Home
Class 12
CHEMISTRY
Why Sm^(2+), Eu^(2+) and Yb^(2+) ions in...

Why `Sm^(2+), Eu^(2+)` and `Yb^(2+)` ions in solutions are good reducing agents but an aqueous solution of `Ce^(4+)`is a good oxidizing agent

Text Solution

Verified by Experts

The most stable oxidation state of lanthanides is +3, hence ions in +2 state tend to change to +3 state by loss of electron and those in +4 state tend to change to +3 state by gain of electron
Promotional Banner

Topper's Solved these Questions

  • TRANSITION ELEMENTS & COORDINATION COMPOUNDS

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (SUBJECTIVE)(Level- I)|15 Videos

  • TRANSITION ELEMENTS & COORDINATION COMPOUNDS

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (SUBJECTIVE)(Level- II)|1 Videos

  • TRANSITION ELEMENTS & COORDINATION COMPOUNDS

    FIITJEE|Exercise MATCHING TYPE SINGLE CORRECT QUESTION|9 Videos

  • THERMODYNAMICS AND THERMOCHEMISTRY

    FIITJEE|Exercise SINGLE INTEGER ANSWER TYPE QUESTIONS|5 Videos

Similar Questions

Explore conceptually related problems

Why Sm^(2+) , Eu^(2+) and yb^(2+) ions in solution are good reducing agent but an aqeius solution of Ce^(4+) is a good oxidising agent?

Why Ce^(4+) is a good oxidising agent

Which is a good reducing agent '?

Why PbCl_(4) is good oxidising agent?

Ca is the good reducing agent, because