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 glucose? (2022)
A.
C6H12O6
B.
C5H10O5
C.
C6H10O5
D.
C6H14O6
Show solution
Solution
Glucose is a simple sugar with the molecular formula C6H12O6.
Correct Answer:
A
— C6H12O6
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Q. What is the molecular formula of glucose? (2022) 2022
A.
C6H12O6
B.
C5H10O5
C.
C6H10O5
D.
C6H14O6
Show solution
Solution
Glucose is a simple sugar with the molecular formula 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
Show solution
Solution
Hydrogen peroxide has the molecular formula H2O2.
Correct Answer:
B
— H2O2
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Q. What is the molecular formula of the simplest alkyne? (2022)
A.
C2H2
B.
C2H4
C.
C3H4
D.
C3H6
Show solution
Solution
The simplest alkyne is ethyne (acetylene) with the molecular formula C2H2.
Correct Answer:
A
— C2H2
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Q. What is the molecular formula of water?
A.
H2O
B.
H2O2
C.
HO
D.
H2
Show solution
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
Show solution
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
Show solution
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
Show solution
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 moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to its plane?
A.
1/2 MR^2
B.
MR^2
C.
1/4 MR^2
D.
2/3 MR^2
Show solution
Solution
The moment of inertia of a disk about an axis through its center is I = 1/2 MR^2.
Correct Answer:
A
— 1/2 MR^2
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Q. What is the moment of inertia of a solid cylinder of mass M and radius R about its central axis?
A.
1/2 MR^2
B.
1/3 MR^2
C.
MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a solid cylinder about its central axis is given by I = 1/2 MR^2.
Correct Answer:
A
— 1/2 MR^2
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Q. What is the moment of inertia of a solid disk about its central axis?
A.
(1/2)MR^2
B.
(1/3)MR^2
C.
(1/4)MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a solid disk about its central axis is (1/2)MR^2.
Correct Answer:
A
— (1/2)MR^2
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Q. What is the moment of inertia of a solid sphere about an axis through its center?
A.
(2/5)mr^2
B.
(1/2)mr^2
C.
(1/3)mr^2
D.
(5/2)mr^2
Show solution
Solution
The moment of inertia of a solid sphere about an axis through its center is given by I = (2/5)mr^2.
Correct Answer:
A
— (2/5)mr^2
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Q. What is the moment of inertia of a solid sphere of mass M and radius R about an axis through its center?
A.
2/5 MR^2
B.
3/5 MR^2
C.
1/2 MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer:
A
— 2/5 MR^2
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Q. What is the moment of inertia of a thin circular hoop of mass M and radius R about an axis through its center?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
Show solution
Solution
The moment of inertia of a thin circular hoop about an axis through its center is I = MR^2.
Correct Answer:
A
— MR^2
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Major Competitive Exams MCQ & Objective Questions
Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.
What You Will Practise Here
Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams
Exam Relevance
The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.
Common Mistakes Students Make
Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam
FAQs
Question: How can I improve my performance in Major Competitive Exams?Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.
Question: What types of questions should I focus on for these exams?Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.
Question: Are there specific strategies for tackling objective questions?Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.
Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!
