(B) The solubility of alkaline earth metal sulphates decreases from $Be$ to $Ba$ because the hydration energy decreases more rapidly than the lattice

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Solubility of sulphates in water

(B) The solubility of alkaline earth metal sulphates decreases from $Be$ to $Ba$ because the hydration energy decreases more rapidly than the lattice energy as the size of the cation increases down the group. Therefore,the correct option is $B$.

Class 11 Chemistry s-Block Elements Alkaline earth metals

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Which of the following decreases on going gradually from $Be$ to $Ba$ in the periodic table?

A
Basic character of hydroxides
B
Solubility of sulphates in water
C
Solubility of hydroxides in water
D
Strength of elements as reducing agent

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Class 11

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s-Block Elements

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Alkaline earth metals

$A$ mixture of $MgCl_2$ and $MgO$ is called

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Predict the feasibility of the given reactions in aqueous solution:
$(i)$ $Be(OH)_2 + 2 OH^{-} \longrightarrow [Be(OH)_4]^{2-}$
$(ii)$ $Be(OH)_2 + 2 H^{+} \longrightarrow Be^{2+} + 2 H_2O$

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In India,green flames are observed during fireworks at weddings. Which of the following elements is responsible for this?

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Statement $A$: Helium and Beryllium have the same outer electronic configuration of the type $ns^2$.
Reason $R$: Both are chemically inert.

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The setting of Plaster of Paris involves:

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Which of the following decreases on going gradually from $Be$ to $Ba$ in the periodic table?

Similar Questions

$A$ mixture of $MgCl_2$ and $MgO$ is called

Predict the feasibility of the given reactions in aqueous solution:$(i)$ $Be(OH)_2 + 2 OH^{-} \longrightarrow [Be(OH)_4]^{2-}$$(ii)$ $Be(OH)_2 + 2 H^{+} \longrightarrow Be^{2+} + 2 H_2O$

In India,green flames are observed during fireworks at weddings. Which of the following elements is responsible for this?

Statement $A$: Helium and Beryllium have the same outer electronic configuration of the type $ns^2$.Reason $R$: Both are chemically inert.

The setting of Plaster of Paris involves:

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Predict the feasibility of the given reactions in aqueous solution:
$(i)$ $Be(OH)_2 + 2 OH^{-} \longrightarrow [Be(OH)_4]^{2-}$
$(ii)$ $Be(OH)_2 + 2 H^{+} \longrightarrow Be^{2+} + 2 H_2O$

Statement $A$: Helium and Beryllium have the same outer electronic configuration of the type $ns^2$.
Reason $R$: Both are chemically inert.