Here are 30 important chemistry questions designed for NEET students, complete with answers and explanations.

NEET Chemistry Practice Questions

Question 1: Which of the following is the most electronegative element?

(A) Chlorine (Cl)

(B) Bromine (Br)

(C) Fluorine (F)

(D) Iodine (I)

Correct Answer: (C)

Reasoning: Fluorine is the most electronegative element in the periodic table due to its small atomic size and high effective nuclear charge, which results in a strong attraction for electrons in a chemical bond. Electronegativity generally increases across a period and decreases down a group.

Question 2: How many moles of water are produced when 16 g of oxygen gas reacts with hydrogen gas?

(A) 0.5 mole

(B) 1.0 mole

(C) 1.5 moles

(D) 2.0 moles

Correct Answer: (B)

Reasoning: The balanced chemical equation for the formation of water is 2H2​(g)+O2​(g)→2H2​O(l). The molar mass of O2​ is 32g/mol. If 16g of O2​ reacts, it is 16/32=0.5 moles of O2​. From the stoichiometry, 1 mole of O2​ produces 2 moles of H2​O. Therefore, 0.5 moles of O2​ will produce 0.5×2=1.0 mole of H2​O.

Question 3: The number of radial nodes for a 3p orbital is:

(A) 0

(B) 1

(C) 2

(D) 3

Correct Answer: (B)

Reasoning: The number of radial nodes for an orbital is given by the formula (n−l−1), where n is the principal quantum number and l is the azimuthal quantum number. For a 3p orbital, n=3 and l=1. So, the number of radial nodes is (3−1−1)=1.

Question 4: Which of the following has the highest bond order?

(A) N2​

(B) O2​

(C) F2​

(D) C2​

Correct Answer: (A)

Reasoning: Using Molecular Orbital Theory: For N2​: configuration is σ1s2σ∗1s2σ2s2σ∗2s2(π2px2​=π2py2​)σ2pz2​. Bond order = (10−4)/2=3. For O2​: configuration is σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​(π2px2​=π2py2​)(π∗2px1​=π∗2py1​). Bond order = (10−6)/2=2. For F2​: configuration is σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​(π2px2​=π2py2​)(π∗2px2​=π∗2py2​). Bond order = (10−8)/2=1. For C2​: configuration is σ1s2σ∗1s2σ2s2σ∗2s2(π2px2​=π2py2​). Bond order = (8−4)/2=2. Thus, N2​ has the highest bond order.

Question 5: Which of the following is an example of a solid solution?

(A) Brass

(B) Smoke

(C) Milk

(D) Fog

Correct Answer: (A)

Reasoning: A solid solution is a solid-state mixture of two or more elements or compounds. Brass is an alloy made primarily of copper and zinc, where zinc atoms are dispersed within the copper lattice, forming a solid solution. Smoke, milk, and fog are examples of colloids or suspensions.

Question 6: The rate constant of a first-order reaction is 1.15×10−3s−1. How long will 5 g of this reactant take to reduce to 3 g?

(A) 244 s

(B) 444 s

(C) 544 s

(D) 644 s

Correct Answer: (B)

Reasoning: For a first-order reaction, the integrated rate law is t=k2.303​log[A][A]0​​. Given k=1.15×10−3s−1, [A]0​=5g, and [A]=3g. t=1.15×10−32.303​log35​ t=1.15×10−32.303​×log(1.666) t=1.15×10−32.303​×0.2219 t≈444s.

Question 7: What is the change in enthalpy (ΔH) for a reaction, given that the heat absorbed is Q and the work done on the system is W?

(A) ΔH=Q+W

(B) ΔH=Q−W

(C) ΔH=Q (at constant pressure)

(D) ΔH=Q−PΔV

Correct Answer: (C)

Reasoning: Enthalpy change (ΔH) is defined as the heat absorbed or released by a system at constant pressure. From the first law of thermodynamics, ΔU=Q+W. At constant pressure, W=−PΔV. Also, enthalpy H=U+PV. So, ΔH=ΔU+PΔV+VΔP. At constant pressure (ΔP=0), ΔH=ΔU+PΔV. Substituting ΔU=Q−PΔV, we get ΔH=Q−PΔV+PΔV=Q.

