Chemistry MCQs for NEET — Practice Questions with Answers

Practice free Chemistry NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology
Register free to filter questions
NEET 2024

Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N.

Choose the correct answer from the options given below:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Be > B (filled 2s); N > C (half-filled 2p). Order: Li < B < Be < C < N.

NEET 2024

Among Group 16 elements, which one does NOT show $-2$ oxidation state?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Po is the most metallic Group 16 element and does not exhibit a stable $-2$ state.

NEET 2024

In which of the following equilibria, $K_p$ and $K_c$ are NOT equal?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$K_p = K_c(RT)^{\Delta n}$. Only (B) has $\Delta n = 1 \ne 0$.

NEET 2024

The reagents with which glucose does not react to give the corresponding tests/products are:

A. Tollen's reagent B. Schiff's reagent C. HCN D. NH₂OH E. NaHSO₃

Choose the correct options from the given below:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Glucose exists largely in cyclic hemiacetal form, so its aldehyde group is masked — it does not give a positive Schiff's test or react with NaHSO₃.

NEET 2024

Identify the correct reagents that would bring about the following transformation.

Ph–CH₂–CH=CH₂ → Ph–CH₂–CH₂–CHO

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Anti-Markovnikov OH placement needs hydroboration-oxidation, then PCC oxidises the primary alcohol selectively to the aldehyde (CrO₃ would over-oxidise to the carboxylic acid).

NEET 2024

The $E^\circ$ value for the $\text{Mn}^{3+}/\text{Mn}^{2+}$ couple is more positive than that of $\text{Cr}^{3+}/\text{Cr}^{2+}$ or $\text{Fe}^{3+}/\text{Fe}^{2+}$ due to change of:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Mn³⁺ ($d^4$) → Mn²⁺ ($d^5$, half-filled, stable) — large favourable change, hence more positive $E^\circ$.

NEET 2024

Given below are two statements:

Statement I: Both $[\text{Co(NH}_3)_6]^{3+}$ and $[\text{CoF}_6]^{3-}$ complexes are octahedral but differ in their magnetic behaviour.

Statement II: $[\text{Co(NH}_3)_6]^{3+}$ is diamagnetic whereas $[\text{CoF}_6]^{3-}$ is paramagnetic.

In the light of the above statements, choose the correct answer from the options given below:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

NH₃ is strong field → low-spin $d^6$ (no unpaired e⁻, diamagnetic). F⁻ is weak field → high-spin $d^6$ (4 unpaired e⁻, paramagnetic).

NEET 2024

Fehling's solution 'A' is

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Fehling A is aqueous CuSO₄; Fehling B is alkaline sodium-potassium tartrate.

NEET 2024

In which of the following processes entropy increases?

A. A liquid evaporates to vapour.

B. Temperature of a crystalline solid lowered from 130 K to 0 K.

C. $2\,\text{NaHCO}_3{}_{(s)} \to \text{Na}_2\text{CO}_3{}_{(s)} + \text{CO}_2{}_{(g)} + \text{H}_2\text{O}_\text{(g)}$

D. $\text{Cl}_2{}_{(g)} \to 2\,\text{Cl}_{(g)}$

Choose the correct answer from the options given below:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Evaporation (A), decomposition producing gases (C) and atomisation (D) all increase entropy; cooling a crystal (B) decreases entropy.

NEET 2024

The energy of an electron in the ground state ($n = 1$) for $\text{He}^+$ ion is $-x$ J, then that for an electron in $n = 2$ state for $\text{Be}^{3+}$ ion in J is:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$E \propto Z^2/n^2$. He⁺: $4/1 = 4$. Be³⁺ at $n=2$: $16/4 = 4$. Ratio is 1, so energy is $-x$ J.

Ready to ace NEET?

Free access · No credit card required

Frequently Asked Questions

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.