CBSE Class 11th Chemistry Solved Practice Paper : Set-1

Oct 12, 2015, 16:28 IST

Find the CBSE Class 11 Chemistry Solved Practice Paper for the coming pre-board exam. These questions will not only help the students to prepare for exams in a better manner, but will also help them in understanding the depth with which a topic should be studied.

Find the CBSE Class 11 Chemistry Solved Practice Paper for the coming pre-board exam. These questions will not only help the students to prepare for exams in a better manner, but will also help them in understanding the depth with which a topic should be studied.

Few sample questions from the Question Paper are given below:

Q. Arrange the elements B, C, N, F, and Si, in increasing order of their non-metallic character.

Ans. In a periodic table, from left to right in a period the non-metallic character increases and from top to bottom in a group the non-metallic character decreases. Therefore, the increasing order of non-metallic character for the given elements is:

                                            B < Si < C < N < F

Q. Calculate the molarity of NaOH in the solution prepared by dissolving its 8 g in enough water to form 500 mL of the solution.

Ans.


Q. Among NH3, H2O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?

Ans. H-bonding in a molecule depends upon the electronegativity and the number of hydrogen atoms. The increasing order of electronegativity among nitrogen, oxygen and fluorine is N < O < F. But the extent of H-bonding follows the order: NH3 < HF < H2O. This is so because water molecule has two H-atoms and the O-atom has two lone pairs through which it can form extensive H-bonding. But in case of HF, it has only one H-atom so HF make long chains of Hf molecules and in ammonia molecule in spite of having three H-atoms it does not shows extensive H-bonding because N-atom has only one lone pair.

Q. The order of reactivity among group 1 elements is Li < Na < K < Rb <Cs while that of group 17 elements is F > CI > Br > I. Explain.

Ans. The outermost cofiguration of the alkali metal is ns1 therefore, they lose an electron to form monopositive ion. Atomic radius is inversely proportional to Ionization enthalpy. Since in group 1 Cs has highest atomic radius hence it has highest reactivity while Li has smallest radius therefore, it has lowest reactivity. In case of halogens, the outermost configuration is ns2np5 therefore; they gain one electron to attain a stable configuration. If the atomic radius is less the nuclear attraction is more and greater is the tendency to accept an electron. In group 17 from top to bottom atomic radius increases thus the tendency to accept an electron decreases. Hence, the reactivity also decreases.

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