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NEET-XI-Chemistry

01: Redox Reactions

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  • #
    Chapter 1 - Redox Reactions
  • Qstn #1
    Assign oxidation numbers to the underlined elements in each of the following species:
  • #1-a
    NaH2PO4
    Ans :

    Let the oxidation number of P be x.

    We know that,

    Oxidation number of Na = +1

    Oxidation number of H = +1

    Oxidation number of O = -2



    Then, we have



    Hence, the oxidation number of P is +5.
  • #1-b
    NaHSO4
    Ans :



    Then, we have



    Hence, the oxidation number of S is + 6.
  • #1-c
    H4P2O7
    Ans :



    Then, we have



    Hence, the oxidation number of P is + 5.
  • #1-d
    K2MnO4
    Ans :



    Then, we have



    Hence, the oxidation number of Mn is + 6.
  • #1-e
    CaO2
    Ans :



    Then, we have



    Hence, the oxidation number of O is - 1.
  • #1-f
    NaBH4
    Ans :



    Then, we have



    Hence, the oxidation number of B is + 3.
  • #1-g
    H2S2O7
    Ans :



    Then, we have



    Hence, the oxidation number of S is + 6.
  • #1-h
    KAl(SO4)2.12 H2O
    Ans :



    Then, we have



    Or,

    We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have



    Hence, the oxidation number of S is + 6.
  • Qstn #2
    What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
  • #2-a
    KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
    Ans : KI3

    In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3to find the oxidation states.

    In a KI3molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.



    Hence, in a KI3molecule, the O.N. of the two I atoms forming the I2molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.
    (b) H2S4O6



    However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.



    The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.
    (c)

    On taking the O.N. of O as –2, the O.N. of Fe is found to be. However, O.N. cannot be fractional.

    Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.


    (d)

    2 (x) + 6 (+1) + 1 (-2) = 0

    or, 2x + 4 = 0

    or, x = -2

    Hence, the O.N. of C is –2.
    (e)

    2 (x) + 4 (+1) + 2 (-2) = 0

    or, 2x = 0

    or, x = 0

    However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.


  • #2-b
    H2S4O6
    Ans : H2S4O6



    However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.



    The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.
  • #2-c
    Fe3O4
    Ans :

    On taking the O.N. of O as –2, the O.N. of Fe is found to be. However, O.N. cannot be fractional.

    Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.


  • #2-d
    CH3CH2OH
    Ans :

    2 (x) + 6 (+1) + 1 (-2) = 0

    or, 2x + 4 = 0

    or, x = -2

    Hence, the O.N. of C is –2.
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