Challenges Faced by Students in Chemistry:

• Conceptual Understanding: The subject involves concepts like electrochemical reactions, reaction mechanisms and others, which can be difficult for the students to visualise and grasp.
• Memorisation vs Applications: Students often focus on rote learning, but in chemistry, some topics require application-based concepts. So, the students must focus on understanding the applications rather than rote learning.
• Numerical Problems: Solving numerical problems, especially those involving calculations, requires practice, precision, and accuracy.
• Organic Reaction Mechanism: Getting a clear understanding of organic reactions can be tricky as it requires proper conceptual clarity.

Challenging Problems from Difficult Chapters in Chemistry:

Electrochemistry:

This chapter is considered as one of the most Challenging Class 12th Chemistry Problems. Many students find it difficult to solve questions related to this topic.

Below, we are mentioning the types of questions asked in the exam from this chapter:

• Calculating the cell potential using the Nernst Equation.
• Deriving and solving problems related to Gibbs free energy and equilibrium constant.
• Numericals on conductance and molar conductivity
• Faraday’s Laws of Electrolysis

Problem:

State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mole Cu2+ ions to Cu?

Solution:

When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte.

Let W be the mass of the substance deposited by passing Q coulomb of charge. Then, according to Faraday’s law, we have

W∝Q

Coulomb is the quantity of charge when a current of 1 ampere is passed for 1 sec. Thus, the amount of charge in coulombs

Q = current in amperes x time in seconds

Q = I*t

W∝I*t

W = Z*I*t

Here, Z is a constant, known as electrochemical equivalent, and is characteristic of the substance deposited.

When a current of 1 ampere is passed for 1 sec, i.e. 1 coulomb (Q = 1), then

W = Z

The electrode reaction is Cu2+ + 2e- —> Cu

therefore, the quantity of charge required for the reduction of 1 mole of Cu2+ = 2F

Cu2+ = 2F = 2*96500 = 193000C

Tips to Solve:

• It is important to remember all the equations and their derivations.
• Students are advised to memorise the key formulas.
• Another crucial thing to remember is to pay attention to the unit conversions.

Chemical Kinetics:

Another important yet difficult chapter of Class 12 physical chemistry. This chapter deals with helping students understand the different aspects of reaction rates and mechanisms.

Below, we are mentioning the types of questions asked in the exam from this chapter:

• Determining rate constants and reaction orders.
• Problems related to half-life reactions (first and second order).
• Activation energy calculation using the Arrhenius equation.

Problem:

A first-order gas phase reaction: A2B2(g) –> 2A(g) + 2B(g) at the temperature 400°C has the rate constant k = 2*10-4 sec-1. What percentage of A2B2 is decomposed on heating for 900 seconds? (Antilog 0.0781 = 1.197)

Solution:

Since the reaction is of the first-order

therefore: k = 2.303/t log a/a-x

log 100/x = 0.781

Given

k= 2.0*10-4 sec-1

t = 900 sec

Substituting these values, we get

2*10-4 sec-1 = 2.303/900 log a/a-x

log a/a-x = 2*10-4 *900/ 2.303

log a/a-x = 0.0781

a/a-x = antilog (0.0781) = 1.197

a = 1.197 a – 1.197x

0.197a = 1.197x

therefore: x/a = 0.197/ 1.197 = 0.1645

therefore: %decomposed = 0/1645*100 = 16.45%

Tips to Solve:

• Understand the integrated rate law expressions for different reaction orders.
• Use logarithmic properties to simplify calculations.
• Expert advise- always read and understand the question carefully before attempting it.

The d-and f-block Elements:

This chapter may be confusing for many students. However, understanding the properties of the d and f-block elements, including radioactive behaviour, similar electronic configuration, and others, would make it easy for the students to score good marks in this chapter.

Below, we are mentioning the types of questions asked in the exam from this chapter:

• Explaining trends in oxidation states, magnetic properties, and colour of transition metal ions.
• Questions on the electronic configuration were asked in the exam.
• The exam’s question paper also focused questions on applications and various properties of d-block and f-block elements.

Problem:

When a white crystalline compound A is heated, K₂Cr₂O₂ and conc. H₂SO₄, a reddish brown gas B, evolved, which gave a yellow-coloured solution C when passed through NaOH. A yellow-coloured ppt D is obtained when adding CH₃COOH and (CH₃COO)Pb to solution C. Also, on heating A with NaOH and passing the evolved gas through the K₂HgI₄ solution, a reddish-brown precipitate E is formed. Identify A, B, C, D and E and write the chemical equations for the reaction involved.

