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Periodic Table and Periodicity question

2002 · Shift 0 · Q27
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Periodic Table and Periodicity question

2002 · Shift 0 · Q27

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
Ce+3,  La+3,  Pm+3  C{e^{ + 3}},\,\,L{a^{ + 3}},\,\,P{m^{ + 3}}\,\,Ce+3,La+3,Pm+3 and Yb+3  Y{b^{ + 3}}\,\,Yb+3 have ionic radial in the increasing order as
  1. A
    La+3  <Ce+3  <Pm+3  <Yb+3L{a^{ + 3}}\,\, \lt C{e^{ + 3}}\,\, \lt P{m^{ + 3}}\,\, \lt Y{b^{ + 3}}La+3<Ce+3<Pm+3<Yb+3
  2. B
    Yb+3  <Pm+3  <Ce+3 < La+3Y{b^{ + 3}}\,\, \lt P{m^{ + 3}}\,\, \lt C{e^{ + 3}}\,\lt \,L{a^{ + 3}}Yb+3<Pm+3<Ce+3<La+3
  3. C
    La+3  =Ce+3  <Pm+3  <Yb+3L{a^{ + 3}}\,\, = C{e^{ + 3}}\,\, \lt P{m^{ + 3}}\,\, \lt Y{b^{ + 3}}La+3=Ce+3<Pm+3<Yb+3
  4. D
    Yb+3  <Pm+3  <La+3  <Ce+3Y{b^{ + 3}}\,\, \lt P{m^{ + 3}}\,\, \lt L{a^{ + 3}}\,\, \lt C{e^{ + 3}}Yb+3<Pm+3<La+3<Ce+3
View written solutionFree

Correct answer: B

  1. We need the increasing order of ionic radii of Ce3+, La3+, Pm3+, Yb3+.\mathrm{Ce^{3+},\ La^{3+},\ Pm^{3+},\ Yb^{3+}}.Ce3+, La3+, Pm3+, Yb3+.

  2. These are all lanthanoid ions in the common oxidation state +3+3+3.

  3. Across the lanthanoid series, the ionic radii of Ln3+\mathrm{Ln^{3+}}Ln3+ ions decrease from La to Lu. This is due to lanthanoid contraction.

    So, as atomic number increases: La3+>Ce3+>Pm3+>Yb3+\mathrm{La^{3+} > Ce^{3+} > Pm^{3+} > Yb^{3+}}La3+>Ce3+>Pm3+>Yb3+ in size.

  4. Therefore, in increasing order of ionic radius (smallest to largest), we reverse this: Yb3+<Pm3+<Ce3+<La3+\mathrm{Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}}Yb3+<Pm3+<Ce3+<La3+

  5. Now compare with the options:

    • A: wrong, it is decreasing order reversed incorrectly.
    • B: Yb3+<Pm3+<Ce3+<La3+\mathrm{Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}}Yb3+<Pm3+<Ce3+<La3+ ✓
    • C: wrong, La3+\mathrm{La^{3+}}La3+ and Ce3+\mathrm{Ce^{3+}}Ce3+ are not equal.
    • D: wrong, Ce3+\mathrm{Ce^{3+}}Ce3+ is not larger than La3+\mathrm{La^{3+}}La3+.
  6. Hence, the correct option is B.

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