CBSE Class 11 students can check their preparedness with help of the Chemistry Sample Paper given in this article. CBSE Class 11 Chemistry Sample Paper is prepared by subject experts after doing the detailed analysis of the latest syllabus and paper format. Students can also check the detailed solutions if they face any difficulty while attempting paper.
Some salient features of the CBSE Class 11 Chemistry Sample Paper, are:
1. Based on the latest CBSE Class 11 Syllabus.
2. Questions are included from each and every chapter.
3. The paper design is exactly similar to CBSE Class 11 Question Paper.
4. Detail solution for all the questions.
1. All the questions are compulsory.
2. The questions paper consists of 27 questions divided into 4 sections A, B, C and D.
3. Section A comprises of 5 questions of 1 mark each. Section B comprises of 7 questions of 2 marks each. Section C comprises of 12 questions of 3 marks each. Section D comprises of 3 questions of 5 marks each.
4. There is no overall choice. However, there is an internal choice in one question of 2 marks weightage, one question of 3 marks weightage and all the three questions of 5 marks weightage.
5. Use of calculators is not permitted.
Some sample questions from CBSE Class 11 Chemistry Solved Sample Paper 2019 are given below:
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Presence of delocalized 6-electrons above and below the benzene ring, make it a rich source of electrons. As a result, it attracts the electron deficient species, i.e., electrophilic reagents with much ease as compared to the nucleophiles that are electron-rich sources. That's why benzene gives electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty.
[SiF6]2– is known whereas [SiCl6]2– not. Give possible reasons.
The main reasons are:
(i) Six large chloride ions cannot be accommodated around Si4+ due to limitation of its size.
(ii) Interaction between lone pair of chloride ion and Si4+ is not very strong.
Can phosphorus with outer electronic configuration 3s23p3 form PH5?
Although phosphorus exhibits +3 and +5 oxidation states, it cannot form PH5. Besides some other considerations, high ∆a H value of dihydrogen and ∆eg H value of hydrogen do not favour to exhibit the highest oxidation state of P, and consequently the formation of PH5.
Question: Give reason for the following:
(a) Sodium less reactive than potassium.
(b) KO2 is paramagnetic.
(c) Alkali metals are not found in nature.
(a) The ionisation enthalpy of the alkali metals decreases down the group. Hence the ionisation enthalpy of potassium is lower than that of sodium which makes it more electropositive and stronger reducing agent as compared to sodium. Therefore, sodium is less reactive than potassium.
(b) The superoxide O2– is paramagnetic because of one unpaired electron in π*2p molecular orbital.
(c) The alkali metals have only one electron in their valence shell, which they can lose easily to complete their octet. Thus the low ionization energies and high electropositive character cause these metals to be highly reactive hence, preventing them to exist in free state in nature. They are found in the earth’s crust in the form of halide, sulphate, carbonate, silicate, borate, oxide ores, etc.
Write IUPAC names of the following compounds:
(i) 2, 2, 4, 4-Tetramethylpentane
(ii) 3, 3-Dimethylpentane
(iii) 3,3-Di-tert-butyl-2, 2, 4, 4–tetramethylpentane
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