Question

Write the condensed electron configuration and partial orbital diagram for the following.
A) Se2-
B)Fe2+
C)Hf
D)Mo
E)Ga3+

Answer 1

Here are the condensed electron configurations and partial orbital diagrams for the specified ions and elements:

### A) Se²⁻ (Selenium ion)
- **Condensed Electron Configuration**: [Ar] 4s² 3d¹⁰ 4p⁶
- **Partial Orbital Diagram**:
```
4p: ↑↑
4s: ↑↓
3d: ↑↑↑↑↑
```
(Note: Se normally has 34 electrons, but with 2 additional electrons for the -2 charge, it has 36 total electrons.)

### B) Fe²⁺ (Iron ion)
- **Condensed Electron Configuration**: [Ar] 4s² 3d⁶ → [Ar] 3d⁶ (after losing 2 electrons from 4s)
- **Partial Orbital Diagram**:
```
3d: ↑↑↑↑↓
4s:
```
(Note: Fe normally has 26 electrons, but loses 2 electrons from the 4s orbital for the +2 charge, leaving 24 electrons.)

### C) Hf (Hafnium)
- **Condensed Electron Configuration**: [Xe] 6s² 4f¹⁴ 5d²
- **Partial Orbital Diagram**:
```
5d: ↑↑
6s: ↑↓
4f: ↑↑↑↑↑↑↑↑↑↑
```
(Note: Hf has a total of 72 electrons.)

### D) Mo (Molybdenum)
- **Condensed Electron Configuration**: [Kr] 5s¹ 4d⁵
- **Partial Orbital Diagram**:
```
4d: ↑↑↑↑↑
5s: ↑
```
(Note: Mo has a total of 42 electrons; it has an unusual configuration that minimizes energy. It loses one electron from 5s to half-fill 4d.)

### E) Ga³⁺ (Gallium ion)
- **Condensed Electron Configuration**: [Ar] 4s² 3d¹⁰ 4p¹ → [Ar] 3d¹⁰ (after losing 3 electrons)
- **Partial Orbital Diagram**:
```
3d: ↑↑↑↑↑↑↑↑↑↑
4s:
4p:
```
(Note: Ga normally has 31 electrons, but loses 3 electrons (2 from 4s and 1 from 4p) for the +3 charge, leaving 28 electrons remaining.)

These representations assume filling of orbitals according to the Aufbau principle, Pauli's exclusion principle, and Hund's rule where applicable.