Practice free Chemistry NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Which reaction is NOT a redox reaction?
(A) is a simple double-displacement reaction; oxidation states do not change.
'Spin only' magnetic moment is same for which of the following ions?
A. $Ti^{3+}$ B. $Cr^{2+}$ C. $Mn^{2+}$ D. $Fe^{2+}$ E. $Sc^{3+}$
Unpaired e⁻: Ti³⁺ ($d^1$, 1), Cr²⁺ ($d^4$, 4), Mn²⁺ ($d^5$, 5), Fe²⁺ ($d^6$, 4), Sc³⁺ ($d^0$, 0). B and D both have 4 unpaired e⁻.
The highest number of helium atoms is in
Atoms: (A) 0.1 mol $\approx 6\times 10^{22}$; (B) $4N_A$; (C) 1 atom; (D) $N_A$. (B) is the largest.
Match List I with List II.
| List I (Complex) | List II (Type of isomerism) |
|---|---|
| A. $[\text{Co(NH}_3)_5(\text{NO}_2)]\text{Cl}_2$ | I. Solvate isomerism |
| B. $[\text{Co(NH}_3)_5(\text{SO}_4)]\text{Br}$ | II. Linkage isomerism |
| C. $[\text{Co(NH}_3)_6][\text{Cr(CN)}_6]$ | III. Ionization isomerism |
| D. $[\text{Co(H}_2\text{O})_6]\text{Cl}_3$ | IV. Coordination isomerism |
Choose the correct answer from the options given below:
NO₂ binds via N or O (linkage); SO₄/Br outside vs. inside (ionization); both ions complex (coordination); H₂O exchange (solvate).
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follow the order $H_2O > H_2Te > H_2Se > H_2S$.
Statement II: On the basis of molecular mass, $H_2O$ is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in $H_2O$, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
Anomalous H-bonding makes H₂O's bp the highest in the family.
A compound with a molecular formula of $C_6H_{14}$ has two tertiary carbons. Its IUPAC name is:
$(CH_3)_2CH\text{-}CH(CH_3)_2$ has two adjacent tertiary carbons.
Match List I with List II.
| List I (Conversion) | List II (Number of Faraday required) |
|---|---|
| A. 1 mol of $H_2O$ to $O_2$ | I. 3F |
| B. 1 mol of $MnO_4^{-}$ to $Mn^{2+}$ | II. 2F |
| C. 1.5 mol of Ca from molten $CaCl_2$ | III. 1F |
| D. 1 mol of FeO to $Fe_2O_3$ | IV. 5F |
Choose the correct answer from the options given below:
A: $2H_2O \to O_2 + 4H^+ + 4e^-$, 1 mol → 2F. B: Mn $+7 \to +2$ → 5F. C: $Ca^{2+} + 2e^- \to Ca$, 1.5 mol → 3F. D: $Fe^{2+} \to Fe^{3+}$, 1 mol → 1F.
The compound that will undergo $S_N^1$ reaction with the fastest rate is
(A) forms a stable 2° benzylic carbocation (resonance-stabilised).
Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si
Choose the correct answer from the options given below:
Standard Pauling values: Si 1.9, C 2.55, N 3.04, O 3.44, F 3.98.
Major products A and B formed in the following reaction sequence are:
2-methyl-1-cyclohexanol (i.e. 1-methylcyclohexanol — OH and CH₃ on the same ring carbon)
$\xrightarrow{PBr_3}$ A (major) $\xrightarrow[\Delta]{alc.\ KOH}$ B (major)
PBr₃ replaces OH with Br at the same (tertiary) carbon → 1-bromo-1-methylcyclohexane. Alcoholic KOH gives E2 to the Saytzeff alkene → 1-methylcyclohexene.
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.