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. What is the electron configuration of chlorine (Cl)?
A.
1s2 2s2 2p6 3s2 3p5
B.
1s2 2s2 2p6 3s2 3p6
C.
1s2 2s2 2p6 3s2 3p4
D.
1s2 2s2 2p6 3s2 3p3
Show solution
Solution
The electron configuration of chlorine (atomic number 17) is 1s2 2s2 2p6 3s2 3p5.
Correct Answer:
A
— 1s2 2s2 2p6 3s2 3p5
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Q. What is the electron configuration of the element with atomic number 30?
A.
[Ar] 4s2 3d10
B.
[Kr] 5s2 4d10
C.
[Ar] 4s2 3d8
D.
[Ar] 4s2 3d9
Show solution
Solution
The element with atomic number 30 is Zn. Its electron configuration is [Ar] 4s2 3d10.
Correct Answer:
A
— [Ar] 4s2 3d10
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Q. What is the electron configuration of the ion Fe3+?
A.
[Ar] 4s2 3d5
B.
[Ar] 3d5
C.
[Ar] 4s2 3d3
D.
[Ar] 3d6
Show solution
Solution
Iron (Fe) has an atomic number of 26, and the electron configuration for Fe3+ is [Ar] 3d5 after losing 3 electrons.
Correct Answer:
B
— [Ar] 3d5
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Q. What is the electron configuration of the sulfur ion (S^2-)?
A.
[Ne] 3s² 3p⁴
B.
[Ne] 3s² 3p⁶
C.
[Ne] 3s² 3p³
D.
[He] 2s² 2p⁶
Show solution
Solution
Sulfur (S) has 16 electrons. S^2- has 18 electrons, resulting in the configuration [Ne] 3s² 3p⁶.
Correct Answer:
B
— [Ne] 3s² 3p⁶
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Q. What is the electron geometry of a molecule with four electron groups around the central atom?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
The electron geometry for a molecule with four electron groups is tetrahedral, as the groups will arrange themselves to minimize repulsion.
Correct Answer:
C
— Tetrahedral
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Q. What is the electronegativity difference between sodium (Na) and chlorine (Cl)?
A.
0.8
B.
1.0
C.
2.1
D.
3.0
Show solution
Solution
Electronegativity of Na ≈ 0.9 and Cl ≈ 3.0. Difference = 3.0 - 0.9 = 2.1.
Correct Answer:
C
— 2.1
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Q. What is the electronegativity of fluorine on the Pauling scale?
A.
3.0
B.
3.5
C.
4.0
D.
4.5
Show solution
Solution
The electronegativity of fluorine is 4.0 on the Pauling scale.
Correct Answer:
C
— 4.0
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Q. What is the endpoint of a titration?
A.
The point where the reaction is complete
B.
The point where the solution is neutral
C.
The point where the indicator changes color
D.
The point where the titrant is exhausted
Show solution
Solution
The endpoint of a titration is indicated by a color change of the indicator, signaling that the reaction is complete.
Correct Answer:
C
— The point where the indicator changes color
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Q. What is the enthalpy change (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if ΔHf for H2O(g) is -241.8 kJ/mol?
A.
-483.6 kJ
B.
241.8 kJ
C.
0 kJ
D.
483.6 kJ
Show solution
Solution
The enthalpy change for the reaction is 2 times the standard enthalpy of formation of H2O(g), which is -241.8 kJ/mol, giving -483.6 kJ.
Correct Answer:
A
— -483.6 kJ
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Q. What is the enthalpy change for the reaction: C(s) + O2(g) → CO2(g) if the standard enthalpy of formation of CO2 is -393.5 kJ/mol?
A.
-393.5 kJ/mol
B.
393.5 kJ/mol
C.
0 kJ/mol
D.
Not enough information
Show solution
Solution
The enthalpy change for the formation of CO2 from its elements in their standard states is equal to the standard enthalpy of formation, which is -393.5 kJ/mol.
Correct Answer:
A
— -393.5 kJ/mol
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Q. What is the enthalpy change for the reaction: H2(g) + 1/2 O2(g) → H2O(l) at standard conditions?
A.
-285.8 kJ/mol
B.
0 kJ/mol
C.
285.8 kJ/mol
D.
-241.8 kJ/mol
Show solution
Solution
The standard enthalpy change of formation for water (H2O) is -285.8 kJ/mol.
Correct Answer:
A
— -285.8 kJ/mol
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Q. What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g) if the standard enthalpy of formation of NH3(g) is -45.9 kJ/mol?
A.
-91.8 kJ
B.
45.9 kJ
C.
0 kJ
D.
91.8 kJ
Show solution
Solution
The enthalpy change for the reaction is 2 times the standard enthalpy of formation of NH3(g), which is -45.9 kJ/mol, giving -91.8 kJ.
Correct Answer:
A
— -91.8 kJ
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Q. What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g) if ΔHf° for NH3 is -45.9 kJ/mol?
A.
-137.7 kJ
B.
-91.8 kJ
C.
0 kJ
D.
45.9 kJ
Show solution
Solution
The enthalpy change for the reaction is calculated as ΔH = 2 * ΔHf°(NH3) = 2 * (-45.9 kJ) = -91.8 kJ.
Correct Answer:
A
— -137.7 kJ
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Q. What is the enthalpy change when 2 moles of water vapor condense to liquid water at constant temperature?
A.
-2ΔH_vap
B.
2ΔH_vap
C.
0
D.
ΔH_vap
Show solution
Solution
The enthalpy change is -2ΔH_vap, as condensation releases heat equal to twice the enthalpy of vaporization.
Correct Answer:
A
— -2ΔH_vap
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Q. What is the enthalpy change when 4 moles of a substance release 800 J of heat?
