Chemistry Syllabus (JEE Main)

Inorganic Chemistry Organic Chemistry Physical Chemistry

Q. What is the oxidation state of hydrogen in most of its compounds?

A. -1
B. 0
C. +1
D. +2

Solution

In most of its compounds, hydrogen has an oxidation state of +1.

Correct Answer: C — +1

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Q. What is the oxidation state of manganese in KMnO4?

A. +2
B. +4
C. +6
D. +7

Solution

In KMnO4, the oxidation state of manganese is +7.

Correct Answer: D — +7

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Q. What is the oxidation state of nitrogen in NH4+?

A. -3
B. +1
C. +3
D. +5

Solution

In NH4+, nitrogen has an oxidation state of -3.

Correct Answer: A — -3

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Q. What is the oxidation state of nitrogen in NO3^-?

A. +3
B. +5
C. 0
D. -1

Solution

In NO3^-, nitrogen has an oxidation state of +5.

Correct Answer: B — +5

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Q. What is the oxidation state of phosphorus in H3PO4?

A. +5
B. +3
C. 0
D. -3

Solution

In H3PO4, phosphorus has an oxidation state of +5.

Correct Answer: A — +5

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Q. What is the oxidation state of phosphorus in phosphoric acid (H3PO4)?

A. +3
B. +5
C. 0
D. -3

Solution

In phosphoric acid, phosphorus has an oxidation state of +5.

Correct Answer: B — +5

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Q. What is the oxidation state of sulfur in H2SO4?

A. 0
B. +4
C. +6
D. -2

Solution

In H2SO4, the oxidation state of sulfur is +6.

Correct Answer: C — +6

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Q. What is the oxidation state of Uranium in Uranium Hexafluoride (UF6)?

A. +4
B. +6
C. +5
D. 0

Solution

In UF6, Uranium is in the +6 oxidation state as it forms six bonds with fluorine.

Correct Answer: B — +6

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Q. What is the percentage by mass of a solution containing 20 grams of solute in 180 grams of solution?

A. 10%
B. 20%
C. 15%
D. 25%

Solution

Percentage by mass = (mass of solute / mass of solution) * 100 = (20 g / (20 g + 180 g)) * 100 = 20%.

Correct Answer: B — 20%

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Q. What is the percentage by mass of a solution containing 5 g of NaCl in 95 g of water?

A. 5%
B. 10%
C. 15%
D. 20%

Solution

Percentage by mass = (mass of solute / total mass) × 100 = (5 g / (5 g + 95 g)) × 100 = 5%.

Correct Answer: A — 5%

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Q. What is the percentage composition of carbon in glucose (C6H12O6)?

A. 40%
B. 50%
C. 60%
D. 30%

Solution

Molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 g. Percentage of carbon = (72/180) x 100 = 40%.

Correct Answer: A — 40%

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Q. What is the period number of the element with atomic number 20?

A. 2
B. 3
C. 4
D. 5

Solution

The element with atomic number 20 is Calcium, which is in period 4.

Correct Answer: C — 4

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Q. What is the periodic trend for electronegativity across a period?

A. Increases
B. Decreases
C. Remains the same
D. Varies irregularly

Solution

Electronegativity increases across a period from left to right.

Correct Answer: A — Increases

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Q. What is the pH of a 0.01 M HCl solution?

A. 1
B. 2
C. 3
D. 4

Solution

The pH of a 0.01 M HCl solution is 2, as HCl is a strong acid and fully dissociates.

Correct Answer: B — 2

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Q. What is the pH of a 0.01 M NaOH solution?

A. 12
B. 2
C. 10
D. 1

Solution

For NaOH, which is a strong base, pOH = -log[OH-]. [OH-] = 0.01 M, so pOH = 2. Therefore, pH = 14 - pOH = 14 - 2 = 12.

Correct Answer: A — 12

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Q. What is the pH of a 0.01 M solution of sodium acetate (CH3COONa)?

A. 4.76
B. 7.00
C. 9.24
D. 10.00

Solution

Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]), where pKa of acetic acid is 4.76.

