Q. What is the molecular formula of fructose?
A.
C6H12O6
B.
C5H10O5
C.
C6H10O5
D.
C5H12O6
Show solution
Solution
Fructose has the same molecular formula as glucose, which is C6H12O6.
Correct Answer:
A
— C6H12O6
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Q. What is the molecular formula of hydrogen peroxide?
A.
H2O
B.
H2O2
C.
H2
D.
O2
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Solution
Hydrogen peroxide has the molecular formula H2O2.
Correct Answer:
B
— H2O2
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Q. What is the molecular formula of water?
A.
H2O
B.
H2O2
C.
HO
D.
H2
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Solution
The molecular formula of water is H2O.
Correct Answer:
A
— H2O
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Q. What is the molecular geometry of ammonia (NH3)?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Trigonal pyramidal
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Solution
Ammonia has a trigonal pyramidal geometry due to the presence of a lone pair on nitrogen.
Correct Answer:
D
— Trigonal pyramidal
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Q. What is the molecular geometry of BF3 according to VSEPR theory?
A.
Trigonal planar
B.
Tetrahedral
C.
Octahedral
D.
Linear
Show solution
Solution
BF3 has three bonding pairs and no lone pairs, resulting in a trigonal planar geometry.
Correct Answer:
A
— Trigonal planar
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Q. What is the molecular geometry of BF3?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Bent
Show solution
Solution
BF3 has a trigonal planar geometry due to the three bonding pairs and no lone pairs on the central atom.
Correct Answer:
B
— Trigonal planar
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Q. What is the molecular geometry of CH4 according to VSEPR theory?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
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Solution
According to VSEPR theory, CH4 has four bonding pairs and no lone pairs, resulting in a tetrahedral geometry.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular geometry of CH4?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
CH4 has a tetrahedral geometry due to four bonding pairs around the central carbon atom.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular geometry of methane (CH4)?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
Methane has a tetrahedral molecular geometry due to sp3 hybridization of the carbon atom.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular geometry of NH3 according to VSEPR theory?
A.
Trigonal planar
B.
Tetrahedral
C.
Bent
D.
Trigonal pyramidal
Show solution
Solution
NH3 has three bonding pairs and one lone pair, resulting in a trigonal pyramidal geometry.
Correct Answer:
D
— Trigonal pyramidal
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Q. What is the molecular geometry of SF4?
A.
Tetrahedral
B.
Trigonal bipyramidal
C.
Seesaw
D.
Square planar
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Solution
SF4 has four bonding pairs and one lone pair, resulting in a seesaw molecular geometry.
Correct Answer:
C
— Seesaw
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Q. What is the molecular geometry of SO2?
A.
Linear
B.
Trigonal planar
C.
Bent
D.
Tetrahedral
Show solution
Solution
SO2 has two bonding pairs and one lone pair, resulting in a bent molecular geometry.
Correct Answer:
C
— Bent
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Q. What is the molecular geometry of the molecule with the electronic configuration of 1s2 2s2 2p2?
A.
Linear
B.
Trigonal Planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
The electronic configuration corresponds to C2, which has a tetrahedral geometry due to sp3 hybridization.
Correct Answer:
C
— Tetrahedral
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Q. What is the molecular orbital configuration of F2?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(π2p)⁴(π*2p)²
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³(π*2p)²
Show solution
Solution
The correct configuration for F2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)².
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
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Q. What is the molecular orbital configuration of O2?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)²
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)⁴
Show solution
Solution
The correct configuration for O2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹.
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
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Q. What is the molecular orbital configuration of the F2 molecule?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)¹
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³(π*2p)²
Show solution
Solution
The correct configuration for F2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)².
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
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Q. What is the molecular orbital configuration of the O2 molecule?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)²
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)¹
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)²
Show solution
Solution
The correct configuration for O2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹.
Correct Answer:
A
— (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
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Q. What is the molecular shape of a molecule with the formula AX3E?
A.
Trigonal planar
B.
Tetrahedral
C.
Trigonal pyramidal
D.
Bent
Show solution
Solution
AX3E indicates three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.
Correct Answer:
C
— Trigonal pyramidal
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Q. What is the molecular shape of BF3 according to VSEPR theory?
A.
Bent
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
BF3 has three bonding pairs and no lone pairs, resulting in a trigonal planar shape.
Correct Answer:
B
— Trigonal planar
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Q. What is the molecular shape of NH3 according to VSEPR theory?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Trigonal pyramidal
Show solution
Solution
NH3 has three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.
Correct Answer:
D
— Trigonal pyramidal
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Q. What is the molecular weight of water (H2O)?
A.
16 g/mol
B.
18 g/mol
C.
20 g/mol
D.
