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Consider the isoelectronic species, Na^{+} , Mg^{2+}, F^{-} and\; O^{2-}.

The correct order of increasing length of their radii is _________.

(i) F^{-} < O^{2-} <Mg^{2+}<Na^{+}

(ii) Mg^{2+}<Na^{+} <F^{-} < O^{2-}

(iii) O^{2-} <F^{-} < Na^{+} <Mg^{2+}

(iv) O^{2-} <F^{-} < Mg^{2+}<Na^{+}


Answers (1)

The answer is the option (ii) Mg^{2+}<Na^{+} <F^{-} < O^{2-}

Explanation: All of them are isoelectronic species because they have the same number of valence electrons and moreover they even have the same number of electrons on the whole, which is 10. So, their radii would be different because of their different nuclear charges. The greater the charge in the cation greater is the force with which the nucleus pulls the electrons or greater is the force of attraction of electrons to the nucleus and thus will have a lesser radius. Whereas in case of anions greater is the charge lesser is the force of attraction thus will have the larger radius, and in this very case the net repulsion between the electrons will outweigh the nuclear charge, and the ion will thus expand in size.

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