A) \[Y{{b}^{3+}}<P{{m}^{3+}}<L{{a}^{3+}}<C{{e}^{3+}}\]?
B) \[Y{{b}^{3+}}<P{{m}^{3+}}<C{{e}^{3+}}<L{{a}^{3+}}\]
C) \[P{{m}^{3+}}<L{{a}^{3+}}<C{{e}^{3+}}<Y{{b}^{3+}}\]
D) \[C{{e}^{3+}}<Y{{b}^{3+}}<P{{m}^{3+}}<L{{a}^{3+}}\]
Correct Answer: B
Solution :
The electronic configuration of the given elements is as [a] \[_{58}C{{e}^{3+}}=[Xe]\,4{{f}^{1}},5{{d}^{10}},6{{s}^{0}}\] [b] \[_{57}L{{a}^{3+}}=[Xe]\,4{{f}^{0}},5{{d}^{0}},6{{s}^{0}}\] [c] \[_{61}P{{m}^{3+}}=[Xe]\,4{{f}^{4}},5{{d}^{0}},6{{s}^{0}}\] [d] \[_{70}P{{m}^{3+}}=[Xe]\,4{{f}^{13}},5{{d}^{0}},6{{s}^{0}}\] Since, the electrons are filling in the same shell (ie, \[4f\]-orbitals) and /-electrons, due to diffused \[f\]orbital, do not shield the other \[f\]-electrons. Hence, as the atomic number increases, the ionic size decreases, (lanthanide contraction). Thus, the correct order of ionic radii is \[Y{{b}^{3+}}<P{{m}^{3+}}<C{{e}^{3+}}<L{{a}^{3+}}\]
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