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The ionic radii `(Å)" of "C^(4+) and O^(2-)` respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion `N^(3-)` would beA. 2.6B. 1.71C. 1.4D. 0.95 |
Answer» Correct Answer - B The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is `C^(4-)gtN^(3-)gtO^(2-)`. |
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