A) \[{{H}^{-}}>,{{H}^{+}}>H\]
B) \[N{{a}^{+}}>{{F}^{-}}>{{O}^{2-}}\]
C) \[{{F}^{-}}>{{O}^{2-}}>N{{a}^{+}}\]
D) None of these
Correct Answer: D
Solution :
| (No option is correct) |
| [a] \[{{H}^{-}}>{{H}^{+}}>H\] |
| 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 as they contain same number of electrons. For isoelectronic species, |
| \[ionic\text{ }radii\text{ }\propto \frac{1}{atomic\text{ }number}\] |
| Ion: \[N{{a}^{+}}{{F}^{-}}{{O}^{2-}}\] |
| Atomic number: 11 9 8 |
| Hence, the correct order of ionic radii is |
| \[{{O}^{2-}}>{{F}^{-}}>N{{a}^{+}}\] |
| [c] Similarly, the correct option is |
| \[{{O}^{2-}}>{{F}^{-}}\,>N{{a}^{+}},\] |
| [d] Ion : \[A{{l}^{3+}}\] \[M{{g}^{2+}}\] \[{{N}^{3-}}\] |
| Atomic number: 13 12 7 |
| Hence, the correct order is\[{{N}^{3-}}\,>M{{g}^{2+}}\,>A{{l}^{3+}}\]. |
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