node. The orbital and the energy needed to ionise the electron from it are respectively

  1. A.4p and 0.85 eV
  2. B.3p and 1.51 eV
  3. C.4d and 0.85 eV
  4. D.3d and 1.51 eV

Correct answer: A

Explanation: Angular nodes =l=1=l=1 (a p orbital). Radial nodes n−l−1=2⇒n=4n-l-1=2\Rightarrow n=4: a 4p orbital. For hydrogen E4=−13.616=−0.85E_4=-\frac{13.6}{16}=-0.85 eV, so ionisation energy =0.85=0.85 eV. (4d has 1 radial and 2 angular nodes; 3p has 1 radial node.)

Question 7

Chemistry · 4 marks · −1

The total number of electrons with azimuthal quantum number l=1l=1 in a ground-state iron atom (Z=26Z=26) is ______.

Correct answer: 12

Explanation: Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s21s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2. Electrons with l=1l=1 are the p electrons: 2p6+3p6=122p^6+3p^6=12.

Question 8

Chemistry · 4 marks · −1

The energy required to ionise one mole of hydrogen atoms from their ground state, in kJ mol−1^{-1} (nearest integer), is ______. (Ionisation energy of one H atom =13.6=13.6 eV; 11 eV per particle =96.49=96.49 kJ mol−1^{-1}.)

Correct answer: 1312

Explanation: 13.6 eV×96.49 kJ mol−1 eV−1=1312.3≈131213.6\ \mathrm{eV}\times96.49\ \mathrm{kJ\,mol^{-1}\,eV^{-1}}=1312.3\approx1312 kJ mol−1^{-1}.

Question 9

Chemistry · 4 marks · −1

The IUPAC name of CH3−CH(CH3)−CH2−CH2−OH\mathrm{CH_3{-}CH(CH_3){-}CH_2{-}CH_2{-}OH} is

  1. A.2-methylbutan-4-ol
  2. B.3-methylbutan-1-ol
  3. C.isopentanol-1
  4. D.3-methylbutan-4-ol

Correct answer: B

Explanation: Longest chain containing the carbon bearing OH has 4 carbons; number from the OH end so that OH gets locant 1. The methyl is then on C3: 3-methylbutan-1-ol.

Question 10

Chemistry · 4 marks · −1

The hybridisation states of the three carbon atoms in CH2=CH−C≡N\mathrm{CH_2{=}CH{-}C{\equiv}N} (from left to right) are

  1. A.sp2,sp2,spsp^2, sp^2, sp
  2. B.sp2,sp,spsp^2, sp, sp
  3. C.sp3,sp2,spsp^3, sp^2, sp
  4. D.sp2,sp2,sp2sp^2, sp^2, sp^2

Correct answer: A

Explanation: The two carbons of the C=C double bond are sp2sp^2; the nitrile carbon (triple bond) is spsp.

Question 11

Chemistry · 4 marks · −1

The correct order of acidic strength of (I) 4-nitrophenol, (II) phenol, (III) 4-methylphenol and (IV) ethanol is

  1. A.II > I > III > IV
  2. B.I > III > II > IV
  3. C.IV > III > II > I
  4. D.I > II > III > IV

Correct answer: D

Explanation: The −NO2-\mathrm{NO_2} group (−M-M, −I-I) stabilises the phenoxide ion, increasing acidity; −CH3-\mathrm{CH_3} (+I, hyperconjugation) destabilises it, decreasing acidity. Ethoxide has no resonance stabilisation, so ethanol is least acidic: I > II > III > IV.

Question 12

Chemistry · 4 marks · −1

The number of hydrogen atoms available for hyperconjugation with the double bond in 2-methylbut-2-ene, (CH3)2C=CH−CH3\mathrm{(CH_3)_2C{=}CH{-}CH_3}, is

  1. A.10
  2. B.6
  3. C.9
  4. D.12

Correct answer: C

Explanation: Hyperconjugation uses C–H bonds on carbons attached to the sp2sp^2 carbons (α\alpha-H). Two CH3\mathrm{CH_3} on C2 give 6 and one CH3\mathrm{CH_3} on C3 gives 3: total 9. The H on C3 is vinylic, not α\alpha.

Question 13

Chemistry · 4 marks · −1

In Lassaigne's test, the sodium fusion extract of an organic compound gives a blood-red colour with iron(III) chloride. This shows that the compound contains

  1. A.nitrogen only
  2. B.sulfur only
  3. C.both nitrogen and sulfur
  4. D.a halogen

Correct answer: C

Explanation: When N and S are both present, sodium fusion forms NaSCN; Fe3+\mathrm{Fe^{3+}} gives blood-red [Fe(SCN)]2+\mathrm{[Fe(SCN)]^{2+}}. N alone gives Prussian blue; S alone gives a violet colour with sodium nitroprusside.

Question 14

Chemistry · 4 marks · −1

The total number of stereoisomers of 4-bromopent-2-ene, CH3−CH=CH−CH(Br)−CH3\mathrm{CH_3{-}CH{=}CH{-}CH(Br){-}CH_3}, is

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

Correct answer: A

Explanation: The C2=C3 double bond shows cis–trans isomerism (C2 carries H and CH3\mathrm{CH_3}; C3 carries H and CHBrCH3\mathrm{CHBrCH_3}). C4 is a stereocentre (H, Br, CH3\mathrm{CH_3}, CH=CHCH3\mathrm{CH{=}CHCH_3}). The two are independent and there is no internal symmetry, so 2×2=42\times2=4 stereoisomers (E/Z each as R and S).

