Which one of the following lanthanoids does not form $$MO_2$$? [M is lanthanoid metal]

The formation of $$MO_2$$ requires the lanthanoid to exhibit a +4 oxidation state. This is possible only when the resulting $$M^{4+}$$ ion has a particularly stable electronic configuration.

The lanthanoids known to form $$MO_2$$ are those where the +4 state leads to an empty ($$4f^0$$), half-filled ($$4f^7$$), or nearly half-filled 4f subshell. Cerium (Ce), praseodymium (Pr), and terbium (Tb) are well-known examples. Additionally, compounds like $$PrO_2$$ and $$NdO_2$$ have been characterized.

For the given options:

Pr ($$[Xe]4f^3$$): $$Pr^{4+}$$ is $$[Xe]4f^1$$. $$PrO_2$$ exists as a stable fluorite-type oxide.

Nd ($$[Xe]4f^4$$): $$Nd^{4+}$$ is $$[Xe]4f^2$$. $$NdO_2$$ has been prepared under high oxygen pressure.

Dy ($$[Xe]4f^{10}$$): $$Dy^{4+}$$ is $$[Xe]4f^8$$. $$DyO_2$$ has been reported under special conditions.

Yb ($$[Xe]4f^{14}$$): Ytterbium has a completely filled 4f subshell, making the +2 oxidation state ($$4f^{14}$$) very stable. Achieving the +4 state would require $$[Xe]4f^{12}$$, which disrupts the stable fully filled configuration. The very high fourth ionization energy makes this oxidation state energetically prohibitive. Hence, $$YbO_2$$ does not form.

The correct answer is Option 1: Yb.