Question 8: Which of the following species has a trigonal planar geometry?

(A) NH3​

(B) BF3​

(C) CH4​

(D) H2​O

Correct Answer: (B)

Reasoning: To determine the geometry, we look at the central atom and its lone pairs. For BF3​: Boron (B) is the central atom with 3 valence electrons. It forms 3 single bonds with Fluorine (F) atoms. There are no lone pairs on B. According to VSEPR theory, three bond pairs and no lone pairs result in a trigonal planar geometry. NH3​ is trigonal pyramidal, CH4​ is tetrahedral, and H2​O is bent.

Question 9: The pH of a 10−8M HCl solution is approximately:

(A) 8

(B) 7

(C) 6.96

(D) 7.04

Correct Answer: (C)

Reasoning: When the concentration of acid or base is very low (e.g., 10−7M or less), the contribution of water’s autoionization (H2​O⇌H++OH−) to the H+ concentration becomes significant. For 10−8M HCl, [H+]total​=[H+]HCl​+[H+]water​. Let [H+]water​=x. Then [OH−]water​=x. Kw​=[H+][OH−]=(10−8+x)(x)=10−14. 10−8x+x2=10−14. Solving the quadratic equation x2+10−8x−10−14=0, we get x≈0.95×10−7. So, [H+]total​=10−8+0.95×10−7=0.1×10−7+0.95×10−7=1.05×10−7M. pH=−log(1.05×10−7)≈6.978. (Rounding to two decimal places, it’s 6.98 or sometimes closer to 6.96 depending on rounding during calculation). The option 6.96 is the closest reasonable answer.

Question 10: Which of the following is a redox reaction?

(A) NaOH+HCl→NaCl+H2​O

(B) BaCl2​+Na2​SO4​→BaSO4​+2NaCl

(C) 2Na+Cl2​→2NaCl

(D) CaCO3​→CaO+CO2​

Correct Answer: (C)

Reasoning: A redox reaction involves a change in the oxidation states of atoms. In 2Na+Cl2​→2NaCl: Sodium (Na) changes oxidation state from 0 to +1 (oxidation). Chlorine (Cl2​) changes oxidation state from 0 to −1 (reduction). The other reactions are acid-base neutralization, precipitation, and decomposition respectively, which do not involve changes in oxidation states of the key elements.

Question 11: The correct order of ionic radii of N3−, O2−, F− is:

(A) N3−<O2−<F−

(B) N3−>O2−>F−

(C) N3−=O2−=F−

(D) O2−<F−<N3−

Correct Answer: (B)

Reasoning: These are isoelectronic species, meaning they all have the same number of electrons (10 electrons, like Neon). For isoelectronic species, the ionic radius decreases with an increase in nuclear charge (atomic number). Nitrogen (N) has atomic number 7. Oxygen (O) has atomic number 8. Fluorine (F) has atomic number 9. Therefore, N3− has the lowest nuclear charge per electron, resulting in the largest size, while F− has the highest nuclear charge per electron, resulting in the smallest size.

Question 12: Which of the following is not a characteristic of physisorption?

(A) Non-specific nature

(B) Reversible nature

(C) High enthalpy of adsorption

(D) Formation of multimolecular layers

Correct Answer: (C)

Reasoning: Physisorption (physical adsorption) is characterized by weak van der Waals forces, which leads to low enthalpy of adsorption (typically 20-40kJ/mol). It is non-specific, reversible, and forms multimolecular layers. Chemisorption, on the other hand, involves chemical bond formation and has a high enthalpy of adsorption.

