Arrange $N a^{+}, M g^{2 +}$ and $A l^{3 +}$ in increasing order of energy of hydration:

N a^{+} < M g^{2 +} < A l^{3 +}

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A l^{3 +} < M g^{2 +} < N a^{+}

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A l^{3 +} < N a^{+} < M g^{2 +}

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M g^{2 +} < A l^{3 +} < N a^{+}

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Solution

The correct option is D $N a^{+} < M g^{2 +} < A l^{3 +}$
The hydration enthalpies (ie, energies of hydration) of metal ions decreases with increase in ionic radii. Because of the smaller size of $A l^{3 +}$ than $M g^{2 +}$ and $N a^{+}$ , its energy of hydration larger than both these ions. The order of heat of hydration is $A l^{3 +} > M g^{2 +} > N a^{+}$

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Similar questions

$N a^{+} . M g^{2 +}, A l^{3 +}$ and $S i^{4 +}$ are isoelectronic. The order of their ionic size is

N a^{+}

M g^{2 +}

A l^{3 +}

S i^{4 +}

{N a}^{+}, {C a}^{2 +}, {M g}^{2 +}, {A l}^{3 +}

{C a}^{2 +} < {N a}^{+} < {M g}^{2 +} < {A l}^{3 +}

A l^{3 +}, N a^{+}, C a^{2 +}

The hydration energy of $M g^{2 +}$ is larger than that of:

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