
Which one of the following statements related to lanthanons is incorrect:
A. all the lanthanons are much more reactive than aluminium.
B. Ce ( + 4) solutions are widely used as oxidizing agents in volumetric analysis.
C. Europium show + 2 oxidation state.
D. The basicity decreases as the ionic radius decreases from Pr to Lu.
Answer
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Hint: Lanthanides have deeply buried f-orbitals. 4 f orbitals are more attracted to the nucleus and aluminium is a p-block element. The most stable oxidation state shown by lanthanides is + 3. From Pr to Lu effective nuclear charge increases.
Complete step by step answer:
The valence orbitals that are 4 f orbitals, are deep buried so they are not affected by the ligand field, so, lanthanides are less reactive and aluminium is a p-block element that shows the oxidation state + 3. Lanthanides and aluminium are different block elements, so simply their reactivity cannot be compared, so it cannot be said that all the lanthanons are much more reactive than aluminium. So, option (A) is correct.
Lanthanides show few oxidation states. The most stable oxidation state shown by lanthanides is + 3 so, Ce (+ 4) easily reduces into Ce (+ 3). So, Ce (+ 4) solutions are widely used as oxidizing agents in volumetric analysis. So, option (B) is incorrect.
Lanthanides also show + 2 oxidation state. Mostly europium, samarium, ytterbium, and thulium are stable in + 2 oxidation state. So, option (C) is incorrect.
From Pr to Lu size of the lanthanides decreases so, the effective nuclear charge increases which makes the Ln-O bond strong so, the tendency to lose the ${\text{O}}{{\text{H}}^ - }$ decreases from Pr to Lu. So, the statement, The basicity decreases as the ionic radius decreases from Pr to Lu, is true and the option (D) is incorrect.
Therefore option (A) all the lanthanons are much more reactive than aluminium, is correct.
Note: All lanthanides have lower energy valence shells, so all lanthanides show similar physical and chemical properties. On going, La to Lu the size decreases, basicity decreases, and the acidity increases.
Complete step by step answer:
The valence orbitals that are 4 f orbitals, are deep buried so they are not affected by the ligand field, so, lanthanides are less reactive and aluminium is a p-block element that shows the oxidation state + 3. Lanthanides and aluminium are different block elements, so simply their reactivity cannot be compared, so it cannot be said that all the lanthanons are much more reactive than aluminium. So, option (A) is correct.
Lanthanides show few oxidation states. The most stable oxidation state shown by lanthanides is + 3 so, Ce (+ 4) easily reduces into Ce (+ 3). So, Ce (+ 4) solutions are widely used as oxidizing agents in volumetric analysis. So, option (B) is incorrect.
Lanthanides also show + 2 oxidation state. Mostly europium, samarium, ytterbium, and thulium are stable in + 2 oxidation state. So, option (C) is incorrect.
From Pr to Lu size of the lanthanides decreases so, the effective nuclear charge increases which makes the Ln-O bond strong so, the tendency to lose the ${\text{O}}{{\text{H}}^ - }$ decreases from Pr to Lu. So, the statement, The basicity decreases as the ionic radius decreases from Pr to Lu, is true and the option (D) is incorrect.
Therefore option (A) all the lanthanons are much more reactive than aluminium, is correct.
Note: All lanthanides have lower energy valence shells, so all lanthanides show similar physical and chemical properties. On going, La to Lu the size decreases, basicity decreases, and the acidity increases.
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