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 \[{{N}^{3-}}\,>M{{g}^{2+}}\,>A{{l}^{3+}}\].
You need to login to perform this action.
You will be redirected in
3 sec
