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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