JEE MAIN - Chemistry (2009 - No. 10)
The set representing the correct order of ionic radius is :
$$N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}$$
$$L{i^ + } > N{a^ + } > M{g^{2 + }} > B{e^{2 + }}$$
$$M{g^{2 + }} > B{e^{2 + }} > L{i^ + } > N{a^ + }$$
$$L{i^ + } > B{e^{2 + }} > N{a^ + } > M{g^{2 + }}$$
Explanation
In periodic table $$Li, Na, Mg, Be$$ are present in following ways :-
Note :
$$(1)\,\,\,$$ In periodic table from left to right the size of element decreases.
$$(2)\,\,\,$$ In periodic table, from top to bottom size of element increases.
So, $$B{e^{ + 2}}$$ will be the smallest in size.
Among $$L{i^ + }$$ and $$N{a^ + },$$ size of $$N{a^ + } > L{i^ + }$$ and among $$B{e^{ + 2}}$$ and $$M{g^{ + 2}},$$ size of $$M{g^{ + 2}} > B{e^{ + 2}}$$.
Note : Size of $$L{i^ + } > M{g^{ + 2}},$$ this is a exception case.
So, the correct order will be, $$N{a^ + } > L{i^ + } > M{g^{ + 2}} > B{e^{ + 2}}$$
| Group 1 | Group 2 |
|---|---|
| Li | Be |
| Na | Mg |
Note :
$$(1)\,\,\,$$ In periodic table from left to right the size of element decreases.
$$(2)\,\,\,$$ In periodic table, from top to bottom size of element increases.
So, $$B{e^{ + 2}}$$ will be the smallest in size.
Among $$L{i^ + }$$ and $$N{a^ + },$$ size of $$N{a^ + } > L{i^ + }$$ and among $$B{e^{ + 2}}$$ and $$M{g^{ + 2}},$$ size of $$M{g^{ + 2}} > B{e^{ + 2}}$$.
Note : Size of $$L{i^ + } > M{g^{ + 2}},$$ this is a exception case.
So, the correct order will be, $$N{a^ + } > L{i^ + } > M{g^{ + 2}} > B{e^{ + 2}}$$
Comments (0)
