Question

The incorrect hydrated radius order is :
(A). $Li_{\left( {aq} \right)}^ + < Be_{\left( {aq} \right)}^{2 + }$
(B). $Na_{\left( {aq} \right)}^ + < Al_{\left( {aq} \right)}^{3 + }$
(C). $I_{\left( {aq} \right)}^ - > Cl_{\left( {aq} \right)}^ -$
(D). $Ba_{\left( {aq} \right)}^{2 + } < Ca_{\left( {aq} \right)}^{2 + }$

Answer 1

Hydrated radius is defined as radius of ions and closely bonded water molecules. It is inversely related to atomic radius of an element. As atomic size decreases, hydrated radius increases and vice-versa.

Complete step by step answer:
We know the hydrated radius is defined as radius of ions and closely bonded water molecules. This means that the hydrated radius is the radius of ions when surrounded by water molecules. The energy released when an ion is hydrated i.e. accepts water molecules is known as hydration energy. Smaller ion in size accepts a large number of water molecules so its hydration energy is more and they greater will be its hydrated radius and vice – versa . This means that as the ionic size decreases, its hydrated radius increases. The ionic size decreases along a period and the ionic / atomic size decreases down a group. This means that along a period hydrated radius increases and down a group hydrated radius decreases.
$L{i^ + }$ and $B{e^{2 + }}$ are members of second period and size atomic size of $B{e^{2 + }}$ is less than that of $L{i^ + }$ . Thus hydrated radius of $B{e^ + }$ is greater than that of $L{i^ + }$ . i.e. $B{e^{2 + }} > L{i^ + }$
Thus option (A) is the correct statement.
Out $A{l^{3 + }}$ and $N{a^ + }$ , atomic size of $N{a^ + }$ is greater than that of $A{l^{3 + }}$ . Thus the hydrated radius of $N{a^ + }\left( {aq} \right)$ is less than that of $A{l^{3 + }}\left( {aq} \right)$ . Thus option (B) i.e. $N{a^ + }\left( {aq} \right) < A{l^{3 + }}\left( {aq} \right)$ is the correct statement.
$Ba$and $Ca$ are the elements of group $1s$ and the atomic radius of $B{a^{2 + }}$ is greater than that of $C{a^{2 + }}$ . Thus $B{a^{2 + }}\left( {aq} \right)$ is less than that of $C{a^{2 + }}\left( {aq} \right)$ that is, $B{a^{2 + }}\left( {aq} \right)$ is less than that of $C{a^{2 + }}\left( {aq} \right)$ that is, $B{a^{2 + }}\left( {aq} \right) < C{a^{2 + }}\left( {aq} \right)$ . Thus option (D) is also a correct statement . $I$ and $Cl$ are the elements of a seventeenth group and the atomic radius of $I$ is more than that of $Cl$ . Thus the hydrated radius of $I\left( {aq} \right)$ should be less than that of $C{l^ - }\left( {aq} \right)$ . But according to option (C), $I\left( {aq} \right)$ is greater than $C{l^ - }\left( {aq} \right)$ which is contrary to our reason. Thus option (C) is correct.
Thus option (C) is the required answer.

Note: Smaller is the atomic size of the atom, more will be its hydrated radius and more will be its hydration energy (the amount of energy liberated when an ion accepts a water molecule or we can say when an ion gets hydrated).