NEET Chemistry Previous Year Questions (2024–2025)
NEET Chemistry PYQs 2025: 95 Solved Questions - Online Practice — Practice Questions
1. Ratio of wavelengths of light absorbed by H atom for n=2->n=3 and n=4->n=6 transitions is:
- (A) 1/36
- (B) 1/16
- (C) 1/9
- (D) 1/4
Answer: (D) 1/4
Delta E prop (1/n1^2-1/n2^2); for 2->3, dE1=RH*5/36; for 4->6, dE2=RH*5/144; lambda prop 1/dE so lambda1/lambda2=dE2/dE1=1/4.
2. (Orig Q47) Which statements are true? A: Cs has very low melting point (unlike Ga). B: Electronegativity of N and Cl (Pauling) are not same. C: Ar,K+,Cl-,Ca2+,S2- are isoelectronic. D: First IE order Si>Al>Mg>Na. E: Atomic radius of Cs > Li and Rb.
- (A) A, B and E only
- (B) C and E only
- (C) C and D only
- (D) A, C and E only
Answer: (B) C and E only
N and Cl have equal electronegativity (3.0) so B is false; Ga also has high melting point contrary to A being framed as unlike, so A is false; D's stated order is wrong (correct is Si>Mg>Al>Na); C and E are true.
3. Match ions (Co2+, Mg2+, Pb2+, Al3+) with their cation-analysis group number (Group I, III, IV, VI).
- (A) A-III,B-IV,C-II,D-I
- (B) A-III,B-IV,C-I,D-II
- (C) A-III,B-II,C-IV,D-I
- (D) A-III,B-II,C-I,D-IV
Answer: (B) A-III,B-IV,C-I,D-II
Standard qualitative analysis groups: Pb2+ is Group I, Al3+ is Group III, Co2+ is Group IV, Mg2+ is Group VI, matching A-III,B-IV,C-I,D-II.
4. Predict major product P: cyclopentane with CH3 substituent, treated with (i) HBr/benzoyl peroxide (ii) KCN (iii) Na(Hg)/C2H5OH.
- (A) Cyclopentane ring with CH3 and CH2NH2 (isomer A)
- (B) Cyclopentane ring with CH3 and CH2NH2 (isomer B)
- (C) Cyclopentane ring with CH3 and NC group (isomer A)
- (D) Cyclopentane ring with CH3 and NC group (isomer B)
Answer: (A) Cyclopentane ring with CH3 and CH2NH2 (isomer A)
Anti-Markovnikov HBr addition (peroxide) puts Br on terminal carbon; KCN substitutes Br to nitrile; Na(Hg)/EtOH reduces nitrile to CH2NH2, giving the amine product.
5. Energy and radius of first Bohr orbit of He+ and Li2+ (RH=2.18e-18 J, a0=52.9 pm) are given by which option?
- (A) En(Li2+)=-19.62e-18 J, rn(Li2+)=17.6 pm; En(He+)=-8.72e-18 J, rn(He+)=26.4 pm
- (B) En(Li2+)=-8.72e-18 J, rn(Li2+)=26.4 pm; En(He+)=-19.62e-18 J, rn(He+)=17.6 pm
- (C) En(Li2+)=-19.62e-16 J, rn(Li2+)=17.6 pm; En(He+)=-8.72e-16 J, rn(He+)=26.4 pm
- (D) En(Li2+)=-8.72e-16 J, rn(Li2+)=17.6 pm; En(He2+)=-19.62e-16 J, rn(He+)=17.6 pm
Answer: (A) En(Li2+)=-19.62e-18 J, rn(Li2+)=17.6 pm; En(He+)=-8.72e-18 J, rn(He+)=26.4 pm
En=-RH Z^2/n^2 and rn=a0 n^2/Z; for Li2+ (Z=3): E=-19.62e-18 J, r=17.6 pm; for He+ (Z=2): E=-8.72e-18 J, r=26.4 pm.
6. Which of the following are paramagnetic? A:[NiCl4]2-, B:Ni(CO)4, C:[Ni(CN)4]2-, D:[Ni(H2O)6]2+, E:Ni(PPh3)4.
- (A) A and C only
- (B) B and E only
- (C) A and D only
- (D) A, D and E only
Answer: (C) A and D only
Weak-field Cl- and H2O keep Ni2+ high-spin with unpaired electrons (paramagnetic) in A and D; CO and PPh3 give Ni(0) d10 diamagnetic (B,E); strong-field CN- gives diamagnetic square planar d8 (C). So only A and D are paramagnetic.
