The sum of the first three ionization energies of the lanthanoids Ce, Eu, Gd, Yb and Lu is

Yb > Eu > Lu > Gd > Ce

The correct answer is option (3) Yb > Eu > Lu > Gd > Ce.

The first ionization energy of an element is the energy required to remove one electron from a neutral atom. The second ionization energy is the energy required to remove a second electron from a positively charged ion, and so on.

The lanthanide contraction is a phenomenon in which the atomic radii of the lanthanides decrease as the atomic number increases. This is due to the increasing nuclear charge of the lanthanides, which pulls the electrons closer to the nucleus.

The sum of the first three ionization energies of the lanthanoids is lowest for Yb because it has the largest atomic radius. The atomic radius of Yb is larger than the atomic radii of the other lanthanides because it has a filled 4f subshell. The filled 4f subshell shields the outer electrons from the nuclear charge, which makes the atomic radius larger.

The sum of the first three ionization energies of the lanthanoids increases as the atomic number increases from Ce to Lu. This is because the nuclear charge increases as the atomic number increases, which pulls the electrons closer to the nucleus and makes it more difficult to remove them.