The Chemistry (School & UG) category covers the fundamental to advanced concepts of Chemistry , structured to support Class 11–12 students, competitive exam aspirants, and undergraduate learners . The content emphasizes conceptual clarity, problem-solving skills, and strong alignment with NCERT and university syllabi .
In this category, you will study:
Physical Chemistry – atomic structure, thermodynamics, chemical kinetics, equilibrium, electrochemistry, and solutions
Organic Chemistry – basic principles, reaction mechanisms, hydrocarbons, functional groups, biomolecules, and polymers
Inorganic Chemistry – periodic trends, chemical bonding, coordination compounds, metallurgy, and s-, p-, d-, and f-block elements
Numerical problem-solving – formula-based calculations, mole concept, and graph-based questions
Reaction mechanisms and derivations with step-by-step explanations
Comparisons, trends, and exceptions important for exams
NCERT-based theory , supported by diagrams, tables, MCQs, assertion–reason questions, numericals, and PYQs
The content is designed to build strong foundations , improve analytical and quantitative skills , and prepare students for school examinations, competitive exams, and undergraduate assessments .
Develop a clear, logical, and application-oriented understanding of Chemistry to excel academically and confidently tackle problem-based and conceptual questions at both school and UG levels .
Q. A chemical reaction releases 250 J of heat. If the reaction occurs at constant pressure, what is the change in enthalpy?
A.
-250 J
B.
250 J
C.
0 J
D.
500 J
Show solution
Solution
At constant pressure, ΔH = q = -250 J.
Correct Answer:
A
— -250 J
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Q. A process absorbs 300 J of heat and does 100 J of work. What is the change in internal energy (ΔU)?
A.
200 J
B.
300 J
C.
400 J
D.
100 J
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Solution
ΔU = q - W = 300 J - 100 J = 200 J.
Correct Answer:
A
— 200 J
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Q. A reaction has an enthalpy change of 200 kJ for the formation of 1 mole of product. What is the enthalpy change for the formation of 0.5 moles of product?
A.
100 kJ
B.
200 kJ
C.
50 kJ
D.
400 kJ
Show solution
Solution
ΔH for 0.5 moles = 200 kJ * 0.5 = 100 kJ.
Correct Answer:
A
— 100 kJ
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Q. According to Graham's Law, how does the rate of effusion of a gas relate to its molar mass?
A.
Inversely proportional
B.
Directly proportional
C.
Equal
D.
Unrelated
Show solution
Solution
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Correct Answer:
A
— Inversely proportional
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Q. According to Graham's Law, the rate of effusion of a gas is inversely proportional to what?
A.
Its molar mass
B.
Its temperature
C.
Its pressure
D.
Its volume
Show solution
Solution
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Correct Answer:
A
— Its molar mass
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Q. According to the Kinetic Molecular Theory, which of the following is NOT a postulate?
A.
Gas particles are in constant random motion.
B.
Gas particles occupy a definite volume.
C.
Collisions between gas particles are elastic.
D.
The average kinetic energy is proportional to temperature.
Show solution
Solution
Gas particles do not occupy a definite volume; they fill the container they are in.
Correct Answer:
B
— Gas particles occupy a definite volume.
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Q. According to Werner's theory, what is the coordination number of a complex with six ligands?
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Solution
The coordination number is defined as the number of ligands surrounding the central metal atom. In this case, it is 6.
Correct Answer:
C
— 6
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Q. According to Werner's theory, what type of isomerism is primarily observed in coordination compounds?
A.
Geometric isomerism
B.
Optical isomerism
C.
Structural isomerism
D.
All of the above
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Solution
Werner's theory accounts for various types of isomerism, including geometric, optical, and structural isomerism in coordination compounds.
Correct Answer:
D
— All of the above
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Q. At constant temperature, what happens to the pressure of a gas if its volume is halved?
A.
Pressure doubles
B.
Pressure halves
C.
Pressure remains the same
D.
Pressure quadruples
Show solution
Solution
According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. Halving the volume doubles the pressure.
Correct Answer:
A
— Pressure doubles
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Q. At constant temperature, what happens to the pressure of a gas if its volume is doubled?