7. Statement I: Arsenic can form arsine like nitrogen forms ammonia. Statement II: Antimony cannot form antimony pentoxide. Choose the correct option.
- (A) Both Statement I and II are correct
- (B) Both Statement I and II are incorrect
- (C) Statement I correct, Statement II incorrect
- (D) Statement I incorrect, Statement II correct
Answer: (C) Statement I correct, Statement II incorrect
Arsenic does form arsine (AsH3), so Statement I is correct; antimony does form Sb2O5 (antimony pentoxide), so Statement II is incorrect.
8. Which electronic configurations belong to main group elements? A:[Ne]3s1, B:[Ar]3d3 4s2, C:[Kr]4d10 5s2 5p5, D:[Ar]3d10 4s1, E:[Rn]5f0 6d2 7s2.
- (A) B and E only
- (B) A and C only
- (C) D and E only
- (D) A, C and D only
Answer: (B) A and C only
Main group elements have valence in only s or p subshells; A ([Ne]3s1, Na) and C ([Kr]4d10 5s2 5p5, I) qualify; B, D, E involve d/f subshell valence electrons (transition/inner-transition).
9. Dalton's Atomic theory could not explain which of the following?
- (A) Law of conservation of mass
- (B) Law of constant proportion
- (C) Law of multiple proportion
- (D) Law of gaseous volume
Answer: (D) Law of gaseous volume
Dalton's theory explained conservation of mass, constant and multiple proportions, but could not explain Gay-Lussac's law of gaseous volumes.
10. Oxidation states of underlined elements in KO2, H2O2, H2SO4 are, respectively:
- (A) +1, -1, and +6
- (B) +2, -2, and +6
- (C) +1, -2, and +4
- (D) +4, -4, and +6
Answer: (A) +1, -1, and +6
In KO2 (superoxide), K is +1; in H2O2 (peroxide), O is -1; in H2SO4, S is +6.
11. If half-life of a first order reaction is 1 minute, time required for 99.9% completion is closest to:
- (A) 2 minutes
- (B) 4 minutes
- (C) 5 minutes
- (D) 10 minutes
Answer: (D) 10 minutes
kt=2.303 log(a/(a-x)); with t1/2=1 min, t=(2.303/0.693) log(1000) approx 10 minutes.
12. Correct order of wavelength of light absorbed by complexes [Co(NH3)6]3+, [Co(CN)6]3-, [Cu(H2O)4]2+, [Ti(H2O)6]3+ is:
- (A) B<D<A<C
- (B) B<A<D<C
- (C) C<D<A<B
- (D) C<A<D<B
Answer: (B) B<A<D<C
Crystal field splitting order CN->NH3>H2O and charge effects give splitting order [Co(CN)6]3-<[Ti(H2O)6]3+<[Co(NH3)6]3+<[Cu(H2O)4]2+; since wavelength is inversely proportional to splitting, order is B<A<D<C.
13. Which compound can exist as cis-trans isomers?
- (A) Pent-1-ene
- (B) 2-Methylhex-2-ene
- (C) 1,1-Dimethylcyclopropane
- (D) 1,2-Dimethylcyclohexane
Answer: (D) 1,2-Dimethylcyclohexane
Cis-trans isomerism needs two different substituents on each stereogenic carbon; only 1,2-dimethylcyclohexane satisfies this among the options.
14. Phosphoric acid ionizes in three steps (Ka1,Ka2,Ka3); K is overall constant. Which statements are true? A: logK=logKa1+logKa2+logKa3. B: H3PO4 stronger than H2PO4- and HPO4^2-. C: Ka1>Ka2>Ka3. D: Ka1=(Ka3+Ka2)/2.
- (A) A and B only
- (B) A and C only
- (C) B, C and D only
- (D) A, B and C only
Answer: (D) A, B and C only
Overall K is the product of stepwise constants so log K is their sum (A true); acidity decreases with each successive ionization from a more negatively charged species (B, C true); D is not a valid general relation.
15. Which reaction does NOT give benzene as the product? (i) sodium benzoate + sodalime/heat, (ii) n-hexane with Mo2O3 catalyst, (iii) acetylene through red hot iron tube, (iv) benzenediazonium chloride with H2O/warm.
- (A) 1
- (B) 2
- (C) 3
- (D) 4
Answer: (D) 4
Reactions (1)-(3) all give benzene (decarboxylation, dehydrocyclization, cyclic trimerization); reaction (4) with water gives phenol, not benzene.