Choose the stable oxidation states of cerium (Ce).

Options

A. `+1 and +2`
B. `+3 and +4`
C. `+2 and+3`
D. `+2 and +7`

Answer

To determine the stable oxidation states of cerium (Ce), we can follow these steps:

### Step 1: Determine the Atomic Number and Electronic Configuration
- The atomic number of cerium (Ce) is 58.
- The electronic configuration of cerium can be written as:
\[ \text{Ce: } [Xe] 6s^2 4f^1 5d^1 \]
Here, [Xe] represents the electron configuration of xenon, which is the nearest noble gas.

### Step 2: Identify the Valence Electrons
- The valence electrons of cerium are in the 6s, 4f, and 5d orbitals. Specifically, cerium has:
- 2 electrons in the 6s orbital
- 1 electron in the 4f orbital
- 1 electron in the 5d orbital

### Step 3: Consider Possible Oxidation States
- Cerium can lose electrons from its outermost shells (6s, 4f, and 5d) to form positive oxidation states.
- If cerium loses 1 electron from the 4f orbital and 1 electron from the 5d orbital, it can achieve a +2 oxidation state.
- If cerium loses all 4 outer electrons (2 from 6s, 1 from 4f, and 1 from 5d), it can achieve a +4 oxidation state.

### Step 4: Determine Stability of Oxidation States
- The +3 oxidation state can be achieved by losing 3 electrons (2 from 6s and 1 from 4f).
- The +4 oxidation state is stable because it achieves a noble gas configuration, which is energetically favorable.

### Conclusion
- The stable oxidation states of cerium are +3 and +4.

### Final Answer
The stable oxidation states of cerium (Ce) are +3 and +4.

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