Most basic hydroxide among the following is:
A. $Lu{(OH)_3}$
B. $Eu{(OH)_3}$
C. $Yb{(OH)_3}$
D. $Ce{(OH)_3}$
Answer
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Hint: The basicity of metal hydroxide increases as we move down the group. $La{(OH)_3}$ is most basic and $Lu{(OH)_3}$ is least basic and due to lanthanide contraction, as the size of lanthanide ions decreases from \[L{a^{3 + }}\] to \[L{u^{3 + }}\], the covalent character of the hydroxides often increases and thus the basic strength decreases.
Complete step by step answer:
The lanthanides, also known as lanthanons are a group of 15 elements of atomic numbers starting from 57 to 71 in which scandium (atomic number 21) and yttrium (atomic number 39) are sometimes included. Due to lanthanide contraction, the size of \[{M^{3 + }}\] ions (that is, \[L{u^{3 + }}\], \[E{u^{3 + }}\], \[Y{b^{3 + }}\] and \[C{e^{3 + }}\] ) decreases and thus, the basic strength of their hydroxides decreases. The order of size of given \[{M^{3 + }}\;\] ions is \[C{e^{3 + }} > E{u^{3 + }} > Y{b^{3 + }} > L{u^{3 + }}\] and as the size of lanthanide ions decreases from \[L{a^{3 + }}\] to \[L{u^{3 + }}\] and the covalent character of the hydroxides most often increases and thus the basic strength of hydroxides decreases. Therefore, $La{(OH)_3}$, being the most basic and $Lu{(OH)_3}$ being the least basic of all.
As we know that, the order of basic strength of hydroxides from the least basic hydroxide to the most basic is $Ce{(OH)_3} > Eu{(OH)_3} > Yb{(OH)_3} > Lu{(OH)_3}$.
Therefore, the correct answer is option (D).
Note: Lanthanides are difficult to separate from each other because of similarities in their physical and chemical properties. Many of the separation processes take the advantage of a small decrease in the ionic radius that happens across the lanthanide series. The extracted liquid contains arms, called ligands, that grab the lanthanide.
Complete step by step answer:
The lanthanides, also known as lanthanons are a group of 15 elements of atomic numbers starting from 57 to 71 in which scandium (atomic number 21) and yttrium (atomic number 39) are sometimes included. Due to lanthanide contraction, the size of \[{M^{3 + }}\] ions (that is, \[L{u^{3 + }}\], \[E{u^{3 + }}\], \[Y{b^{3 + }}\] and \[C{e^{3 + }}\] ) decreases and thus, the basic strength of their hydroxides decreases. The order of size of given \[{M^{3 + }}\;\] ions is \[C{e^{3 + }} > E{u^{3 + }} > Y{b^{3 + }} > L{u^{3 + }}\] and as the size of lanthanide ions decreases from \[L{a^{3 + }}\] to \[L{u^{3 + }}\] and the covalent character of the hydroxides most often increases and thus the basic strength of hydroxides decreases. Therefore, $La{(OH)_3}$, being the most basic and $Lu{(OH)_3}$ being the least basic of all.
As we know that, the order of basic strength of hydroxides from the least basic hydroxide to the most basic is $Ce{(OH)_3} > Eu{(OH)_3} > Yb{(OH)_3} > Lu{(OH)_3}$.
Therefore, the correct answer is option (D).
Note: Lanthanides are difficult to separate from each other because of similarities in their physical and chemical properties. Many of the separation processes take the advantage of a small decrease in the ionic radius that happens across the lanthanide series. The extracted liquid contains arms, called ligands, that grab the lanthanide.
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