In lanthanoids, the dioxide formula $MO_2$ corresponds to the metal being in the +4 oxidation state, because oxygen is $-2$ and $M + 2(-2)=0 \Rightarrow M=+4.$

So the question reduces to: Which lanthanoid does not commonly show the +4 oxidation state?

Among lanthanoids, the more stable +4 oxidation state is shown mainly by a few elements such as:

• Ce: forms $CeO_2$
• Pr: can form $PrO_2$
• Tb: can form $TbO_2$
• Nd can also form dioxide under suitable conditions

Now check the given options:

• A: Nd — can form $NdO_2$.
• B: Dy — does not commonly form a stable dioxide $DyO_2$.
• C: Yb — does not form $YbO_2$; however Yb is well known mainly for the +2 and +3 oxidation states.
• D: Pr — forms $PrO_2$.

In standard JEE chemistry, the lanthanoids known to form dioxides are typically Ce, Pr, Tb, and Nd. Therefore among the given options, Yb is taken as the correct choice for not forming $MO_2$.

Hence, the correct answer is: $\boxed{\text{C: Yb}}$