Question 13: The oxidation state of sulfur in H2​SO4​ is:

(A) +2

(B) +4

(C) +6

(D) +8

Correct Answer: (C)

Reasoning: Let the oxidation state of sulfur be x. In H2​SO4​: Hydrogen (H) has an oxidation state of +1. Oxygen (O) has an oxidation state of −2. The sum of oxidation states in a neutral molecule is zero. 2(+1)+x+4(−2)=0 2+x−8=0 x−6=0 x=+6.

Question 14: Which of the following is the strongest acid?

(A) HClO4​

(B) HClO3​

(C) HClO2​

(D) HClO

Correct Answer: (A)

Reasoning: The strength of oxyacids of the same central atom increases with an increase in the number of oxygen atoms attached to the central atom (or with an increase in the oxidation state of the central atom). More oxygen atoms lead to greater resonance stabilization of the conjugate base and stronger electron-withdrawing effects, making the O−H bond more polar and easier to break. In HClO4​, Chlorine is in the highest oxidation state (+7), making it the strongest acid among the given options.

Question 15: Which gas is responsible for the depletion of the ozone layer?

(A) Carbon dioxide (CO2​)

(B) Methane (CH4​)

(C) Chlorofluorocarbons (CFCs)

(D) Sulfur dioxide (SO2​)

Correct Answer: (C)

Reasoning: Chlorofluorocarbons (CFCs) are highly stable compounds that, when released into the atmosphere, eventually reach the stratosphere. There, UV radiation breaks them down, releasing chlorine atoms. These chlorine atoms act as catalysts, repeatedly reacting with and destroying ozone molecules (O3​), leading to the depletion of the ozone layer.

Question 16: The total number of atomic orbitals in the third main energy level (n=3) is:

(A) 3

(B) 6

(C) 9

(D) 12

Correct Answer: (C)

Reasoning: For a given principal quantum number n, the total number of orbitals is n2. For n=3, the possible values for the azimuthal quantum number l are 0,1,2 (corresponding to s,p,d subshells). Number of orbitals in s subshell (l=0) = 2l+1=1. Number of orbitals in p subshell (l=1) = 2l+1=3. Number of orbitals in d subshell (l=2) = 2l+1=5. Total number of orbitals = 1+3+5=9. Alternatively, using the formula n2, for n=3, total orbitals = 32=9.

Question 17: Which of the following compounds exhibits hydrogen bonding?

(A) CH4​

(B) H2​S

(C) NH3​

(D) HCl

Correct Answer: (C)

Reasoning: Hydrogen bonding occurs when hydrogen is directly bonded to a highly electronegative atom like Nitrogen (N), Oxygen (O), or Fluorine (F). In NH3​ (ammonia), hydrogen is bonded to nitrogen, allowing for the formation of intermolecular hydrogen bonds. In CH4​, H2​S, and HCl, hydrogen is not bonded to N, O, or F, so they do not exhibit significant hydrogen bonding.

Question 18: The strongest reducing agent among the following is:

(A) Li

(B) Na

(C) K

(D) Cs

Correct Answer: (D)

Reasoning: Reducing agents are substances that readily lose electrons (get oxidized). Down a group in the alkali metals, the ionization energy decreases, meaning the outermost electron is held less tightly and is easier to remove. Therefore, caesium (Cs) has the lowest ionization energy among the given alkali metals and is the strongest reducing agent.

Question 19: In organic chemistry, the delocalization of σ electrons is known as:

(A) Inductive effect

(B) Resonance effect

(C) Hyperconjugation

(D) Electromeric effect

Correct Answer: (C)

Reasoning: Hyperconjugation (also known as no-bond resonance) involves the delocalization of σ (sigma) electrons of a C-H bond (or C-C bond) with an adjacent π bond or a vacant p-orbital. This effect stabilizes carbocations, free radicals, and alkenes. Inductive effect involves the displacement of σ electrons through a chain, resonance involves delocalization of π electrons, and electromeric effect is a temporary electron transfer in the presence of an attacking reagent.

Question 20: Which of the following is an example of a disproportionation reaction?

