The outer electrons configuration of `Lu` (Atomic No.,71)` is :

Options

A. `4f^(3) 5d^(5) 6s^(2)`
B. `4f^(8)5d^(0) 6s^(2)`
C. `4f^(4)5d^(4) s^(2)`
D. `4f^(14)5d^(1)6s^(2)`

Explanation

To determine the outer electron configuration of Lutetium (Lu), which has an atomic number of 71, we can follow these steps:

### Step 1: Identify the Atomic Number
Lutetium has an atomic number of 71, which means it has 71 electrons.

### Step 2: Determine the Electron Configuration
The electron configuration is built by filling the atomic orbitals according to the Aufbau principle, Hund's rule, and the Pauli exclusion principle.

1. **Fill the 1s orbital**: 2 electrons
2. **Fill the 2s orbital**: 2 electrons
3. **Fill the 2p orbital**: 6 electrons
4. **Fill the 3s orbital**: 2 electrons
5. **Fill the 3p orbital**: 6 electrons
6. **Fill the 4s orbital**: 2 electrons
7. **Fill the 3d orbital**: 10 electrons
8. **Fill the 4p orbital**: 6 electrons
9. **Fill the 5s orbital**: 2 electrons
10. **Fill the 4d orbital**: 10 electrons
11. **Fill the 5p orbital**: 6 electrons
12. **Fill the 6s orbital**: 2 electrons
13. **Fill the 4f orbital**: 14 electrons (completely filled)
14. **Fill the 5d orbital**: 1 electron

### Step 3: Write the Complete Electron Configuration
Combining all the filled orbitals, the complete electron configuration for Lutetium is:
\[ \text{Xe} \, 4f^{14} \, 5d^1 \, 6s^2 \]

### Step 4: Identify the Outer Electron Configuration
The outermost electrons are those in the highest energy levels, which are the 5d and 6s orbitals. Thus, the outer electron configuration of Lutetium is:
\[ 5d^1 \, 6s^2 \]

### Final Answer
The outer electron configuration of Lutetium (Lu) is:
\[ 5d^1 \, 6s^2 \]

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