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Question

Which of the following is a correct order of atomic sizes: Ce(58), Sn(50), Yb(70), and Lu(71)?


A

Sn>Lu>Yb>Ce

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B

Ce>Yb>Sn>Lu

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C

Ce>Sn>Yb>Lu

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D

Lu>Ce>Sn>Yb

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Solution

The correct option is C

Ce>Sn>Yb>Lu


Explanation for the correct option:

C. Ce>Sn>Yb>Lu

Trend of atomic size in periodic table:

  • On moving from left to right in a periodic table, the atomic radius of an atom decreases due to the addition of an extra proton which leads to a stronger pull of electrons while the extra electron goes to the same principal energy level.
  • While going down in a group, an extra energy level adds to the structure which leads to an increase in atomic radius. Thus, on moving down the group, the atomic radius increases.
  • Cerium (Ce) belongs to Lanthanoids, Tin (Sn) belongs to Group 16 and period 4, Ytterbium (Yb), and Lutetium (Lu) also belong to lanthanoids.
  • Lanthanoids are considered in the sixth period.
  • There is a sharp decrease in the atomic size of lanthanoids from Ce>Yb>Lu as a result of lanthanoid contraction. The size of Yb and Lu is smaller than Sn.
  • Thus, the correct order of atomic size is Ce>Sn>Yb>Lu.

Explanation of incorrect options:

The correct order of atomic size is Ce>Sn>Yb>Lu. Thus, options A, B, and D are incorrect.

Hence, the correct option is C i.e. Ce>Sn>Yb>Lu


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