(A) 2NaOH+H2​SO4​→Na2​SO4​+2H2​O

(B) 2H2​O2​→2H2​O+O2​

(C) CuO+H2​→Cu+H2​O

(D) AgNO3​+NaCl→AgCl+NaNO3​

Correct Answer: (B)

Reasoning: A disproportionation reaction is a type of redox reaction in which a single element is simultaneously oxidized and reduced. In 2H2​O2​→2H2​O+O2​: The oxidation state of Oxygen in H2​O2​ is −1. In H2​O, Oxygen’s oxidation state is −2 (reduction). In O2​, Oxygen’s oxidation state is 0 (oxidation). Here, oxygen is both oxidized and reduced.

Question 21: The IUPAC name of CH3​CH2​CH(CH3​)CH2​OH is:

(A) 2-methylbutan-1-ol

(B) 3-methylbutan-1-ol

(C) 2-methylbutan-4-ol

(D) 3-methylbutan-4-ol

Correct Answer: (A)

Reasoning:

1. Identify the longest carbon chain containing the -OH group. This is a 4-carbon chain (butane).
2. Number the chain from the end closest to the -OH group. So, the -OH is on carbon 1.
3. Identify substituents. There is a methyl group on carbon 2. The name is 2-methylbutan-1-ol.

Question 22: The maximum number of electrons in a subshell is given by the formula:

(A) n2

(B) 2n2

(C) 2(2l+1)

(D) 2l+1

Correct Answer: (C)

Reasoning: For a given azimuthal quantum number l, the number of orbitals in that subshell is (2l+1). Each orbital can accommodate a maximum of two electrons (Pauli exclusion principle). Therefore, the maximum number of electrons in a subshell is 2×(2l+1). n2 gives total orbitals in a shell, 2n2 gives total electrons in a shell, and 2l+1 gives the number of orbitals in a subshell.

Question 23: Which of the following is a pseudo first-order reaction?

(A) Hydrolysis of ester in acidic medium

(B) Decomposition of N2​O5​

(C) Formation of HI

(D) Photosynthesis

Correct Answer: (A)

Reasoning: A pseudo first-order reaction is a second-order reaction (or higher) that behaves like a first-order reaction because one of the reactants is present in a very large excess, so its concentration remains essentially constant during the reaction. The hydrolysis of an ester in the presence of acid or base (e.g., CH3​COOC2​H5​(l)+H2​O(l)H+​CH3​COOH(l)+C2​H5​OH(l)) is a pseudo first-order reaction because water is typically present in large excess, making its concentration effectively constant.

Question 24: The solubility product of AgCl is 1.6×10−10. If 0.1M NaCl solution is added to a saturated solution of AgCl, what will be the concentration of Ag+ ions?

(A) 1.6×10−10M

(B) 1.6×10−9M

(C) 1.6×10−11M

(D) 1.6×10−5M

Correct Answer: (B)

Reasoning: The solubility product constant (Ksp​) for AgCl is given by Ksp​=[Ag+][Cl−]. In a 0.1M NaCl solution, the concentration of Cl− ions is 0.1M. Since NaCl is a strong electrolyte, it fully dissociates. Due to the common ion effect, the solubility of AgCl will decrease significantly. 1.6×10−10=[Ag+]×0.1 [Ag+]=0.11.6×10−10​=1.6×10−9M.

Question 25: Which of the following is used as a moderator in nuclear reactors?

(A) Heavy water (D2​O)

(B) Liquid sodium

(C) Boron rods

(D) Uranium

Correct Answer: (A)

Reasoning: Moderators are substances used in nuclear reactors to slow down the fast-moving neutrons produced during fission, making them thermal neutrons that are more likely to cause further fission. Heavy water (D2​O) and graphite are common moderators because they have a low neutron absorption cross-section and can effectively slow down neutrons through elastic collisions. Liquid sodium is a coolant, boron rods are control rods, and uranium is the fuel.

