To determine the principal (n) and azimuthal (l) quantum numbers of the valence electrons in tripositive Lutetium (Lu3+), we need to first determine the electronic configuration of neutral Lutetium (Lu). Neutral Lutetium has an atomic number of 71 . Its electron configuration is: Lu:[Xe]4f145d16s2 In the tripositive state (Lu3+), three electrons are removed. The electrons are generally removed from the outermost orbitals first. Thus, removing three electrons from neutral Lutetium involves the removal of two 6 s electrons and one 5 d electron: Lu3+:[Xe]4f14
Therefore, the valence electrons in Lu3+ are in the 4f subshell. The principal quantum number n for the 4 f subshell is 4 , and the azimuthal (angular momentum) quantum number l for the f subshell is 3 . So, the correct set of quantum numbers for the valence electrons in tripositive Lutetium is: n=4‌‌&‌‌l=3 Therefore, the correct option is: Option D