A.
Pressure doubles
B.
Pressure halves
C.
Pressure remains the same
D.
Pressure quadruples
Show solution
Solution
According to Boyle's Law, at constant temperature, pressure is inversely proportional to volume. If the volume is doubled, the pressure is halved.
Correct Answer:
B
— Pressure halves
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Q. At what temperature does the Kelvin scale start?
A.
0°C
B.
100°C
C.
273.15°C
D.
0 K
Show solution
Solution
The Kelvin scale starts at absolute zero, which is 0 K or -273.15°C.
Correct Answer:
D
— 0 K
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Q. At which conditions do real gases behave most like ideal gases?
A.
High pressure and low temperature
B.
Low pressure and high temperature
C.
High pressure and high temperature
D.
Low pressure and low temperature
Show solution
Solution
Real gases behave most like ideal gases at low pressure and high temperature, where intermolecular forces are minimized.
Correct Answer:
B
— Low pressure and high temperature
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Q. Calculate the change in enthalpy (ΔH) when 2 moles of a substance absorb 500 J of heat at constant pressure.
A.
250 J/mol
B.
500 J/mol
C.
1000 J/mol
D.
125 J/mol
Show solution
Solution
ΔH = q/n = 500 J / 2 mol = 250 J/mol.
Correct Answer:
A
— 250 J/mol
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Q. Calculate the dipole moment of a molecule with a charge of 1.6 x 10^-19 C and a bond length of 1.5 Å.
A.
2.4 x 10^-29 C·m
B.
2.4 x 10^-28 C·m
C.
2.4 x 10^-30 C·m
D.
2.4 x 10^-31 C·m
Show solution
Solution
Dipole moment (p) = charge (q) × distance (d) = (1.6 x 10^-19 C) × (1.5 x 10^-10 m) = 2.4 x 10^-29 C·m.
Correct Answer:
A
— 2.4 x 10^-29 C·m
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Q. Calculate the dipole moment of HCl given that the bond length is 1.27 Å and the charge separation is 0.33 e.
A.
1.1 D
B.
0.4 D
C.
0.8 D
D.
0.2 D
Show solution
Solution
Dipole moment (μ) = charge (q) × distance (d). μ = 0.33 e × 1.27 × 10^-10 m = 4.2 × 10^-29 C·m = 1.1 D.
Correct Answer:
A
— 1.1 D
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Q. Calculate the enthalpy change (ΔH) for the reaction: 2NO(g) + O2(g) → 2NO2(g) given the following bond enthalpies: N≡N = 941 kJ/mol, O=O = 498 kJ/mol, N=O = 201 kJ/mol.
A.
-180 kJ
B.
-200 kJ
C.
-220 kJ
D.
-240 kJ
Show solution
Solution
ΔH = [2(201) + 498] - [2(941)] = -220 kJ.
Correct Answer:
C
— -220 kJ
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Q. Calculate the formal charge on a nitrogen atom in NH3.
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Solution
Formal charge = Valence electrons - (Non-bonding electrons + 1/2 Bonding electrons) = 5 - (0 + 1/2*6) = 0.
Correct Answer:
A
— 0
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Q. Calculate the formal charge on the nitrogen atom in NO3-.
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Solution
Formal charge = valence electrons - (non-bonding electrons + 1/2 bonding electrons). For nitrogen in NO3-, it is 5 - (0 + 1/2(6)) = 0.
Correct Answer:
A
— 0
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Q. Calculate the Gibbs free energy change (ΔG) for a cell with E° = 0.75 V and n = 2 moles of electrons transferred.
A.
-150 kJ
B.
-75 kJ
C.
75 kJ
D.
150 kJ
Show solution
Solution
ΔG = -nFE° = -2 * 96485 C/mol * 0.75 V = -144.73 kJ.
Correct Answer:
A
— -150 kJ
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Q. Calculate the ionization energy of hydrogen in eV if the energy level is -13.6 eV.
A.
13.6 eV
B.
1.24 eV
C.
3.4 eV
D.