A.
200 J/mol
B.
400 J/mol
C.
100 J/mol
D.
300 J/mol
Show solution
Solution
ΔH = q/n = -800 J / 4 mol = -200 J/mol.
Correct Answer:
B
— 400 J/mol
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Q. What is the equilibrium constant (K) for a reaction with ΔG° = -40 kJ at 298 K?
A.
10
B.
100
C.
1000
D.
10000
Show solution
Solution
ΔG° = -RTln(K); K = e^(-ΔG°/RT) = e^(40,000/(8.314*298)) ≈ 1000.
Correct Answer:
C
— 1000
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Q. What is the equilibrium constant expression for the reaction A + B ⇌ C?
A.
[C]/([A][B])
B.
[A][B]/[C]
C.
[C]^2/([A][B])
D.
[C]/[A] + [B]
Show solution
Solution
The equilibrium constant K is given by K = [C]/([A][B]) at equilibrium.
Correct Answer:
A
— [C]/([A][B])
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Q. What is the equivalent weight of sulfuric acid (H2SO4) for the purpose of neutralization?
A.
49 g
B.
98 g
C.
24.5 g
D.
196 g
Show solution
Solution
Equivalent weight = molar mass / n, where n = number of H+ ions. Molar mass of H2SO4 = 98 g/mol, n = 2. Therefore, equivalent weight = 98 g / 2 = 49 g.
Correct Answer:
C
— 24.5 g
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Q. What is the expected bond angle in a molecule with sp2 hybridization?
A.
90 degrees
B.
120 degrees
C.
180 degrees
D.
109.5 degrees
Show solution
Solution
Molecules with sp2 hybridization have bond angles of approximately 120 degrees due to trigonal planar geometry.
Correct Answer:
B
— 120 degrees
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Q. What is the expected bond angle in a molecule with tetrahedral geometry?
A.
90 degrees
B.
109.5 degrees
C.
120 degrees
D.
180 degrees
Show solution
Solution
The bond angle in a tetrahedral geometry is approximately 109.5 degrees.
Correct Answer:
B
— 109.5 degrees
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Q. What is the expected bond length of a C-C single bond?
A.
1.54 Å
B.
1.34 Å
C.
1.20 Å
D.
1.45 Å
Show solution
Solution
The typical bond length for a C-C single bond is approximately 1.54 Å.
Correct Answer:
A
— 1.54 Å
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Q. What is the expected major product when 1-chloropropane reacts with potassium tert-butoxide?
A.
Propene
B.
2-propanol
C.
1-butene
D.
2-bromopropane
Show solution
Solution
The reaction of 1-chloropropane with potassium tert-butoxide favors elimination, leading to the formation of propene as the major product.
Correct Answer:
A
— Propene
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Q. What is the expected oxidation state of phosphorus in phosphoric acid (H3PO4)?
Show solution
Solution
In phosphoric acid, phosphorus has an oxidation state of +5.
Correct Answer:
C
— 5
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Q. What is the expected product when 1-butyne undergoes hydroboration-oxidation?
A.
1-butanol
B.
2-butanol
C.
butanal
D.
butyric acid
Show solution
Solution
Hydroboration-oxidation of 1-butyne leads to the formation of 2-butanol due to anti-Markovnikov addition.
Correct Answer:
B
— 2-butanol
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Q. What is the expected product when 2-bromo-2-methylpropane reacts with a strong base in a bimolecular elimination reaction?
A.
2-methylpropene
B.
2-bromopropane
C.
2-methyl-2-bromopropane
D.
Propane
Show solution
Solution
The strong base facilitates the E2 elimination, resulting in the formation of 2-methylpropene.
Correct Answer:
A
— 2-methylpropene
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Q. What is the expected product when 2-methylpropene undergoes hydroboration-oxidation?
A.
2-methylpropan-1-ol
B.
2-methylpropan-2-ol
C.
2-methylpropanoic acid
D.
2-methylpropene
Show solution
Solution
Hydroboration-oxidation of 2-methylpropene leads to the formation of 2-methylpropan-1-ol due to anti-Markovnikov addition of water.
Correct Answer:
A
— 2-methylpropan-1-ol
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Q. What is the expected product when anisole is treated with chlorosulfonic acid?
A.
Anisole sulfonic acid
B.
p-Anisole sulfonic acid
C.
o-Anisole sulfonic acid
D.
No reaction
Show solution
Solution
Chlorosulfonic acid introduces a sulfonyl group at the para position due to the activating effect of the methoxy group.
Correct Answer:
B
— p-Anisole sulfonic acid
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Q. What is the expected product when anisole undergoes bromination?
A.
Bromobenzene
B.
Ortho-bromoanisole
C.
Para-bromoanisole
D.
No reaction
Show solution
Solution
The expected product is para-bromoanisole, as the methoxy group is an electron-donating group that directs electrophilic substitution to the para position.
Correct Answer:
C
— Para-bromoanisole
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Q. What is the expected product when phenol is treated with bromine water?
A.
Bromobenzene
B.
2,4,6-Tribromophenol
C.
Benzene
D.
Bromophenol
Show solution
Solution
The expected product is 2,4,6-tribromophenol, as phenol is highly reactive towards electrophilic substitution due to the hydroxyl group.
Correct Answer:
B
— 2,4,6-Tribromophenol
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Q. What is the Faraday constant approximately equal to?
A.
96485 C/mol
B.
96500 C/mol
C.
100000 C/mol
D.
90000 C/mol
Show solution
Solution
The Faraday constant is approximately equal to 96485 C/mol, representing the charge of one mole of electrons.
Correct Answer:
A
— 96485 C/mol
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