22 g/mol
Show solution
Solution
The molecular weight of water is calculated as (2*1) + (16) = 18 g/mol.
Correct Answer:
B
— 18 g/mol
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Q. What is the name of the reaction where benzene is converted to phenol?
A.
Hydrogenation
B.
Nitration
C.
Sulfonation
D.
Hydroxylation
Show solution
Solution
The conversion of benzene to phenol is known as hydroxylation, typically involving the addition of a hydroxyl group (-OH) to the aromatic ring.
Correct Answer:
D
— Hydroxylation
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Q. What is the Nernst equation used for?
A.
Calculating pH
B.
Determining cell potential
C.
Finding molarity
D.
Measuring temperature
Show solution
Solution
The Nernst equation is used to determine the cell potential under non-standard conditions.
Correct Answer:
B
— Determining cell potential
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Q. What is the normality of a solution containing 1 mole of H2SO4 in 1 liter of solution?
A.
1 N
B.
2 N
C.
0.5 N
D.
4 N
Show solution
Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so N = 2 moles / 1 L = 2 N.
Correct Answer:
B
— 2 N
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Q. What is the normality of a solution containing 2 moles of H2SO4 in 1 liter of solution?
A.
2 N
B.
4 N
C.
1 N
D.
0.5 N
Show solution
Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so 2 moles × 2 = 4 N.
Correct Answer:
B
— 4 N
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Q. What is the normality of a solution containing 3 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)
A.
3 N
B.
6 N
C.
1.5 N
D.
1 N
Show solution
Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents per mole, so 3 moles = 6 equivalents. Normality = 6 equivalents / 2 L = 3 N.
Correct Answer:
B
— 6 N
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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)
A.
4 N
B.
2 N
C.
8 N
D.
1 N
Show solution
Solution
Normality (N) = number of equivalents / liters of solution. H2SO4 has 2 equivalents per mole, so 4 moles = 8 equivalents. Normality = 8 equivalents / 2 L = 4 N.
Correct Answer:
C
— 8 N
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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution?
A.
4 N
B.
8 N
C.
2 N
D.
1 N
Show solution
Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents, so 4 moles = 8 equivalents. N = 8 eq / 2 L = 4 N.
Correct Answer:
B
— 8 N
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Q. What is the normality of a solution that is 1 M in H2SO4?
A.
1 N
B.
2 N
C.
0.5 N
D.
4 N
Show solution
Solution
Normality (N) = Molarity (M) x number of equivalents. H2SO4 has 2 acidic protons, so 1 M x 2 = 2 N.
Correct Answer:
B
— 2 N
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Q. What is the number of atoms in 2 moles of aluminum (Al)? (2044)
A.
6.022 x 10^23
B.
1.2044 x 10^24
C.
3.011 x 10^23
D.
12.044 x 10^24
Show solution
Solution
Number of atoms = moles x Avogadro's number = 2 moles x 6.022 x 10^23 atoms/mole = 1.2044 x 10^24 atoms.
Correct Answer:
B
— 1.2044 x 10^24
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Chemistry Syllabus (JEE Main) MCQ & Objective Questions
The Chemistry Syllabus for JEE Main is crucial for students aiming to excel in their exams. Understanding this syllabus not only helps in grasping fundamental concepts but also enhances performance in objective questions and MCQs. Regular practice with these types of questions is essential for scoring better and mastering important topics.
What You Will Practise Here
Basic Concepts of Chemistry
Atomic Structure and Chemical Bonding
States of Matter: Gases and Liquids
Thermodynamics and Thermochemistry
Equilibrium: Chemical and Ionic
Redox Reactions and Electrochemistry
Hydrocarbons and Environmental Chemistry
Exam Relevance
The Chemistry syllabus is a significant part of CBSE, State Boards, NEET, and JEE exams. Questions from this syllabus often appear in various formats, including multiple-choice questions, assertion-reason type questions, and numerical problems. Familiarity with the common question patterns can greatly enhance your exam preparation and confidence.
Common Mistakes Students Make
Misunderstanding the periodic trends and their implications.
Confusing different types of chemical bonds and their properties.
Neglecting to balance redox reactions properly.
Overlooking the significance of units in thermodynamic calculations.
Failing to apply concepts of equilibrium in problem-solving.
FAQs
Question: What are the key topics I should focus on in the Chemistry syllabus for JEE Main?Answer: Focus on atomic structure, chemical bonding, thermodynamics, and equilibrium as they are frequently tested.
Question: How can I improve my performance in Chemistry MCQs?Answer: Regular practice with past papers and understanding concepts deeply will help you tackle MCQs effectively.
Start your journey towards mastering the Chemistry Syllabus (JEE Main) by solving practice MCQs today. Test your understanding and build confidence for your exams!
