JEE Main · 2021 · 26 Feb 2021 · Shift 1 · d and f Block Elements

Forming $\mathrm{M}{\mathrm{O}}_2$ needs the lanthanoid in the $+4$ state, and among the lanthanoids that state is confined to cerium, praseodymium, neodymium, terbium and dysprosium.
Ytterbium is $[\mathrm{Xe}]4{\mathrm{f}}^{14}6{\mathrm{s}}^2$. After the two $6s$ electrons and one $4f$ electron are removed it is already at $+3$, and a fourth electron would have to be torn from the filled $4f$ shell; the ionisation enthalpy demanded is far beyond what oxide formation can repay, so ytterbium is held to the $+2$ and $+3$ states.
Ytterbium therefore does not form $\mathrm{M}{\mathrm{O}}_2$.