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The correct order of ionic radii of O^{2-},N^{3-},F^-,Mg^{2+},Na^+,Al^{3+}
Option: 1 N^{3-}<F^-<O^{2-}<Mg^{2+}<Na^+<Al^{3+}
Option: 2 Al^{3+}<Na^+<Mg^{2+}<O^{2-}<F^-<N^{3-}
Option: 3 N^{3-}<O^{2-}<F^-<Na^+<Mg^{2+}<Al^{3+}
Option: 4 Al^{3+}<Mg^{2+}<Na^+<F^-<O^{2-}<N^{3-}

Answers (1)

best_answer

The charge to mass ratio(z/e) of the given species is given as follows:

O2- = z/e = 8/10 = 0.8

N3- = 7/10 = 0.7

F- = 9/10 = 0.9

Mg2+ = 12/10 = 1.2

Na+ = 11/10 = 1.2

Al3+ = 13/10 = 1.3

Now, higher the z/e, smaller is the atomic radii.

Thus, the atomic radii follows the order given below:

Al3+ < Mg2+ < Na+ < F- < O2- < N3-

Therefore, Optioin(4) is correct.

Posted by

Kuldeep Maurya

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