Chemical Bonding And Molecular Structure

  • Question
    CBSEENCH11008055

    The ionic radii (in Å) of N3–, O2– and F are respectively:

    • 1.36, 1.40 and 1.71

    • 1.36, 1.71 and 1.40

    • 1.71, 1.40 and 1.36

    • 1.71, 1.36 and 1.40

    Solution

    C.

    1.71, 1.40 and 1.36

    Number of electrons in N3- = 7+3 = 10
    Number of electrons in O2- = 8+2 = 10
    Number of electrons in F- = 9+1 = 10
    Since, all the three species have each 10 electrons hence they are isoelectronic species.
    It is considered that, in case of isoelectronic species as the negative charge increase, ionic radii increase and therefore the value of ionic radii are
    N3- = 1.71 (highest among the three)
    O2- = 1.40
    F- = 1.36 (lowest among the three)
    N3- > O2-> F-

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