Question 15

Chemistry · 4 marks · −1

In a Kjeldahl estimation, the ammonia evolved from 0.5 g of an organic compound exactly neutralised 10 mL of 1 M H2SO4\mathrm{H_2SO_4}. The percentage of nitrogen in the compound is ______.

Correct answer: 56

Explanation: 10 mL of 1 M H2SO4\mathrm{H_2SO_4} = 20 mmol H+\mathrm{H^+} = 20 mmol NH3\mathrm{NH_3} = 20 mmol N =0.28=0.28 g. %N =0.280.5×100=56%=\frac{0.28}{0.5}\times100=56\%.

Question 16

Chemistry · 4 marks · −1

In a Carius determination, 0.2 g of an organic compound gave 0.376 g of AgBr. The percentage of bromine in the compound is ______. (Ag = 108, Br = 80)

Correct answer: 80

Explanation: Mass of Br =80188×0.376=0.16=\frac{80}{188}\times0.376=0.16 g. %Br =0.160.2×100=80%=\frac{0.16}{0.2}\times100=80\%.

Question 17

Chemistry · 1 marks

Which of the following is a reducing sugar?

  1. B.Glucose
  2. C.Cellulose
  3. D.Starch
  4. A.Sucrose

Correct answer: B

Explanation: Glucose is a reducing sugar because it has a free aldehyde group (-CHO) that can reduce Tollens' reagent and Fehling's solution. Sucrose is a non-reducing sugar.

Question 18

Chemistry · 4 marks · −1

Deficiency of which vitamin causes scurvy?

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

Correct answer: C

Explanation: Scurvy (bleeding gums) is caused by deficiency of vitamin C (ascorbic acid). Vitamin A: night blindness; D: rickets; K: delayed blood clotting.

Question 19

Chemistry · 4 marks · −1

Prolonged heating of glucose with HI gives n-hexane. This observation shows that

  1. A.glucose contains an aldehyde group
  2. B.glucose has five hydroxyl groups
  3. C.glucose exists in a cyclic form
  4. D.the six carbon atoms of glucose are linked in a straight chain

Correct answer: D

Explanation: HI/heat reduces all functional groups to C–H, leaving the carbon skeleton; obtaining n-hexane (not a branched hexane) proves an unbranched six-carbon chain. Aldehyde and OH groups are shown by other reactions (oxidation to gluconic acid, acetylation).

Question 20

Chemistry · 4 marks · −1

How many different dipeptides can be formed from glycine and alanine if each dipeptide contains two amino-acid residues and a residue may be repeated?

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

Correct answer: B

Explanation: Order matters (N-terminal vs C-terminal): Gly-Gly, Gly-Ala, Ala-Gly, Ala-Ala = 2×2=42\times2=4.

Question 21

Chemistry · 4 marks · −1

Which statement about nucleic acids is correct?

  1. A.DNA contains ribose and uracil
  2. B.RNA contains deoxyribose and thymine
  3. C.DNA contains deoxyribose and thymine, while RNA contains ribose and uracil
  4. D.Both DNA and RNA contain thymine

Correct answer: C

Explanation: DNA: 2-deoxyribose with bases A, G, C, T. RNA: ribose with A, G, C, U (uracil replaces thymine).

Question 22

Chemistry · 4 marks · −1

The specific rotations of sucrose, D-glucose and D-fructose are +66.5∘+66.5^\circ, +52.5∘+52.5^\circ and −92.4∘-92.4^\circ respectively. Sucrose is completely hydrolysed in dilute acid. The specific rotation of the resulting mixture (taking specific rotation of a mixture as the mass-weighted average) is

  1. A.−39.9∘-39.9^\circ
  2. B.+19.95∘+19.95^\circ
  3. C.+66.5∘+66.5^\circ
  4. D.−19.95∘-19.95^\circ

Correct answer: D

Explanation: Hydrolysis gives equimolar glucose and fructose; both have molar mass 180, so the masses are equal. [α]=52.5+(−92.4)2=−19.95∘[\alpha]=\frac{52.5+(-92.4)}{2}=-19.95^\circ. The sign changes from + to −, hence the name invert sugar.

Question 23

Chemistry · 4 marks · −1

The number of chiral carbon atoms in the open-chain structure of D-glucose is ______.

Correct answer: 4

Explanation: OHC−(CHOH)4−CH2OH\mathrm{OHC{-}(CHOH)_4{-}CH_2OH}: C2, C3, C4 and C5 each carry four different groups; C1 (CHO) and C6 (CH2OH\mathrm{CH_2OH}) are not stereocentres. Hence 4.

Question 24

Chemistry · 4 marks · −1

Glucose reacts with acetic anhydride to form a pentaacetate. The molar mass of glucose pentaacetate (in g mol−1^{-1}) is ______. (C = 12, H = 1, O = 16)

Correct answer: 390

Explanation: Each of the five –OH groups becomes –OCOCH3_3, adding C2H2O=42\mathrm{C_2H_2O}=42 per group. 180+5×42=390180+5\times42=390 g mol−1^{-1} (C16H22O11\mathrm{C_{16}H_{22}O_{11}}).

Question 25

Chemistry · 4 marks · −1

The shape of the SF4\mathrm{SF_4} molecule is

  1. A.tetrahedral
  2. B.square planar
  3. C.see-saw
  4. D.trigonal pyramidal

Correct answer: C

Explanation: S has 6 valence electrons: 4 bond pairs + 1 lone pair = 5 electron pairs (sp3dsp^3d, trigonal bipyramidal). The lone pair occupies an equatorial position, giving a see-saw shape.

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