TIFR GS 2021 Question with detailed solution| Chemistry

TIFR GS2021 (Tata Institute of Fundamental Research) Graduate School 2021 exam Question with a detailed solution for chemical science.

TIFR GS 2021 Question with detailed solution paper

Hello everyone today we will discuss the TIFR GS2021 (Tata Institute of  Fundamental Research) Graduate School Admission 2021 exam Question with a detailed solution for chemical science. Today we have to solve TIFR's previous year's question detailed solution for Chemistry students.

TIFR GS2021 Chemistry questions solve step by step. This is the first article of the TIFR GS 2021 question-solving section. We will go forward with this TIFR previous year's questions Answer solution section. Every week we will discuss the question-answer detailed solution.

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Question with solution paper TIFR GS 2021

Chem article provides only chemistry details solution paper with pdf, our team provided as soon as possible. This is the Question-Answer series of the TIFR examination Aspirant. you can follow the below question with detailed solutions including: 

Q.1. Indicate the stereochemical relationship between the two molecules. 

1) Diastereomers

2) Enantiomers

3) Identical and achiral

4) Identical and chiral



Answer: (2) Enantiomers

Hints: Two give compound mirror images to each other, therefore two compounds are enantiomers to each other.

If two same compounds are given mirror images to each other then we called enantiomers. We see that the two compounds are mirror images of each other.

Q.2. What is the correct C-H bond (shown in below molecules) dissociation energy?

1) A>B>C

2) C>B>A

3) C>A>B

4) A>C>B



Answer: 3) C>A>B

Hints: we know the bond strength of the hybridization order: sp > sp2 > sp3. Compound A attach C-H bond (which carbon sp2 hybridized), compound B attach C-H bond (which carbon atom is sp3 hybridized) and compound C attach C-H bond (which carbon atom sp hybridized) are involved in Bond dissociation energy. According to the bond strength of hybridization of C-H Bond: Compound C > A > B.   

Q.3. For a particle in a 1-D box of length 2L the normalization constant A for its wavefunction πœ“ (x, t) =A sin (π‘›πœ‹π‘₯/2𝐿) (where n = 1,2,3…) is given by:

1) A=√(1/4𝐿)

2) A=√(2/𝐿)

3) A=√(1/2𝐿)

4) A=√(1/𝐿)


 

Answer: (4) A=√(1/𝐿)

Hints:

For 1D box particle wavefunction:

πœ“ (π‘₯, 𝑑) = √(2/𝐿) sin (π‘›πœ‹π‘₯/𝐿)

At, Length L= 2L, then we get,

πœ“ (π‘₯, 𝑑) = √(2/2𝐿) sin (π‘›πœ‹π‘₯/2𝐿)

πœ“ (π‘₯, 𝑑) = √(1/𝐿) sin (π‘›πœ‹π‘₯/2𝐿)

Compered the give equation then we get the normalization constant A = √(1/𝐿) for its wavefunction πœ“.



Q.4. Charge polarization of the C-F bond is much larger compared to the C-Br bond. Which of the following statements on the dipole moments of CH3Br and CH3F is true?

1) The dipole moment of CH3F is greater than CH3Br

2) The dipole moment of CH3Br is greater than CH3F

3) Both molecules have zero dipole moment

4) The dipole moments of CH3F and CH3Br are the same



Answer: 4) The dipole moments of CH3Br and CH3Br are the same

Hints: The polarization of the C-F bond is much larger than the C-Br bond.

We know that dipole moment is equal to the separation charge into the distance between two atoms. Dipole moment = q x d. where q = Separation charge and d = Distance between two atoms (bond length).   

C-F bonds have less dipole moment due to Florine lone pair electrons repletion with carbon atom is high compared to C-Br. And on the other hand, C-Br has a high dipole moment due to bromine atom lone pair electrons repulsion comparatively less.

The size of the carbon atom and halogen atom (F, Br) matches better to the Br atom compared to the F atom. Because Br atom size is larger than F atom. And on the other hand, the partial charge of the halogen atom Florine has more than Bromine. So, all combined negate the Dipole moments of the CH3F and CH3Br. We can call the dipole moments of the CH3F and CH3Br are same.


Q.5. Quantum confinement results in

1) Energy gap in a semiconductor is proportional to the inverse of the size

2) Energy gap in a semiconductor is proportional to the inverse of the square of size

3) Energy gap in a semiconductor is proportional to the square of size

4) Energy gap in a semiconductor is proportional to the inverse of the square root of the size



Answer: 2) Energy gap in a semiconductor is proportional to the inverse of the square of size


Q.6. An instrument measures the position of two independent atoms concerning the origin along a line as 20 ∓ 3 Γ… and 40 ∓ 4 Γ…. The distance between the two atoms on the line is:

1) 20∓πŸ• Γ…

2) 20∓𝟏 Γ…

3) 20∓πŸ“ Γ…

4) 20∓πŸ’ Γ…



Answer: 3) 20∓πŸ“ Γ…

Hints: Distance d=x_2-x_1 for independent variables error in d is given by ∆d=√(〖(∆x_1)〗^2-〖(∆x_2)〗^2 )

Q.7. Predict the product of the following reaction [PCC = Pyridinium chlorochromate].

Options:

(A)

(B)

(C)

(D)



Answer: (D) No Reaction

Hints: 

[PCC = Pyridinium chlorochromate] is an oxidising reagent. In case alcohol is converted into aldehyde/acid. PCC has oxidized only primary and secondary alcohol into aldehyde and acid respectively. But PCC reagent does not oxidize tertiary alcohol.

you can see that the given reactant compound is tertiary alcohol, so they have no reaction.

Conclusion:

Test question for mixing topics like stereochemistry, C-H bond dissociation energy, Quantum chemistry 1-D box problem, normalization constant value, dipole moments, Energy gap in a semiconductor, etc. This is for entrance examinations such as TIFR GS 2021. We are hopeful that this question-answer series helpful for beginners. Thank you for reading. Do not forget to like, share and comment. Give your feedback in below comment section.