A) \[{{H}^{-}}>{{H}^{+}}>H\]
B) \[N{{a}^{+}}>{{F}^{-}}>{{O}^{2-}}\]
C) \[{{F}^{-}}>{{O}^{2-}}>N{{a}^{+}}\]
D) \[A{{l}^{3+}}>M{{g}^{2+}}>{{N}^{3-}}\]
Correct Answer: A
Solution :
Options b, c and d are incorrect. \[{{H}^{-}}>H>{{H}^{+}}\] It is known that radius of a cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Whereas the radius of anion is always greater than a cation due to decrease in effective nuclear charge. Hence the correct order is \[{{H}^{-}}>{{H}^{+}}>H.\] [b] \[N{{a}^{+}}>{{F}^{-}}>{{O}^{2-}}\] The given species are isoelectronic and they are in contact with the same number of electrons. For isoelectronic species, \[\text{Ionic}\,\text{radii}\propto \frac{\text{1}}{\text{atomic}\,\text{number}}\] \[\therefore \] The correct order of ionic radii is \[{{O}^{2-}}>{{F}^{-}}>N{{a}^{+}}\] [c] Similarly, the correct option is \[{{O}^{2-}}>{{F}^{-}}>N{{a}^{+}}\] [d] The correct order is \[{{N}^{3-}}>M{{g}^{2+}}>A{{l}^{3+}}\]
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