
“The enthalpy of hydration decreases in going from \[L{i^ + }\] to \[C{s^ + }\]“. Answer whether the above statement is true or false. If true enter 1, else enter 0.
Answer
232.8k+ views
Hint: The smaller the ion, the stronger the attractions are between the molecules of water and the ion. Hydration enthalpy will be more for smaller ions due to stronger bonding and vice versa. It also depends on ionic potential i.e. charge/size ratio.
Complete step-by-step answer:
The energy released as a result of the formation of new bonds between ions and the water molecules is called enthalpy of hydrogen. In alkali metals, it decreases with the increase in ionic sizes.
\[L{i^ + } > N{a^ + } > {K^ + } > R{b^ + } > C{s^ + }\]
High hydration enthalpy means the hydrated ion is very stable. Hydration occurs because of attraction of lone pairs of electrons on oxygen atoms of water molecules with a positive ion. Thus, more attraction will have more hydration and so more will be the hydration enthalpy. Also, more the positive charge, more the enthalpy of hydration.
From this we conclude that hydration enthalpy is directly proportional to the charge of ions and inversely proportional to the size of an ion. That is, it depends on a charge/size ratio called ionic potential.
Lithium ion has the greatest reducing nature in aqueous medium because of its maximum hydration energy and high polarizing power. And as caesium ion has less extent of hydration, its electrode potential is less compared to all.
For group 1 elements, charge remains constant +1 and so the hydration energy depends only on size. Size of alkali metal ions increases on going down the group due to addition of shells. Hence. We can say that the enthalpy of hydration decreases in going from \[L{i^ + }\] to \[C{s^ + }\].
The statement is true, 1.
Note: It can also be understood by surface charge density on which the magnitude of hydration enthalpy depends. The surface charge density is more for smaller ions as they have high force of attraction between the ion and the water polar end and thus smaller ions have higher hydration enthalpy.
Complete step-by-step answer:
The energy released as a result of the formation of new bonds between ions and the water molecules is called enthalpy of hydrogen. In alkali metals, it decreases with the increase in ionic sizes.
\[L{i^ + } > N{a^ + } > {K^ + } > R{b^ + } > C{s^ + }\]
High hydration enthalpy means the hydrated ion is very stable. Hydration occurs because of attraction of lone pairs of electrons on oxygen atoms of water molecules with a positive ion. Thus, more attraction will have more hydration and so more will be the hydration enthalpy. Also, more the positive charge, more the enthalpy of hydration.
From this we conclude that hydration enthalpy is directly proportional to the charge of ions and inversely proportional to the size of an ion. That is, it depends on a charge/size ratio called ionic potential.
Lithium ion has the greatest reducing nature in aqueous medium because of its maximum hydration energy and high polarizing power. And as caesium ion has less extent of hydration, its electrode potential is less compared to all.
For group 1 elements, charge remains constant +1 and so the hydration energy depends only on size. Size of alkali metal ions increases on going down the group due to addition of shells. Hence. We can say that the enthalpy of hydration decreases in going from \[L{i^ + }\] to \[C{s^ + }\].
The statement is true, 1.
Note: It can also be understood by surface charge density on which the magnitude of hydration enthalpy depends. The surface charge density is more for smaller ions as they have high force of attraction between the ion and the water polar end and thus smaller ions have higher hydration enthalpy.
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