The solubility of $Fe{{(OH)}_{2}}$ is greater than $Fe{{(OH)}_{3}}$ in water. If true, write $1$ , else write $0$ .

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Hint: Before talking about the answer, we should know what solubility is. Solubility is defined as the ability of a compound to get soluble. Solubility is expressed in terms of molarity. You should also know about solubility products, that is, ${{K}_{sp}}$ .

Formula used:For the following reaction: $[AB]\rightleftharpoons a[{{A}^{+}}]+b[{{B}^{-}}]$
Solubility product, ${{K}_{sp}}={{[{{A}^{+}}]}^{a}}{{[{{B}^{-}}]}^{b}}$
where, ${{K}_{sp}}$ is solubility product, ${{A}^{+}}$ is the cation, ${{B}^{-}}$ is the anion and $a,b$ are concentration of $A$ and $B$ .

Complete step by step answer:
Firstly, we will calculate the solubility product of $Fe{{(OH)}_{2}}$
The reaction can be written as follows:
\[Fe{{(OH)}_{2}}\rightleftharpoons F{{e}^{2+}}+2O{{H}^{-}}\]
Here, the solubility of $F{{e}^{2+}}$ is $'s'$ and solubility of $O{{H}^{-}}$ ions is $'2s'$
Therefore, \[{{K}_{sp}}=(s){{(2s)}^{2}}\]
\[{{K}_{sp}}=4{{s}^{3}}\]
The ${{K}_{sp}}$ value of $Fe{{(OH)}_{2}}$ is $2\times {{10}^{-15}}$
\[
 \Rightarrow 2\times {{10}^{-15}}=4{{s}^{3}} \\
 \Rightarrow s={{(\dfrac{2\times {{10}^{-15}}}{4})}^{{}^{1}/{}_{3}}}=7.94\times {{10}^{-6}} \\
\]
Similarly, for $Fe{{(OH)}_{3}}$, the reaction will be written as:
\[Fe{{(OH)}_{3}}\rightleftharpoons F{{e}^{3+}}+3O{{H}^{-}}\]
and the solubility of $F{{e}^{3+}}$ is $'s'$ and solubility of $O{{H}^{-}}$ ions is $'3s'$
Therefore, \[{{K}_{sp}}=(s){{(3s)}^{3}}\]
\[{{K}_{sp}}=27{{s}^{4}}\]
${{K}_{sp}}$ value of $Fe{{(OH)}_{3}}$ is $6\times {{10}^{-38}}$
\[
 \Rightarrow 6\times {{10}^{-38}}=27{{s}^{4}} \\
 \Rightarrow s={{(\dfrac{6\times {{10}^{-38}}}{27})}^{{}^{1}/{}_{4}}}=2.17\times {{10}^{-10}} \\
\]
Now as we can see, the solubility of $Fe{{(OH)}_{2}}$ is greater than $Fe{{(OH)}_{3}}$
So, the correct answer is $1$ .
Here, ${{K}_{sp}}$ is a solubility product which is defined as the equilibrium constant for solid to dissolve in an aqueous solution. It tells us about the level at which the solute is dissolved in the solution. It also tells us about the slightly ionic compound.

Additional Information:
If the ionic product is more than the saturated product, precipitation will take place.
If the ionic product is less than the saturated product, precipitation will not take place. Solution is diluted.
If the ionic product is equal to the saturated product, the solution is saturated and there will be no precipitation.
Solubility products depend only on temperature.
Factors affecting solubility are common ion effect and simultaneous solubility.


Note:
As we increase the solubility product, there will be an increase in solubility.
${{K}_{sp}}$ of $Fe{{(OH)}_{2}}$ is more than the ${{K}_{sp}}$ of $Fe{{(OH)}_{3}}$ and hence solubility of $Fe{{(OH)}_{2}}$ is greater than that of $Fe{{(OH)}_{3}}$ .