JEE Main 26 Feb 2021 Shift 1 Solved Paper

In oxides, MO2 ( M is lanthanoid metal) only four lanthanoids exhibit +4 oxidation state. These lanthanoids are praseodymium (Pr,Z=59), neodymium (Nd,Z=60) terbium (Tb,Z=65) dysprosium (Dy,Z=66)
So, Yb (ytterbium) option (d) does not form MO2 type of oxide.
Note The common and predominant oxidation state of lanthanoids is +3. Consequently M4+ compounds are strong oxidising agents which changes to the common +3 state. Similarly, lanthanoid compounds of +2 state have a tendency to show reducing property as they get changed to +3 state easily.