The most common oxidation states of cerium are

Options

A. `+3,+4`
B. `+2,+3`
C. `+2,+4`
D. `+3,+5`

Explanation

To determine the most common oxidation states of cerium, we can follow these steps:

### Step 1: Write the Electronic Configuration of Cerium
Cerium (Ce) has an atomic number of 58. Its electronic configuration can be written as:
\[ \text{Ce: } [Xe] \, 4f^1 \, 5d^1 \, 6s^2 \]

### Step 2: Identify Possible Oxidation States
To find the common oxidation states, we need to consider how cerium can lose electrons to achieve stability. The electrons in the outermost shells (6s and 5d) are the most likely to be lost.

### Step 3: Determine the +3 Oxidation State
When cerium loses three electrons, it can achieve the +3 oxidation state. The electronic configuration after losing three electrons would be:
\[ \text{Ce}^{+3}: [Xe] \, 4f^1 \, 5d^0 \, 6s^0 \]
This means two electrons are lost from the 6s subshell and one from the 5d subshell.

### Step 4: Determine the +4 Oxidation State
For the +4 oxidation state, cerium loses four electrons. The electronic configuration would be:
\[ \text{Ce}^{+4}: [Xe] \, 4f^0 \, 5d^0 \, 6s^0 \]
In this case, cerium has lost all of its 6s and 5d electrons, achieving a stable noble gas configuration.

### Step 5: Conclusion
The most common oxidation states of cerium are +3 and +4.

### Final Answer
The most common oxidation states of cerium are +3 and +4.

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