Solution:

The following reactions are:

4NH₄Cl + K₂Cr₂O₂ + 6H₂SO₄ → 2CrO₂Cl₂ + 4NH₄HSO₄ + 2KHSO₄ + 3H₂O

CrOCl₂ + 2NaOH → Na2Cr₄ + 2HCl

Na2CrO₄ + (CH₃COO)2Pb → PbCrO₄ ↓ + 2CH₃COONa

NH₄Cl + NaOH → NaCl + H₂O + NH₃

2K₂HgI₂ + NH₃ + 3KOH → NH₂(HgO)HgI + 7KI +2H₂O

According to above

• Compound A is NH₄Cl
• Compound B is CrO₂Cl₂
• Compound C is Na₂CrO₄
• Compound D is PbCrO₄
• Compound E is NH₂(HgO)HgI

Tips to Solve:

• Relate the problem to electronic configurations and stability trends in oxidation states.
• Practice balancing reactions systematically.
• Learn and understand the properties of d-and f-block elements.

Coordination Compounds:

This chapter focuses on the chemical species formed by transition metals and ligands, which students often find to be the most challenging class 12th chemistry problem. It focuses mainly on understanding different compounds’ bonding, properties, and structures.

Below, we are mentioning the types of questions asked in the exam from this chapter:

• Determining hybridisation and geometry of complexes.
• Writing IUPAC names of coordination compounds.
• Crystal field splitting energy and magnetic moment calculations.

Problem:

Using crystal field theory, draw an energy level diagram, write the electronic configuration of the central metal atom/ ion and determine the magnetic moment value in the following:

[CoF₆]³¯, [Co(H₂O)₂]² +, [Co(CN)₆]³¯

Solution:

[CoF₆]³¯

Co³+ = 3d₆

F¯ ion is a weak field ligand. The value of ∆₀ is small (P>∆₀). Electron pairing is not possible.

Number of unpaired electrons = 4

Magnetic moment = √n(n+2) = √4(4+2) = √24 = 4.9BM

[Co(H₂O)₆]² +

Co²+ = 3d⁷

H₂O is a weak field ligand. The value of ∆₀ is small. Electron pairing is not possible.

Number of unpaired electrons = 3

Magnetic moment = √n(n+2) = √3(3+2) = √15 = 3.87BM

[Co(CN)₆]³¯

Co³+ = 3d⁶

CN¯ is a strong field ligand.w The value of ∆₀ is large. Electron pairing is not possible.

Number of unpaired electrons = Zero

Since there are no unpaired electrons present, the complex is diamagnetic in nature.

Tips to Solve:

• Familiarise yourself with Crystal Field Theory (CFT) and its application in predicting the magnetic properties of complexes.
• Use ligand strength to identify high-spin or low-spin configurations.

Aldehydes, Ketones and Carboxylic Acids:

One of the most important chapters of organic chemistry emphasises the reactivity and chemical behaviour of carbonyl and carboxylic acid functional groups. It covers essential reactions and their mechanisms frequently appearing in the exam.

Below, we are mentioning the types of questions asked in the exam from this chapter:

• Writing reaction mechanisms for Aldol condensation, Cannizzaro reaction, and Clemmensen reduction.
• Identifying products in reactions involving nucleophilic addition to carbonyl compounds.
• Acidic strength comparison of carboxylic acids.

Problem:

Ethylbenzene is generally prepared by acetylation of benzene followed by reduction and not by direct alkylation. Think of a possible reason.

Solution:

Preparation of ethylbenzene from acylation of benzene and reduction is shown as:

Direct alkylation is not performed because a polysubstituted product is formed. Due to the disadvantage of poly substitution, Friedel-Craft’s alkylation reaction is not used to prepare alkylbenzenes. Instead, Friedel-Craft’s acylation is used.

Tips to Solve:

• Understand the reactivity of functional groups and common reaction mechanisms.
• Practice identifying intermediates in multi-step reactions.
• Learn and understand the properties and names of reactions.

In the blog above, we discussed the Most Challenging Class 12th Chemistry Problems commonly asked from various chapters that are often considered difficult. We hope this will help you become familiar with the type of questions. The board exams are crucial in a student’s academic journey, so it is essential for students to prepare for these exams with full dedication and sincerity.

Best of luck to all the students!

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