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

<p>The stability of a divalent state can often be attributed to the stability of the electron configuration of that state. Among the options given, we can evaluate the electron configurations of each element in their divalent state to find out which is the most stable. The electronic configurations of the neutral atoms are :</p> <ul> <li>Ce (Atomic Number 58) : $[Xe] 4f^1 5d^1 6s^2$</li> <li>Sm (Atomic Number 62) : $[Xe] 4f^6 6s^2$</li> <li>Eu (Atomic Number 63) : $[Xe] 4f^7 6s^2$</li> <li>Yb (Atomic Number 70) : $[Xe] 4f^{14} 6s^2$</li> </ul> <p>In their divalent state (+2 oxidation state), the electron configurations would be :</p> <ul> <li>Ce : $[Xe] 4f^1 5d^1$ (removal of the 2 electrons from the 6s orbital)</li> <li>Sm : $[Xe] 4f^6$ (removal of the 2 electrons from the 6s orbital)</li> <li>Eu : $[Xe] 4f^7$ (removal of the 2 electrons from the 6s orbital)</li> <li>Yb : $[Xe] 4f^{14}$ (removal of the 2 electrons from the 6s orbital)</li> </ul> <p>Among these, Eu in its divalent state has a half-filled (4f) shell, which offers extra stability due to symmetrical electron distribution. Therefore, Eu (Atomic Number 63) is the most stable in the divalent form. </p> <p>So the correct answer is Option C : Eu (Atomic Number 63).</p>