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WBJEE: Important Questions and Preparation Tips –Periodic Classification & Periodicity in Properties

Dec 21, 2017 17:20 IST
    WBJEE 2018: Periodicity in Properties
    WBJEE 2018: Periodicity in Properties

    In this article, the engineering section of Jagranjosh after the detailed analysis of the pattern and the syllabus of the examination brings to you the complete chapter notes of chapter Periodic Classification & Periodicity in Properties including important concepts, formulae and some previous year solved questions for coming WBJEE entrance examination 2018. Students always get 1-2 questions from this topic in the examination.

    In these chapter notes all important topics related to Classification & Periodicity in Properties like need to classify elements, Law of Triads, Law of Octaves, Mendeleev's Periodic Table, Modern Periodic Table, Characteristics of Modern Periodic table, Periodic Trends in Properties of Elements etc.

    WBJEE: Important Questions and Preparation Tips – States of Matter

    Important Concepts:

    WBJEE Important Concepts

    WBJEE: Important Questions and Preparation Tips – Electrochemistry

    Some previous year solved questions are given below:

    Question 1:

    Which of the following statements regarding Lanthanides is false?

    (a) All lanthanides are solid at room temperature.

    (b) Their usual oxidation state is +3

    (c) They can be separated from one another by ion-exchange method.

    (d) Ionic radii of trivalent lanthanides steadily increases with increase in atomic number. 

    Solution 1:

    Lanthanide are reffered as rare earth metals. they are found in 6th period in periodic table. At room temperature all lanthanides are solid. They have usual oxidation state of +3. Also, they can be separated by ion exchange method.

    Ionic radii of trivalent lanthanides steadily decreases with increase in atomic number and the phenomenon is known as Lanthanoid contraction.

    Hence, the correct option is (d).

    Question 2:

    Which of the following arrangements is correct in respect to solubility in water?

    (a) CaSO4 > BaSO4 > BeSO4 > MgSO4 > SrSO4

    (b) BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4

    (c) BaSO4 > SrSO4 > CaSO4 > MgSO4 > BeSO4

    (d) BeSO4 > CaSO4 > MgSO4 > SrSO4 > BaSO4

    Solution 2:

    In a group the solubility in water decreases down the group due to increase in ionic size.

    So, the correct order of solubility in water is:

    Be2+ < Mg2+ < Ca2+ < Sr2+ < Ba2+

    Hence, the correct option is (b).

    WBJEE: Important Questions and Preparation Tips – Solutions

    Question 3:

    An element X belongs to fourth period and fifteenth group of the periodic table. Which of the following statements is true?

    (a) It has a completely filled s-orbital and a partially filled d-orbital.

    (b) It has completely filled s-and p-orbitals and a partially filled d-orbital.

    (c) It has completely filled s-and p-orbitals and a half filled d-orbital.

    (d) It has a half filled p-orbital, and completely filled  s- and d-orbitals.

    Solution 3:

    The element that belong to fourth period and fifteenth group of the periodic table is Arsenic.

    The electronic configuration of Arsenic is:

     As33: [Ar]4s23d104p3

    Hence, the correct option is (d).

    WBJEE: Important Questions and Preparation Tips – Equilibrium

    Question 4:

    Which one has the highest boiling point?

    (a) He

    (b) Ne

    (c) Kr

    (d) Xe

    Solution 4:

    As we move down the group of noble gases, molecular mass increases by which dipole produced for a moment and hence London forces increases from He to Xe.

    Therefore, more amount of energy is required to break these forces, thus boiling point also increases from He and Xe.

    Hence, the correct option is (d).

    WBJEE: Important Questions and Preparation Tips – Basic Concepts of Chemistry

    Question 5:

    Which one of the following arrangements represents the correct order of electron gain enthalpy

    (with negative sign) of the given atomic species?

    (a) CI < F < S < CO

    (b) CO < S < F < CI

    (C) S  < CO < CI < F

    (d) F < CI < CO < S

    Solution 5:

    Election Gain enthalpy: It is the energy released when one mole of electrons is added to gaseous atoms of an element. Generally electron gain enthalpy increases in a period from left to right but decreases in a group on moving down.

    Fluorine due to its smaller size has unexpectedly low electron gain enthalpy than Cl.

    Similar is shown in case of CO and S. Thus, the order of electron gain enthalpy is

    CO < S < F < CI

    Hence, the correct option is (b).

    WBJEE: Important Questions and Preparation Tips – Solid State

    WBJEE 2018: Notification, Application, Dates, Eligibility, Exam Pattern, Syllabus

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