A) \[L{{a}^{3+}}<C{{e}^{3+}}<P{{m}^{3+}}<Y{{b}^{3+}}\]
B) \[Y{{b}^{3+}}<P{{m}^{3+}}<C{{e}^{3+}}<L{{a}^{3+}}\]
C) \[L{{a}^{3+}}\,=C{{e}^{3+}}\,<P{{m}^{3+}}\,<Y{{b}^{3+}}\]
D) \[Y{{b}^{3+}}<P{{m}^{3+}}<L{{a}^{3+}}<C{{e}^{3+}}\]
Correct Answer: B
Solution :
Atomic radii decreases across a series because of increase in nuclear charge i.e., in a series it decreases with increase in atomic number. \[{{r}_{n}}\] (radius ) \[\propto \frac{1}{Z}\] Atomic number of the given species are La = 57 Ce = 58 Pm = 61 Yb = 70 All these species belong to the same series, so number of shell are constant for all of these species but nuclear charge per electron increases with increase in atomic number (number of protons). Thus, nuclear charge felt by \[{{e}^{-}}\] of \[Y{{b}^{3+}}\] is highest and therefore size is smallest and the ionic radii in the increasing order is \[Y{{b}^{3+}}<P{{m}^{3+}}<C{{e}^{3+}}<L{{a}^{3+}}\]
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