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11 August, 05:45

Which of the following is correctly ranked in order of increasing bond polarity? O-O < O-F Cl-S < Cl-Br Al-O < P-O H-H < N-N B-F < C-F F-O < O-O Li-F < C-O

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  1. 11 August, 06:14
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    O-O
    option 1 is correct.

    Explanation: The bond polarity can be explained on the basis electronegative difference between the atoms.

    Greater the electronegativity difference between the two atoms in a bond greater is the polarity of the bond.

    Electronegativity difference of a bond between two atoms = EN of atom having higher electronegativity - EN of atom having lower electronegativity

    Likewise we can calculate the electronegativity difference for the following bonds and estimate there bond polarities.

    1. EN[O-O]=EN (O) - EN (O) = (3.44) - (3.44) = 0

    EN[O-F]=EN (F) - EN (O) = (3.98) - (3.44) = 0.54

    So the bond polarity of O-F is more than O-O as it has more electronegativity difference. These two species are correctly arranged as O-O
    2. EN[Cl-S]=EN (S) - EN (Cl) = (3.16) - (2.58) = 0.58

    EN[Cl-Br]=EN (Cl) - EN (Br) = (3.16) - (2.96) = 0.20

    So the bond polarity of Cl-S is more than Cl-Br as it has more electronegativity difference. These two species are incorrectly arranged as Cl-S>Cl-Br.

    3. EN[Al-O]=EN (O) - EN (Al) = (3.44) - (1.61) = 1.83

    EN[P-O]=EN (O) - EN (P) = (3.44) - (2.19) = 1.25

    So the bond polarity of Al-O is more than P-O as it has more electronegativity difference. These two species are incorrectly arranged as Al-O>P-O.

    4. EN[H-H]=EN (H) - EN (H) = (2.2) - (2.2) = 0

    EN[N-N]=EN (H) - EN (N) = (3.04) - (3.04) = 0

    The electronegativity difference for both the bonds is 0 hence both the bonds would not have any polarity. Hence the mentioned order is incorrect.

    5. EN[B--]=EN (F) - EN (B) = (3.98) - (2.04) = 1.94

    EN[C-F]=EN (F) - EN (C) = (3.98) - (2.55) = 1.43

    So the bond polarity of B- - is more than C-F as it has more electronegativity difference. These two species are incorrectly arranged as B-->C-F

    6. EN[Li-F]=EN (F) - EN (Li) = (3.98) - (0.98) = 3.0

    EN[C-O]=EN (O) - EN (C) = (3.44) - (2.55) = 0.89.

    So the bond polarity of Li-F is more than C-O as it has more electronegativity difference. These two species are incorrectly arranged as Li-F>C-O

    7. EN[F-O]=EN (F) - EN (O) = (3.98) - (3.44) = 0.54

    EN[O-O]=EN (O) - EN (O) = (3.44) - (3.44) = 0

    So the bond polarity of F-O is more than O-O as it has more

    electronegativity difference. So it is incorrectly arranged.
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