Question 26: The oxidation state of K in KO2​ is:

(A) +1

(B) +2

(C) +0.5

(D) -1

Correct Answer: (A)

Reasoning: KO2​ is Potassium superoxide. In ionic compounds, alkali metals (Group 1 elements like K) always exhibit an oxidation state of +1. Therefore, in KO2​, the oxidation state of Potassium (K) is +1. For the overall compound to be neutral, the O2​ unit must have a −1 charge, making it a superoxide ion (O2−​), where the oxidation state of each oxygen atom is −0.5.

Question 27: What is the bond angle in H2​O molecule?

(A) 180∘

(B) 109.5∘

(C) 120∘

(D) 104.5∘

Correct Answer: (D)

Reasoning: In the H2​O molecule, oxygen is the central atom and has two bond pairs and two lone pairs of electrons. According to VSEPR theory, these four electron pairs arrange themselves in a tetrahedral geometry around the oxygen atom. However, the presence of two lone pairs causes greater repulsion than bond pairs, compressing the bond angle between the two O-H bonds from the ideal tetrahedral angle of 109.5∘ to approximately 104.5∘.

Question 28: Which of the following elements is a metalloid?

(A) Aluminum (Al)

(B) Silicon (Si)

(C) Calcium (Ca)

(D) Sulfur (S)

Correct Answer: (B)

Reasoning: Metalloids are elements that have properties intermediate between metals and non-metals. They often show properties of both. Silicon (Si) is a classic example of a metalloid; it is a semiconductor, which is a characteristic property of metalloids. Aluminum is a metal, calcium is an alkaline earth metal, and sulfur is a non-metal.

Question 29: The correct increasing order of acidity of the following compounds is:

(A) Water < Alcohol < Phenol < Carboxylic acid

(B) Alcohol < Water < Phenol < Carboxylic acid

(C) Phenol < Alcohol < Water < Carboxylic acid

(D) Carboxylic acid < Phenol < Alcohol < Water

Correct Answer: (B)

Reasoning: The acidity of compounds is related to the stability of their conjugate base.

• Alcohols (ROH) are very weak acids; their conjugate base (RO−) is unstable.
• Water (H2​O) is slightly more acidic than alcohols due to the smaller size of oxygen.
• Phenols (C6​H5​OH) are more acidic than alcohols and water because the phenoxide ion (C6​H5​O−) is stabilized by resonance with the benzene ring.
• Carboxylic acids (RCOOH) are the most acidic among these because their conjugate base (carboxylate ion, RCOO−) is highly stabilized by two equivalent resonance structures, leading to delocalization of the negative charge over two oxygen atoms. Therefore, the increasing order of acidity is Alcohol < Water < Phenol < Carboxylic acid.

Question 30: Which of the following is an example of homogeneous catalysis?

(A) Haber’s process for ammonia synthesis

(B) Contact process for sulfuric acid

(C) Hydrolysis of ester catalyzed by H2​SO4​

(D) Decomposition of KClO3​ in the presence of MnO2​

Correct Answer: (C)

Reasoning: In homogeneous catalysis, the catalyst and the reactants are in the same physical phase. (A) Haber’s process (N2​(g)+3H2​(g)Fe(s)​2NH3​(g)) involves a solid catalyst (Fe) and gaseous reactants, so it is heterogeneous. (B) Contact process (2SO2​(g)+O2​(g)V2​O5​(s)​2SO3​(g)) involves a solid catalyst (V2​O5​) and gaseous reactants, so it is heterogeneous. (C) Hydrolysis of ester catalyzed by H2​SO4​ (CH3​COOC2​H5​(l)+H2​O(l)H2​SO4​(l)​CH3​COOH(l)+C2​H5​OH(l)) involves all reactants and the catalyst in the liquid phase, making it homogeneous. (D) Decomposition of KClO3​ (2KClO3​(s)MnO2​(s)​2KCl(s)+3O2​(g)) involves a solid catalyst (MnO2​) and a solid reactant, so it is heterogeneous.