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

Options

A. Pr
B. Dy
C. Nd
D. Yb

Answer

To determine which lanthanoid does not form the oxide of the type \( MO_2 \) (where \( M \) is a lanthanoid metal), we need to analyze the oxidation states of the lanthanoids and their ability to form such oxides.

### Step-by-Step Solution:

1. **Identify the Lanthanoids**: The lanthanoid series includes the elements from Lanthanum (La, atomic number 57) to Lutetium (Lu, atomic number 71). The relevant lanthanoids for this question are:
- La (57)
- Ce (58)
- Pr (59)
- Nd (60)
- Pm (61)
- Sm (62)
- Eu (63)
- Gd (64)
- Tb (65)
- Dy (66)
- Ho (67)
- Er (68)
- Tm (69)
- Yb (70)
- Lu (71)

2. **Oxidation States**: The lanthanoids typically exhibit oxidation states of +3 and +4. However, not all lanthanoids can form \( MO_2 \) type oxides, which require a +4 oxidation state.

3. **Identify the Lanthanoids that Form \( MO_2 \)**: The lanthanoids that can exhibit a +4 oxidation state and form \( MO_2 \) include:
- Ce (Cerium)
- Pr (Praseodymium)
- Nd (Neodymium)
- Tb (Terbium)
- Dy (Dysprosium)

4. **Identify the Exception**: Among the lanthanoids, Samarium (Sm, atomic number 62) does not form \( MO_2 \) because it primarily exhibits a +3 oxidation state and does not stabilize the +4 oxidation state in its oxides.

5. **Conclusion**: Therefore, the lanthanoid that does not form \( MO_2 \) is Samarium (Sm).

### Final Answer:
The lanthanoid that does not form \( MO_2 \) is **Samarium (Sm)**.

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