Correct Answer: C — 9.24

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Q. What is the pH of a 0.01 M solution of sodium hydroxide (NaOH)?

A. 12
B. 2
C. 10
D. 14

Solution

pH = 14 - pOH; pOH = -log[OH-] = -log(0.01) = 2; pH = 14 - 2 = 12.

Correct Answer: A — 12

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Q. What is the pH of a 0.01 M solution of sulfuric acid (H2SO4)?

A. 1
B. 2
C. 3
D. 0.5

Solution

H2SO4 is a strong acid and dissociates completely, so pH = -log(0.01) = 2.

Correct Answer: A — 1

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Q. What is the pH of a 0.05 M solution of sulfuric acid (H2SO4)?

A. 1
B. 1.3
C. 1.7
D. 2

Solution

H2SO4 is a strong acid and dissociates completely; [H+] = 0.05 M, so pH = -log(0.05) ≈ 1.3

Correct Answer: B — 1.3

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Q. What is the pH of a 0.1 M NaOH solution?

A. 10
B. 11
C. 12
D. 13

Solution

pOH = -log[OH-] = -log(0.1) = 1; pH = 14 - pOH = 14 - 1 = 13

Correct Answer: C — 12

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Q. What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8 x 10^-5?

A. 2.87
B. 4.76
C. 3.87
D. 5.00

Solution

Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.

Correct Answer: B — 4.76

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Q. What is the pH of a 0.1 M solution of ammonium chloride (NH4Cl)?

A. 5.1
B. 5.5
C. 6.1
D. 6.5

Solution

NH4Cl is a salt of a weak base and a strong acid; it hydrolyzes to give H+, resulting in a pH < 7.

Correct Answer: C — 6.1

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Q. What is the pH of a 0.1 M solution of hydrochloric acid (HCl)?

A. 1
B. 0.1
C. 10
D. 0

Solution

The pH of a strong acid like HCl is calculated using the formula pH = -log[H+]. For 0.1 M HCl, [H+] = 0.1 M, so pH = -log(0.1) = 1.

Correct Answer: A — 1

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Q. What is the pH of a 0.1 M solution of potassium hydrogen phthalate (KHP)?

A. 4.0
B. 5.0
C. 6.0
D. 7.0

Solution

KHP is a weak acid; its pKa is approximately 5.0, so the pH of a 0.1 M solution is around 5.0.

Correct Answer: B — 5.0

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Q. What is the pH of a 0.1 M solution of sodium acetate (CH3COONa)?

A. 4.76
B. 7
C. 9.24
D. 10

Solution

pH = 7 + 0.5(pKa + log[C]) = 7 + 0.5(4.76 + log(0.1)) = 9.24

Correct Answer: C — 9.24

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Q. What is the pH of a 0.1 M solution of sodium bicarbonate (NaHCO3)?

A. 7.5
B. 8.4
C. 9.0
D. 6.0

Solution

NaHCO3 is a weak base; its pH is around 8.4.

Correct Answer: B — 8.4

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Q. What is the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate?

A. 4.74
B. 5.74
C. 6.74
D. 7.74

Solution

Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]). Here, pKa ≈ 4.74, so pH = 4.74 + log(1) = 4.74.

Correct Answer: A — 4.74

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Q. What is the pH of a buffer solution containing 0.2 M acetic acid and 0.1 M sodium acetate?

A. 4.76
B. 5.00
C. 5.74
D. 6.00

Solution

Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = 4.76 + log(0.1/0.2) = 5.74

Correct Answer: C — 5.74

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Q. What is the pH of a buffer solution made from 0.2 M acetic acid and 0.2 M sodium acetate?

A. 4.76
B. 5.76
C. 6.76
D. 7.76

Solution

Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]) = 4.76.

Correct Answer: A — 4.76

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Q. What is the pH of a neutral solution at 25°C?

A. 0
B. 7
C. 14
D. 1

Solution

At 25°C, the pH of a neutral solution is 7.

Correct Answer: B — 7

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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!

Chemistry Syllabus (JEE Main)

Chemistry Syllabus (JEE Main) MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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