A) \[L{{u}^{3+}}<E{{u}^{3+}}<L{{a}^{3+}}<{{Y}^{3+}}\]
B) \[L{{a}^{3+}}<E{{u}^{3+}}<L{{u}^{3+}}<{{Y}^{3+}}\]
C) \[{{Y}^{3+}}<L{{a}^{3+}}<E{{u}^{3+}}<L{{u}^{3+}}\]
D) \[{{Y}^{3+}}<L{{u}^{3+}}<E{{u}^{3+}}<L{{a}^{3+}}\]
Correct Answer: D
Solution :
The correct order of ionic radii of \[{{Y}^{3+}},L{{a}^{3+}}E{{u}^{3+}}\] and \[L{{u}^{3+}}\], is \[{{Y}^{3+}}<L{{u}^{3+}}<E{{u}^{3+}}<L{{a}^{3+}}\] because Eu and Lu are the members of lanthanide series (so they show lanthanide contraction) and La is the representative element of all elements of such series and \[{{Y}^{3+}}\] ion has lower radii as comparison to \[L{{a}^{3+}}\] because it lies immediately above it in periodic table. (Atomic no. Y = 39, La = 57, Eu = 63, Lu = 71)
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