Among Ce(4f^{1} 5d^{1} 6s^{2}),Nd(4f^{3} 6s^{ | vedclass

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(B) For the $3^{rd}$ ionisation energy,the electron must be removed from the $+2$ oxidation state of the given metals.The electronic configuration of

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$Eu$ and $Ce$

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(B) For the $3^{rd}$ ionisation energy,the electron must be removed from the $+2$ oxidation state of the given metals.The electronic configuration of $Ce^{2+}$ is $4f^{1} 5d^{1}$.The electronic configuration of $Nd^{2+}$ is $4f^{3}$.The electronic configuration of $Eu^{2+}$ is $4f^{7}$.The electronic configuration of $Dy^{2+}$ is $4f^{10}$.$Eu$ will have the maximum $3^{rd}$ ionisation enthalpy because the electron has to be removed from the stable half-filled $4f^{7}$ configuration of $Eu^{2+}$.$Ce$ will have the lowest $3^{rd}$ ionisation enthalpy because the electron has to be removed from the $5d$-orbital,which is more shielded and further from the nucleus compared to the $4f$-orbitals.

Among $Ce(4f^{1} 5d^{1} 6s^{2})$,$Nd(4f^{3} 6s^{2})$,$Eu(4f^{7} 6s^{2})$ and $Dy(4f^{10} 6s^{2})$,the elements having highest and lowest $3^{rd}$ ionisation energies,respectively are

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