The outer electronic configuration of two members of the lanthanoid series are as follows:
`4f^(1)5d^(1)6s^(2)` and `4f^(7) 5d^(0) 6s^(2)`
What are their atomic number ? Predict the oxidation states exhibited by elements in their compounds.

Answer

To solve the question, we will analyze the given outer electronic configurations of the two lanthanoid elements, determine their atomic numbers, and predict the oxidation states exhibited by these elements in their compounds.

### Step-by-Step Solution:

1. **Identify the first electronic configuration: `4f^(1) 5d^(1) 6s^(2)`**
- The outer electronic configuration indicates that there is 1 electron in the 4f subshell, 1 electron in the 5d subshell, and 2 electrons in the 6s subshell.
- The element with this configuration is **Cerium (Ce)**.

2. **Determine the atomic number of Cerium:**
- The atomic number of Cerium is **58**. This can be confirmed by checking the periodic table.

3. **Predict the oxidation states of Cerium:**
- Cerium typically exhibits oxidation states of **+3** and **+4**.
- In the +3 oxidation state, Cerium loses three electrons: one from the 4f subshell and two from the 6s subshell, leading to the configuration `4f^(1) 5d^(0) 6s^(0)`.
- In the +4 oxidation state, Cerium loses one additional electron from the 5d subshell, leading to the configuration `4f^(0) 5d^(0) 6s^(0)`.

4. **Identify the second electronic configuration: `4f^(7) 5d^(0) 6s^(2)`**
- This configuration shows that there are 7 electrons in the 4f subshell, 0 electrons in the 5d subshell, and 2 electrons in the 6s subshell.
- The element with this configuration is **Europium (Eu)**.

5. **Determine the atomic number of Europium:**
- The atomic number of Europium is **63**. This can also be confirmed by checking the periodic table.

6. **Predict the oxidation states of Europium:**
- Europium typically exhibits oxidation states of **+2** and **+3**.
- In the +2 oxidation state, Europium loses two electrons from the 6s subshell, leading to the configuration `4f^(7) 5d^(0) 6s^(0)`.
- In the +3 oxidation state, Europium loses one additional electron from the 4f subshell, leading to the configuration `4f^(6) 5d^(0) 6s^(0)`.

### Summary of Results:
- **Cerium (Ce)**: Atomic number = 58, Oxidation states = +3, +4
- **Europium (Eu)**: Atomic number = 63, Oxidation states = +2, +3