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Chemical Bonding and Molecular Structure Test 3

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Chemical Bonding and Molecular Structure Test 3
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  • Question 1
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    Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

    Q.  Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is 

    Solution

    Molecular orbital configuration of B2 as per the condition of violation of hund's rule will be .

  • Question 2
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    Which of the following statement is incorrect regarding NO2 molecule?

    Solution

    NO2 molecule has unpaired electrons which are responsible for its brown colour and paramagnetic behavior.

  • Question 3
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    Molecule MX3 (atomic number M < 21) has zero dipole moment, the sigma bonding orbitals used by M are

    Solution

    Molecule has zero dipole moment, thus it does not have lone pair. Hence, non-polar.

    Thus, (M — X) bonds indicate sp2-hybridisation.

  • Question 4
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    bond in between (C2— C3) vinyl acetylene is formed by ...... overlapping

    Solution

    Vinyl acetylene is

    bond are at equal position, numbering of C-chains based on IUPAC nomenclature is done from side.

  • Question 5
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    Hydrogen bonds are formed in many compounds e.g., H2O, HF, NH3 . The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

    Solution

    H2​O>HF>NH3​

    Strength of hydrogen bonding depends on the size and electronegativity of the atom. 

    Smaller the size of the atom, greater is the electronegativity and hence stronger is the H−bonding. Thus, the order of strength of H-bonding is H...F>H...O>H...N. 

    But each HF molecule is linked only to two other HF molecules while each H2O molecule is linked to four other H2​O molecules through H−bonding.

    Hence, the decreasing order of boiling points is H2​O>HF>NH3​.

  • Question 6
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    Considering the state of hybridisation of C-atoms,select the molecule among the following which is linear

    Solution

    Molecules with  have bond angle 180°. In (d) there is also and is not symmetrical.
    (c) is symmetrical w.r.t.

  • Question 7
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    In which one of the following species the central atom has the type of hybridisation which si not the same as that present in the other three? 

    Solution

    Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid oebitals.

    where,

    V= number of valance electrons of central atom

    X = number of monovalent atoms

    C= charge on cation

    A = charge on anion
     

  • Question 8
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    BF3 is electron deficient (Lewis acid) and NH3 is electron-rich (Lewis acid). In 1 : 1 complex of BF3 and NH3, then coordination geometry and hybridisation of N and B atoms are :




    Solution


    After 1 :1 complex is formed in which NH3 is electron donor and BF3 is electron acceptor

    Each atom (N, B) both are sp3 hybridised. Since, lone pairs of N-atoms are also involved in bonding, hence,both are sp3 hybridized and tetrahedral.

  • Question 9
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    If there are five electron pairs in outer shell, then structure and bond angle as predicted by Sidgwick-Powell theory is

    Solution

    Three electron pairs are inclined at 120°.
    Two electron-pairs are at 90° with the first three.

    Structure is trigonal bipyramidal.

  • Question 10
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    Hybridisation of Acetylene is

    Solution

    Since acetylene is made up of triple bond. So the hybridization of carbon in acetylene is sp.

  • Question 11
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    Based on VSEPR model, AX4E representation is for

    Solution

  • Question 12
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    Direction (Q. Nos. 12 and 14) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.

    Q. 

    Statement I : BF3 molecule is planar but NF3 is pyramidal.

    Statement II : N atom is smaller than B.

    Solution


    (Ip-bp) repulsion make it pyramidal.
    Thus, Statement I and II both are correct but Statement II is not the correct explanation of Statement I.

  • Question 13
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    Statement I : B atom is sp2-hybridised in B2H6.
    Statement II : There is no lone pair or unpaired electron in B2H6.

    Solution

    (d) B2H6 has following types of bonding

    Bridging H-atoms and B-atoms are electron deficient. Each B-atom is however, joined to four H-atoms thus sp3-hybridised.
    There is no lone pair or unpaired electron.
    Thus, Statement I is incorrect but Statement II is correct.

  • Question 14
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    Statement I : BCI3 exists as monomer while AICI3 exists as dimer.

    Statement II : Boron atom is too small to form the stable 4-membered ring system

    Solution


    Size of B is too small to form stable 4-membered ring system. (Also delocalised n-bonding is disturbed). Hence, dimerofBCI3is not formed.
    Dimer AI2CI6 is stable due to stable ring

    Thus, Statements I and II both are correct and Statement II is the correct explanation of Statement 

  • Question 15
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    Direction (Q. Nos. 15-16) This section contains  a paragraph, each describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).

    In certain polar solvents, PCI5 undergoes an ionisation reaction as shown

     

    Q. Select the incorrect statement (s).

    Solution


  • Question 16
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    In certain polar solvents, PCI5 undergoes an ionisation reaction as shown

     

    Q. Number of lone pairs on P in I, il and III are

    Solution

     Lone pair on each phosphorus atom = 0

  • Question 17
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    Direction (Q. Nos. 17-18) This section contains 2 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE  is correct.

    Q. Match the following compounds given in Column I with their geometry given is Column II
     


    Solution





  • Question 18
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    Match the processes in Column I with type of changes in hybridisation in Column II.
     

    Solution






     

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