Which one of the lanthanoids given below is t | vedclass

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(C) The stability of the divalent state in lanthanoids is related to the electronic configuration and the reduction potential $E^0_{M^{3+}/M^{2+}}$.$E

Vedclass · Accepted Answer

$Eu$ (Atomic Number $63$)

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(C) The stability of the divalent state in lanthanoids is related to the electronic configuration and the reduction potential $E^0_{M^{3+}/M^{2+}}$.$Eu^{2+}$ has a stable half-filled $f$-orbital configuration $([Xe] 4f^7)$.$Yb^{2+}$ has a stable fully-filled $f$-orbital configuration $([Xe] 4f^{14})$.However,comparing the standard reduction potentials: $E^0_{Eu^{3+}/Eu^{2+}} = -0.35 \ V$ and $E^0_{Yb^{3+}/Yb^{2+}} = -1.05 \ V$.$A$ less negative reduction potential indicates that $Eu^{3+}$ is more easily reduced to $Eu^{2+}$ than $Yb^{3+}$ is to $Yb^{2+}$,making $Eu^{2+}$ the most stable among the given options.

Which one of the lanthanoids given below is the most stable in divalent form?

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