0.85 eV
Show solution
Solution
Ionization energy is the energy required to remove an electron from the ground state. For hydrogen, it is 13.6 eV.
Correct Answer:
A
— 13.6 eV
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Q. Calculate the ionization energy of hydrogen in eV if the energy of the electron in the ground state is -13.6 eV.
A.
13.6 eV
B.
1.24 eV
C.
3.4 eV
D.
27.2 eV
Show solution
Solution
Ionization energy is the energy required to remove the electron from the ground state. For hydrogen, it is 13.6 eV.
Correct Answer:
A
— 13.6 eV
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Q. Calculate the molar mass of water (H2O).
A.
16 g/mol
B.
18 g/mol
C.
20 g/mol
D.
22 g/mol
Show solution
Solution
The molar mass of water is calculated as follows: H (1 g/mol) x 2 + O (16 g/mol) = 18 g/mol.
Correct Answer:
B
— 18 g/mol
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Q. Determine the bond angle in a tetrahedral molecule like CH4.
A.
90°
B.
109.5°
C.
120°
D.
180°
Show solution
Solution
In a tetrahedral geometry, the bond angles are approximately 109.5° due to the arrangement of four electron pairs.
Correct Answer:
B
— 109.5°
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Q. For a first-order reaction, if the half-life is 10 minutes, what is the rate constant?
A.
0.0693 min^-1
B.
0.1 min^-1
C.
0.693 min^-1
D.
0.5 min^-1
Show solution
Solution
The half-life (t1/2) of a first-order reaction is given by t1/2 = 0.693/k. Rearranging gives k = 0.693/t1/2 = 0.693/10 min = 0.0693 min^-1.
Correct Answer:
A
— 0.0693 min^-1
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Q. For a first-order reaction, if the half-life is 10 minutes, what will be the half-life if the concentration is doubled?
A.
10 minutes
B.
20 minutes
C.
5 minutes
D.
It cannot be determined
Show solution
Solution
The half-life of a first-order reaction is independent of concentration; it remains 10 minutes.
Correct Answer:
A
— 10 minutes
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Q. For a molecule with a bond length of 1.2 Å and a bond order of 2, what is the bond energy in kJ/mol?
A.
500
B.
600
C.
700
D.
800
Show solution
Solution
The bond energy for a double bond (bond order = 2) is typically around 600 kJ/mol for C=C.
Correct Answer:
B
— 600
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Q. For a reaction A → B, if the rate constant k is 0.1 s^-1, what is the time required for the concentration of A to decrease to 25% of its initial value?
A.
10 seconds
B.
20 seconds
C.
30 seconds
D.
40 seconds
Show solution
Solution
For a first-order reaction, t = (ln(2)/k) * ln([A]0/[A]t). To reach 25%, t = (ln(2)/0.1) * ln(4) = 20 seconds.
Correct Answer:
B
— 20 seconds
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Q. For a reaction at equilibrium, if the concentration of products is increased, what will happen to the position of equilibrium according to Le Chatelier's principle?
A.
Shift to the right
B.
Shift to the left
C.
No change
D.
Depends on temperature
Show solution
Solution
According to Le Chatelier's principle, increasing the concentration of products will shift the equilibrium position to the left to counteract the change.
Correct Answer:
B
— Shift to the left
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Q. For a reaction at equilibrium, if the temperature is increased and ΔH is positive, what will happen to the position of equilibrium?
A.
Shift to the right
B.
Shift to the left
C.
No change
D.
Depends on concentration
Show solution
Solution
According to Le Chatelier's principle, increasing temperature shifts the equilibrium to favor the endothermic direction (right).
Correct Answer:
A
— Shift to the right
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Q. For a reaction at equilibrium, what happens if the concentration of a reactant is increased?
A.
The equilibrium shifts to the right
B.
The equilibrium shifts to the left
C.
The equilibrium remains unchanged
D.
The reaction stops
Show solution
Solution
According to Le Chatelier's principle, increasing the concentration of a reactant will shift the equilibrium position to the right, favoring the formation of products.
Correct Answer:
A
— The equilibrium shifts to the right
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