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The d-And-f-Block Elements
Question
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Which of the following lanthanoid ions is diamagnetic?
(At. number; Ce =58, Sm =62, Eu =63, Yb = 70)

  • Ce2+

  • Sm2+

  • Eu2+

  • Yb2+

Answer

D.

Yb2+

Lanthanoid ion with no unpaired electron is diamagnetic in nature.
Ce58 = [Xe]4f2 5d0 6s2
Ce2+ = [Xe]4f2 (two unpaired electrons)
Sm62 =  [Xe]4f6 5d0 6s2
Sm2+ = [Xe] 4f6 (six unpaired electron)

Eu63 =  [Xe]4f7 5d0 6s2
Eu2+ = [Xe] 4f7 (seven unpaired electron)

Yb70 =  [Xe]4f14 5d0 6s2
Yb2+ = [Xe] 4f14 (No unpaired electron)
Because of the absence of unpaired electrons, Yb2+ is diamagnetic.

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Some More Questions From The d-And-f-Block Elements Chapter

Explain why Cu+ ion is not stable in aqueous solutions?

Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

The outer electronic configuration of copper is 3d10 4s1, yet it is considered transition element. Why?

Why the melting points of transition elements are high?

Why Zn, Cd and Hg are not regarded as transition elements?

Name the first element of 3rd transition series.

Though copper, silver and gold have completely filled sets of d-orbitals yet they are considered as transition metals. Why?

Why are transition elements so named?

Why are transition elements known as d-block elements?